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Chairperson, CBSE
Contents S NO TOPIC PAGE
NUMBER 1. Ellipse - I 1 3 2. Ellipse - II 9 3. Ellipse - III 11 15 4. Ellipse - IV 17 20 5. Ellipse - V 24 28 6. Hyperbola - I 30 34 7. Hyperbola - II 35 38 8. Hyperbola - III 40 44 9. Hyperbola - IV 45 48 10. Hyperbola - V 50 55 11. Hyperbola - VI 57 61 12. Hyperbola - VII 63 68 13. Hyperbola - VIII 70 73 14. Complex Number - I 75 78 15. Complex Number 80 85 16. Complex Number - III 87 17. Complex Number - IV 90 96 18. Complex Number - V 98 104 19. Complex Number - VI 106 112 20. Complex Number - VII 114 122 21. Complex Number And Quadratic Equations - I 125 131 22. Complex Number And Quadratic Equations - II 133 135
5 Practice Questions
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- II
23. Complex Number And Quadratic Equations 139 141 24. Complex Number And Quadratic Equations - IV 144 147 25. Sequences And Series - I 149 154 26 Sequences And Series - II 156 161 27. Sequences And Series - III 164 169 28. Sequences And Series - IV 171 176 29. Sequences And Series - V 179 188 30. Sequences And Series - VI 191 198 31. Sequences And Series - VII 200 206 32. Sequences And Series - VIII 209 215 33. Permutation And Combinations - I 217 220 34. Permutation And Combinations -II 222 225 35. Permutation And Combinations - III 227 230 36. Permutation And Combinations - IV 233 235 37. Binomial Theorem - I 237 245 38. Binomial Theorem - II 247 254 39. Binomial Theorem - III 256 260 40. Binomial Theorem - IV 262 266 41. Trigonometry - I 268 276 42. Trigonometry - II 282 2843. Trigonometry - III 28 29644. Trigonometry Equations - I 300 345. 31 3146. 32
6 9
Trigonometry Equations - II 08 3
Trigonometry Equation - III 6 3
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- III
32
47. 332 3
48. Trigonometry Triangles - II 34 34
49. Trigonometry Triangles - III 3 3
50. Trigonometry Functions - I 3 3
51. Trigonometry Functions - II 3 3
52. Trigonometry Functions - III 3 3
53.
4
Trigonometry Equations - IV 6
Trigonometry Triangles - I 40
3 7
51 61 64 76 79 84 86 98
MathematicalReasoning 401409 Statistics
16
Practice Questions 329
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54Practice Questions
1
ELLIPSE - I
Topics covered1. Definition of an Ellipse2. Standard Equation of an ellipse3 Basic terminology of ellipse4 Comparison of standard equation of an ellipse when a>b and b<a.1 Definition of an Ellipse
An ellipse is the set of all points in a plane, the sum of whosedistances from two fixed points in the plane is a constant. Fixedpoints are called focus.
PF1+PF2 = constant2 Standard equation of an ellipse.
2
2
ax
+ 2
2
by
= 1 with a>b.
Where b2=a2(1–e2)(i) The line containing the two fixed points (foci)
is called focal axis (majoraxis) and points ofintersection of the curve with focal axis arecalled the vertices of the ellipse ie. A(a,0) andA (–a,0).The distance between F1 and F2 is called thefocal length. F1F2 = 2ae.The distance between the vertices is A A =2a is called major axis.The distance BB1 = 2b is called minor axis
(ii) Point of intersection of the major and minor axis is called the centre of the ellipse. Anychord of the ellipse passing through it gets bisected by it and is called diameter.
(iii) Any chord through focus is called a focal chord and any chord perpendicular to the focalaxis is called double ordinate DE.
(iv) A particular double ordinate through focus and perpendicular to focal axis is called its
latus rectum (L L ). Length of latus rectum = ab2 2
(v) M M and N N are two directrices of the ellipse and their equations are x = ea
and x
= – ea
respectively
(vi) A chord of the ellipse passing through its focus is called a focal chord.(vii) An ellipse is the locus of point which moves in a plane such that the ratio of its distance
from a fixed point (focus) to the fixed line (directrix) is less than 1. This ratio is calledeccentricity and is denoted by e. For an ellipse e<1.
F2 F1
F2
DB L
X
YMN
L1
(ae, 0)(–ae,0)A
(a, 0)E
F1
L1L B
)0,a( A
MN
eax
P
2
(viii) The equation to the ellipse whose focus is (h.k) and directrix is x+my+n = 0 and whose
ecc tricity e<1, is (x–h)2 +(y–k)2 = e2
2
22 m
nmyx.
(ix) Special form If the centre of the ellipse is at point (h, k) and the directions of the axis are
parallel to the co-ordinate axes, then its equation is 2
2
a)h–x(
+ 2
2
b)k–y(
= 1
If we shift the origin at (h,k) rotating totaling co-ordinate axes then equation of the
ellipse with respect to new origin becomes 2
2
aX
+ 2
2
bY
= 1
EXAMPLES1 The eccentricity of the ellipse 9x2+5y2 – 30y = 0 is equal to
(a) 1/3 (b) 2/3 (c) 3/4 (d) None of these.Solution : Here equation of ellipse is
9x2+5y2–30y = 09x2+5(y2–6y+9–9) = 09x2+5(y–3) – 45 = 09x2+5(y –3)2 = 45
5x2
+ 9
)3–y( 2 = 1
9>5 b>aa = 5 b = 3
Hence e = 2
2
ba–1
= 95–1
= 94
= 2/32 P is any point on the ellipse 81x2+144y2=1944 whose foci are S and S . Then SP+S P equals.
(a) (b) 3 6 (c) 36 (d) 324Solution : Here equation of ellipse is
81x2+144y2 =1944
811944x2
+ 144
1944y2
= 1
en
2
4 6
3
24x2
+ 2
27y2
= 1
24> 227
a>b a = 2 6SP+S P = 2a =4 6
1 The eccentricity of ellipse if length of latus rectum is one-third of major axis
(a) 2/3 (b) 32 (c) (d)
4
43
2 The curve represented by x = 3(cost+sint) y = 4(cost–sint) is(a) Ellipse (b) Parabola (c) Hyperbola (d) Circle
3 The foci of the ellipse 25(x+1)2+9(y+2)2 = 225 are at(a) (–1,2)&(–1,6) (b) (–2,1)&(–2,6)(c) (–1,–2)&(–2,–1) (d) (–1,–2)&(–1,–6)
4 The equation –2x2
+ 5–
2
+1 = 0 represents an ellipse, if
(a) <5 (b) <2 (c) 2< <5 (d) <2or >5
5 The sum of the focal distances of any point on the ellipse 9x2+16y2 = 144 is(a) 32 (b) 18 (c) 16 (d) 8
6 The latus rectum of the conic 3x2+4y2–6x+8y–5 = 0
(a) 3 (b)23
(c) 32
(d) None of these
7 The centre of the ellipse 9
)2–yx( 2
+ 16
)y–x( 2
= 1 is
(a) (0,0) (b) (1,1) (c) (1,0) (d) (0,1)
8 In an ellipse the distance between its foci is 6 and its minor axis 8. Then its eccentricity is
(a) 4/5 (b) 521
(c) 3/5 (d) 1/2
y
56
PRACTICE QUESTIONS
4
9 For the ellipse x2+4y2 = 9
(a) The eccentricity is 21
(b) The latus rectum is 23
(c) a focus is 0,33 (d) a directrix is x 32–
10 The difference between the lengths of the major axis and the latus rectum of an ellipse is(a) ae (b) 2ae (c) ae2 (d) 2ae2
Answers1. b 2. a 3. a 4. c 5. d6. a 7. b 8. c 9. b 10. d
5
1 The equation a–10
x 2 +
a–4y2
= 1 represents an ellipse if
(a) a<4 (b) a>4 (c) 4<a<10 (d) a>10Solution : Here equation of ellipse is
a–10x 2
+ a–4
y2 = 1
a2 = 10–a and b2 = 4–a10–a>0 and 4–a>0
10>a and 4>a a<4.
2 The radius of the circle passing through the foci of the ellipse 16x2
+ 9
y2 = 1 and having its
centre (0,3) is(a) 4 (b) 3 (c) 12 (d) 7/2Solution ;
16x2
+ 9
y2 = 1
a2 = 16 b2 = 9a = 4 b = 3 a>b
e = 2
2
ab–1
= 169–1
= 167
= 47
Foci = ( ae,0) foci is 0,7
Radius of the circle through foci & centre (0,3) is 22 3)7( = 97 = 16 = 4.
3 The equation of the ellipse whose focus is (1,–1) directrix x–y–3 = 0 and eccentricity 21
is
(a) 7x2+2xy+7y2–10x+10y+7=0 (b) 7x2+2xy+7y2+7=0(c) 7x2+2xy+7y2+10x–10y–7=0 (d) None of these
ELLIPSE - II
6
Solution : Let P(x,y) be any point on the ellipseThen by definition SP = ePM.
22 )1y()1–x( = 21
23–y–x
(x–1)2+(y+1)2 = 81
(x2+y2+9–2xy+6y–6x)
8x2–16x+8+8y2+16y+8 = x2+y2–2xy+6y–6x+97x2+7y2+2xy–10x+10y+7 = 0
4 The equation (5x–1)2+(5y–2)2 = ( 2–2 +1) (3x+4y–1)2 represents an ellipse if .(a) (0,1) (b) (0,2) –{1} (c) (1,2) (d) (–1,0)Solution :(5x–1)2+(5y–2)2 = ( 2– +1)(3x+4y–1)2
252
51–x + 25
2
52–y = ( –1)2 (3x+4y–1)2
2
51–x +
2
52–y = ( –1)2
25)1–y4x3( 2
Sp2 = e2 (PM)2
e = | –1|In ellipse 0<e<1
0<| –1|<10< <2 –{1}
(0,2) –{1}
5 The eccentricity of an ellipse 2
2
ax
+ 2
2
by
= 1 whose latus rectum is half of its major axis is
(a) 21
(b)32
(c)23
(d) None of these
Solution:
2
2
ax
+ 2
2
by
= 1 a>b
major axis = 2a
latus rectum is =ab2 2
= a
7
According to question ; ab2 2
= a
2b2 = a2
eccentricity e = 2
2
ab–1
= 21–1
= 21
6 If (5,12) and (24,7) are the foci of an ellipse passing through the origin, then the eccentricity of theconic is
(a)12386
(b)13386
(c)25386
(d)38386
Solution :S(5,12), S (24,7)
SP = 22 125 = 13
S P = 22 724 = 25
SS = 2ae = 22 519
= 25361= 386
ae = 2386
SP+S P = 2a = 13+252a= 38a = 19
e = PSSPSS
= a2ae2
= 38386
7 Locus of the point which divides double ordinate of the ellipse 2
2
ax
+ 2
2
by
= 1 in the ratio 1:2
internally is
S
P(0,0)
(5,12) (24,7)S
8
(a) 2
2
ax
+ 2
2
by9
= 1 (b) 2
2
ax
+ 2
2
by9
= 91
(c) 2
2
ax9
+ 2
2
by9
= 1 (d) None of these.
Solution :Let P(h,k) be a point divides double ordinate in the ratio 1:2 internally
Let coordinates of ends of double ordinate (h,y1) and (h,–y1).
By section formula k = 3y2y– 11 =
3y1
y1 = 3kNow the point (h,y1) = (h,3k) lies on the ellipse
2
2
ah
+ 2
2
b9
= 1 or 2
2
ax
+ 2
2
by9
= 1 ( (h,k) are arbitrary)
8 If C is the centre of the ellipse 9x2+16y2 = 144 and S is one focus. The ratio of CS to semi majoraxis is(a) 7 :16 (b) 7 : 4 (c) 5 : 7 (d) None of theseSolution : Here equation of ellipse is9x2+16y2 = 144
16x2
+ 9
y2 = 1
16>9 a>b
e = 2
2
ab–1
= 169–1
= 167
= 47
O
B
Y
XP(h, k)
(h , )1 –y1
(h , )1 1y
h2
k
e
9
Foci is ( ae,0) = ( 7 ,0)
Semi major axis is = 2a2
= a = 4
CS = 7 and semi major axis is 4.
Required ratio is 7 :4.
1 The equation of the circle drawn with the two foci of 2
2
ax
+ 2
2
by
= 1 as the end points of a
diameter is(a) x2+y2=a2+b2 (b) x2+y2 = a2 (c) x2+y2=2a2 (d) x2+y2 = a2–b2
2 The radius of the circle passing through the foci of the ellipse 16x2
+ 7y2
= 1 and having its
centre (0,3) is
(a) (b) 3 (c) 12 (d) 27
36–r–r
x2
2
+ 5r6–r
y2
2
= 1 will represents the ellipse, if r lies in the interval
(a) (– ,2) (b) (3, ) (c) (5, ) (d) (1, )
4 The semi latus rectum of an ellipse is(a) The AM of the segments of its focal chord.(b) The GM of the segments of its focal chord(c) The HM of the segments of its focal chord(d) None of these
5 The following equation represents an ellipse 25(x2–6x+9) + 16y2 = 400. How should the axes be
transformed so that the ellipse is represented by the equation 16x2
+ 25y2
= 1______
6 Let P be a variable point on the ellipse 16x2
+ 25y2
= 1 with foci S1 and S2. It A be area of the
triangle PS1S2 then the maximum value of A is __________
7 In an ellipse, if the lines joining a focus to the extremities of the minor axis make an equilateraltriangle with the minor axis, the eccentricity of the ellipse is(a) 3/4 (b) 2/3 (c) 1/2 (d) 2/3
3 2
PRACTICE QUESTIONS
10
8 Column Matching :
For the ellipse 5x2
+ 4y2
= 1
Column I Column II1 x = 0 a a directrix2 y = 0 b a latus rectum3 x = 1 c minor axis4 x = 5 d major axis
The centre of the ellipse 14x2–4xy+11y2–44x–58y+71 = 0 is _______
Answer
1. d 2. a 3. c 4. c 5. (3,0) 6. 127. b 8. (1) c (2) d (3) b (4) a
9
9. (3,1)
11
ELLIPSE -III
Topics covered1. Auxilliary circle2. Eccentric angle3 Equation of chord4 Position of a point with respect to an ellipse.
Auxiliary CircleThe circle described on the major axis of an ellipse as diameter is called an auxiliary circle ofthe ellipse
If 2
2
ax
+ 2
2
by
= 1 is an ellipse then its auxiliary circle is x2+y2=a2
Eccentric angle of a point
Let P be any point on the ellipse 2
2
ax
+ 2
2
by
= 1
Draw PM perpendicular to major axis from P andproduce MP to meet the auxiliary circle at Q. Join CQ.
QCA = is called eccentric angle of point PNote that the angle ACP is not eccentric angle.i.e. eccentric angle of P on an ellipse is the angle whichthe radius through the corresponding point on, theauxiliary circle makes with the major axis
Q (acos , asin ) x-coordinate of P is acos
2
22
acosa
+ 2
2
by
= 1
2
2
by
= 1– cos2
y2 = b2 sin2
y = b sinCoordinate of P is (acos , bsin )
i.e. x = acos and y = bsin is the parameter equations of the ellipse.(acos ,bsin ) is also called the point ‘ ’
Equation of the chord
O
Q Auxiliarycircle
ellipse
M A
x +y =a22 2
C
B
Y
X
P(x, y)
B
Y
AX
1by
ax
2
2
2
2
P
12
Let P (acos ,bsin ) and Q(acos ,bsin ) be any two points of the ellipse 2
2
ax
+ 2
2
by
= 1 then the
equation of the chord joining these two points is
y– bsin = cosa–cosasinb–sinb
(x–acos )
Simplifying the equation we get
2–cos
2sin
by
2cos
ax
& are eccentric angle of points P and Q of ellipsePosition of a point (h,k) with respect to an ellipse
Let ellipse be 2
2
ax
+ 2
2
by
= 1
Now P will lie outside, on or inside the ellipse 2
2
ax
+ 2
2
by
= 1 according as
2
2
ah
+ 2
2
bk
–1>,=,<0
EXAMPLES1 Find the equation of the curve whose parametric equation are x=1+4cos , y = 2+3sin
RSolution: We havex=1+4cos , y = 2+3sin
41–x
= cos and 32–y
= sin
Squaring and adding we get2
41–x
+ 2
32–y
= cos2 + sin2
16)1–x( 2
+ 9
)2–y( 2
= 1
Which is an ellipse.
2 Find the eccentric angle of a point on the ellipse 6x2
+ 2y2
= 1 whose distance from the
centre of the ellipse is 5Solution :
We have 6x2
+ 2y2
= 1
13
a2 = 6 b2 = 2a = 6 b = 2
any point on the ellipse with as eccentric angle is P sin2,cos6Here centre is origin
CP = 22 sin2cos6 = 5 6cos2 +2sin2 = 5 4cos2 = 3
cos2 = 43
cos =23
= 6 , 65
, 67
, 63 If and are the eccentric angles of the extremities of a focal chord of an ellipse, then the
eccentricity of the ellipse is
(a) )–cos(coscos
(b) )–sin(sin–sin
(c) )–cos(cos–cos
(d) )sin(sinsin
Solution : Equation of chord joining points having eccentric angles and is
ax
cos 2 + by
sin 2 = cos 2–
Since these points are extremities of focal chord so it passes through focus (ae,0) then
e cos 2 = cos 2–
e =2
cos
2–cos
Multiply & divide by 2sin 2 on right side
e =2
cos2
sin2
2–cos
2sin2
11
14
e = )sin(sinsin
4 An ellipse passes through the point (4,–1) and touches the line x+4y–10 = 0. Find its equationof its axes coincide with coordinate axes.
Solution: Let the equation of ellipse be 2
2
ax
+ 2
2
by
= 1
It passes through (4,–1)
2a16
+ 2b1
= 1 or a2+16b2 = a2b2......................(1)
x+4y–10 = 0 is a tangent to the ellipse.
y = – 41
x + 410
y= mx+c
m = – 41
, c = 410
c = 222 bma is a condition for tangent
410
= 22 b161a
16100
= 16a 2
+ b2
16b2 = 100–a2
a2+16b2 = 100From (1) we get
100 = a2b2
b2 = 2a100
a2+ 2a1600
= 100
a4–100a2+1600 = 0a4–80a2–20a2+1600 = 0a2(a2–80)–20(a2–80) = 0(a2–80)(a2–20) = 0
b2 = 810
= 45
or b2 20100
= 5
15
Equation of ellipse is80x2
+5y4 2
= 1
or20x2
+5y2
= 1
5 If ax
+ by
= 2 touches the ellipse 2
2
ax
+ 2
2
by
= 1, then find its eccentric angle of point of
contact.Solution : Let be the eccentric angle of the point of contact :
coordinates of the point is (acos , bsin )Equation of tangent at this point is
acosx
+ bsiny
–1 = 0 ...................(1)
Given that ax
+ by
– 2 = 0 .....................(2) is tangent
Comparing (1) and (2) as these two are identical, we get
a1a
cos
=
b1b
sin
= 2–1–
cos = 21
= sin
= 4
1 The sum of the squares of the reciprocals of two perpendicular diameter of an ellipse is
(a) 41
22 b1
a1
(b) 21
22 b1
a1
(c) 22 b1
a1
(d) None of these
2 Prove that any point on the ellipse whose foci are (–1,0) and (7,0) and eccentcicity 21
is
(3+8cos , 4 3 sin ), R. Also nd the eq of the ellipse
3 Let E be the ellipse 9x2
+ 4y2
= 1 and C be the circle x2+y2 = 9. Let P and Q be the points
(1,2) and (2,1) respectively. Then(a) Q lies inside C but outside E(b) Q lies outside both C and E
fi n
PRACTICE QUESTIONS
16
(c) P lies inside both C and E(d) P lies inside C but outside E
4 P is a variable on the ellipse 2
2
ax
+ 2
2
by
= 1 with AA A as the major axis. Then the maximum
area of the triangle AP A is(a) ab (b) 2ab (c) ab/2 (d) None of these
5 A man running round a race course notes that the sum of the distances of two flag-posts from himis always 10m and the distance between the flag-posts is 8m. The area of the path he encloses insquare meters is(a) 15 (b) 12 (c) 18 (d) 8
6 If the line x+my+n = 0 cuts the ellipse 2
2
ax
+ 25y2
= 1 in points whose eccentric angles
differ by 2 then 2
2222
nmba
(a) 1 (b) 2 (c) 4 (d) 3/2
7 If PSQ is a focal chord if the ellipse 16x2+25y2 = 400 such that SP = 8, then SQ =(a) 1 (b) 2 (c) 3 (d) 4
8 If equation of the ellipse is 2x2+3y2–8x+6y+5 = 0 then which of the following are true?(a) equation of director circle is x2+y2–4x+2y = 10(b) director circle will pass through (4, –1)(c) equation of auxillary circle is x2+y2–4x+2y+2 = 0(d) None of these
9 The foci of ellipse 2
5x
+ 2
3y
= 1 are S and S . P is a point on ellipse whose eccentric
angle is /3. The incentre of triangle SPS is
(a) 3,2 (b) 32,2 (c) 2
3,2 (d) 2,3
Answers
1. a 2. 64
)3–x( 2
+ 48y2
= 1 3. d 4. a 5. a 6. b 7. b 8. c 9. b
17
ELLIPSE - IV
Intersection of a line and an ellipse
Line y = mx+c_______(1) and ellipse 2
2
ax
+ 2
2
by
= 1 _________(2)
Solving equations (1) & (2) we get(a2m2+b2)x2 +2a2|cmx+a2(c2–b2) = 0If D>0 then y = mx+c is a secant
D = 0 then y = mx+c is a tangentD<0 y = mx+c does not meet ellipse
Point formEquation of tangent to the ellipse at point (x1, y1)
Let the equation of ellipse be 2
2
ax
+ 2
2
by
= 1
Then equation of tangent in point form is 21
axx
+ 21
byy
= 1
Parametric form
Equation of tangent at point (acos ,bsin ) to the ellipse is acosx
+ bsiny
= 1
Slope form
y = mx 222 bma is a tangent to an ellipse 2
2
ax
+ 2
2
by
= 1 and point of contact is
222
2
222
2
bmab,
bmama
Number of tangents through a given point P (h,k)
y = mx+ 222 bma is any tangent to the ellipse 2
2
ax
+ 2
2
by
= 1
If it passes through P (h,k) then
k = mh+ 222 bma
k–mh = 222 bma(k–mh)2 = a2m2+b2
m2(h2–a2) – 2hkm+(k2–b2) = 0It is a quadratic in m and will give two values of m hence there are two tangents.
18
Examples1 If the line 3x+4y = 7 touches the ellipse 3x2+4y2 = 1, then the point of contact is
(a) 71,
71
(b) 31–,
31
(c) 71–,
71
(d) None of these
Solution : (a)Let P (x1,y1) be point of contact the equation of tangent to the ellipse
31
x2 +
41
y2
= 1 is
31
xx1 +
41
yy1 = 1
3xx1+4yy1 1=0 __________(1)Given that 3x+4x– 7 = 0 ________________(2) touches the ellipse
(1) and (2) are sameBy comparing we get
3x3 1 =
4y4 1 = 7
1–
x = 71
, y1 = 71
71,
71
is the point of contact
Example 2: The number of values of c such that the line y = 4x+c touches the curve 4
x2 + y2 = 1 is
(a) 0 (b) 1 (c) 2 (d) infinite.
Solution : Given ellipse is 4
x2 +
1y2
a2 = 4 b2 = 1and a line y = 4x+c is a tangent
m = 4
c = 222 bma
= 1164= 65
c has 2 values c = 65 or 65–
3 If 3 bx+ay =2ab touches the ellipse 2
2
ax
+ 2
2
by
= 1 then the eccentric angle of the point of contact
–
–
19
is
(a) 6 (b) 4 (c) 3 (d) 2Solution : (a)
Equation of tangent ax
23
+ by
21
= 1____________(1)
and equation of tangent at the point (acos , bsin ) is ax
cos + by
sin = 1________(2)
comparing (1) & (2) we get
cos = 23
and sin = 21
tan = 31
= tan 6
= 6
4 A tangent having slope of 34–
to the ellipse 18x2
+ 32y2
= 1 intersects the major and minor axes
at points A and B respectively. If C is the centre of the ellipses, then the area of the triangleABC is(a) 12sq.u (b) 24sq.u (c) 36sq.u (d) 48sq.uSolution : (b)
Equation of tangent to the ellipse 2
2
ax
+ 2
2
by
= 1 is
y = mx+ 222 bma (b>a)
Here m = 34–
, a2 = 18, b2 = 32
y = 34–
x + 329
1618
y = 34–
x + 8
Then points on the axis where tangents meet are A(6,0) and B(0,8)
Then area of ABC is 21
×6×8 = 24sq.u
20
5 If the tangents to the ellipse 2
2
ax
+ 2
2
by
= 1 make angle and with the major axis such that
tan +tan = , then the locus of their point intersection is(a) x2+y2 = a2 (b) x2+y2 = b2 (c) x2–a2 = 2 xy (d) (x2–a2) = 2xySolution : (d)Equation of tangent to the ellipse with slope m is
y = mx+ 222 bmaIf it is passes through the point P (h,k) then
k = mh+ 222 bma
or (k–mh)2 = 222 bmak2+m2h2–2mkh = a2m2+b2
m2(h2–a2)–2mkh+k2–b2 = 0It is a quadratic in m having two roots m1 & m2
m1+m2 = 22 a–hkh2
and m1m2 = 22
22
a–hb–k
Given that tan +tan = m1+m2 =
22 a–hkh2
=
2kh = (h2–a2) locus of point P(h,k) is
(x2–a2) = 2xy
1 If P(x,y), F1(3,0), F2(–3,0) and 16x2+25y2 = 400, then PF1+PF2 equals(a) 8 (b) 6 (c) 10 (d) 12
2 The length of the major axis of the ellipse
(5x–10)2 + (5y+15)2 = 4
)7y4–x3( 2 is
(a) 10 (b) 320
(c) 720
(d) 4
3 Angle subtended by common tangents of two ellipses 4(x–4)2+25y2 = 100 and 4(x+1)2+y2 = 4at origin is
(a) 3 (b) 4 (c) 6 (d) 2
PRACTICE QUESTIONS
21
4 The distance of a point on the ellipse 6
x2 +
2y2
= 1 from the centre is 2. Then the eccentric angle
of the point is
(a) 4 (b) 43
(c) 65
(d) 6
5 If the chord through the points whose eccentric angles are and on the ellipse 25x2
+ 9
y2=1
passes through a focus, then the value of tan 2 tan 2 is
(a) 91
(b) –9 (c) 91–
(d) 9
6 In an ellipse the distance between its foci is 6 and its minor axis is 8, the eccentricity of theellipse is
(a) 54
(b) 53
(c) 521
(d) 21
7 The number of values of C such that the straight line y = 4x+c touches the curve 4
x2 + y2 = 1,
is(a) 0 (b) 2 (c) 1 (d)
8 The line 3x+5y = 15 2 is a tangent to the ellipse 25x2
+ 9
y2 = 1, at a point whose eccentric
angle is
(a) 6 (b) 4 (c) 3 (d) 32
9 Tangents are drawn to the ellipse 3x2+5y2 = 32 and 25x2+9y2 = 450 passing through the point(3,5). The number of such tangents are(a) 2 (b) 3 (c) 4 (d) 0
10 Tangents are drawn to the ellipse 9
x2 +
5y2
= 1 at ends of latus rectum. The area of quadrilateral
so formed is
(a) 27 (b) 227
(c) 427
(d) 5527
11 An ellipse passes through the point (4,–1) and its axes are along the axes of coordinates. If theline x+4y–10 = 0 is a tangent to it then its equation is
(a)100x2
+ 5y2
=1 (b)8x2
+4/5
y2
=1 (c)20x2
+ 5y2
=1 (d) None of these
22
12 Prove that the line 2x+3y = 12 touches the ellipse9x2
+ 4y2
= 2
13 The tangent at the point sin11
16,cos4 to the ellipse 16x2+11y2 = 256 is also a tangent to the
circle x2+y2 –2x = 15, find the value of .
14 Find the equations of tangents to the ellipse 9x2+16y2 = 144 which pass through the point (2,3).
15 The angle between pair of tangents drawn to the ellipse 3x2+2y2 = 5 from the point (1,2) is tan1(12/
5 )
16 Prove that the portion of the tangent to the ellipse intercepted between the curve and the directrixsubtends a right angle at the corresponding focus.
17 Linked Comprehension Type.
For all real p, the line 2px+y 2p–1 = 1 touches a fixed ellipse whose axes are coordinateaxes.(i) The eccentricity of the ellipse is
(a) 32
(b)23
(c) 31
(d) 21
(ii) The foci of ellipse are
(a) 3,0 (b) 32,0 (c) 0,2
3 (d) None of these
(iii) The locus of point of intersection of perpendicular tangents is(a) x2+y2 = 5/4 (b) x2+y2 = 3/2 (c) x2+y2 = 2 (d) None of these
18 C1 : x2+y2 = r2 and C2 =
16x2
+ 9y2
= 1 intersect at four distinct points A, B, C and D, Their
common tangents form a parallelogram A B C D .(i) If ABCD is a square then r is equal to
(a) 25
12(b) 5
12(c) 55
12(d) None of these
(ii) If A B C D is a square then r is equal to
(a) 20 (b) 12 (c) 15 (d) None of these(iii) If A B C D is a square, then the ratio of area of the circle C1 to the area of the
circumcircle of A B C is
(a) 169
(b) 43
(c) 21
(d) None of these
23
19 The ellipse 2
2
ax
+ 2
2
by
= 1 is such that it has the least area but contains the circle (x–1)2+y2 = 1
(i) The eccentricity of the ellipse is
(a)32
(b) 31
(c) 21
(d) None of these
(ii) Equation of auxilliary circle of ellipse is(a) x2+y4 = 6.5 (b) x2+y4 = 5 (c) x2+y4 = 45 (d) None of these(iii) Length of latus rectum of the ellipse is(a) 2 units (b) 1unit (c) 3units (d) 2.5 units
20 The equation of the straight lines joining the foci of the ellipse 25x2
+ 16y2
= 1 to the foci of the
ellipse 24x2
+ 49y2
= 1 forms a parallelogram. Then the area of the figure formed by the foci of
these two ellipse.(a) 15 (b) 30 (c) 20 (d) 18
ANSWERS
1. c 2. b 3. b 4. a,b 5. c,d 6. b 7. b 8. b
9. b 10. a 11. b,c 13. 6 14. y = 3, x+y = 5
17. (i) a (ii) d (iii) a 18. (i) a (ii) d (iii) c 19. (i) a (ii) c (iii) b20. b
24
ELLIPSE - V
Equations of Normals in different forms(i) Point form
Let equation of ellipse be 2
2
ax
+ 2
2
by
= 1
Equation of normal is 1
2
xxa
– 1
2
yyb
= a2–b2
(ii) Parametric form
Let equation of ellipse be 2
2
ax
+ 2
2
by
= 1
Equation of normal is axsec – bycosec = a2–b2
(acos , bsin ) is parametric coordinates of P(x1,y1), i.e. P(acos(iii) Slope form
For an ellipse 2
2
ax
+ 2
2
by
= 1
Equation of normal in terms of slope (m) is
y = mx 222
22
mba
)b–a(m
Condition of normality when y = mx+c is the normal of 2
2
ax
+ 2
2
by
= 1 is
c2 = 222
2222
mba)b–a(m
or c = 222
22
mba
)b–a(m
Examples
1 If the normal at the point P( ) to the ellipse 14x2
+ 5y
= 1 intersects it again at the point Q (2 )
then cos is equal to
(a) 32
(b) – 32
(c) 23
(d) – 23
Solution :
, bsin )
25
The normal at (asin , bsin ) is cosax
– sinby
= a2 –b2
Here a2 = 14 & b2 = 5a = 14 b = 5
cosx14
– sin
y5 = 14 – 5
cosx14
– sin
y5 = 9 _______________(1)
It meets the curve again at (2 ) (acos2 ,bsin2 )
i.e. 2sin5,2cos14 equation (1) satisfy this point
cos2cos14
– sin2sin5
= 9
cos)1–cos2(14 2
– sin = 9
28cos2 – 14 –10cos2 = 9cos18cos2 –9cos –14 = 0(6cos –7) (3cos +2) = 0
cos = 67
> 1not possible ( cos <1)
cos = 32–
option (b) is correct.
2 The equation of the normal to the ellipse 2
2
ax
+ 2
2
by
= 1 at the end of the latus rectum in the first
quadrant is(a) x+ey–ae3 = 0 (b) x–ey+ae3 = 0 (c) x–ey–ae3 = 0 (d) None of theseSolution :
The end of the latus rectum in the first quadrantal is ab,ae
2
Equation of normal at ab,ae
2
is
( cossin5 2 )
26
aexa 2
– ab
yb2
2
= a2–b2 22
1
2
1
2
b–ayyb–x
xa
eax
– ay = a2–b2
ax–eay = ea2–eb2
= ea2–e(a2–a2e2)=ea2–ea2+a2e3
ax–aey–a2e3 = 0 x–ey–ae3 = 0
Correct option is c
3 The condition that the line xcos +ysin = p may be a normal to the ellipse 2
2
ax
+ 2
2
by
= 1then
(a) (a2–b2)2 = p2 (a2sec2 +b2cosec2 ) (b) (a2–b2)2 = p2 (a2cosec2 +b2sec2 )(c) (a2+b2)2 = p2 (a2sec2 +b2cosec2 ) (d) None of theseSolution :
The equation of normal to 2
2
ax
+ 2
2
by
= 1 at ‘ ’ is
cosaxa2
– sinb
yb2
– (a2–b2) = 0 ____________(1)
Given that xcos +ysin –p = 0 is normal to the ellipse comparing (1) & (2) we get
coscos/a
= sinb–
= pb–a 22
or coscosa
= sinsinb–
= pb–a 22
or acoscos
= – bsinsin
= 22 b–ap
cos = cos)b–a(ap
22 sin = sin)b–a(bp–22
Squaring and adding we get
cos2 +sin2 = 2222
22
cos)b–a(pa
+ 2222
22
sin)b–a(pb
(a2–b2)2 = p2(a2sec2 +b2cosc2 ) correct option is a
4 If the normals at P(x ,y ), Q(x ,y ) and R(x ,y ) to the ellipse are concurrent, then1 1 2 2 3 3
/sin
,
,
27
(a)3333
1111
2222
yxyxyxyxyxyx
= –1 (b)3333
2222
1111
yxyxyxyxyxyx
= 0
(c)1111
3333
2222
yxyxyxyxyxyx
= 1 (d) None of these
Solution :
The equations of the normals at P(x1,y1),Q(x2,y2) and R(x3,y3) to the ellipse 2
2
ax
+ 2
2
by
= 1 are
1
2
xxa
– 1
2
yyb
= a2 – b2 ______________(1)
2
2
xxa
– 2
2
yyb
= a2 – b2 ______________(2)
3
2
xxa
– 3
2
yyb
= a2 – b2 ______________(3)
respectivelyThese lines are concurrent, if
22
3
2
3
2
22
2
2
2
2
22
1
2
1
2
b–ayb–
xa
b–ayb–
xa
b–ayb–
xa
= 0
a2b2(a2–b2)1
y1–
x1
1y
1–x1
1y
1–x1
33
22
11
= 0
R1 x1y1R1 ; R2 x2y2R2 ; R3 x3y3 we get
28
3333
2222
1111
yxx–yyxx–yyxx–y
= 0
OR3333
2222
1111
yxyxyxyxyxyx
= 0
your correct option is ‘b’
1 In the normal at the end of latus rectum of the ellipse 2
2
ax
+ 2
2
by
= 1 with eccentricity e, passes
through one end of the minor axis, then :(a) e2(1+e2) = 0 (b) e2(1+e2) = 1 (c) e2(1+e2) = – 1 (d) e2(1+e2) = 2
2 If the normals to 2
2
ax
+ 2
2
by
= 1 at the ends of the chords 1 x+m1y =1 and 2 x+m2y =1 are
concurrent, then :(a) a2
1 2+b2m1m2 =1 (b) a21 2+b2m1m2 = –1
(c) a21 2–b2m1m2 = – 1 (d) None of these
3 If the normal at an end of a latus rectum of an ellipse passes through one extremity of the minoraxis, then the eccentricity of the ellipse is given by(a) e4+e2–1 = 0 (b) e4+e2–5 = 0 (c) e3 = 5 (d) None of these
4 The number of normals that can be dr wn from a point to a given ellipse is(a) 2 (b) 3 (c) 4 (d) 1
5 If the normal at any point P on the ellipse 2
2
ax
+ 2
2
by
= 1 meets the axes in G and g respectively,,
then PG :Pg is equal to(a) a : b (b) a2 : b2 (c) b2 : a2 (d) b : a
6 If normal to ellipse 2
2
ax
+ 2
2
by
= 1 at ab,ae
2
is passing through (0,–2b), then value of
eccentricity is
(a) 2 –1 (b) 2 1–2 (c) 1–22 (d) None of these
7 If normal at any point P to the ellipse 2
2
ax
+ 2
2
by
= 1 , a>b meet the axes at M and N so that PNPM
a
PRACTICE QUESTIONS
29
= 32
, then value of eccentricity e is
(a) 21
(b)32
(c) 31
(d) None of these
8 If the tangent drown at point (t2,2t) on the parabola y2 = 4x is same as the normal drawn at point( 5 cos , 2sin ) on the ellipse 4x2+5y2 = 20. Then the values of t and are
(a) = cos–151–
& t = 51–
(b) = cos–15
1 & t = 5
1
(c) = cos–152–
& t = 52–
(d) None of these
9 The normals at four points on the ellipse 2
2
ax
+ 2
2
by
= 1 meet in the point (h,k). Then the mean
position of the four points is
(a) )ba(2kb,
)ba(2ha
22
2
22
2
(b) )ba(2kb,
)ba(2ha
22
3
22
3
(c) )b–a(2bk,
)b–a(2ah
2222 (d) )b–a(2kb,
)b–a(2ha
22
2
22
2
10 The equation of the normal at the point (2,3) on the ellipse 9x2+16y2 = 180 is(a) 3y = 8x–10 (b) 3y–8x+7 = 0 (c) 8y+3x+7 = 0 (d) 3x+2y+7 = 0
11 Number of distinct normal lines that can be drawn to the ellipse 169x2
+ 25y2
= 1 from the point P
(0,6) is(a) one (b) two (c) three (d) four
12 Any ordinate MP of the ellipse 25x2
+ 9y2
= 1 meets the auxiliary circle at Q, then locus of the point
of intersection of normals at P and Q to the respective curve is(a) x2+y2 = 8 (b) x2+y2 = 34 (c) x2+y2 = 64 (d) x2+y2 = 15
13 If the normals at P ( ) and Q 2 to the ellipse 2
2
ax
+ 2
2
by
= 1 meet the major axis at G and g
respectively, then PG2+Qg2 =(a) b2(1–e2)(2–e2) (b) a2(e4–e2+2) (c) a2(1+e2)(2+e2) (d) b2(1+e2)(2+e2)
Answers1. b 2 b 3. a 4. c 5. c 6. c 7. c 8. a9. d 10. b 11. c 12. c 13. b
30
HYPERBOLA - IEquation of Hyperbola
Definition 1 : The locus of a point in a plane, the difference of whose distance from two fixed pointsin the plane is constant.According to definition
PF2–PF1 = constant
F2(Focus) Vertex Vertex
Centre
Conjugate axis
Transverse axis
F1
A
Two fixed points are known as Foci of the hyperbola.The mid point of the line segment joining the foci is called the centre. The line joining the vertices isknown as transverse axis and the line through the centre and perpendicular to transverse axis isknown as conjugate axis. The point at which the hyperbola intersect the transverse axis is known asvertices of the hyperbola.Let distance between two foci be 2c and distance between two vertices be 2a and a quantity ‘b’ beb2=c2–a2.
Let a point P on the hyperbola be (x, y), then according to definition of hyperbola 22 y)cx( –
22 y)c–x( =2a
22 y)cx( =2a + 22 y)c–x(
(x+c)2+y2 = 4a2 + (x–c)2+y2+4a 22 y)c–x( cac2a2
4cx–4a2=4a 22 y)c–x(
acx
–a= 22 y)c–x(
2
22
axc
+a2–2cx=x2+c2–2cx+y2
2
22
aa–c
x2–y2=(c2–a2)
22
2
2
2
a–cy–
ax
=1 2
2
2
2
by–
ax
=1
2
2
2
2
by–
ax
=1
F2(–c, 0) ( c, 0)
(x, y)
XF1O
P
X
31
y= 22 a–xab
y is real when x2 a2
y is real when x a or x –aDefinition 2 : The locus of a point which moves in a plane such that the ratio of its distance from afixed point to its perpendicular distance from a fixed straight line (not passing through given fixedpoint) is always constant and greater than 1.
PMPS
=e > 1
The equation of hyperbola whose focus is the point (h, k) anddirectrix is ax+by+c=0 and whose eccentricity is e, is
(x–h)2+(y–k)2=e222
2
bacbyax
Example 1.
Equation of hyperbola whose focus is (1,2) equations of directrix is x+y+1=0 and eccentricity is 23
is
(a) x2+y2–18xy–y–31=0(b) x2+y2+18xy+34x+50y–31=0(c) x2+y2–18xy–x+y–31=0(d) x2+y2+18xy+x+y+31=0SolutionLet variable point P(x,y) then PS=ePM
22 )2–y()1–x( = 21yx
23
x2+y2–2x–4y+5= 89
(x2+y2+2xy+2x+2y+1)
x2+y2+18xy+34x+50y–31=0Ans. bEccentricity :
e = rticesbetween ve Distancefocibetween Distance
= a2c2
= ac
e2 = 2
2
ac
= 2
22
aba
=1+ 2
2
ab
a2e2=a2+b2
)1–e(ay–
ax
22
2
2
2
=1
M
ax+by+c=0
P(x,y)
S(x ,y )1 1
32
Some terms related to Hyperbola
Equation of hyperbola 2
2
2
2
by–
ax
=1
1. Centre : All chords passing through a point and bisected at that point is known as centre ofhyperbola C(0,0)
2. Eccentricity : e= rticesbetween ve Distancefocibetween Distance
e= 2
2
axis) e(transversaxis) (conjugate1
e= PMPS
, where S is focus, P is any point on the hyperbola, PM is distance from directrix.
3. Foci : S and S are foci whose coordinates are S(ae,0) and S (–ae, 0)
XS
Z1Z2
Y
ACX
S A
1Z2ZY
4. Directrices : Z1 1Z and Z2 2Z are the directrices whose equations are x= ea
and x=– ea
5. Vertices : A and A are the vertices of hyperbola . A(a,0) and A (–a,0).
6. Axes : The line A A’ is called transverse axis and the line perpendicular to its through the centreof the hyperbola is called conjugate axis. Equations of transverse axis is y=0 and equation ofconjugate axis is x=0.Length of transverse axis =2aLength of conjugate axis =2b
7. Double Ordinate : A chord of hyperbola which is perpendicular to transverse axis is known asdouble ordinate Q Q , Q(h, k), Q (h,–k).
N
Q
Q
8. Latus rectum : The double ordinates passing through focus is known as latus rectum.
33
L1(ae,ab2
), ab,–aeL
2
1
L2(–ae,ab2
), ab,–ae–L
2
2
9. Focal Chord : A chord passing through focus isknown as focal chord.
10. Focal Distance :
PS1=ePM1=e(x1– ea
)
= ex1–a
PS2=e(x1+ ea
)
= ex1+a PS2–PS1=2a
Rectangular or Equilateral HyperbolaIf a=b, then equation of hyperbola isx2–y2=a2 known as rectangular or equilateral hyperbola.The eccentricity of rectangular hyperbola is
e2=1+ 2
2
ab
=1+1
e= 2Equation of hyperbola if centre is (h, k) and axes are parallel to coordinate axes is
2
2
2
2
bk)–(y–
ah)–(x
=1
XS
L1L2
Y
XS
1L2L
Y
X
Z1Z2
S1(ae,o)
P(x ,y1 1)
S2
M2 M1
Y
OX
1Z2ZY
34
1. The eccentricity of the conic 9x2–16y2=144 is
a. 34
b. 54
c. 45
d. 17
2. The equation of the conic with focus at (1, –1) directrix along x–y+1=0 and eccentricity 2 isa. xy=1 b. 2xy+4x–4y–1=0c. x2–y2=1 d. 2xy–4x+4y+1=0
3. Equation of the hyperbola whose vertices are ( 3,0) and foci at ( 5,0) isa. 9x2–25y2=144 b. 16x2–9y2=144 c. 9x2–16y2=144 d. 25x2–9y2=225
4. The eccentricity of the hyperbola whose latus rectum is half of its transverse axis, is
a.23
b. 23
c. 32
d. 32
5. The differences of the focal distances of any point on the hyperbola is equal toa. eccentricity b. latus-rectumc. length of conjugate axis d. length of transverse axis
6. The distance between the foci of a hyperbola is 16 and its eccentricity is 2 , then equation ofthe hyperbola isa. x2–y2=32 b. x2–y2=16 c. y2–x2=16 d. y2–x2=32
7. The length of latus rectum of the hyperbola 3x2– 6y2
a. 24 b. 2 3 c. 3 2 d. 4 38. The eccentricity of the hyperbola 9x2–16y2–18x+32y–151=0 is
a. 35
b. 45
c. 25
d. 5
9. The foci of a hyperbola coincide with the foci of the ellipse 9y
25x 22
=1. If its eccentricity is 2
then equation of hyperbola is
a.4y–
12x 22
=1 b.12y–
8x 22
=1 c.12y–
4x 22
=1 d. x2–y2=4
10. The eccentricity of rectangular hyperbola isa. 2 b. 1 c. 3 d. 2
Answers1. c 2. d 3. b 4. a 5. d6. a 7. d 8. b 9. c 10 d
PRACTICE QUESTIONS
35
HYPERBOLA - IIConjugate Hyperbola
Position of a point : Let a point P(x1,y1) and equation of hyperbola be 2
2
2
2
by–
ax
–1=0
S1= 2
21
2
21
by–
ax
–1
If S1 > 0, point is inside the hyperbola.
If S1 = 0, point is on the hyperbola.
If S1 < 0, point is outside the hyperbola.
Conjugate Hyperbola : Corresponding to every hyperbola there exists a hyperbola such that thetransverse axis and conjugate axis of one is equal to the conjugate axis and transverse axis of theother. Such hyperbolas are known as conjugate to each other.
Therefore for the hyperbola 2
2
2
2
by–
ax
=1
Conjugate hyperbola is 2
2
2
2
by–
ax
Let e1 be the eccentricity of 2
2
2
2
by–
ax
=1 and e2 be
the eccentricity of 2
2
2
2
by–
ax
=–1 then e21=1+ 2
22
2
2
aba
ab
and e22=1+ 2
22
2
2
bab
ba
22
21 e
1e1
=1
The foci of a hyperbola and its conjugate hyperbola are concyclic and form the vertices of a square.Auxiliary circle and eccentric angleA circle drawn with centre O and transverse axis as diameteris known as auxiliary circle. Equation of auxiliary circle isx2+y2=a2
A is any point on the circle whose coordinates are (acos ,a sin ), where is known as eccentric angle. Now, In
OAB, OA=a
cos = OBa
OB=asec
X
R(x ,y )3 3
Q(x ,y )2 2
P(x ,y )1 1
Y
X
Y
XS2 S1O (ae,0)
(0, be)
(0,–be)
(–ae,0)
S3
S4
Y
X
Y
XB
P
O
(asec )btan
(acos ,asin )A
36
P point lies on hyperbola, so 2
2
2
22
by–
aseca
=1 tanby
P(asec , btan )0 < 2
The equations x=asec and y=btan represents a hyperbola. So, the parametric form of the hyperbola
2
2
2
2
by–
ax
=1 can be represented as x=asec ,y=btan
For the hyperbola 2
2
2
2
b)k–y(–
a)h–x(
=1, parametric form is x=h+asec ; y=k+btan .
Hyperbola Conjugate Hyperbola
Equation 1by–
ax
2
2
2
2
1by
ax– 2
2
2
2
Centre (0, 0) (0, 0)Vertice (a, 0) & (–a, 0) (0, b) & (0,–b)Foci (ae, 0) & (–ae, 0) (o, be) & (0, –be)Length of transverse axis 2a 2bLength of conjugate axis 2b 2a
Length of latus rectumab2 2
ba2 2
Equation of transverse axis y=0 x=0Equation of conjugate axis x=0 y=0
Equation of directrices x= ea
y= eb
Eccentricity e= 2
2
ab1 or b2=a2(e2–1) e= 2
2
ba1 or a2=b2(e2–1)
Example 1.
The point t2b–
2bt,
ata
2at
lies on the, [ for all values of t(t 0)]
(a) circle (b) parabola (c) ellipse (d) hyperbolaSolution
Let x= t2a
2at
and y= t2b–
2bt
t1t
ax2
and t1–t
by2
Squaring and subtracting, we get
37
2
2
2
2
by4–
ax4
=4
2
2
2
2
by–
ax
=1
Ans. dExample 2.The position of the point (5, –4) relative to the hyperbola 9x2–y2=1 is(a) on the hyperbola (b) outside the hyperbola(c) Inside the hyperbola (d) can not saySolution
S=9x2–y2–1S1=9(5)2–(–4)2–1=225–16–1=208>0
point (5, –4) lies inside the hyperbola.Ans c.
Example 3.Two circles are given such that they neither intersect nor touch. The locus of centre of variable circlewhich touches both the circles externally is(a) a circle (b) a parabola(c) an ellipse (d) a hyperbolaSolutionLet radii of the fixed circles be r1 and r2 and radius ofvariable circle be rLet variable circles with centre C touches two fixed circles with centre C1 and C2.Then CC1=r+r1 and CC2=r+r2 CC1–CC2=r1–r2=constantLocus of C is hyperbola whose foci are C1 and C2Ans. dExample 4:If the latus rectum subtends a right angle at the centre of the hyperbola
2
2
2
2
by–
ax
=1, then its eccentricity is
(a)2
15(b) 3
(c)25
(d)2
35
SolutionmAC×mBC= –1
ae.ab–
ae.ab 22
= –1
b4=a4e2
C2C1
C
XSC(0, 0)
(ae, 0)
A
B
ab,ae
2
ab,–ae
2
38
4
4
ab
=e2
(e2–1)2=e2
e4–3e2+1=0
e2=2
53e2=
253 e=
253
[if e = 3 52
<1 but ecentricity of hyperbola >1 so neglecting this value of e]
And. d
1. The equation of hyperbola whose foci are (8,3), (0,3) and eccentricity is 34
is
a. 17
)3–y(–9
4)–(x 22
b. 19
)3–y(–7
4)–(x 22
c. 17
)4–y(–9
3)–(x 22
d. none of these
2. If S and S be the foci, C the centre and P be a point on a rectangular hyperbola then SP× S Pis equal toa. 2.SP b. (SP)2 c. (CP)2 d. 2.CP
3. If e and e be the eccentricity of a hyperbola and its conjugate, then
a. 1e1–
e1
22 b. e2+ e 2=1 c. e2– e 2=1 d. 1e1
e1
22
4. The foci of a hyperbola coincide with the foci of the ellipse 19y
25x 22
, The equation of hyperbola
if its eccentricity is 2 isa. 3x2–y2=12 b. 4x2–y2=12 c. x2–3y2=12 d. x2–4y2=12
5. The equation 22 )2–y()4–x( + 22 )2–y()4x( =8 representsa. an ellipse b. a parabolac. a pair of coincident line segment d. hyperbola
6. For hyperbola 2
2
2
2
siny–
cosx
=1, which of the following remains constant with change in .
a. abscissae of vertices b. abscissae of focic. eccentricity d. directrix
PRACTICE QUESTIONS
39
7. Two rods are rotating about two fixed points in opposite directions. If they start from theirposition of co-incidence and one rotates at the rate double that of the other, then locus of pointof intersections of two rods isa. a parabola b. a circle c. an ellipse d. a hyperbola
8. The equations r1
y–r–1
x 22
=1 , r > 1 represents
a. an ellipse b. a circle c. a hyperbola d. None of these
9. The equation 2x2+3y2–8x–18y+35=k representsa. no locus if k > 0 b. an ellipse if k < 0c. a point if k = 0 d. a hyperbola if k > 0
10. A hyperbola having the transverse axis of length 2sin is confocal with the ellipse 3x2+4y2=12.Then its equation isa. x2cosec2 – y2sec2 =1 b. x2sec2 – y2cosec2 =1c. x2sin2 – y2cos2 =1 d. x2cos2 – y2sin2 =1
Answers
1. a 2. c 3. d 4. a 5. c6. b 7. d 8. d 9. c 10. a
40
HYPERBOLA - IIIEquation of Hyperbola
Equation of a Hyperbola referred to two perpendicular linesLet equation of hyperbola be
2
2
2
2
by–
ax
=1
From diagram PM=y and PN=x
2
2
2
2
bPM–
aPN
=1
ie. if perpendicular distance of a point P(x,y) from two mutually perpendicular lines say 1 1x+b1y+c1=0and 2 2x+b2y+c2=0 then
2
22
22
222
2
21
21
111
bba
cybxa
–a
bacybxa
=1
then the locus of point P denotes a hyperbola* centre of the hyperbola , we get after solving l1=0 and l2=0* Transverse axis : l2=0* Conjugate axis : l1=0
* Foci : The foci of the hyperbola is the point of intersection of the lines aeba
cybxa21
21
111and
l2=0
* Directrix : ea
bacybxa
21
21
111
* Length of transverse axis = 2a* Length of conjugate axis = 2b
* Length of latus Rectum = ab2 2
Example 1 :Find the eccentricity and centre of the hyperbola
22512–y3x4–
10012–y4–x3 22
=1
Solution :l1=3x–4y–12, l2=4x+3y–12a=10, b=15
e = 213
1002251
XO
(x, y)P
Y
N
=a =a
ll
41
3x–4y–12=04x+3y–12=0
x= 2584
, y=– 2512
Centre 2512,–
2584
Example 2:Find the eccentricity of the conic 4(2y–x–3)2–9(2x+y–1)2=80
Solution :
3 )–x–y 21 )–yx
=
3 )–x–y 2 21 )–yx
e = = 313
Example 3:Find the coordinates of the centre, foci and vertices, length of axes and latus rectum, equationof axes and directries, and eccentricity of the conic 9x2–16y2–18x+32y–151=0
Solution :9x2–16y2–18x+32y–151=09(x2–2x+1–1) –16(y2–2y+1–1)–151=09(x–1)2–16(y–1)2=144
91–y–
161–x 22
=1
Let x–1=X, y–1=Y, a=4, b=3Centre : X=0, Y=0 x=1, y=1 i.e. (1, 1)
Eccentricity : e= 1691 = 4
5
Foci : X= ae, Y=0 x=1 5, y=1 i.e. (6, 1) and (–4, 1)Vertices : X= a, Y=0 x=1 4, y=1 i.e. (5, 1) and (–3, 1)Length of transverse axis = 2a = 8Length of conjugate axis = 2b = 6
Length of latus rectums = ab2 2
= 29
Equation of transverse axis : Y=0 y–1=0Equation of conjugate axis : X=0 x–1=0
Equation of directries X = ea
x–1= 516
i.e. 5x–21=0 and 5x+11=01=0
4 ( 2 ( 2
( 2 ( 2
ba
2
– 9 8 08 08 08 0
–2 0 =1
a = 20, b = 2 2 8 09
91x =
2
2
1 2
20 91 = 4
8 0 / 9
80
42
Example 4:The equation of the transverse and conjugate axes of a hyperbola are respectively 3x+4y–7=0,4x–3y+8=0 and their respective lengths are 4 and 6. The equation of the hyperbola isa. 17x2+312xy+108y2–634x–312y–715=0b. 108x2+312xy+17y2–312x–634y–715=0c. 108x2–312xy+17y2–312x–634y–715=0d. none of these
Solution :The equations of hyperbola is
2
2
22
2
2
22
26
348y3–x4
–
24
437–y4x3
=1
2258y3–x4–
1007–y4x3 22
=1
9(9x2+16y2+49+24xy–42x–56y)–4(16x2+9y2+64–24xy+64x–48y)=90017x2+312xy+108y2–634x–312y–715=0Ans (a)
Line and Hyperbola
Let equation of line be y=mx+c and equation of hyperbola be 1by–
ax
2
2
2
2
1b
)cmx(–ax
2
2
2
2
xbmc2–
bc–
bxm–
ax
22
2
2
22
2
2
–1=0
0b
bc–xbmc2–
bm–
a1x 2
22
22
2
22
0)bc(–mcx2–m–abx 222
2
22
43
D=4m2c22
222
ama–b
(c2+b2)
i. D < 0 i.e. c2–a2m2+b2 < 0 line do not intersect hyperbola.ii. D = 0 i.e. c2–a2m2+b2 = 0 line touches the hyperbola.iii. D > 0 i.e. c2–a2m2+b2 > 0 line intersect hyperbola at two points.
Hence y = mx 222 b–ma is a tangent to hyperbola.
Let this tangent passes through a point (h, k) then k = mh 222 b–ma(k–mh)2= a2m2–b2
m2(h2–a2)–2mkh + k2+b2=0Hence maximum two tangents can be drawn through a point P.
Now m1 + m2 = 22 a–hkh2
m1 . m2 = 22
22
a–hbk
If is the angle between the two tangents, then
tan 21
21
mm1m–m
tan2 221
212
21
)1(4–)(
mmmmmm
= 2
22
22
22
222
22
–1
–4–
–2
ahbk
ahbk
ahkh
tan2 22222
222222
)ba–kh()bk)(a–h(4–hk4
tan2 22222
222222
)ba–kh()bh–kaba(4
If =90° then h2+k2–a2+b2=0i.e. h2+k2=a2–b2
Locus of (h,k) is x2+y2=a2–b2
Hence, Locus of point of intersection of two perpendicular tangents is known as Director Circle. Itsequation is x2+y2=a2–b2
If a=b, director circle is a point circle.If a < b, no real director circle is possible.
For equation of hyperbola 2
2
2
2
b)h–y(–
a)h–x(
equation of tangent in slope from is y–k=m(x–
h) 222 b–ma .
O
YP(h, k)
X
+4. = 4 222
2
am c 2+ 2 c - m c + - m b b 22a
b 4 2 2
2
222
a= 4b2 c +b a m– 2
44
1. If the foci of the ellipse 1byx
2
22
and the hyperbola 181y–
144x 22
coincide then b2 =
a. 3 b. 5 c. 7 d. 9
2. If PQ is a double ordinate of the hyperbola 1by–
ax
2
2
2
2
such that OPQ is an equilateral triangle,
O being the centre of the hyperbola, the range of eccentricity is
a. 32,0 b. ,
32
c. 34,0 d. ,
34
3. An ellipse and hyperbola are confocal and the conjugate axis of the hyperbola is equal to theminor axis of the ellipse. If e1 and e2 are the eccentricities of the ellipse and hyperbola then
a. e12+e2
2=2 b. e1+e2=2 c. 2e1
e1
21d. 2
e1
e1
22
21
4. The centre of a hyperbola 100
7–y4x3 2
–225
8y3–x4 2
=1 is
a. 2552,
2511–
b. 2552–,
2511
c. (0, 0) d. (10, 15)
5. The equations of the transverse and conjugate axes of a hyperbola are x+2y–3=0 and
2x–y+4=0 respectively and their respective lengths are 2 and 32
, equation of hyperbola is
a. 2(2x–y+4)2–3(x+2y–3)2=1 b. 2(x+2y–3)2–3(2x–y+4)2=1c. 2(2x–y+4)2–3(x+2y–3)2=5 d. 2(x+2y–3)2–3(2x–y+4)2=5
6. For all real values of m the straight line y=mx+ 4–m9 2 is a tangent to the hyperbolaa. 4x2–9y2=36 b. 9x2–4y2=36 c. x2–36y2=9 d. 36x2–y2=36
7. The equation of tangents to the curve 4x2–9y2=1 which is parallel to 4y=5x+7 isa. 4y=5x–30 b. 4y=5x+24c. 24y–30x= 161 d. 30x–24y– 161 =0
8. If the line 5x+12y=9 touches the hyperbola x2–9y2=9 then point of contact is
a. (5, 4) b. 34–,5 c. 3
4,5 d. (5, –4)
9. If y=mx+ 51 is a tangent to the hyperbola 100x2
–49y2
=1, then m =
a. 1 b. 17 c. –1 d. 2
10. The locus of the point of intersection of perpendicular tangents to 25x2
–16y2
=1 is
a. x2+y2=25 b. x2+y2=16 c. x2+y2=41 d. x2+y2=9Answers1. c 2. b 3. d 4. a 5. c 6.a 7. c 8. b 9. c 10. d
232
PRACTICE QUESTIONS
45
HYPERBOLA - IVEquation of Tangent
Equation of Tangent
i. Point Form : 2
2
2
2
by–
ax
=1
Differentiate w.r.t. x
dxdy
ay
ax
22
2–2=0
ybxb
dxdy
2
2
Slope of tangent = 1
21
2
yaxb
Equation of tangent, y–y1=1
21
2
yaxb
(x–x1)
2
21
2
21
21
21
by–
ax
byy–
axx
21
21
byy–
axx
=1 (But P(x1,y1) lies on hyperbola)
or T = 21
21
byy–
axx
–1
Equation of tangent is T=0
Equation of tangent at point P(x 1,y1) to the hyperbola 2
2
2
2
b)k–y(–
a)h–x(
=1 is
21
21
b)k–y)(k–y(–
a)h–x)(h–x(
=1
(ii) Parametric Form :Parametric equation of hyperbola is x=asec , y=btanEquation of tangent is
ax
sec – by
tan =1
(iii) Slope Form :
y = mx 222 b–ma
O
Y
P(x , y )1 1
X
46
Hyperbola 1by–
ax
2
2
2
2
Point of contact.
Point form : 21
21
byy–
axx
=1 (x1,y1)
Parametric Form : ax
sec – by
tan =1 (asec ,btan )
Slope form: y = mx 222 b–ma 222
2
222
2
b–mab,
b–mama
Example 1 :The equation of a tangent to the hyperbola 16x2–25y2–96x+100y–356=0, which makes an
angle 4 with the transverse axis, is
a. y=x+2 b. y=x–5 c. y=x+3 d. x=y+2Solution :
The equation of the hyperbola is16(x2–6x) –25(y2–4y)= 356
253–x 2
–16
2– 2y=1
The equation of tangent of slope m = tan 4 =1 to this hyperbola are
y–2=1(x–3) 16–125y–2=x–3 3
y=x+2 or y=x–4Ans. (a)
Example 2 :
The point of intersection of two tangents to the hyperbola 2
2
ax
– 2
2
by
=1, the product of whose
slopes is c2, lies on the curvea. y2–b2=c2(x2+a2) b. y2+a2=c2(x2–b2)c. y2+b2=c2(x2–a2) d. y2–a2=c2(x2+b2)
Solution :
Let P(h,k) be the point of intersection of two tangents to the hyperbola 2
2
ax
– 2
2
by
=1
The equation of tangent to hyperbola is
47
y=mx ± 222 b–maIf it passes through (h, k) then
k=mh ± 222 b–ma (k–mh)2=a2m2–b2
m2(h2–a2)–2mkh+k2+b2=0
Let m1 and m2 be the slopes of the tangents passing throgh P. Then m1.m2= 22
22
a–hbk
c2= 22
22
a–hbk
Hence locus of P(h, k) is y2+b2=(x2–a2)c2
Ans. (c)
Example 3 :
If the tangents drawn from a point on the hyperbola x2–y2=a2–b2 to the ellipse 2
2
ax
+ 2
2
by
=1
make angles and with the transverse axis of the hyperbola, thena. tan –tan =1 b. tan +tan =1 c. tan .tan =1 d. tan .tan =–1
Solution :Let P(h,k) be the point on the hyperbola x2–y2=a2–b2, then h2–k2=a2–b2 .........(i)
The equation of tangent to the ellipse is 2
2
ax
+ 2
2
by
=1 is
y=mx ± 222 bmaIf it passes through (h, k) then
k=mh ± 222 bma(k–mh)2=a2m2+b2
m2(h2–a2)–2mkh+k2–b2=0Let m1 and m2 be the roots of this equation, then
m1.m2= 22
22
a–hb–k
=1 (from equation (i))
tan .tan =1Ans (c)
Example 4 :If the line 2x+ 6 y=2 touches the hyperbola x2–2y2=4, then the point of contact is
a. 6,2– b. 62,5– c. 61,
21
d. 6,–4
48
Solution :
Equation of hyperbola is 2
2
2x
– 2
2
2
y=1
Equation of tangent is y = 32x
32–
m= –32
, a=2, b = 2 , c=32
and c2=a2m2–b2
Point of contact is 222
2
222
2
b–mab,
b–mama
i.e. 6,4
Ans. (d)
Example 5 :
Number of real tangents can be drawn from the point (5, 0) to the hyperbola 16x2
–9y2
=1 is
a. 2 b. 1 c. 0 d. 4Solution :
S1= 1625
– 90
–1 > 0
Point (5, 0) lies inside the hyperbola. Hence no tangents can be drawn.Ans. (c)
1. P is a point on the hyperbola 2
2
ax
– 2
2
by
=1, N is the foot of the perpendicuilar from P on the
transverse axis. The tangent to the hyperbola at P meets the transverse axis at T. If O is thecentre of the hyperbola, then OT×ON isa. b2 b. a2 c. ab d. none of these
2. Tangents drawn from the point (c,d) to the hyperbola 2
2
ax
– 2
2
by
=1make angles and with
the x-axis. If tan tan =1, then c2–d2=a. a2–b2 b. a2.b2 c. a2+b2 d. none of these
3. The common tangents to the two hyperbolas 2
2
ax
– 2
2
by
=1 and 2
2
ay
– 2
2
bx
=1 is
a. y=x+ 22 b–a b. y=x+ 22 ba c. y=2x+ 22 b–a d. none of these
PRACTICE QUESTIONS
49
4. The equation of the tangent to the curve 4x2–9y2=1 which is parallel to 4y=5x+7 is
a. 30x+24y=720 b. 7y=6x–15 c. y=372
x+ 715
d. none of these
6. The locus of a point P(h, k) moving under the condition that the line y=hx+k is a tangent to the
hyperbola 2
2
ax
– 2
2
by
=1 is
a. Parabola b. circle c. ellipse d. hyperbola
Answers1. b 2. c 3. a 4. d5. c 6. d
50
Hyperbola - VNormals
Equation of Pair of Tangents
Let equation of hyperbola be 2
2
2
2
by–
ax
=1 and a point P(x1,y1) then the combined equation of
tangents PA and PB is SS1 = T2 where
S= 2
2
2
2
by–
ax
–1
S1= 2
21
2
21
by–
ax
–1
T= 21
21
byy–
axx
–1
Example 1 :
A pair of tangents drawn from the point (4, 3) to the hyperbola 16x2
–9y2
=1 then angle between
the tangents is
a. tan–1 43
b. 2 c. tan–1 34
d. tan–1 21
Solution :
S = 16x2
–9y2
–1, S1 = 1616
– 99
–1=–1, T= 16x4
– 9y3
–1
Equation of pair of tangents is SS1=T2
222
1–3y–
4x1–
9y–
16x1–
9y
16x–
22
+1=9y
16x 22
+1– 3y2
2x–
6xy
3x2–4xy–12x+16y=03x–4y=0, x–4=0
m1= 43
m2=
= 2 –tan–1
43
=cot–1
43
=tan–1
34
Ans (c)
O
Y
P
A
B
(x , y )1 1
X
Y
51
Equation of Normal to the Hyperbla :i. Point Form:
2
2
2
2
by–
ax
=1
Slope to tangent = +1
21
2
yaxb
Slope to normal = –1
21
2
xbya
Equation of normal y–y1= –1
21
2
xbya
(x–x1)
1
2
xxa
+1
2
yyb
=a2+b2
ii. Parametric Form: P(asec , btan )
secax
+ tanby
=a2+b2
or axcos +bycot = a2 + b2
iii. Slope Form:
y = mx 222
22
bm–a)ba(m
Point of contanct is 222
2
222
2
bm–amb,
bm–aa
Exercise 1
A normal to the hyperbola 2
2
ax
– 2
2
by
=1 meets the axes in M and N and lines MP and NP are
drawn perpendiculars to the axes meeting at P. The locus of P isa. a2x2–b2y2=(a2+b2)2 b. a2x2–b2y2=a2+b2
c. b2x2–a2y2=a2+b2 d. a2x2–b2y2=a2–b2
Solution :Equation of normal is axcos +bycot =a2+b2
Normat meets the axes at M 0,cosa
ba 22
and N cotbba,0
22
Now equation of PM is x = cosa
ba 22
O
Y
PNormal
Tangent
(x , y )1 1
X
Y
X
X O
Y
Y
P
M
N
X
52
sec = 22 baax
and equation of PN is y = cotb
ba 22
tan = 22 baby
sec2 –tan2 =1
222
22
222
22
bayb–
baxa
=1
a2x2–b2y2=(a2+b2)2
Ans. (a)Exercise 2
The line lx+my–n=0 will be normal to the hyperbola 2
2
ax
– 2
2
by
=1 if
a. l2a2+b2m2=n
)ba( 22
b. 2
222
2
2
2
2
nba
mb
la
c. 2
222
2
2
2
2
nba
mb–
la
d. 2
22
2
2
2
2
nba
mb–
la
Solution :Equation of normal is axcos +bycot =a2+b2
This equation compare with lx+my=n
i.e. n
bamcotb
lcosa 22
sec = )ba(lan
22
tan = )ba(mbn
22
sec2 –tan2 =1
2
222
2
2
2
2
n)ba(
mb–
la
Ans. (c)Exercise 3
The line xcos +ysin = p touches the hyperbola 2
2
ax
– 2
2
by
=1 if
a. a2cos2 +b2sin2 =p b. a2cos2 –b2sin2 =p2
53
c. a2cos2 +b2sin2 =p2 d. a2cos2 –b2sin2 =pSolution :
The line y=mx+c touches the hyperbola 2
2
ax
– 2
2
by
=1, then c2=a2m2–b2
Given equation is y = – cot +pcosec p2cosec2 =a2(–cot )2–b2
p2 = a2cos2 –b2sin2
Ans. (b)Example 4 :
The normal at P to a hyperbola of eccentricity e, intersects its transverse and conjugate axes atL and M respectiely. The locus of the mid point of LM is a hyperbola then eccentricity of thehyperbola is
a. 1–e1e
b.1–e
e2 c. e d. none of these
Solution :
The equation of the normal at P (asec ,btan ) of the hyperbola 2
2
ax
– 2
2
by
=1 is
axcos +bycot =a2+b2
This intersect the transverse and conjugate axes at L 0,seca
ba 22
and M tanb
ba,022
respectively.Let N(h,k) be the mid point of LM, then
h=a2ba 22
sec and k = b2ba 22
tan
sec = 22 baah2
and tan = 22 bakb2
sec2 –tan2 =1
4a h –4b k =(a +b )2 2 2 2 2 2 2
Thus locus of (h,k) is 4a2x2–4b2y2=(a2+b2)2
Let e1 be the eccentricity of this hyperbola. Then
e12=1+ 2
2
ba
= 2
22
bba
= )1–e(aea22
22
=1–e
e2
2
e1= 1–ee2
Ans. (b)
y = – x cot + p cosec
54
Some Important Results1. Normal other than transverse axis never passes through the focus.
2. Locus of the feet of the perpendicular drawn from focus of the hyperbola 2
2
ax
– 2
2
by
=1 upon any
tangent is its auxiliary circle i.e. x2+y2=a2.
XO
Y
Y
PT
S X
T
3. The product of the feet of these perpendiculars is b2.4. The portion of the tangent between the point of contact and the directrix subtends a right angle
at the corresponding focus.
X O
Y Z
Y
P
ST
X
Z
5. The tangent and normal at any point of a hyperbola bisect the angle between the focal radii.This spells the reflection property of the hyperbola as an incoming light ray aimed towards onefocus is reflected from the outer surface of the hyperbola towards the other focus.
XO
Y
Y
P
S XS
55
6. If an ellipse and a hyperbola have the same foci, they cut at right anlges at any of their commonpoints.
XO
Y
90°
S1S2
X
7. The ellipse 2
2
ax
+ 2
2
by
=1 and the hyperbola 22
2
k–ax
– 22
2
b–ky
=1 (a>k>b>0) are confocal and
therefore orthogonal.8. The foci of the hyperbola and the points P and Q in which any tangent meets the tangents at the
vertices are concyclic with PQ as diameter of the circle.
XO
P
Q
Y
S1S2
X
1. Normal is drawn at one of the extremities of the latus rectum of hyperbola 2
2
ax
– 2
2
by
=1, which
meets the axis at points A and B. Then area of triangle OAB (O being the origin) is
a. a5e2 b. a2e5 c. 21
a2e5 d. none of these
2. If the normal at P( ) on the hyperbola 2
2
ax
– 2
2
a2y
=1meets the conjugate axis at G then AG× A G
is (Where A and A are the vertices of the hyperbola)a. a2e2sec2 –a2 b. a4e2sec2 –a4 c. a2e4sec2 –a2 d. none of these
3. Normals are drawn to the hyperbola 2
2
ax
– 2
2
by
=1 at points 1 and 2 meeting the conjugate axis
at G1 and G2 respectively. If 1+ 2= 2 then CG1 × CG2 is (where C is centre of hyperbola and
e is eccentricity of hyperbola)
Y’
PRACTICE QUESTIONS
56
a.1–e
ea2
42
b.1–e
ea2
22
c.1–e
ea2
44
d. none of these
4. Let P(asec , btan ) and Q(asec , btan ), where + = 2 , be two points on the hyperbola
2
2
ax
– 2
2
by
=1. If (h, k) is the point of intersection of the normals at P and Q, then k is equal to
a.a
ba 22
b. aba–
22
c.b
ba 22
d. bba–
22
5. If the normal at the point P to the rectangular hyperbola x2–y2=a2 meet the axes in N and Nand C(0, 0) is the centre of the hyperbola thena. PN=P N =PC b. PC=PN P N c. PN=P N PC d. none of these
Answers1. c 2. c 3. a 4. d5. a
57
HYPERBOLA - VIAsymptotes
Equation of Chord of Contact
Let equation of hyperbola be 2
2
2
2
by–
ax
=1,
Equation of Chord of contact is
21
21
byy–
axx
–1=0
or T=0
where T = 21
21
byy–
axx
–1
Example 1 :
If tangents of the parabola y2=4ax intersect the hyperbola 2
2
ax
– 2
2
by
=1 at P and Q, then locus of
point of intersection of tangents at P and Q isa. a3y2+b4x2=0 b. a3y2+b4x=0 c. a2y2+b2x2=0 d. none of these
Solution :Let P1(h, k) be the point of intersections of tangents at P and Q. Therefore, the equation ofchord of contact PQ is
22 byk–
axh
=1
y=kb–
kahxb 2
2
2
which touches the parabola y2=4ax
kahb
akb–
2
2
2
–hbka
kb
2
32
Locus of (h, k) is y2=– 3
4
ab
x
a3y2+b4x=0Ans. (b)
Y
P
RQ
(x , y )1 1
X
58
Equation of the chord of the hyperbola whose mid point is given :A is the mid point of PQ, then equation of chord is
21
21
byy–
axx
= 2
21
2
21
by–
ax
ie. T=S1 where
T = 21
21
byy–
axx
–1
S1 = 2
21
2
21
by–
ax
–1
Example 2:The locus of the middle points of the chord of hyperbola 3x2–2y2+4x–6y=0 parallel to y=2x isa. 3x+4y=4 b. 4x+3y=12 c. 3x–4y=4 d. none of these
Solution :Let mid point of the chord be (h, k), equation of the chord is T=S1 i.e.
3xh–2yk+2(x+h)–3(y+k)=3h2–2k2+4h–6k x(3h+2) –y(2k+3) –3h2+2k2–2h+3k=0
Its slope is 3k22h3
=2 (slope of y=2x)
3h–4k=4Locus of (h, k) is 3x–4y=4
Ans. (c)Asymptotes of Hyperbola
An asymptotes of any hyperbola is a straight line which touches in it two points at infinity.ORIf the length of the perpendicular let fall from a point on a hyperbolato a straight line tends to zero as the point on the hyperbola movesto infinity along the hyperbola, then the straight line is calledasymptote of the hyperbola.
The equation of two asymptotes of the hyparbola 2
2
2
2
by–
ax
=1
are y= xab
or by
ax
=0
Pair of asymptotes : 2
2
2
2
by–
ax
=0
1. If b=a, then 2
2
2
2
by–
ax
=1 reduces to x2–y2=a2. The asymptotes of rectangular hyperbola
x2–y2=a2 are y= x which are at right angles.2. A hyperbola and its conjugate hyperbola have the same asymptotes.
Y
P
A
Q
(x , y )1 1
X
Y
OX
xab y x
aby–
59
3. The angle between the asymptotes of 2
2
2
2
by–
ax
=1 is 2 tan–1 ab
4. The asymptotes pass through the centre of the yperbola.5. The bisectors of the angle between the asymptotes are the coordinate axes.
6. Let H = 2
2
2
2
by–
ax
–1=0
A = 2
2
2
2
by–
ax
=0
and C = 01by–
ax
2
2
2
2
be the equation of the hyperbola, asymptotes and the conjugate hyperbola respectively, thenclearlyC+H=2A
Example 3:The asymptotes of the hyperbola xy–3y–2x=0 area. x–3=0 b. x+y=0 c. y–2=0 d. x–y=0
Solution :Since equation of a hyperbola and its asymptotes differ in constant terms only. Pair of asymptotesis given by
xy–3y–2x+c=0It represents a pair of straight line
c
23–1–
23–0
21
1–210
= 0
43––
23–
2c
21– =0
43
43
4c– =0
c=6 xy–3y–2x+6=0
(x–3) (y–2)=0Asymptotes are x–3=0 and y–2 0
Ans : a, c
h
60
Example 4:The equation of the hyperbola which has 3x–4y+7=0 and 4x+3y+1=0 as its asymptotes andwhich passes through the origin isa. x2–y2=12xy b. 12x2–7xy–12y2+31x+17y=0c. 12x2–12y2=7xy d. 12x2+7xy–12y2+25x–19y=0
Solution :Combined equation of the asymptotes is
(3x–4y+7) (4x+3y+1)=012x2–7xy–12y2+31x+17y+7=0
Since equation of hyperbola and combined equation of its asymptotes differ by a constant,therefore equation of hyperola may be
12x2–7xy–12y2+31x+17y+c=0But it passes through the origin. So c=0Hence equation of hyperbola is12x2–7xy–12y2+31x+17y=0
Ans. (b)Example 5:
The product of the lengths of perpendiculars drawn from any point on the hyperbola x2–2y2=2to its asymptotes is
a. 23
b. 1 c. 2 d. 32
Solution :Given hyperbola is
2x2
–1y2
=1
Let P ( 2 sec , tan ) a point on the hyperbola.
Equation of asymptotes is 0y–2
x and 0y
2x
Product of the lengths of perpendiculars = 121
tansec.1
21
tan–sec
= 23
tan–sec 22
= 32
Ans (d)
61
Exercise 6The angle between the asymptotes of the hyperbola
16x2
–9y2
=1
a. tan–143
b. tan–1724
c. 2tan–134
d. 2tan–154
Solution :
Equations of asymptotes are 4x
– 3y
=0 and 4x
+ 3y
=0
slope of first asymptote is tan = 43
= tan–143
Angle between the asymptotes is 2 =2tan–143
=tan–1
169–1432
=tan–1724
Ans. (b)
1. If angle between asymptotes of hyperbola 2
2
ax
– 2
2
by
=1 is 120° and product of perpendiculars
drawn from foci upon its any tangent is 9, then locus of point of intersection of perpendiculartangents of the hyperbola can bea. x2+y2=6 b. x2+y2=9 c. x2+y2=3 d. x2+y2=18
2. For a hyperbola whose centre is at (1,2) and asymptotes are parallel to lines 2x+3y=0 andx+2y=1, then equation of hyperbola passing through (2, 4) isa. (2x+3y–5) (x+2y–8)=40 b. (2x+3y–8) (x+2y–5)=40c. (2x+3y–8) (x+2y–5)=30 d. none of these
3. Asymptotes of the hyperbola 21
2
ax
– 21
2
by
=1and 22
2
ax
– 22
2
by
=1are perpendicular to each other then,
a.2
1
aa
= 2
1
bb
b. a1a2=b1b2 c. a1–a2=b1–b2 d. a1a2+b1b2=0
PRACTICE QUESTIONS
62
4. If S=0 be the equation of the hyperbola x2+4xy+3y2–4x+2y+1=0, then the value of k for whichS+k=0 represents its asymptotes isa. –22 b. 18 c. –16 d. 20
5. A hyperbola passes through (2, 3) and has asymptotes 3x–4y+5=0 and 12x+5y–40=0, thenthe equation of its transverse axis isa. 77x–21y–265=0 b. 21x–77y–265=0c. 21x–77y–265=0 d. 21x+77y–265=0
6. The combined equation of the asymptotes of the hyperbola 2x2+5xy+2y2+4x+5y=0 isa. 2x2+5xy+2y2+7=0 b. 2x2+5xy+2y2+4x+5y+7=0c. 2x2+5xy+2y2+4x+5y+2=0 d. None of these
7. The asymptotes of the hyperbola xy+hx+ky=0 area. x+h=0 and y+k=0 b. x+h=0 and y–k=0c. x–k=0 and y–h=0 b. x+k=0 and y+h=0
8. If foci of hyperbola lie on y=x and one of the asymptote is y=2x, then equation of the hyperbola,given that it passes through (5, 4) isa. 2x2–2y+5xy+5=0 b. 2x2+2y2–5xy+10=0c. x2–y2–xy+5=0 d. None of these
Linked comprehension type (for problems 9 - 11)In hyperbola portion of tangent intercept between asymptotes is bisected at the point of contact.Consider a hyperbola whose centre is at origin. A line x+y=2 touches this hyperbola at P(1, 1)and intersects the asymptotes at A and B such that AB=6 2 units.
9. Equation of asymptotes area. 2x2+2y2–5xy=0 b. 2x2+5xy+2y2=0c. 3x2+6xy+4y2=0 d. None of these
10. Equation of tangent to the hyperbola at 27,1– is
a. 3x+2y=4 b. 3x+4y=11 c. 5x+2y=2 d. none of these11. Angle subtended by AB at centre of the hyperbola is
a. tan–1 34
b. tan–1 32
c. tan–1 43
d. none of these
Answers1. d 2. b 3. d 4. a5. d 6. c 7. c 8- b9. b 10.. a 11. c
63
HYPERBOLA - VIIRectangular Hyperbola
Rectangular Hyperbola:A hyperbola whose asymptotes include a right angle is said to be rectangular hyperbola.ORIf the lengths of transverse and conjugate axes of any hyperbola be equal it is called rectangularor equilateral hyperbola.According to first definition
2tan–1ab
= 2
tan–1
ab
= 4
ab
= tan 4b = a
then 2
2
ax
– 2
2
by
= 1 becomes x2–y2 = a2
According to second definition
When a = b, 2
2
ax
– 2
2
by
= 1 becomes x2–y2 = a2
Eccentricity e = 2
2
ab1 = 2
Then asymptotes of x2–y2 = a2 are x+y = 0 and x–y = 0 . Each of these two asymptotes isinclined at an angle of 45° with the transverse axis. So, if we rotate the coordinate axes through
an angle of 4 keeping the origin fixed, then the axes coincide with the asymptotes of the
hyperbola.Now equation of asymptotes of new hyperbola is x y = 0Then equation of hyperbola is xy = k (constant)
The hyperbola passes through the point 2a,
2a
k = 2a 2
Then equation of hyperbola is xy = 2a 2
or xy = c2 where c2 = 2a 2
If the asymptotes of a rectangular hyperbola are x = a, y = b, then its equation is (x–a)(y–b) = c2
xy = c2
Y
y=–x y=xx –y =a22 2
xy=c2
A
(a, 0)OX
2a,
2a
64
1 Asymptotes : x = 0, y = 02 Transverse axis : y = x
Conjugate axis : y = – x3 Vertices A(c,c), A (–c,–c)
4 Foci : S 2c,2c , S 2c,–2c–
5 Length of transverse axis = A A = c226 Equation of auxiliary circle x2+y2 = 2c2
7 Equation of director circle x2+y2 = 08 x2–y2 = a2 and xy = c2 intersect at right anglesProperties of Rectangular Hyperbola1 Eccentricity of rectangular hyperbola is 2 .
2 Since x = ct, y = tc
satisfies xy = c2
(x,y) = tc,ct (t 0) is called a ‘t’ point on the rectangular hyperbola. The x = ct, y = t
c
represents its parametric equation with parameter ‘t’
3 Equation of chord joining P1
1 tc,ct and Q
22 t
c,ct is
x+y t1 t2 – c(t1+t2) = 0
Slope of chord = – 21tt
1
4 Equation of tangent at (x1,y1) is xy1+yx1 = 2c2
5 Equation of tangent at t is tx
+yt = 2c
Slope of tangent = – 2t1
6 Equation of normal at (x1,y1) is xx1–yy1 = x12–y1
2
Equation of normal at t is xt3–yt–ct4+c = 0Slope of normal = t2
7 Point of intersection of tangents at t1 and t2 is
2121
21
ttc2,
tttct2
8 Point of intersection of normal at t1 and t2 is
)tt(t.t)tttt(ctct
,)tt(t.t
c–)tttt(tct
2121
2221
21
32
31
2121
2221
2121
65
Example 1If the normal at the point t1 to the rectangular hyperbola xy = c2 meets it again at the point t2 then(a) t1t2 = – 1 (b) t1
3t2 = – 1 (c) t1t23 = – 1 (d) t1
2t22 = – 1
Solution
Equation of normal at 1
1 tc,ct to the hyperbola xy = c2 is
xt13–yt1–ct1
4+c = 0
But this passes through 2
2 tc,ct then
ct2t13–
2tc
t1–ct14+c = 0
t2t13–
2
1
tt
–t14+1 = 0
t22t1
3–t1–t14t2+t2 = 0
t13t2(t2–t1)+1(t2–t1) = 0
(t2–t1) (t13t2+1) = 0
t13t2 = – 1 ( t1 t2)
Ans (b)Example 2
A triangle has its vertices on a rectangular hyperbola xy = c2
The orthocenter of the triangle lies on(a) x2+y2 = c2 (b) x2–y2 = c2 (c) xy = c2 (d) None of theseSolution
Let A1
1 tc,ct B
22 t
c,ct C3
3 tc,ct be the vertices of atriangle lies on xy = c2
Now slope of BC = 23
23
ctcttc
tc
= – 32tt
1
Hence slope of AD = t2t3Equation of AD is
y–1tc
= t2t3 (x–ct1)
t1y–c = xt1t2t3–ct12t2t3______________(1)
Similarly
slope of AC = – 31tt
1
Y
DC
BE
A
X
66
slope of BE = t1t3Equation of BE is t2y–c = xt1t2t3–ct2
2t1t3____________(2)
Solving (1) and (2) we get y = –ct1t2t3 & x = – 321 ttt
c
Which satisfies xy = c2
Therefore orthocentre lies on xy = c2
Ans (c)Example 3
If PN is the perpendicular from a point on a rectangular hyperbola xy = c2 to its asymptotes thenlocus of the midpoint of PN is
(a) xy = 2c2
(b) xy = 4c2
(c) xy = 2c2 (d) None of these
SolutionLet P(x1, y1) be a point on xy = c2
Le Q(h,k) be the Midpoint of PN then x1 = h and k = 2y
(x , y ) lies on xy = c1 12 h(2k) = c2 Locus of
(h,k) is 2xy = c2
Ans (a)Example 4
PQ and RS are two perpendicular chords of the rectangular hyperbola xy = c2. If O is the centre ofthe rectangular hyperbola, then product of the slopes of OP, OQ, OR and OS is(a) –1 (b) –2 (c) 1 (d) 2Solution
Let coordinates of P, Q, R, S be i
i tc,ct , i = 1, 2, 3, 4
Now PQ RS RSPQ mm = –1
34
34
12
12
ct–cttc–
tc
ct–cttc–
tc
= – 1
– 4321 tt1–
tt1
= – 1
t1t2t3t4 = – 1
Now slope of OP = 1
1
cttc
= 21t1
Y
N
PQ
X1
67
Similarly slope of OQ = 22t1
Similarly slope of OR = 23t1
Similarly slope of OS = 24t1
Product of their slopes = 24
23
22
21 t.t.t.t
1 = 2)1(–
1 = 1
Ans (c)Example 5
The angle between the rectangular hyperbolas(y–mx) (my+x) = a2 and (m2–1) (y2–x2) + 4mxy = b2 is
(a) 2 (b) 3 (c) 4 (d) tan–1ab
Solution(y–mx) (my+x) = a2
m–dxdy
(my+x) + 1dxdym (y–mx) = 0
(my+x) dxdy
+ m(y–mx) dxdy
= m(my+x) – (y–mx)
dxdy
= xm–myxmymxy–mxym
2
2
= my2xxm–y–mx2ym
2
2
= m1________(1)
For another hyperbola(m2–1) (y2–x2) + 4mxy = b2
(m2–1) x2–dxdyy2 + 4m dx
dyxy = 0
(m2–1) y dxdy
– x (m2–1)+2my+2mx dxdy
= 0
(m2y–y+2mx) dxdy
= m2x–x–2my
dxdy
= y–mx2ymmy2–x–xm
2
2
= m2 _________(2)
Now m1×m2 = – 1
68
angle = 2Ans (a)
Example 6The family of the curves which intersect the family of rectangular hyperbola xy = c2 orthogonally is(a) family of circle b) family of parabola(c) family of ellipse (d) family of hyperbolaSolutionxy = c2
Differentiate w.r.t. x
y+x dxdy
= 0
Replace dxdy
by – dydx
, we get
y – x dydx
= 0
y dy – xdx = 0y2–x2 = k (where k is paramenter)
family of hyperbola.Ans (d)
1 The coordinates of the foci of the rectangular hyperbola xy = c2 is
(a) ( c, c) (b) 2c,
2c
(c) 2c,2c (d) 2c,
2c
2 The equation of directories of the rectangular hyperbola xy = c2 is
(a) x+y = c (b) x+y = c c (c) x+y = cc
(d) x y = 0
3 If the line ax+by+c = 0 is a normal to the hyperbola xy = 1, then(a) a>0, b>0 (b) a<0, b<0 (c) a>o, b<0 (d) a<o, b>0
4 Consider the set of hyperbolas xy = k, k R. Let e1, be the eccentricity when k = 4 and e2 be theeccentricity when k= 9, then e1–e2 is equal to
(a) 0 (b) 1 (c) 23
(d) 2
5 If chords of the hyperbola x2–y2 = a2 touch the parabola y2 = 4ax then the locus of the middle pointsof these chords in the crane(a) y2 = x3 (b) y2(x–a) = x2 (c) y3(x–a)=x2 (d) y2(x–a) = x3
6 If the tangent and normal to a rectangular hyperbola xy = 4 cut off intercepts a1 and a2 on one axisand b1 and b2 on the other, then a1a2+b1b2 is(a) –4 (b) 0 (c) 1 (d) 4
PRACTICE QUESTIONS
69
7 The points of intersection of the cranes whose parametric equations are x = t2+1, y = 2t and x = 25,
y = 52
is given by
(a) (1,2) (b) (2,2) (c) (1,–3) (d) (–2,4)8 Number of maximum tangents from any point to the hyperbola xy = c2 is
(a) 1 (b) 2 (c) 3 (d) 49 The length of the latus rectum of the hyperbola xy–3x–3y+7 = 0 is
(a) 2 (b) 4 (c) 2 2 (d) None of these10 If x = 9 is the chord of contact of the hyperbola x2–y2 = 9, then the equation of the corresponding
pair of tangent is(a) 9x2–8y2–18x+9 = 0 (b) 9x2–8y2–18x–9 = 0(c) 9x2–8y2+18x+9 = 0 (d) 9x2–8y2+18x–9 = 0
11 The equation of the chord joining two points (x1,y1) and (x2,y2) on the rectangular hyperbola xy =c2 is
(a)21 x–x
x +
21 y–yy
= 1 (b)21 yy
x +
21 xxy
= 1
(c)21 xx
x +
21 yyy
= 1 (d)21 y–y
x +
21 x–xy
= 1
Answers :1. c 2. b 3. c, d 4. a 5. d6. b 7. b 8. b 9. b 10. a 11. c
70
HYPERBOLA - VIIIPractice Problems
Practice Problems1 Equation of conjugate axis of hyperbola xy–3y–4x+7 = 0 is
(a) x+y = 7 (b) x+y = 3 (c) x–y = 7 (d) None of theseSolution:
xy–3y–4x+7 = 0xy–3y–4x+12 = 5(x–3) (y–4) = 5
Equation of asymptotes are x–3 = 0 and y-4 = 0Since the hyperbola is rectangular hyperbola, axesare bisectors of asymptotesHence their slopes are 1
Equation of conjugate axis isy–4 = – 1(x–3)x+y = 7
Ans (A)2 If S1 and S2 are the foci of the hyperbola whose transverse axis length is 4 and conjugate axis
length is 6, S3 and S4 are the foci of the conjugate hyperbola, then the area of the quadrilateralS1S3S2S4 is(a) 156 (b) 36 (c) 26 (d) None of theseSolution :
S1S3S2S4 forms a square.
So required area = 4×area of S1 O S3 = 4×21 ae × be1
= 2 abee1 = 2. 2. 3. e e1= 12ee1
Now e = 491 =
213
& e1 = 491 =
213
Hence area = 12×213
×213
= 26 sq.units
Ans (c)3 The ellipse 4x2+9y2 = 36 and the hyperbola a2x2–y2 = 4 intersect at right angles then the equation
of the circle through the points of intersection of two conic is(a) x2+y2 = 25 (b) 5(x2+y2 )+3x+4y = 0(c) 5(x2+y2)–3x–4y = 0 (d) (x2+y2) = 5Solution:Since ellipse and hyperbola intersect orthogonally, they are confocal.
e = 94–1 =
35
S4
S1
S3
S2 o
Y
X
(3, 4)y=4
x=3
71
foci of ellipse 0,5
(ae)2 = a2+b2
5 = 2a
4 + 4 a = 2
Let point of intersection in the first quadrant be P(x1, y1).P lies on both the curves.
4x12+9y1
2 = 36 and 4x12–y1
2 = 4Adding these two, we get 8x1
2+8y12 = 40
x12+y1
2 = 5Equation of circle is x2+y2 = 5
4 If e is the eccentricity of the hyperbola 2
2
a
x – 2
2
b
y = 1 and 2 is angle between the asymptotes
then cos =
(a)e1
(b)e
e–1(c)
ee1
(d) None of these
Solution:
e = 2
2
a
b1
we know 2 = 2tan–1ab
tan = ab
e = 2tan1 = sec cos = e1
.
Ans (a)5 From a point p (1,2) pair of tangents are drawn to a hyperbola in which one tangent to each
arm of hyperbola. Equation of asymptotes of hyperbola are 3 x–y+5 = 0 and 3 x+y–1 = 0then eccentricity of hyperbola is
(a) 3 (b)3
2(c) 2 (d) None of these
Solution:Equation of asymptotes are
3 x–y+5 = 0
– 3 x–y+1 = 0a1a2+b1b2 = –3+1<0origin lies in acute angle and P(1,2) lies in obtuse angle.e = sec where 2 is the angle between asymptotes.
2 = 3
=6
XO
P
Y
X
X
P(1,2)
Y
72
e = sec6
= 3
2
Ans (b)
6 If a variable line has its intercepts on the coordinate axes e, e where 2e
, 2e
are the eccentricities
of a hyperbola and its conjugate hyperbola, then the line always touches the circle x2+y2 = r2,where r =(a) 4 (b) 3(c) 2 (d) Can not be decidedSolution
Now 2e
4 + 2)e(
4 = 1 4 = 22
22
)e(e
)e(e
Line passing through the points (e,0) and (0,e ) is e x+ey = eeIt is a tangent to the circle x2+y2 = r2
22 )e(e
ee = r
2 = rAns (c)
7 If angle between asymptotes of hyperbola 2
2
a
x – 2
2
b
y = 1 is 120° and product of perpendiculars
dr wn from foci upon its any tangent is 9, then locus of point of intersection of perpendiculartangents of the hyperbola can be(a) x2+y2 = 18 (b) x2+y2 = 6 (c) x2+y2 = 9 (d) x2+y2 = 3Solution
2tan–1ab
= 60° ab
= 3
1
b2 = 9 a2 = 27Required locus is director circle i.e. x2+y2 = 27 – 9
x2+y2 = 18
If ab
= tan 60° = 3
a2 = 3Then equation of director circle is x2+y2 = 3 – 9 = – 6 which is not possible.Ans (a)
8 The equation of the transverse axis of the hyperbola(x–3)2+(y+1)2 = (4x+3y)2 is(a) 3x–4y = 0 (b) 4x+3y = 0 (c) 3x–4y = 13 (d) 4x+3y = 9
a
73
Solution(x–3)2+(y+1)2 = (4x+3y)2
(x–3)2+(y+1)2= 25 2
5y3x4
PS = 5 PMDirectrix is 4x+3y = 0 and focus is (3,–1)
Equation of transverse axis is y+1 = 43
(x–3)
3x–4y = 13.Ans (c).
1 The equation of common tangents to the parabola y2 = 8x and hyperbola 3x2–y2 = 3 is(a) x 2y–1 = 0 (b) x 2y+1 = 0 (c) 2x y+1 = 0 (d) 2x y–1 = 0
2 A tangent to the hyperbola y = 5x9x
passing through the origin is
(a) x–2y =0 (b) 5x–y = 0 (c) 5x+y = 0 (d) x+225y =03 The equation of the common tangent to the curves y2 = 8x and xy = – 1 is
(a) y = x+2 (b) y = 2x+1 (c) 2y = x+8 (d) 3y = 9x+2
4 Let PQ be a double ordinate of the hyperbola 2
2
ax
– 2
2
by
= 1. If O be the centre of the
hyperbola and OPQ is an equilateral triangle, then eccentricity e is
(a) > 3 (b) >2 (c) > 32
(d) None of these
5 The difference between the length 2a of the transverse axis of a hyperbola of eccentricity eand the length of its latus rectum is(a) a(2e2–1) (b) 2a(e2–1) (c) 2a|3–e2| (d) 2a|2–e2|
6 The slopes of common tangents to the hyperbolas 9x2
– 16y2
= 1 and 9y2
– 16x2
= 1 are
(a) 2 (b) 2 (c) 1 (d) None of these
7 The two conics 2
2
by
– 2
2
ax
= 1 and y2 = – ab
x intersect ifff
(a) 0<b 21
(b) 0<a 21
(c) b2 < a2 (d) b2 > a2
8 The point on the hyperbola 24x2
– 18y2
= 1 which is nearest to the line 3x+2y+1 = 0 is
(a) (–6,3) (b) (3,–6) (c) (–6,–3) (d) (6,3)
PRACTICE QUESTIONS
74
9 If (asec , btan ) and (asec , btan ) be the coordinates of the ends of a focal chord of the
hyperbola 2
2
ax
– 2
2
by
= 1, then tan 2 tan 2 =
(a) e–1e1
(b) e1e–1
(c) 1e1–e
(d) None of these
10 If the latus rectum of a hyperbola through one focus subtends 60° angle at the other focus, then itseccentricity e is(a) 2 (b) 3 (c) 5 (d) 6
Answers
1. c 2. b 3. a 4. c
5. d 6. c 7. a 8. d
9. b 10. b
75
COMPLEX NUMBER - IAlgebra of Complex Number
Our motive for the introduction of complex numbers is to make every algebraic equation solvable.Let us consider the equation z2+4 = 0. This equation has no solution in the set of real numbers.There is no real no. x whose square is –4. In order to remedy this situation, a new kind ofnumbers were introduced and were given the name complex nos. Eular was the first to introduce
the symbol i for 1– with the property i2 = –1.i is also called the symbol as the imaginary unit.i is called an imaginary number.
Powers of i(i) ii1 1–i2 i–i.ii 23
1i4 ii.ii 45 1–i.ii 246 i–i.ii 347
1i.ii 448 ii.ii 89 1–i10 i–i11
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1i p4 ii 1p4 1–i 2p4 i–i 3p4
1i 4– iii
i1i 43
3– 1–i1i 2
2– i–i1i 1–
Complex numbersAn expression of the form x+iy, where x & y are real numbers and i is a symbol, is called acomplex number and usually denoted by z.
z = (x) + i (y)
Real part of z or Re (z) Imaginary part z or Im (z)Note: If Re z = 0, then z is Note: If Imz = 0, then z iscalled purely imaginary called purely real.
* Every real number is a complex number.* 0 is both purely real and purely imaginary number.* A complex is imaginary no. if and only if its imaginary part is non-zero. Here real part
may or may not be zero. 7+11i is an imaginary number but not purely imaginary.
Addition of complex numberLet z1 = a1+ib1 & z2 = a2 + ib2be two complex numberThen their sum z1 +z2 is defined as the complex number (a1+a2) + i(b1+b2)
an
i0 1
76
Multiplication of Complex NumberLet z1 = a1+ ib1 & z2 = a2 + ib2 be two complex number. Then the multiplication of z1 with z2 isdenoted by z1 z2 and is defined as (a1a2–b
1b
2) + i(a
1a
2+ b
1b
2).
Division of Complex NumberThe division of a complex number z1 by a non-zero complex z2 is defined as the multiplicaiton of
z1 by the multiplication inverse of z2 and is denoted by2
1
zz
22
11
2
1
ibaiba
zz
22
22
22
11
ib–aib–a
ibaiba
= 22
22
2121
babbaa
+ 22
22
2112
babb–aai
Conjugate of ZLet z = a + ib be a complex number. Then the conjugate of z is denoted byz and is equal to a–ibthus z = a + ib
z = a – ibe.g. if z = 3 +4i then z = 3 – 4i
Modulus of Z
The modulus of a complex number x + iy is denoted by iyx and is denoted as iyx =
22 yx = non-negative square root of x2 + y2 e.g. z = 3 – 4i, then z = 54–3 22 .
Equality of complex numberTwo complex nos z1 = x1 + iy1 & z2 = x2 + iy2 are said to be equal if and only if x1 = x2 & y1 = y2i.e. Z1 = Z2 Re(z1) = Re (z2) & Im (z1) = Im (z2).
Example: 1 Evaluate 53 1 2 3n n n n
G.E =i–(4x13+1) + [in (1+i+i2+i3) +4] 21
.i
11 2
11 2
11 2
1
2
1 1 4 .
(0 ) 4 .
( 4 ) .
2
1 2 2 2
n
n
i i i i i
i i i
i i
i i
ii i i i iii
77
Example: 2 Show that the polynomial x sx rx qx p 3424144 is divisible byx3+x2+x+1 ,where p,q,r,s N.
Solution: Let f(x) = x sx rx qx p 3424144Now x3+x2+x+1 = (x2+1)(x+1) =(x+i)(x-i)(x+1)
f(i) = i si ri qi p 3424144 = 1+ i+i2+i3 = 1+ i -1- i =0
f(-i) = )()()() )34()24()14(4( iiii srqp= +1+(– ) (+1) + (–i) 2 + (–i) 3 = 1-. –1+i=0
f(–1) = 0
Thus by division theorem x3 +x2 +x +1 is factor of x sx rx qx p 3424144 .
Exercise: 3 Express sin2cos–1
1i
in the form x+iy
sin2cos–11
i = 2
cos2
4sin2
sin 22
1
= 2 2
sin 2 cos sin 2 cos2 2 2 2
2 sin sin 2 cos sin 2 cos 2 sin sin 4 cos2 2 2 2 2 2 2 2
i i
i i
= 2 2
sin 2 cos2 2
sin 2 sin 8 cos2 2 2
i
= sin 2 cos
2 2
sin 1 cos 4 4 cos2
i
= sin 2 cos
2 2
sin 5 3 cos2
i
i i
i
78
= 2 cot1 2
5 3 cos 5 3 cos
Exercise: 4 Find the multiplicative inverse of complex no 3+2iLet z = 3+2i
3 2 3 2 3 23 2 3 2 9 4 13 13
i i ii i
1. If a < 0, b > 0, then b.a is equal to
a. ba– b. ib.a– c. ba d. none of these
2. The value of the sum , where 1–i ,is
a. i b. i –1 c. – i d. 0
3. The smallest positive integral value of n of which n
ii
11
is purely imaginary with positive
imaginary part, isa. 1 b. 3 c. 5 d. none of these
4. If n is an odd integer, 1–i , then (1+i)6n +(1–i)6n is equal toa . 0 b. 2 c. –2 d. none of these
5. If sini2–1sini23
is a real number and 0 < < 2 , then =
a. b. 2 c. 2 d. 6
6. If b+ic=(1+a)z and a2 + b2 + c2 =1, then iziz
11
a. ciba
1 b. ciba
1 c. ciba
1 d. ciba
1
7. If 31
iyx = a + ib then by
ax
=
a. 0 b. 1 c. –1 d. none of these
i
PRACTICE QUESTIONS
79
8. If (a + ib)5 = i , then (b+ai)5 is equal toa. i b. i– c. i– d. i––
9. The set of values of Ra for which x2 + i(a–1)x+5 = 0 will have a pair of conjugate complexroots isa. R b. {1}
c. d. none of these
10. The relation between the real numbers a and b, which satisfy the equation ix1ix–1
=a – ib, for some
real value of x, is
a. (a – b) (a + b) = 1 b. bab–a
c. a2+b2 = 1 d. none of these
Answers :1.b 2.b 3.b 4.a 5.a 6.d 7.d 8.a 9.b 10.c
}0a2–a{ 2: a 21.
80
COMPLEX NUMBER - IISquare Root & Polar Form
Square root of a complex number1. If the square root of a+ib is to evaluated let iy , x , y Rxiba2. Square both sides and equate real and imaginary part which will give value of (x2–y2) and .3. Find x2+y2 by 4. From x2 – y2 & x2 + y2, we get the value of x.5. Put x in xy, we obtained corresponding value of y.6. Now iyxiba
Direct FormulaThe square root of z = a + ib are
2a–z
i2
azfor b > 0
& 2
a–zi–
2az
for b 0
Note:
i. abba is ture only when at least-one of a & b is non-negative.
ii. The square root of are 2
iii. The square root of are
iv. The square root of i are
v. The square root of – i are
Example: Find the square root of –7–24iSolution: Here a =-7 & b = -24 0
2524–7–z 22
Now by using formula,
i24–7– = i4–32
725i–2
7–252
a–zi–
2az
2xy
81
Geometrical Representation of a Point is:Coordinate Geometry Vector Complex Number
a
b
P(a,b)
Cartesian Plane
y
xa
b
P(a,b)
Cartesian Plane
y
xCartesian Plane
r
jb
rjbiaP
Complex plane, Gaussian planeor Argand plane
P (z=x+iy)
yx
22 yxzr
Modulus
The modulus of a complex number x+iy is denoted by iyx = 22 yx = non negative squareroot of x2+y2
e.g. z = 3 – 4i, then z = 54–3 22
Argument or Amplitude of a complex number
z = x+iy
x
y
P(x,y)22 yxr
22 yxxcos & 22 yx
ysin
rxcos & r
xsin
x = rcos & x = rsinThe argument of a complex number z = x+iy is the value of which satisfies the two equations
22 yxxcos and
22 yxysin
Argument of z is denoted by argument z or amplitude z.There will be infinite number of values of satisfying the above equations and all these valueswill be the argument of z but usually we take only that value of to which 20Example: z = –1 –i here x = –1, y = –1
21–
yxxcos
22 45,
43
82
21–
yxysin
22 47,
45
since4
5 satisfies both the equation
argument z = 45
and general value of argument z = 45n2 , where n = 0, 2,1 .....
Another way of finding argument of a complex numberWorking Rule
i. Take xytan and from this find the value of lying between 0 and 2
ii. Then find in which quadrant the point z lies.iii. Argument of z will be , ,– or 2 according as the point z lies in the 1st, 2nd,
3rd or 4th quadrants.Exercise: Let z = –1 –i here x = –1, y = –1
4tan1
1–1–
xytan
4 (between 0 and 2 )
Since the point z = –1 –i (–1,–1) lies in 3rd quadrant
argument = = 45
4Principal vaule of the argument: There are infinite values of satisfying the equation
22 yxxcos and
22 yxysin
But there will be a unique value of such that – . The value of argument satisfying theinequality – is called principal value of argument.For above example
Principal value of argument = 43–2–
45
Note: If argument , subtract 2 from it to get the principal value of argument and if ar-gument – , add 2 to it, to get the principal value of argument.
Polar form of a complex Numberz = x+iy= sinicosr is called the polar form of complex Number..
83
For above example
2zr
Polar form of z is 4
5sini4
5cos2
or43–sini
43–cos2
Complex Number
Cartesian Representation Polar Repersentationz = a+ib z = sinicosr
= sinicosr
= ire (using euler’s formula)Algebraic operations
Cartesian form Polar form
Complex No. z = a + ib, w= c + id z = ire ,w= ise
Addition z +w= (a + c) + i (b + d) z +w= cosscosr +sinssinri
Subtraction z –w= (a – c) + i (b – d) z – w= coss–cosr +
sins–sinri
Multiplicaiton zw= (ac – bd) + i zw = rsei( + )
(ad + bc)
Division id–cidcid–cidaz –ie
sr
wz
= 2222 dcad–bci
dcbdac
Conjugate of a Complex Number
The conjugate of a complex number z =x + iy is denoted by z = iyx and is defined as
z = iyx and if z = eir (Eular’s form), then z = e ir
84
Equality of Complex NumberTwo complex numbers z1 =x1 + iy1 and z2 = x2 +iy2 are said to be equal if and only if x1 = x2 & y1 =+y2i.e. z1 = z2 , then Re(z1) =Re(z2) & Im(z1) = Im(z2)
Properties of Conjugate1. z–zz–z2. zz
3. zz arg–arg = arg (–z) – arg z– = 2
4. 2zzz
5. 2121 zzzz
6. nn zz
7. 2121 z.zz.z
8.2
1
2
1
zz
zz
Example-Write the following in Polar form(1)z =1 + 3 i
(2) z = i31(3) z = i31(4) z = i31
Solution : z = 1 + 3 i (1, 3 )
r = z = 3)1(22 =2
Then tan = 3tan3
13
xy
(1) As z lies in 1st quadrant
so, argz = = = 3Therefore polar form of z is
2(cos 3 + i sin 3 )
85
(2) z lies in 2nd quadrant so argz = = = 32
Therefore polar form of z is
2(cos 32
+ i sin 32
)2 22 cos sin3 3
i
(3) z lies in 3rd quadrant so argz = = = 34
Therefore polar form of z is
2(cos 34
+ i sin 34
)
(4) lies in 4th quadrant so argz = = = 3Therefore polar form of z is
2(cos( 3 ) + i sin( 3 ))
1. Im(z) is equal to
a. izz21
b.zz
c. izz21
d. none of these
2. If z1 = 9y2–4–10ix, z2 = 8y2–20i where 21 zz then z = x+iy is equal toa. –2+2i b. i22– c. i2– d. none of these
3. If z is a complex number satisfying the relation i12z1z , then z is
a. i4121
b. i4321
c. i4–121
d. i4–321
4. For a complex number z, the minimum value of zz isa. 1 b. 2 c. 3 d. none of these
5. If 1z , then z1z1
is equal to
a. z b. z c. z+ z d. none of these6. If 11–z1 , 22–z2 , 33–z3 , then 321 zzz
a. is less than 6 b. is more than 3c. is less than 12 d. is between 6 and 12
7. If z1,z2 are two non-zero complex numbers such that 2121 zzzz , then complex arg 2
1zz
equal to
a. b. – c. 0 d. 2
2 –
i2
( is
PRACTICE QUESTIONS
86
8. If z =i–3i3
, then the fundamental argument of z is
a. 3–
b. 3 c. 6 d. none of these
9. If z = x+iy satisfies amp(z–1) = amp (z+3i) then the value of (x–1): y is equal toa. 2:1 b. 1:3 c. –1:3 d. none of these
10. If (1+x)n = a0+a1x+a2x2+.....+anx
n, then (a0– a2+......)2 + (a1– a3+......)2 is equal to
a. 3n b. 2n c. n
n
212–1
d. none of these
11. If x = 2+5i and 2 !7!31
!9!11
+ !2!5
!51
b , then the value of x3–5x2 +33x–19 is equal to
a. a b. b c. a–b d. none of these
12. If x + iy such that 1–z1z and complex 1z1–z
= 4 then
a. x = 12 , y = 0 b. x = 0, y = 12c. x = 0, y = 1–2 d. x = 1–2 , y = 0
13. If the square root of 1631
21
2
2
2
2
yx
xy
ixy
yx is
mi–
xy
yx
, then m is
a. 2 b. 3 c. 4 d. none of these
14.If i = 1– , then 4+5334
23i
21– +
365
33i
21–3 is equal to
a. 3i–1 b. 3i–1– c. 3i d. 3i–
15. If z1, z2 and z3 are complex numbers such that 321 zzz = 1z1
z1
z1
321,
then 321 zzz isa. equal to 1 b. less than 1 c. greater than 3 d. equal to 3
16. The complex number sinx + icosx and cosx – isinx are conjugate to each other for
a. nx b. x = 0 c.21nx d.
Answers :1.c 2.b 3.c 4.b 5.a 6.c 7.c 8.b 9.b 10.b11.b 12.b 13.c 14.c 15.a 16.d
nx4
or x4
87
COMPLEX NUMBER - IIIArguments and Modulus
Props of Argarg( z ) = – arg zz = x+iy= r(cos + isin ) ___ polar form= rei ____Euler’s formarg z = z = r (cos – isin )= re– i
= r (cos(– )+isin(– )arg z = – = – arg z.arg(z1z2) = arg z1+argz2
z1= r1 (cos 1+isin 1) arg z1 = 1
z2= r2 (cos 2+isin 2) arg z2 = 2
z1z2 = r1r2)( 21ie
= r1r2[cos[( 1+ 2)+isin( 1+ 2)]arg(z1z2) = 1 2
= arg z1+arg z2In general , we write arg(z1z2) = arg z1+argz2+2k , wherek is -1 or 0 or 1
arg2
1
zz
= arg (z1) – arg z2 + 2k where k = 0, –1 ,1
2
1
zz
= 2
1
rr
)–(i 21e
= 2
1
rr
[cos( 1– 2)+isin( 1– 2)]
arg2
1
zz
= 1– 2
= arg z1 – arg z2arg(zn) = n arg zz = r (cos + isin )
= r ie
zn = rn ie
+
z n
88
n [cos(n )+isinn ]arg zn = n = n arg z
Q.1 If arg z1 = 170° and arg z2 = 70°, then final the principal argument of z1 z2arg (z1z2) = arg(z1)+arg z2 = 240°Thus z1 z2 lies in 3rd quadrant. Hence its principal argument is –120°.
Q.2 If z1 and z2 are conjugate to each other, then find arg (–z1z2).z1 & z2 are conjugate to each other i.e z2 = 1z
arg (–z1z2) = arg (–z1 1z ) = arg (–|z1|2)
= arg (negative real no.)= .
Q.4 If z1, z2& z3, z4 are two pairs of conjugate complex nos, then find the value of arg 4
1
zz
+ arg 3
2
zz
.
z2 = 1z & z4 = 3z the forez1 z2 = |z1|
2 & z3 z4 = |z3|2
Now arg4
1
zz
+ arg 3
2
zz
= arg 43
21
zzzz
= arg zz
zz
3
1arg
3
12
2
2
=arg (positive real no.)=0
Properties of Modulus
1 |z1z2| = |z1| |z2|Explanation:Let z1 =x1 + iy1 &z2 = x2 +iy2
Now|z1z2|=|(x1+iy1)(x2+iy2)|=|(x1x2-y1y2)+ i(x1y2+x2y1)|= yyxxyyxx 2121212122
= zzyxyxyxyx 21
22222222
22112211
as |z1z2.................zn| = |z1||z2|.......................|zn|so, |zn| = |z.z..............to n factors| = |z||z|.......................to n factors=|z|n.
22
1
zz
= 2
1
zz
(try yourself)
3 |z1+z2|2 = |z1|
2 + |z2|2 + 2 re (z 2z )
4 |z1–z2|2 =|z1|
2 + |z2|2 – 2 re (z 2z )
re
= rZn
89
Explanation of (3) &(4): zzz2
zzzzzzzzzzzzzzzzzz 12121121212121
2
2221
|z1+z2|2 = |z1|
2 + |z2|2 2 re (z1 2z ) ( zzzzzz 212121
)
5 |z1+z2|2 + |z1 –z2|
2 = 2 (|z1|2 + |z2|
2)Explanation : Adding (3) & (4), we get the result.
6 |z1+z2| |z1| + |z2| (Triangle Inequality)Explanation : 2sin2cossincos
11211121 ii rrzz
= 22112211 sinsincoscos rrrr i =
= 22112211 sinsincoscos 22
rrrr
= )sinsincos(cos2221 212121
2
2
22
1
2
1
22
1sincossincos rrrr
= )cos(221 2121
22
rrrr
)1)cos((221 2121
22
rrrr|z1+z2| rr 21
2
or |z1+z2| rr 21
Thus |z1+z2| |z1| + |z2|7 |z1–z2| ||z1| – |z2||
Explanation: 2sin2cossincos11
211121 ii rrzz= 22112211 sinsincoscos rrrr i =
= 22112211 sinsincoscos 22
rrrr
= )sinsincos(cos2221 2121212
2
22
1
2
1
22
1sincossincos rrrr
= )cos(221 2121
22
rrrr)1)cos((221 2121
22
rrrr|z1 2| rr 21
2
or |z1 2| zzrr 2121
Thus |z1 2| |z1 2|
-z
-z
z
| | z
90
COMPLEX NUMBER - IVPractice Problems
1 If , are two different complex numbers such that | | = 1 or | | = 1 then the expression –1–
equals
(a) 21
(b) 1 (c) 2 (d) None of these
Solution
–1–
= ––
( = | |2 = 1)
= )–(–
= |)–(||–|
= |–||||)–)(1(–| = –||
|–||1–| = ||
1 = 1
2 If z = x+iy such that |z+1| = |z–1| and amp 1z1–z
/4, then
(a) x = 2 +1 y = 0 (b) x = 0, y = 2 + 1
(c) x = 0, y = 2 –1 (d) x = 2 –1, y = 0let z = x+iy
|z+1| = |z–1| |(x+1)+iy| = |(x–1)+iy|
22 y)1x( = 22 y)1–x( (x+1)2+y2 = (x–1)2+y2 x2+2x+1= x2–2x+1 4x = 0 x = 0
amp 1z1–z
= /4, amp 1iy1–iy
= /4 ....................(1)
1iy1–iy
× 1–iy1–iy
= )1y(–iy2–y–1
2
2
= 1)(y)y–(1
2
2
+ 2y1iy2
1yy2
2 × )y–1()1y(–
2
2
= tan /4 2y–1y2–
= 1
1–y2 = – 2yy2– 2y–1 = 0
y = 2
442 =
2222
= 1 2
,
=
91
1– 2 can be neglected as it is negative but tan lies in Ist quadrant.3 For any complex number z, the maximum value of |z|–|z–1| is
(a) 21
(b) 1 (c) 23
(d) 2
|z1–z2| ||z1|–|z2|| |z1| – |z2||z|–|z–1| |z–(z–1)| = 1
Maximum value of |z| – |z–1| = 1
4 If z1 & z2 are two complex numbers such that 21
21
z.z–2z2–z
is uni moduler, while z2 is not uni moduler, then
the value of |z1| is(a) 1 (b) 2 (c) 3 (d) 4Solution
21
21zz–2z2–z
= 1 |zz–2||z2–z|
21
21 = 1
221 |z2–z| = 2
21 |zz–2|
( 21 z2–z ) )z2–z( 21 = (2–z1 z 2) )zz–2( 21
(z1–2z2) ( 1z –2 2z ) = )zz–2( 21 )zz–2( 21
z1 1z –2z1 2z –2z2 1z +4 z2 2z= 4–2z1 2z – 1z z2 +z1 1z 2z 2z
|z1|2+4|z2|
2 = 4+|z1|2 |z2|
2
|z1|2(1–|z2|
2)–4(1–|z2|2) = 0
(|z1|2–4) (1–|z2|
2) = 0 |z2| = 1, |z1| = 2
5 If |z–3+2i| 4, then the sum of least and greatest values of |z| is(a) 2 11 (b) 3 11 (c) 2 13 (d) 3 13|z–3+2i| 4 ..........(1)
|z–3+2i| ||z|–|3–2i|||z–3+2i| ||z|– 13 |............(2)
From (1) & (2)4 |z–3+2i| ||z|– 13 |
||z|– 13 | 4
–4 |z|– 13 4
13 –4 |z| 4+ 13greatest value of |z| is 13 +4
&least value of |z| is 13 –4
their sum = 13 +4+ 13 –4
2| = 1 (not unimodular) so, |z1 | = 2since |z 1
92
= 2 136 If z1, z2, z3 are three distinct complex numbers and a, b, c are three positive real numbers such that
|z–z|a
32 = |z–z|
b
13 = |z–z|
c
21, then
32
2
z–za
+ 13
2
z–zb
+ 21
2
z–zc
is
(a) 23
22
21
22
21
23
21
223
22
2
zzz)zz(c)zz(b)zz(a
(b) 0
(c)133221
23
22222
22221
222
zzzzzzz)b–a(cz)a–c(bz)c–b(a
(d) None of these
Let |z–z|a
32 = |z–z|
b
13 = |z–z|
c
21 = k (ray)
a2 = k2 |z2–z3|2 etc
32
2
z–za
= k2 ( 2z – 3z )
13
2
z–zb
= k2 ( 3z – 1z )
21
2
z–zc
= k2 ( 1z – 2z )
32
2
z–za
+ 21
2
13
2
z–zc
z–zb
= k2 | 2z – 3z + 3z – 1z + 1z – 2z ) = 0
Cube root of unityz3 = 1z3 – 1 = 0(z–1) (z2+z+1) = 0z = 1, z2+z+1 = 0
z = 2
i31–
z = 1, z = 2i31–
, z =
2
2i3–1–
1+ + 2 = 03 = 1
93
w19 = w3×6+1 = (w3)2. =
De Mover’s Theorem(cos +isin )n =cosn +isinn .
nth roots of unity1 = cos0+isin011/n = (cos0+isin0)1/n
= n0r2sini
n0r2cos
= cos nr2
+ isin nr2
r = 0, 1/2 .............(n–1)
= nr2i
e where r = 0, 1/2 ............(n–1)
= 1, nr2i
e , n4i
e , n6i
e ........... n)1–n(2i
e
= 1, , 2................. n–1 where = nr2i
e(1) 1+ + 2+.........+ n–1 = 0(2) 1. . 2........... n–1 = (–1)n–1
Important Relations1 x2+x+1 = (x– ) (x– 2)2 x2–x+1 = (x+ ) (x– 2)3 x2+2y+y2 = (x–y ) (x–y 2)4 x2–xy+y2 = (x+y ) (x+y 2)5 x2+y2 = (x+iy) (x–iy)6 x3+y3 = (x+y) (x+y ) (x+y 2)7 x3–y3 = (x–y) (x–y ) (x–y 2)8 x2+y2+z2–xy–yz–zx = (x+y +z 2) (x+y 2+z )9 x3+y3+z3–3xyz = (x+y+z) (x+y +z 2) (x+yw2+z )
1 If w is a cube root of unity, then +...........
12827
329
83
21
is(a) +1 (b) i2 (c) 0 (d) None of these
...........12827
329
83
21
= 21 ...........
6427
169
431 = 2
1
43–1
1 = 2
1× 1
4 = 2
w+w2 = –1 = i2
2 , real are the roots of x3–3x2+3x+7 = 0
(w is the cube root of unity), then 1–1–
1–1–
1–1–
is
94
(a) 3/ (b) 2 (c) 2 2 (d) 3 2
3 we havex3–3x2+3x+7 = 0(x–1)3+8 = 0 (x–1)3 = (–2)3
3
2––x
= 1 2–1–x
= 1, , 2
x = –2+1, x–1 = –2 , x–1, = –2 2
x = –1, x= 1–2 , x = 1–2 2
= –1, = 1–2 , = 1–2 2
–1 = –2, –1 = –2 , –1 = –2 2
G.E = 22
+ 22–2–
+ 2
2 2
= 2+ 2+ 2
= 3 2
3 The common roots of eqin
z3+2z2+2z+1 = 0 & z1985+z100+1 = 0 are(a) 1, (b) 1, 2 (c) , 2 (d) None of these
z3+2z2+2z+1 = 0..................(1)(z+1) (z2+z+1) = 0z = –1, , 2 are the roots of (1)z = –1, z1985+z100+1 = 1 0z = , ( )1985+ 100+1 = 2+ +1 = 0z = 2, 3970+ 200+1 = 0
The common roots are w,w2
(c) is correct.
4 If , , are the cube roots of p, p<0 then for arang x,y & z zyxzyx
is
(a) w2 (b) –1 (c) 333
333 zyx(d) None of these
Let x3 = px = (1.p)1/3
x = p1/3. 11/3
x = p1/3, p1/3 , p1/3 2
Given expression = 3/1p.zp.yp.xzpypp.x
23/13/1
23/13/13/1
= zyxzyx
2
2
= )zyx()zyx(
2
22
= 2
1
(multiply and divide by )2
95
5 The value of the expression
11 2
11 +12 2
12 + 13 2
13 +.........+1n 2
1n
Tk = 1k 2
1k = (k+ 2) (k+ ) = k2–k+1
Sn = n
1kkT =
n
1k
2 )1k–k(
= 1k–k
= 6)1)(1n(n – 2
)1n(n +n
= n 12
1n–6
1n3n2 2
= n 663–n3–1n3n2 2
= 64n2 2
= 3
)2n(n 2 Ans
6 If w and w2 are the cube roots of unity and
a1
+ b1
+ c1
= 2 2 & 2a1
+ 2b1
+ 2c1
= 2 then the value of 1a1
1b1
1c1
is
(a) 1 (b) –1 (c) (d) 2Note that & 2 are roots of
xa1
+ xb1
+ xc1
= x2
.................(1)
)xc)(xb)(xa(abx)cba(2x3 2
= x2
x3–(bc+ca+ab)x–2abc = 0The roots of the equation are & 2
If in the third root of this equation then + + 2 = 0 = 1Put = 1 on equation (1)
1a1
1b1
1c1
= 2.
2n
bc
2
n
+
ca
+ +
+
96
1. The value of 5252 11 ,where and 2 are the complex cube roots of unity,,isa. 0 b. 32 c. -32 d. 32
2. For any two complex numbers z and z and any real numbers a and b,2
212
21 azbzbzaz isequal to
a. 2122 zzba b. 2
22
122 zzba
c. 22
21
22 zzba d. none of these
3. For any two complex numbersz1 and z2 , the relation 2121 zzzz holds,if
a. arg(z1)=arg(z2) b. arg(z1)+arg (z2)= 2c. z1z2 d. |z1|=|z2|
4. If and 2 are the two imaginary cube roots of unity, then the equation whose roots are382317 a anda is
a. x2 +ax -a2 = 0 b. x2 +a2x +a2 = 0 c. x2 +ax +a2 = 0 d. x2 -a2x +a= 05. If 1,a1,a2,..............an-1 are roots of unity, then the value of (1-a1)(1-a2)........(1-an-1) is
a. 0 b. 1 c. n d. n2
6. If 1is a cube root of unity, then the value of
2001002
200100
21002
22w11
11
a. 0 b. 1 c. d. 2
7. If the area of the triangle formed by the points z,z +iz and iz is 50sq. unit , then |z| is equal toa. 5 b. 8 c. 10 d. none of these
8. Let be an imaginary roots of xn = 1 .Then, (5 - )(5 - 2 ).............( 1n5 ) is
a. 1 b. 415n
c. 14n d. 415n
9. If 1)( is a cube root of unity and BA1 7
a. 0,1 b. 1,1 c.1,0 d.-1,110. If 22x then the value of x4 + 3x3 +2x2 - 11x -6 is
a. 1 b.-1 c.2 d. none of these
11. If z = isincos ,then 1z1z
2n
2n
is equal to (n is an integer)
a. i cotn b. i tan n c.tann d. cotn12. The cube root of unity
21
2
PRACTICE QUESTIONS
97
a. lie on the circle |z|=1 b. are collinearc.form an equilateral triangle d. none of these
13. The complex number z satisfying the equation |z| - 4 = |z - i| - |z +5 i| = 0, isa. i3 b. 32 c. 2i32- d. 0
14. If z1 = 8 + 4 i, z2 = 6 + 4 i and 4zzz-zarg
2
1, then z satisfies
a. |z -7 -4 i| =1 b. |z - 7 -5 i|= 2 c. 84i-z d. 187i-z
15. If z =cos + i sin , then
a. 2cosnz1 z n
nb. cosn2
z1 z n
nn
c isinn2z1 z n
nn d. sinn2i)(
z1 z n
nn
ANSWERS
1.d 2.b 3.a 4.c5.c 6.a 7.c 8.d9.b 10.a 11.b 12c13.c 14.b 15.a
98
COMPLEX NUMBER - VLoci in Complex plane
1. Distance formulaDistance between A & Bi.e. AB = 21 z–z
Imz
A(z )1
B(z )2
Rez
2.Section formulaIf P divides AB in the ratiom:n (internally), then P is
nmnzmz 12
A(z )1
B(z )2
m
n
P
In case of external division ,P is nmnzmz 12 .
If P is the mid point of AB then
P is 2zz 21
Straight lineAPOAOP
i.e.t is some suitable real number
Now writing corresponding complex number, we get
99
z = z1 + t (z2–z1) ......(1)Since Value of t is different for different position of P.t is called a parameter & the equation (1) is called the parametric equation of straight
line passing through the fixed points A (z1) & B (z2).Cartesian equation of (1) Put z = x+iy, z1=x1+iy1 & z2 = x2+iy2 in (1)
If we get x+iy = x1+iy1+t((x2–x1)+(y2–y1))x = x1+t(x2–x1) & y = y1+t(y2–y1)
or ty–yy–y
x–xx–x
12
1
12
1
From (1) tz–zz–z
12
1
t is Purely real number.
so, -----------------(2)
12
1
12
1
z–zz–z
z–zz–z
112112 zzzz–zz–zz 111212 zzzz–zz–zz
0zzzzzz111
21
210
1zz1zz1zz
22
11 ........(3)
Slope of this line is21
21
z–zz–z .
Since, 0z.z–z.zz–zz–z–zz 21212121
0z.z–z.zizz–zi–zz–zi 21212121
0bzaza .......(4)where b = 2121 z.z–zzi
= 2121 .– zzzzi
= i.2i Im 21zz
= –2Im 21zz = a real number & 0a as 21 zz
* Points z1,z2,z3 are collinear if and only if 01zz1zz1zz
33
22
11
1z( z ,then z is purely real)
100
* Equation (2) & (3) represent the non parametric form of equation of straight linepassing through the point A(z1) & B(z2)
* Equation (4) is called general equation with slopeEquation of Perpendicular Bisector
We consider two fixed points A(z1) & B(z2) and variable point P(z) moving such that AP = BPi.e. 21 z–zz–z ......(5)
2z z
From Geometry, it is obvious that P lies on the perpendicular bisector of the segment AB.Equaiton (5) is the equation of perpendicular bisector of the segment joining A(z1) & B(z2).
Another form of perpendicular bisector
22
2121 z–zz–zz–zz–z
* If A(z1) and B(z2) are two fixed points and P is a variable moving on the line segment AB thenAP+BP = ABi.e. 2121 z–zz–zz–z .......(6)
above equation represents the line segment AB
i.e. 1121 z–zz–z–z–z .......(7) i.e. 1121 z–z–z–z–z–z .......(8)From equation (7) & (8), it is clear that p is lying on the line passing through A & B but not lyingon the segment AB.
= ––
–
101
If z1 and z2 are two unequal complex numbers represented by points P and Q,
then 21
21
z–zz–z
is called the complex slope of the line joining z1 &z2.
Thus w =21
21
z–zz–z
then =21
21
z–zz–z
= 1z–zz–z
21
21
* Two lines having complex slopes w1 and w2 are parallel if and only if w1 = w2
w1 =21
21
z–zz–z
(complex slope of AB)
w2 =43
43
z–zz–z
(complex slope of CD)
AB||CD angle b/w AB and CD = 0 or
argument 0z–zz–z
34
12 or
is real
43
21
43
21
34
12
34
12
z–zz–z
z–zz–z
z–zz–z
z–zz–z
2143
43
21
21 wwz–zz–z
z–zz–z
* The two lines having complex slope w1 & w2 are perpendicular if and only if w1 = –w2.* The equation of a line parallel to 0bzaza is
0zaza , where * The equation of a line perpendicular to the line 0bzaza is
0za–za , where * The length of perpendicular from a point P(z0) to the line 0bzaza is
a2bzaza 00 .
102
CIRCLE
Consider a fixed point c(z0) and a variablepoint P(z) which is moving, keeping its distance
from the point C a constant RAt any position, P(z) satisfies the
equation Rz–z 0 ......(1)Equation(1) represents the points on a
circle whose centre is c(z0) & radius is R.
Cartesian form22
0 Rz–z .Let z = x+iy &z0 = x0 + iy0
2200 Ry–yix–x
(x–x0)2 + (y–y0)
2 = R2
General equation of a circle is0bzazazz where b is real. The cen-
tre of this circle is –a and radius
is b–ab–aa 2
Explanation: Let z0 be the centre of the circleand P(z) be any point on the circle.
Then rz–z 0 or 220 rz–z
(z–z0) 0z–z = r2 (z–z0) 0z–z = r2
0r–zzz–zz–zz 22000
0bzaza–zz ......(1)
Where a = –z0, b = 0r–z 220 real number clearly centre of the circle (1) is z0 i.e. –a &
radius=
* Equation of the circle whose one diameter is the linesegment joining z1 & z2 is
0z–zz–z
z–zz–z
2
1
2
1
2APB
103
argument2z–z
z–z
1
2
z–zz–z
1
2 is purely imaginary
or z–zz–z–
z–z
z–z
1
2________
1
2
z–zz–z–
z–zz–z
1
2
1
2
z–zz–z–
z–zz–z
1
2
2
1
1
2
1
2
z–zz–z–
z–zz–z or 0
z–zz–z
z–zz–z
2
1
2
1
(z–z1) 2z–z +(z–z2) 1z–z = 0
* 21 z–zkz–z KBAAP Locus of P(z) is a circle 1k* 2121 z–zkz–zz–z PA+PB = k be locus of z is an ellipse where A & B are foci of
the ellipse.* 2121 z–zkz–z–z–z PA–PB = k be locus of z is an ellipse where A & B are foci at
A & B.
* 221
22
21 z–zz–zz–z , locus of z is a circle.
* argument (fixed), locus of z is segment of a circle.
* argument , locus of z is a circle with z1 and z2 as the vertices of diameter..
* argument , locus of z is a straight line passing through z1 & z2.
104
1. If Z = x + iy and = iziz1
,then 1implies that, in the complex plane
a. z lies on the imaginary axis b. z lies on the real axisc. z lies on the unit circle d. none of these
2. The points z1,z2,z3,z4 in the complex plane are the vertices of a parallelogram taken in order ifand only ifa. z1 +z4 = z2 +z3 b. z1 +z3 =z2 +z4 c. z1 +z2 = z3 +z4 d. none of these
3. For all complex numbers z1,z2 satisfying 1z =12 and 5i43z2 , the minimum value of
21 zz isa. 0 b. 2 c. 7 d. 17
4. If 1z , then the point representing the complex number z31 will lie ona. a circle b. a straight line c. a parabola d. a hyperbola
5. If z = 253 i ,then the locus of z isa. ellipse b. semicircle c. parabola d. straight line
6. If z is complex number then the locus of z satisfying the condition 112 zz isa. perpendicular bisector of line segment joining 1/2 and 1 b. circlec. parabola d. none of the above curves
7. If Im 2112
izz then the locus of the point representing z in the complex plane is
a. circle b. a straight line c. a parabola d. none of these
8. If 4321 zzzz then the points representing z1,z2,z3,z4 area. concyclic b. vertices of a squarec. vertices of a rhombus d. none of these
9. The equation 05)34()34( ziizz represents a circle whose radius is
a. 5 b. 52 c.25 d. none of these
10. The equation 0kk,iziz , can represent an ellipse if k2 isa. <1 b. <2 c. >4 d. none of these
11. The equation kiziz represents hyperbola ifa. 22 k b. 2k c. 2k0 d. none of these
12. Let 21 2 ttitz ,where t is a parameter. The locus of z in the Argand plane isa. a hyperbola b. an ellipsec. a straight line d. None of these
z
PRACTICE QUESTIONS
105
13. Let z1 and z2 be two non-real complex cube roots of unity and 22
21 zzzz be the
equation of a circle with z1,z2 as ends of a diameter, then the value of is
a. 4 b. 3 c. 2 d. 2
14. If the equation kzzzz 22
21 represents the equation of a circle, where z = 2 + 3i,
= 4 + 3i are the extremities of a diameter, then the value of k isa. 1/4 b. 4 c. 2 d. none of these
15. Let and be two fixed non-zero complex numbers and z a variable complex number. If the
lines 01zz1andzz are mutually perpendicular, then
a. 0 b. 0 c. 0 d. 0
ANSWERS
1.b 2.b 3.b 4.a5.b 6.b 7.b 8.a9.a 10.c 11.a 12.a13.b 14b 15d
z
1
2z
106
COMPLEX NOMBER -VIProblem Solving and nth Roots of Unity
1 Let x,y R & i = 1– , then z = x+iy is called a complex numberRe(z) = x, Im(z) = yIf x = 0, z is purely imaginaryIf y = 0, z is purely real
2 For two complex numbers z1 = x1+iy1 z2 = x2+iy2(i) z1 = z2 if and only if x1 = x2 and y1=y2Note that no order relation is possible among the set of complex numbers. i.e. it is wrong tosay 1+i<4+3i. But |1+i| < |4+3i|(ii) z1 z2 = (x1 x2)+i(y1 y2)(iii) z1z2 = (x1x2–y1y2)+i(x1y2+x2y1)
(iv)2
1
zz
= 22
22
21122121
yx)yx–yx(i)yyxx(
; z2 0
3 Conjugate of zConjugate of z = x+iy is defined as z = x–iy (i.e. replace i by –i)Properties(i) z = z(ii) z = z z is puraly real(iii) z = – z z is purely imaginary(iv) z+ z = 2Re(z) = 2Re( z )(v) z– z = 2iIm(z)(vi) 21 zz = 1z 2z
(vii) 22zz = 1z 2z (also nz = ( z )n)
(viii)2
1
zz
= 2
1
zz
4 Geometric representationCoordinate Complex number Vectorrepresentation representation representationPoint P :P(x,y) Point P : z = (x+iy) Point P : Position vector
of P OP = x i +y j5 Modulus and amplitude
|z| = r = 22 yx , the modulus of z (distance of z from the origin)
= tan–1xy
= arg z, the inclination of OP with positive direction of x–axis where P is (x,y).
Here – < arg zz = x+iy ; x,y R (algebraic form)
107
= r(cos +isin ) or r cis (polar/trigonometric form)= r ie (Eulers form)
Properties of modulus(i) –|z| Re(z),Im(z) |z|(ii) |z|=| z |+|–z|+|– z |(iii) z z =|z|2(iv) |z1z2|=|z1||z2|
Also |zn|=|z|n
(iv)2
1
zz
= |z||z|
2
1
(vi) |z|–|z| 21 |z1 z2| |z1|+|z2| (Triangle inequality)
i.e.|z1|+|z2| is the maximum and |z|–|z| 21 is the minimum value of |z1 z2|
(vii) |z1 z2|2 =
2211
21212
22
1
212
22
1
zarg&)zarg(where)cos(|z||z|2|z||z|
)zzRe(2|z||z|
(viii) |z1+z2|2+|z1–z2|
2=2(|z1|2+|z2|
2) (parallelogram law)Also |az1–bz2|
2+|bz1+az2|2=(a2+b2)(|z1|
2+|z2|2); a,b R
Note : If z1z = a, the greatest and least value of |z| are respectively
24aa 2
and
24aa– 2
.
Properties of argument(i) arg(z1z2)=arg(z1)+arg(z2)+2k , where k= –1 or 0 or 1
Also arg(zn) = narg(z)+2k
(ii) arg2
1
zz
= arg z1–arg z2 + 2k , where k = –1 or 0 or 1
(iii) arg zz
= 2arg(z)+2k , where k = –1 or 0 or 1
(iv) argument of zero is not definedarg(z) = 0 z is real and positivearg(z) = z is real and negative
arg(z) = 2 z is purely imaginary, Imz>0
108
arg(z) = 2–
z is purely imaginary, Imz<0
(v) |z1+z2| = |z1| + |z2|
argz1 – argz2 = 0, 2
1
zz
>0
0, z1, z2 are collinear and z1, z2 lie on the same side of 0.(vi) |z1–z2| = |z1| + |z2|
argz1–argz2 = ,2
1
zz
<0
0, z1, z2 are collinear and 0 lies between z1 & z2.(vii) |z1–z2| = |z1+z2|
argz1–argz2 = 2
2
1
zz
and 1z z2 are purely imaginary..
6 De Moivres theorem(i) For any rational number n, then
(cos + i sin )n = cosn +i sinni.e. , ( ie )n = ine
(ii) (cos 1+ i sin 1) (cos 2+ i sin 2)..................(cos n + i sin )= cos( 1+ 2+......+ n) + i sin( 1+ 2+...............+ n)
(iii) If z = r(cos + i sin ) and n Z+, then
z1/n = r1/n nr2sini
nr2cos where r = 0, 1, 2, ...........(n–1)
(nth roots of z)7 The nth roots of unity
z = n 1 = 11/n = n/r2ie , r = 0, 1, 2, ............,(n–1)Let z = r where = n/2ie
The nth roots of unity are ( °=) 1, , 2,......... n–1 where = ei2 /n = cos n2
+ i sin n2
Properties(i) nth roots of unity are solutions of the equation z = 11/n
i.e. zn = 1z n–1 = (z–1)(z– ) (z– 2)..........(z– n–1)nth roots of –1 are the solutions of zn +1 = 0
(ii) nth roots of unity lie on a unit circle |z| = 1 and divide the circumference into n equal parts andare the vertices of a regular polygon of n sides inscribed in the circle |z| = 1
n
109
(ii) Product of nth roots of unity = (–1)n–1
(iii) Sum of nth roots of unity is always zero.(v) nth roots of unity form a G.P with common ratio n/2ie(vi) Sum of pth power of nth roots of unity
= 1+ p+( 2)p+( 3)p +..............(+ n–1)p = nofmultipleaispif;nnofmultipleanotispif;0
8 Cube roots of unity
11/3 = cos 3r2
+ i sin 3r2
; r = 0, 1, 2
= 1, 2
3i1–,
23i–1–
If one of the non real complex roots be , then the other non real complex root will be 2.
i.e. if = 2
3i1– , then 2 =
23i–1–
the 3 cube roots of unity are 1, & 2.Properties(i) z3–1 = (z–1) (z– ) (z– 2)(ii) & 2 are roots of z2+z+1 = 0 i.e., z2+z+1 = (z– ) (z– 2)(iii) arg = 2 /3 & arg 2 = 4 /3(iv) cube roots of unity lie on the unit circle |z| = 1 and divide its circumference into three equal
parts(v) If A(1), B( ) & C( 2), then ABC is an equilateral triangle.(vi) 3 = 1 ; 1+ + 2 = 0 ; 3n = 1 ; 3n+1 = ; 3n+2 = 2
(vii) = 1
= 2 ; 2 = 2
1 =
(viii) x2+y2 = (x+iy) (x–iy) x3+y3 = (x+y) (x+ y) ( x+ 2y) x3–y3 = (x–y) (x– y) ( x– 2y) x2+xy+y2 = (x– y) ( x– 2y) x2–xy+y2 = (x+ y) ( x+ 2y) x2+y2+z2–xy–y z–zx = (x+y +z 2) (x+y 2+z ) x3+y3+z3–3xy z = (x+y+z) (x+y +z 2) (x+y 2+z )
Note If a+b+c = 0 = a2+b2+c2 , then a : b : c = 1: : 2
Any complex number for which partimaginarypartreal
= 1 : 3 or 3 : 1, can be expressed
in terms of , 2 & i9 Square root of a complex number
Let z = a+ib
110
iba = ))zRe(–|z(|21i))zRe(|z(|
21
i.e. iba = )a–ba(21i)aba(
21 2222
To find the square root of a–ib, replace i by – i in the above result.10 Logarithm of a complex number
Let z = a+ib = r ie
|z| = r = 22 balog z = log |z| + iarg z
11 Expansions(i) cosn = nC0 cosn –nC2cosn–2 sin2 +nC4cosn–4 sin4 +............(ii) sinn = nC1cosn–1 sin –nC3cosn–3sin3 +nC5cosn–5sin5 ........ (using De-Moivres theo-rem)
(iii) cosn = 1–n21
{nC0cosn +nC1cos(n–2) +nC2cos(n–4) +....................}
(iv) sinn = 1–n
2/n
2)1(–
{cosn –nC1cos(n–2) +nC2cos(n–4) +...........}
sinn = 1–n
21–n
2)1(– {sinn –nC1sin(n–2) +nC2sin(n–4) +...........}
Solved Examples
1 If is a complex cube root of unity, then the value of 2
2
baccba
+ 2
2
acbcba
is equal to
(a) –1 (b) 2 (c) 2 (d) None of theseSolution :
1
caa)cba(
2
32
+ 21
2
432
acb)cba(
= 1
+ 21
2+ = – 1
Ans (a)
2 If x2+x+1 = 0, then the value of 2
x1x +
2
22
x1x + ..............+
2
2727
x1x is
(a) 27 (b) 72 (c) 54 (d) None of theseSolution:
111
Roots of x2+x+1 = 0 are and 2
Put x=21
+ 2
22 1
+ 2
33 1
...........+2
2727 1
= (–1)2 + (–1)2 + (2)2 + .................+(2)2
= 18×1 + 9×4 = 54Ans (c)
3 If , , are roots of x3–3x2+3x+7 = 0 and is a complex cube roots of unity, then
1–1–
+ 1–1–
+ 1–1–
is equal to
(a) (b) 2 (c) 2 2 (d) 3 2
Solution:(x–1)3 = –8x–1 = x –1 = – 2, –2 , –2 2
x= – 1, 1–2 , 1–2 2
Let = –1 = 1–2 , = 1 –2 2
– 1 = – 2 ; –1 = – 2 , =1 –2 2
1–1–
+ 1–1–
+ 1–1–
= 2–2–
22–2–
2–2– 2
= 1
+ 1
+ 2 = 3 2
Ans (d)4 If (x–1)4 –16 = 0, then the sum of non real complex roots of the equation is
(a) 2 (b) 0 (c) 4 (d) None of theseSolution:
(x–1)4 = 16
x–1 = 41
)16(x–1 = 2, 2ix = 1 2, 1 2i
Sum of non-real roots is (1+2i) + (1–2i) = 2Ans (a)
5 If z is a non-real root of 7 1– , then z86 +z175+ z289 is equal to(a) 0 (b) –1 (c) 3 (d) 1Solution :
z = 7 1– z7 = –1z86 + z175 + z289 = (z7)12 z2 + (z7)25+(z7)41.z2
+ +
112
= z2–1–z2 = –1Ans (b)
1 The value of 6
1k 7k2cosi–
7k2sin is
(a) –1 (b) 0 (c) –i (d) i2 For positive integers n1, n2 the value of expression
1n)i1( + 1n3 )i1( + 2n5 )i1( + 2n7 )i1( is a real number if and only if(a) n1 = n2+1 (b) n1 = n2–1 (c) n1 = n2 (d) n1>0, n2>0
3 The minimum value of |a+b +c 2| where a, b & c are all not equal integers and ( 1) is a cuberoot of unity, is
(a) 3 (b) 21
(c) 1 (d) 0
4 If x is a complex root of the equation
1xxx1xxx1
+ x–111
1x–1111x–1
= 0, then 2005x +2005x1 is
(a) 1 (b) –1 (c) i (d)5 If z = i elog (2– 3 ), then cosz =
(a) 0 (b) 1 (c) 2 (d) None of these
6 If is a complex cube roots of unity and a, b, c are such that a1
+ b1
+ c1
= 2 2
and 2a1
+ 2b1
+ 2c1
= 2 , then 1a1
+ 1b1
+ 1c1
=
(a) 1 (b) –1 (c) 2 (d) –27 If ( 1) be a cube root of unity and (1+ 2)m = (1+ 4)m, then the least positive integral value of
m is(a) 2 (b) 5 (c) 1 (d) 3
8 cot iy–xiyxlogi– e
(a) xy2y–x 22
(b) 22 y–xxy2
(c) 22 yxxy2
(d) xy2x–y 22
9 sin 72
+ sin 74
+sin 78
=
(a)27
(b) 21–
(c) 81
(d) None of these
PRACTICE QUESTIONS
113
10 If 0, 1, 2,.............., n–1 be the , nth roots of the unity then the value of 1–n
0i i
i
)–3( is equal
to
(a) 1–33
n (b) 1–31–n
n (c) 1–31n
n (d) 1–32n
n
Answers1 .d 2. d 3. c 4. a 5. c6. c 7. d 8. a 9. a 10. a
114
COMPLEX NUMBERS - VIIRotation Theorem
GEOMETRY OF COMPLEX NUMBERS1. Distance between two complex numbers z1 & z2 is 21 z–z2. Section formula
Two points P & Q have affixes z1 & z2 respectively in the argand plane, then the affix of a pointR dividing PQ is the ratio m:n
i. internally is nmnzmz 12
ii. externally is n–mnz–mz 12
iii. If R is the mid point PQ, then affix of R is 2
zz 21
3. Special points of a triangle
Segment/Line Figure point of concurrency Formula
Perpendicular CircumcentreC2sinB2sinA2sin
C2sinzB2sinzA2sinzz 3213
bisectors (Equidistant from
321
322
1
z–zzz–zz
vertices)
Angle
bisectors Incentre (equidistant z3= cbaczbzaz 321
from the sides) to the sides) ,z–za 32 ,z–zb 31
21 z–zc
Medians Centroid ( distance3
zzzG 321z
from vertex to G is
32
of total median)
Altitudes Orthocentre (can be CtanBtanAtanCtanzBtanzAtanzZ 321
H
inside, outside or on
the right angle)
321
132321
z–zzz–zzz–zzz
A
CB
S
A
A
CB
A
A
CB
H
A
CB
G
115
4. Area of the triangle with vertices z1, z2 & z3 is 321 z–zz41
5. Equilateral triangleIn equilateral ABC , where the vertices are given by z1, z2, z3 and the circumcentre is z0,i. 2
023
22
21 z3zzz
ii. 13322123
22
21 zzzzzzzzz
iii. 0z–z
1z–z
1z–z
1
133221
6. If z1,z2,z3,z4 are vertices of a parallelogram if and only if z1+z3 = z2+z4.7. If z1,z2,z3,z4 are vertices of a square in the same order, then
i. z1+ z3 = z2+ z4
ii. 14433221 z–zz–zz–zz–z
iii. 4231 z–zz–z
iv.42
31
z–zz–z is purely imaginary
8. Equation of the straight line joining z1 & z2 is
i. arg 0orz–zz–z
12
1 i.e.12
1
z–zz–z must be real i.e.
12
1
12
1
z–zz–z
z–zz–z
or
01zz1zz1zz
22
11 (non-parametric form)
ii. z = z1+(1– )z2 (parametric form)iii. General equation of a line is 0bzaza where .Rb
Note: Condition for 3 points z1,z2,z3 to be collinear is that arg or0z–zz–z
23
13
Two points z1 & z2 lie on the same side or opposite side of the line 0bzaza according as
bzaza 11 and bzaza 22 have same sign or opposite sign.
9. Slope of the line segment joining z1 & z2 is21
21
z–zz–z .Two lines with slopes 1 & 2 are
a. perpendicular if 1 + 2 = 0b. parallel if 1 = 2
116
10. Length of the perpendicular from z1 to 0bzaza is a2
bzaza 11
11. Equation of the perpendicular bisector of the line segment joining z1 & z2 is
21 z–zz + 21 z–zz = 22
21 z–z
12. Equation of a circle
i. rz–z 0 or 0r–zzzz–zz–zz 20000 where z0 the centre & r is the radius
ii. 221
22
21 z–zz–zz–z (circle described on the line joining z1 & z2 as diameter)
iii. General equation of a circle is 0bzazazz where Rb . Centre is –a and radius
is b–aa
* Four points z1,z2,z3,z4 are concyclic if and only if 3241
4231
z–zz–zz–zz–z
is purely real.
13. Loci in complex plane
i. 0z–z = rcircle with centre z0 & radius r.
l 0z–z < r : interior of this circle
l 0z–z > r : exterior of this circle
l 1z–z = 2z–zperpendicular bisector of the segment joining z1 & z2
iii. 1z–z + 2z–z = k21
21
z–zkiflinestraightais
z–zkifellipseanis
iv. 1z–z – 2z–z = k21
21
z–zkiflinestraightais
z–zkifhyperbolaais
v. argument az–zz–z
2
1 , a fixed angle is a circle.
14. Complex number as a rotation arrow in the argand plane.Let z = ire
OP = r & XOPize is a complex number whose modulus is r and
argument XOP . To obtain ize , rotate OP through anangle in anticlockwise direction.
P(z)
Y
XO
Q (ze )i
117
Note: Let z1 & z2 be two complex represented by P & Q such
that POQ . i1ez is a vector of magnitude
1zOP along OQ . i
1
1 ezz is a unit vector along QO .
A vector of magnitude OQz2 units is given by i
1
12 e
zzz
i.e.i
11
22 ez
zz
z
Note: general formula for rotationIf AB is rotated to AC but ,ACAB then
r2
r3
A(z )1
B (z )2
C (z )3
i
2
12
3
13 er
z–zr
z–z
l Multiplication of z with i rotates the vector z through a right angle in anticlockwise direction2/ie1i
l Multiplication of z by – 1 rotates the vector z throughan angle of 1800 in anticlockwise direction.
l Let AB & CD intersect at z0. Let P(z1) & Q (z2) betwo points on AB & CD. Then the angle is given by
= arg 01
02
z–zz–z = arg (z2– z0) – arg (z1– z0)
Solved Examples1. If z1,z2,z3 are the vertices of an equilateral triangle having its circumcentre at the origin such that
z1 = 1+i, find z2 and z3.A(z )1
B(z )2 C(z )3
P(z )1
Y
XO
Q (z )2
P(z )1A B
C
D
Q (z )2
z0
3/23/2
3/2
118
Solution: Clearly, OB and OC are obtained by rotatingOA through 3/2 and 3/4 respectively..
OB = OA 3/4ie and, OC = OA 3/4ie
z2 = z13/2ie and, z3 = z1
3/4iez2 =(1+i) 3/2sini3/2cos and, 3/4sini3/4cosi1z3
z2 = (1+i) 2/3i2/1– and 2/3i–2/1–i1z3
z2= –2
1–3i2
13 and z3 = –2
31i–2
3–1
Example: 2 Let z1 and z2 are roots of the equation z2+pz+q = 0, where the coefficients p and q maybe complex numbers. Let A and B represent z1 and z2 in the complex plane. If 0AOB andOA = OB, where O is origin, prove that p2 = 4qcos2 0/ .Solution: Since z1 and z2 are roots of the equation z2+pz+q = 0
z1+z2 = –p and z1z2 = q
Since OA = OB. So, OB is obtained by rotating OA in
anticlockwise sense through angle .
OB = OA iei
12 ezz
sincos1
2 izz
sinicos11zz
1
2
119
/22 1
1
2 cos cos sin 2 cos2 2 2 2
iz z i ez
2/i
1
12 e2
cos2z
zz
i22
1
12 e2
cos4z
zz
1
222
1
12
zz
2cos4
zzz
(z2+z1)2 = 4z1z2cos2
2
(–p)2 = 4qcos2
2
p2 = 4qcos2
2Example: 3 If the points A,B,C represent the complex numbers z1,z2,z3 respectively and the angles
of the ABC at B and C are both 2–
. Prove that (z3 – z2)2 = 4(z3–z1) (z1–z2)sin2
2.
Solution: We have,
2–CB
2–
2––A
Since AC is obtained by rotating AB through angle .
120
AC = AB ie(z3 – z1) = (z2 – z1)
ie
sinicosz–zz–z
12
13
sinicos1–1–z–zz–z
12
13
2cos
2sini2
2sin2–
z–zz–z 2
12
23
2sini
2cos
2sini2
z–zz–z
12
23
2/i
12
23 e2
sini2z–zz–z
22/i222
12
23 e2
sini4z–zz–z
i22
12
23 e2
sin4–z–zz–z
12
1322
12
23
z–zz–z
2sin4–
z–zz–z
2sinz–zz–z4z–z 2
21132
23
Example: 4 Show that the area of the triangle on the Argand plane formed by the complex
numbers z, iz and z + iz is2z
21
Solution: We have, iz = 2/izeThis implies that iz is the vector obtained by rotating vector
121
z in anticlockwise direction through 900. Therefore, .ABOA So,
Area of OBxOA21OAB
=2z
21izz
21
.
Example: 5 Let z1,z2,z3 be the affixes of the vertices A,B and C respectively of a ABC . Provethat the triangle is equilateral if .zzzzzzzzz 133221
23
22
21
Solution: First, let z1,z2,z3 be the affixes of the vertices A,B,C of an equilateral triangle ABC.Then, we have to prove that .zzzzzzzzz 133221
23
22
21
Since ABCis an equilateral triangle.Therefore,AB = BC = AC and .3/CBA
Clearly, AC can be obtained by rotating ABin anticlockwise sense through 600.z3 – z1 = (z2–z1) 3/ie ......(1)
Also BC can be obtained by rotating BA by 3 anticlockwise
z2 – z3 = (z1–z3) 3/ie ......(2)From (1) and (2)
31
32
12
13
z–zz–z
z–zz–z
Solving we get .zzzzzzzzz 13322123
22
21
122
1. The complex number z = x+ iy which satisfy the equation i5zi5–z
=1 lie on
a. the x-axis b. the straight line y = 5c. a circle passing through the origin d. none of these
2. Let z & w be two complex numbers such that 1w,1z and iw–ziwz = 2 then z =a. 1 or i b. i or –i c. 1 or –1 d. i or –1
3. Let z1 & z2 be the nth roots of unity which subtend a right angle at the origin, then n must be ofthe form (where zk )a. 4k+1 b. 4k+2 c. 4k+3 d. 4k
4. The complex number z1,z2 & z3 satisfying2
3i–1z–zz–z
32
31 are the vertices of a triangle
which isa. of area zero b. right angled isoscelesc. equilateral d. obtuse-angled isosceles
5. For all complex numbers z1,z2 satisfying 12z1 and i4–3–z2 =5, the minimum value of
21 z–z isa. 0 b. 2 c. 7 d. 17
6. The shaded region where P = (–1,0), Q=(–1+ 2 , 2 ), R = (–1+ 2 ,– 2 ), S = (1,0) isrepresented by
Y
XP O
Q
S
R
(PQ=PS=PR)
a. 41zarg,21z b. 2
1zarg,21z
c. 41zarg,21z d. 2
1zarg,21–z
7. A man walks a distance of 3 units from the origin towards the North-East (N450 E) direction.Form there, he walks a distance of 4 units towards the North-West (N450 W) direction to
PRACTICE QUESTIONS
123
reach a point P. Then the position of P is the Argand plane isa. i4e3 4/i b. 4/iei4–3 c. 4/iei34 d. 4/iei43
8. A particle P starts from the point z0 = 1+2i. It moves first horizontally away from origin by 5units and then vertically away from origin by 3 units to reach a point z1. From z1 the particle
moves 2 units in the direction of the vector ji and then it moves through an angle 2 in
anticlockwise direction on a circle with centre at origin to reach a point z2. The point z2 is givenbya. 6+7i b. –7+6i c. 7+6i d. –6+7i
9.* Match the followingColumn I Column II
a. The set of points z satisfying p. an ellipse with eccentricity 4/5
zizzi–z is contained inor equal to q. the set of points z satisfying Im(z) = 0
b. The set of points z satisfying r. the set of points z satisfying 1zIm
04–z4z is contained inor equal to
c. If 2w , then the set of points s. the set of points satisfying 2zRe
w1–wz is contained in or equal to t. the set of points z satisfying 3z
d. If 1w , then the set of points
w1wz is contained in or equal to
10. If a,b,c and u,v,w are complex numbers representing the vertices of two triangles such thatc = (1–r) a +rb, w = (1– r) u + rv, where r is a complex number, then the two trianglesa. have the same area b. are similarc. are congruent d. none of these
11. The locus of the centre of a circle which touches the circles 1z–z = a and 2z–z = b externallyisa. an ellipse b. a hyperbola c. circle d. none of these
12.* If one of the vertices of the square circumscribing the circle 21–z is 3i2 , then which ofthe following can be a vertex of it?a. i3–1 b. i3– c. i31 d. none of these
13. Read the passage and answer the following questions:A(z1), B(z2), C(z3) are the vertices a triangle inscribed in the circle 2z . Internal angle bisectorof the angle A meets the circum circle again at D(z4).
124
i. Complex number representing point D is
a.3
214 z
zzz b.1
324 z
zzz c.3
214 z
zzz d. none of these
ii. argument (z4 / (z2–z3) is
a. 4 b. 3 c. 2 ` d. 32
iii. For fixed positions of B(z2) and C(z3) all the bisectors(internal) of A will pass through afixed point which isa. H.M. of z2 and z3 b. A.M. of z2 and z3c. G.M. of z2 and z3 d. none of these
Note:* Questions with more than one option is correct.
ANSWERS
1. a 2. c 3. d 4. c 5. b 6. a 7. d
8. d 9. a q,r; b p; c p,s,t; d q,s,r,t,t 10. b 11. b
12. a,b,c 13. (i) d (ii) c (iii) c
125
COMPLEX NUMBERS AND QUADRATIC EQUATIONS - IQuadratic Equations
Quadratic equationsThe general form of a quadratic equation over real numbers is ax2+bx+c = 0 where a, b,c R & a0. The solution of the quadrati equation ax2+bx+c = 0 is given by
x = a2
ac4–bb– 2. The expression b2–4ac is called the discriminant of the quadratic equation
and is denoted by D.Nature of roots : For the quadratic equation ax2+bx+c = 0, where a, b, c R and a 0, then
D>O D = O D<0Roots are real and Roots are real & Roots are imaginary{If p+iq isunequal (distinct) equal (coincident) one of the roots then the other
must be the conjugate p–iqwhere p,q R & i = 1– }
For the quadratic equation ax2+bx+c = 0 where a, b, c Q and a 0, then :
D>0 and is a perfect square. D>0 and not a perfect square.
Roots are rational and unequal (distinct). Roots are irrational{If p+ q is one root,then the other root must be the conjugate
p – q }ie, Irrational roots occurr inpairs if a, b, c Q.
Note1 : If is a root of ƒ(x) = 0, then the polynomial ƒ(x) is exactly divisible by x– or (x– ) isa factor of ƒ(x) and vice versa.
Note 2 : ax2+bx+c = 0 cannot have three different roots. If it has, then the equation becomes anidentity in x. ie, a = b = c = 0.Relation between roots and coefficients
If 1, 2,......... n are roots of the equation ƒ(x)=anxn+an–1x
n–1+an–2xn–2+....... ..a2x
2+a1x+a0 = 0,then ƒ(x) = an(x– 1)(x– 2)(x– 3)..............(x– n)
anxn+an–1x
n–1+an–2xn–2+.............+a2x
2+a1x+a0 = an (x– 1) (x– 2)............(x– n)Comparing the coefficients of like powers of x an both sides, we get,
S1 = 1+ 2+.............+ n = i = n
1–n
aa–
= n
1–n
xofcoeftxof.coeft–
S2 = 1 2+ 1 3+................ jiji = (–1)2
n
2–n
aa
= (–1)2 n
2–n
xofcoeftxof.coeft
S3 = 1 2 3+ 2 3 4+................kji
kji = (–1)3
n
3–n
aa
= (–1)3 n
3–n
xofcoeftxof.coeft
......................................
.....................................
126
Sn = 1 2 3...................... n = (–1)n
n
0
aa
= (–1)n nconstant termcoeftof x
Here Sk denotes the sum of the products of the roots taken ‘k’ at a time.Particular cases :-
Quadratic equation : If & are the roots of the quadratic equation ax2+bx+c = 0, then
S1 = + = ab–
, & S2 = = ac
Cubic equation : If , , are the roots of the cubic equation ax3+bx2+cx+d = 0, then
S1 = + + = ab–
S2 = + + = (–1)2
ac
= ac
S3 = = (–1)3
ad
= ad–
Biquadratic equation : If , , , are roots of the biquadratic equation ax4+bx3+cx2+dx+e = 0,then
S1 = + + + = ab–
S2 = + + + + + = (–1)2 ac
Or S2=( + ) ( + )+ + = ac
S3 = + + + = (–1)3
ad
Or S3 = ( + )+ ( + ) = ad–
and S4 = = (–1)4
ae
= ae
Formation of a polynomial equation from given rootsIf 1, 2, 3, ................, n are the roots of an nth degree equation, then the equation isxn–S1x
n–1+S2xn–2–S3x
n–3+..............+(–1)n Sn = 0 where Sk denotes the sum of the products of rootstaken k at a time.Particular cases
Quadratic equation : If , are the roots of a quadratic equation, then the equation isx2–S1x+S2 = 0 ie, x2–( + )x+ = 0.
n
cons tant termcoeft of x
127
Cubic equation : If , , are the roots of a cubic equation. Then the equation is ,x3–S1x
2+S2x–S3 = 0 ie, x3–( + + ) x2+( + + ) x– = 0Biquadratic equation : If , , , are the roots of a biquadratic equation, then the equation is
x4–S1x3+S2x
2–S3x+S4 = 0i e , x 4 – ( + + + ) x 3 + ( + + + + + ) x 2 –
( + + + )x+ = 0.Quadratic Expression : An expression of the form ax2+bx+c, where a, b, c R & a 0 is called a
quadratic expression in x. So in general quad atic expression is represented as: ƒ(x)=ax2+bx+c or y =ax2+bx+c.
Graph of a quadratic ExpressionLet y = ax2+bx+c where a 0.
Then y = a acx
abx2
y = a 2
2
2
22
a4b–
ac
a4b
abxx
y +a4
ac4–b2= a
2
a2bx y+ a4
D = a
2
a2bx
Let y+ a4D
= Y & x+ a2b
= X
Y = aX2 or X2 = aY
Clearly it is the equation of a parabola having its vertex at a4D–,
a2b–
.
If a>0, then the parabola open upwards.If a<0, then the parabola open downwards.
Sign of quadratic Expression(1) The parabola will intersect the x–axis in two distinct points if D>0.
(i) a>0 (ii) a<0
X X
Let ƒ(x) = 0 have 2 real roots Let ƒ(x) = 0 have 2 real roots& ( < ).Then ƒ(x)>0 & ( < )Then ƒ(x)<0x (– , ) ( ) and ƒ(x)<0 x (– , ) ( )x ( , ) & ƒ(x)>0 for all x ( , )
r
128
(2) The parabola will touch the x–axis at one point if D = 0(i) a>0 (ii) a<0
X
X
ƒ(x) 0 x R ƒ(x) 0 x R(3) The parabola will not intersect x–axis if D<0.
(i) a>0 (ii) a<0
X
ƒ(x)>0 x R ƒ(x)<0 x RNOTE : Condition that a quadratic function ƒ(x,y) = ax2+2hxy+by2+2gx+2ƒy+c may be resolved
into two linear factions is that
abc+2ƒgh–aƒ2–bg2–ch2 = cfgfbhgha
= 0
NOTE :
(i) For a>0, ƒ(x) = ax2+bx+c has least value at x= a2b–
. This least value is given by a4D–
(ii) For a<0, ƒ(x) = ax2+bx+c has greatest value at x= a2b–
. This greatest value is given by a4D–
Solved Examples1 If , are roots of ax2+bx+c = 0 ; +h and +h are roots of px2+qx+r = 0 and D1, D2 are theirdiscriminants, then D1: D2 =
(a) 2
2
pa
(b) 2
2
qb
(c) 2
2
rc
(d) None of these
Solution : – = ( +h)–( +h) ( – )2 = (( +h) ( +h))2
( + )2 –4 = (( +h)– ( +h))2 – 4( +h)( +h)
2
ab–
– 4 ac
= 2
pq–
– pr4
+
–
129
2
2
aac4–b
= 2
2
p4–q
2
1
DD
= 2
2
pa
Ans : (a)2 If a Z and the equation (x–a)(x–10)+1 = 0 has integral roots, then the values of a are
(a) 10, 8 (b) 12, 10 (c) 12, 8 (d) None of theseSolution : (x–a)(x–10) = – 1
x–a = 1 & x–10 = –1 OR x–a = –1 & x–10=19–a = 1 x= 10 – 1 11 – a = – 1 x = 11a = 8 x = 9 a =12
Ans : (c)3 If , are roots of the equation (x–a)(x–b)+c = 0 (c 0), then then roots of the equation (x–c–
)(x–c– ) = c are(a) a and b+c (b) a+c and b (c) a+c and b+c (d) None of these
Solution: x2–(a+b)x+ab+c = 0+ = a+b and = ab+c
Now (x–c– )(x–c– ) = c(x–c)2 –( + )(x–c)+ –c = 0(x–c)2–(a+b)(x–c) +ab = 0(x–c)2– a(x–c) –b(x–c) +ab = 0(x–c–a)(x–c–b) = 0x = c+a and b+c
Ans (c)4 Let 2 be the discriminant and , be the roots of the equation ax2+bx+c = 0. Then 2a + and
2a – can be roots of the equation(a) x2+2bx+b2 = 0 (b) x2–2bx+b = 0(c) x2+2bx–3b2–16ac = 0 (d) x2+2bx–3b2+16ac = 0
Solution : , = a2
b– 2
= a2b–
and = a2–b–
Or = a2–b–
and = a2b–
2a = –b+ 2a = –b– 2a + = – b & 2a – = – b2a + = –b+2 2a – = –b–2 S = –2b and P = b2
S = – 2b and P = +b2–4 2
= +b2–4(b2–4ac)= –3b2+16ac
pr21 DD
a 22
p =
2
)= - 2b
)= 16ac -3bP = (2a + ). (2a - 2Heve S means sum of the rootsP means product of the roots
S = (2a + )+(2a -
130
quadatic equation is quadratic equation is x2+2bx–3b2+16ac = 0 x2+2bx+b2 = 0Ans : (a) and (d)
5 The polynomial equation (ax2+bx+c)(ax2–dx–c) = 0 , ac 0 has(a) four real roots (b) atleast two real roots(c) atmost two real roots (d) No real rootsSolution : ac 0
ac>0 or ac<0Now D1=b2–4ac & D2 = d2+4acWhen ac> 0 D2>0 but D1 may positive or negativeWhen ac<0 D1>0 but D2 may be positive or negativeIn either case the polynomial has atleast two real rootsAns (b)
6 If , are roots of x2–p(x+1) –q = 0 , then the value of q212
2
2
+ q212
2
2
is
(a) 1 (b) 2 (c) 3 (d) None of theseSolution : x2–px–p–q = 0
+ = p and = – p–qNow ( +1)( +1) = ( + )+ +1
= p–q–p+1 = 1–q
Now q212
2
2
q212
2
2
= 1–q)1()1(
2
2
+ 1–q)1()1(
2
2
= )1)(1(–)1()1(
2
2
+ )1)(1(–)1()1(
2
2
= –1
+ – = –1––1
––
= 1Ans (a)
7 Let , , be the roots of the equation x3+4x+1 = 0, then ( + )–1+( + )–1+( + )–1 equals(a) 2 (b) 3 (c) 4 (d) 5Solution : + + = 0, + + = 4, = –1
1
+1
+1
= – 1
– 1
–1
= – = – 1–4
= 4
Ans : (c)
+
= –1–1
1
( ) + +
1–1
1( )
+ = + 1
( )
+ ( ) +1 1 + ( )–
+1 +1 +1
+ 1 +1
131
1 The minimum value of ƒ(x) = x2+2bx+2c2 is greatest than the maximum value ofg(x)= –x2–2cx+b2, then (x being a real )
(a) |c|> 3|b|
(b) 2|c|
>|b|
(c) –1<c< 2 b (d) Non real value of b &c exist2 If P(x) is a polynomial of degree less than or equal to 2 and S is the set of all such polynomials so that
P(1) = 1, P(0) = 0 and P1(x)>0 x [0,1], then S =(a) (b) {(1–a)x2+ax, 0<a<2}(c) {(1–a)x2+ax,a>0} (d) {(1–a)x2+ax,0<a<1}
3 In the quadratic equation ax2+bx+c = 0 if = b2–4ac and + , 2+ 2 and 3+ 3 are in G.P,.P,where , are the roots of the equation, then(a) 0 (b) b = 0 (c) c = 0 (d) = 0
4 If a, b, c are the sides of a triangle ABC such that x2–2(a+b+c)x+3 (ab+bc+ca) = 0 has real roots,then
(a) < 34
(b) > 35
(c) 35,
34
(d) 35,
31
5 Let & be the roots of x2–6x–2 = 0, with > . If an = n – n for n 1, then the value of
9
810a2
a2–a is
(a) 1 (b) 2 (c) 3 (d) 46 If x2–10ax–11b = 0 have roots c & d. x2–10cx–11d = 0 have roots a & b, then a+b+c+d is
(a) 1210 (b) 1120 (c) 1200 (d) None of these
7 If tn denotes the nth term of an A.P. and tp = q1
and tq = p1
, then which of the following is necessarily
a root of the equation (p+2q–3r)x2+(q+2r–3p)x+(r+2p–3q) = 0 is(a) tp (b) tq (c) tpq (r) tp+q
8 The curve y = ( +1) x2+2 intersect the curve y = x+3 in exactly one point, if equals(a) {–2, 2} (b) {1} (c) {–2} (d) {2}
9 Read the passage and answer the following questions.Consider the equation x4+(1–2k) x2+k2–1 = 0 where k is real. If x2 is imaginary, or x2<0, the equationhas no real roots. If x2>0, the equation has real roots.
(i)* The equation has no real roots if k
(a) (– –1) (b) (–1,1) (c) 45,1 (d) ,
45
(ii) The equation has only two real roots if k(a) (– –1) (b) (0, 1) (c) (1, 2) (d) (–1, 1)
Here, P(x) = b x + ax + c2
PRACTICE QUESTIONS
132
(iii) The equation has four real roots if k
(a) (– ,0) (b) (–1, 1) (c) 45,1 (d) (1, )
10 If , are the roots of the equation ax2+bx+c = 0, then the value of
1coscoscos1)–cos(cos)–cos(1
is
(a) sin( + ) (b) sin sin (c) 1+cos( + ) (d) None of these11* If (1+k) tan2x–4tan x–1+k = 0 has real roots, then
(a) k2 5 (b) tan(x1+x2)=2 (c) for k=2, x1= 4 (d) for k =1, x1 = 0
12 If p,q {1,2,3,4}, the number of equations of the form px2+qx+1 = 0 having real roots is(a) 15 (b) 9 (c) 7 (d) 8
13 In PQR R = 2 . If tan 2p
& tan 2Q
are the roots of the equation ax2+bx+c = 0 (a 0) then
(a) a+b=c (b) b+c = 0 (c) a+c = b (d) b = c
Note : * Questions with more than one option is correct
Answers
1. b 2. b 3. c 4. a 5. c
6. a 7. c 8. c 9 (i) a, d (ii) d (iii) c10. d 11. a, b, c, d 12. c 13. a
133
COMPLEX NUMBERS AND QUADRATIC EQUATIONS - IIQuadratic Equations(Location o f Roots)
Let ƒ(x) = ax2+bx+c, a,b,c R , 0a and , be the roots of ƒ(x)=0. Let k be anyreal number
Cases Figure Conditionk and i. a > 0 i. 0D (roots may be equal)k Both the ii. a.ƒ(k) > 0
roots are ess than k iii. 2k > i.e 2k > sumof
ii. a < 0 roots or k > a2b–
k and k i. a > 0 i. 0D (roots may be equal)Both the roots are ii. a.ƒ(k) > 0greater than k iii. 2k < i.e 2k < sumof
ii. a < 0 roots or k < a2b–
k i. a > 0 i. D > 0k lies between (distinct roots)the roots ii. a ƒ(k) < 0
ii. a < 0
Wavy Curve MethodLet ƒ(x) = 1k
1a–x 2k2a–x …… nk
na–x – (1)
Where iNki & Rai such that a1 < a2 < ……< an. Mark a1,a2…… an on real axis check thesign of ƒ(x) in each interval. The solution of ƒ(x) > 0 is the union of all intervals in which we have putplus sign and the solution of ƒ(x) < 0 is the union of all intervals in which we have put the minus sign.
Exponential EquationsIf we have an equation of the formax = b where a > 0, thenx if 0b ; x = bloga if b > 0, 1a ;
x if a = 1, 1b ; Rx , if a = 1, b = 1
k X
k X
k X
k X
k X
k X
l
134
Lagrange’s IdentityIf a1, a2, a3, b1, b2, b3 R then
23
22
21 aaa 2
322
21 bbb – 2
332211 bababa
= 21221 ba–ba + 2
2332 ba–ba + 23113 ba–ba
Note: If fe
dc
ba
, then each ratio is equal to
i. ....fdb....eca
ii.n/1
....rfqdpb
....reqcpannn
nnn
where p, q, r, Rn
iii. n
n
...bdf...ace
bdac
SOLVED EXAMPLES1. The values of m for which both roots of the equation x2–mx+1=0 are less than unity is
a. 2,–– b. ]2,–(– c. ,2– d. none of theseSolution:0D a.f(1) > 0 2
(–m)2 –4.1.1 0 1.(1–m+1) > 0 m < 2(m –2) (m+2) 0 m < 2 2,–m .......(3)
m ]2,–(– ),2[ .... (1) 2,–m ....(2) Form (1), (2) and (3) we have
]2,–(–m
Ans: b
2. The values of m Rm , for which both roots of the equation x2–6mx+9m2–2m+2 = 0 exceed3 isa. ]1,(– b. 1,– c. ),1[ d. none of these
Solution:0D
(–6m)2 –4.1.(9m2 –2m+2) 0 a.f(3) > 0 68m–8 0 1.(32–18m+9m2–2m+2) > 0 6m > 6
m 1 ........(1) 9m2 – 20m + 11 > 0 m > 1 ...(3)9m2 – 9m – 11m + 11 > 0(9m – 11) (m–1) > 0
911
......(2) <
135
From (1), (2) and (3) ,9
11m
Ans: d
3. The values of p for which 6 lies between the roots of the equation x2+2(p–3)x+9 = 0 is
a. 43,–– b. 4
3,–– c. 1,– d. none of these
Solution: 0D a.ƒ(6) < 0(–2(p–3))2 –4.1.9 > 0 1.(36+12(p–3) +9)) < 0p2–6p > 0 12p+9 < 0p (p – 6) > 0
p > 0, p> 6 .....(1) P < 43–
.....(2)
From (1) and (2) 43–,–p
Ans: a
4. If a, b, c R , and the equation ax2+bx+ c = 0 has no real roots, thena. (a+b+c) > 0 b. a(a+b+c) > 0c. b (a+b+c) > 0 d. c(a+b+c) < 0Solution:a > 0 a < 0ƒ(0) > 0 c > 0 ƒ(0) < 0 c < 0ƒ(1) > 0 a+b+c > 0 ƒ(1) < 0 a+b+c < 0aƒ(1) > 0 & cƒ(1) > 0 aƒ(1) > 0 & c.ƒ(1) > 0a(a+b+c) > 0 and c (a+b+c) > 0 a(a+b+c) > 0 and c (a+b+c) > 0Ans: b
1. If the roots of equation x2–2ax+a2+a–3 are less than 3, thena. a < 2 b. a > 4 c. 3 < a < 4 d. –2 < a < 3
2. Read the following passage and answer the questions:-ƒ(x) = ax2+bx+c = –x–xa , where are the roots of ƒ(x) = 0. If ac4–b 2 isnegative, then its sign is same as that of a, the coefficient of x2. If ƒ(x) = –x–xa , where
, a is positive, then for any number p which lies between ;& ƒ(p) is negative and forany number q or r which do not lie between & , ƒ(q) or ƒ(r) both will be positive. Also if a2
< x2 < b2, then a < x < b or –b < x < –a.
i. If x1–42–x 2 + 7–2–15 2 > 0 Rx , thena. (0,2) b. (1,3)c. (2,4) d. none of these
PRACTICE QUESTIONS
136
ii. Let ƒ(x) be a quadratic expression which is positive for all real x.If g(x) = ƒ(x)+ ),x(ƒ)x(ƒ then for any real x,a. g(x) > 0 b. g(x) 0 c. g(x) 0 d. g(x) < 0
iii. The inequality 2–x–x22–x–x
2
2
> 2 holds only if
a. 32–x1– only b. only for 1x
32
c. –1 < x < 1 d. –1 < x < 1x32or
32
iv* for real x, the functionc–x
b–xa–x will assume all real values, provided
a. a < b < c b. a > b > cc. a > c > b d. a < c < b
3. Values of ‘a’ for which the roots of the equation (a+1)x2–3ax+4a = 0 1–a greater thanunity is
a. 1,–716–a b. 1,–
716–a
c. 1,–716–a d. none of these
4*. If Rx satisfies (log10(100x))2 + (log10(10x))2 + log10x 14 , then the solution set containsthe interval
a. 10,1 b. 1,10 2/9– c. ,0 d. ,1–
5*. If a, b are the real roots of x2+px+1 = 0 and c, d are the real roots of x2+qx+1 = 0, then(a – c) (b – c) (a + d) (b + d) is divisible bya. a – b – c – d b. a + b + c – dc. a + b + c + d d. a + b – c – d
6. Match the following:-Column I Column II
a. The value of x for which loge(x – 3) < 1 is (p) 21–5,0
b. The value of x for which log1/2x log1/3x is (q) (0, 1)c. If log0.3(x – 1) < log0.09(x– 1), then x lies in the interval (r) (2, 8)d. If 2xsinlog xcos and 3,0x then sinx lies in the interval (s) (3, 3 + e)
(t) 215,0
137
7. Read the paragraph and answer the questions that follow:
LetxQ
ba +2–xQ
b–a = A, where RA,N and a2 – b = 1 1b–aba
i.e 11
b–aorbaba
i. If x54 + x
5–4 = 62, then
a. 2,12,–3–x b. 1,2–2,3–x
c. 3,21,–2–x d. 2,1–3,2–x
ii. Solution of 3–243–232
1–x2–x1x2–x 22
are
a. 1,31 b. 1,21
c. 2,31 d. 2,21
iii. The number of real solutions of the equation 30144–1514415tt are where
t = x2–2 xa. 0 b. 2c. 4 d. 6
8. The maximum value of ƒ(x) = is7x9x3
17x9x32
2
5k+1, Then k is
a. 41 b. 40 c. 8 d. none of these
9. If c
xy–zb
zx–ya
yz–x 222
, then (x+y+z) (a+b+c) is
a. ax+by+cz b. a+b+c c. 3zyx
d. none of these
10 The value of x satisfying the equation 79xlog2–2x3log 1–x1–x isa. 3 b. 9 c. 27 d. 81
11. If 20x9–x20x9–x 22 then which is true?
a. x 4 or 5x b. 5x4c. 4 < x < 5 d. none of these
12. If x2+px+1 is a factor of ax3+bx+c, thena. a2+c2 = –ab b. a2– c2 = –abc. a2– c2 = ab d. none of these
13. If and 3–5,3–5 22 , then the equation whose roots one and is
a. x2–5x–3 = 0 b. 3x2+19x+3 = 0c. 3x2–19x+3 = 0 ‘ d. x2+5x–3 = 0
138
14. If the equation (cos p – 1)x2+ x (cos p) + sin p = 0, in the variable x, has real roots then ‘p’can take any value in the interval.a. 2,0 b. 0,–
c. 2,
2–
d. ,0
15. If (cos +isin ) is a root of the equation ax2+bx+c = 0, a, b, c R , thena. 0csinb2cosa b. 0ccosb2cosac. 0csinb2sina d. none of these
Note:* Questions with more than one option is correct.
ANSWERS
1. a 2. (i) c (ii) a (iii) d (iv) c,d 3. a 4. a,b 5. c,d
6. a s; b q; c r; d p 7. (i) c (ii) b (iii) c 8. c 9. a 10. d
11. c 12. c 13. c 14. d 15. b
139
COMPLEX NUMBERS AND QUADRATIC EQUATIONS - IIIQuadratic Equations(Location o f Roots)
Let ƒ(x) = ax2+bx+c, a, b, c 0a,R and , be the roots of ƒ(x) = 0. Let k1, k2 be tworeal numbers such that k1 < k2
Cases Figure Conditions
Exactly one root i. D > 0 (distinct roots)lies is the interval (k1,k2) i. a > 0 ii. ƒ(k1) ƒ(k2) < 0
ii. a < 0
Both the roots lie i. a > 0 i. D 0between k1 & k2 (roots may be equal)k1 < 2k ii. a < 0 ii. a ƒ(k1) > 0 & a ƒ(k2) > 0
iii. k1 < 2ka2b–
k1 & k2 lie between i. a > 0 i. D > 0 (distinct roots)the roots kk 21 ii. a ƒ(k1) < 0 & a ƒ(k2) < 0
ii. a < 0
Logarithmic EquationsIf we have an equation of the form as loga ƒ(x) = b where a > 0, 1a can be written as ƒ(x) = ab
when ƒ(x) > 0.
Logarithmic Inequalities
For a > 1 For 0 < a < 1
0 < x < y and 0 < x < y andylogxlog aa are ylogxlog aa are
equivalent equivalent
pxloga pxloga
0 < x < aP x > aP
pxloga pxloga
x > aP 0 < x < aP
k2
k1 X
k2k1 X
k2k1X
k2k1 X
k2k1X
k2
k1X
140
Descartes Rule of signsThe maximum number of positive real roots of a polynomial equation ƒ(x) = 0 is the number of
changes of signs from positive to negative and negative to positive.The maximum number of negative real roots of a polynomial equation ƒ(x) = 0 is the number of
changed signs from positive to negative and negative to positive in ƒ(– x) = 0Solved examples1. The values of m for which exactly one root of x2 – 2mx + m2 –1= 0 lies in the interval (–2, 4) is
a. (–3, –1) (3, 5) b. (–3, –1) c. (3, 5) d. noneSolution: D > 0 ƒ(–2)ƒ(4) < 0(–2m)2 –4.1.(m2–1) > 0 (4+4m+m2–1) (16 –8m+m2–1) < 04 > 0 (m2 +4m+3) (m2 –8m+3)
Rm ......(1) (m +1) (m + 3) (m – 3) (m – 5) < 0m (–3, –1) (3, 5) ......(2)
From (1) and (2) , m (–3, –1) (3, 5)
Ans: a2. The values of a for which both the roots of the equation 4x2 –2x + a = 0 lie in the interval
(–1, 1) is.
a. ,2– b. 41,– c. 4
1,2– d. none of these
Solution:0D a.ƒ(–1) > 0 a. ƒ(1) > 0
(–2)2 – 4.4.a 0 4.(4+2+a) > 0 4.(4 –2+a) > 04a –1 0 a+6 > 0 a > –2
a41 – (1) a> –6 – (2) a (–2, ) – (3)
From (1),(2) and (3),a41,2–
Ans: c3. The all possible values of a for which one root of the equation (a –5)x2 –2ax+a – 4 = 0 is
smaller than 1 and the other greater than 2 isa. [5, 24) b. (5, 24] c. (5, 24) d. none of theseSolution: D 0 (a –5) ƒ(1) < 0 (a –5) ƒ(2) < 0(–2a)2– 4(a – 5) (a – 4) 0 (a –5) (a–5 –2a+ a –4) < 0 (a –5) (4(a –5) –4a + a –4) < 0
9a –20 0 (a –5) 9 0 (a–5)(–9) < 0 (a – 5) (a – 24) < 0
a 920
– (1) a > 5 – (2) 5 < a < 24 –(3)
From (1), (2), and (3) )24,5(aAns: c
4. If a, b, c R and the equation x2 + (a+b) x + c = 0 has no real roots, then
>
141
a. c (a+b+c) > 0 b. c + (a+b+c) c > 0c. c – (a+b–c) c > 0 d. c (a+b–c) > 0Solution:ƒ(0) > 0 c > 0 ƒ(0) < 0 c > 0ƒ(1) > 0 1+a+b+c > 0 ƒ(1) < 0 1+a+b+c < 0ƒ(–1) > 0 – (a+b) + c > 0 ƒ(–1) < 0
ƒ(0). ƒ(1) > 0 and ƒ(0). ƒ(–1) > 0 1– (a+b) + c < 0gives b and c ƒ(0) ƒ(1) > 0 and ƒ(0). ƒ(–1) > 0
gives (b) and (c)Ans b and c
1. The values of a for which 2x2–2(2a+1)x + a(a+1) = 0 may have one root less than a and otherroot greater than a are given bya. 1 > a > 0 b. –1< a < 0 c. a 0 d. a > 0 & a < –1
2. The value of a for which the equation (1–a2)x2+2ax–1= 0 has roots belonging to (0, 1) is
a.2
51a b. a > 2 c. 2a2
51d. a > 2
3. If a, b, c, x, y, z, R be such that (a+b+c)2=3(ab+bc+ca–x2–y2–z2), thena. a = b = c = 0 = x = y = z b. x = y = z = 0, a = b = cc. a = b = c = 0; x = y = z d. x = y = z = a = b = c
4. Number of positive integers n for which n2+96 is a perfect square isa. 8 b. 12 c. 4 d. infinite
5. The curve y = 2x1 2 intersects the curve y = 3x is exactly one point, if equalsa. {–2, 2} b. {1} c. {–2} d. {2}
6. A quadratic equation whose product of roots x1 & x2 is equal to 4 and satisfying the
relation 21–x
x1–x
x
2
2
1
1is
a. x2–2x+4 = 0 b. x2+2x+4 = 0 c. x2+4x+4 = 0 d. x2–4x+4 = 07. If a,b,c,d R , then the equation (x2+ax–3b) (x2–cx+3b) (x2–dx+2b) = 0 has
a. 6 real roots b. at least 2 real rootsc. 4 real roots d. 3 real roots
8. Suppose P,Q,R are defined as P=a2b+ab2–a2c–ac2, Q=b2c+bc2–a2b–ab2 & R=a2c+ac2–b2c–bc2, where a>b>c and the equation Px2+Qx+R=0 has equal roots, then a,b,c are ina. A.P b. G.P c. H.P d. AGP
9. If a (p+q)2+2abpq+c=0 & a(p+r)2+2abpr+c=0 0a then
a. qr=p2 b. qr = acp2 c. qr= – p2 d. none of these
PRACTICE QUESTIONS
142
10. x2–xy+y2–4x–4y+16=0 representsa. point b. a circlec. a pair of straight line d. none of these
11. If the roots of the equation ax2+bx+c=0 are of the form 1k2k&
k1k
, then (a+b+c)2 is equal
to
a. 2b2– ac b. 2a c. b2–4ac d. b2–2ac12. Read the passage and answer the following questions:-
aƒ( ) < 0 is the necessary and sufficient condition for a particular real number to the betweenthe roots of a quadratic equation ƒ(x)=0, where ƒ(x)= ax2+bx+c. Again if 1ƒ 2ƒ <0, then
exactly one of the roots will lie between 1& 2 .
1. If ,cab thena. One root of ƒ(x)=0 is positive, the other is negative.b. Exactly one of the roots of ƒ(x)=0 lies in (–1, 1).c. 1 lies between the roots of ƒ(x)=0.d. Both the roots of ƒ(x)=0 are less than 1
2. If a(a+b+c) < 0 < (a+b+c) c, thena. one root is less than 0, the other is greater than 1.b. Exactly one of the roots lies in (0, 1)c. Both the roots lie in (0, 1)d. At least one of the roots lies in (0, 1)
3. If (a+b+c) c < 0 < a(a+b+c), thena. one root is less than 0, the other is greater than 1b. one root lies in o,– and the other in (0, 1)c. both roots lie in (0, 1)d. one root lies in (0, 1) and other in ,1
13. Match the following:-Column I Column II(Number of positive integers for which) (p) 0a. One root is positive and the other
is negative for the equation(m–2)x2–(8–2m)x – (8–3m) = 0
b. Exactly one root of the equation (q) infinitex2–m(2x–8)–15=0 lies in the interval(0, 1) (r) 1
c. The equation x2+2(m+1)x+9m–5=0 hasboth roots negative
d. The equation x2+2(m–1)x+m+5=0 has (s) 2both roots lying on either sides of 1
143
14. If , are the roots of 375x2–25x–2=0 & Sn = nn , then the value of n
1rrn
Slim3
1 is.....
15. If x,y,z are distinct positive number such that x1z
z1y
y1x , then xyz = .......
ANSWERS
1. d 2. b 3. a.b 4. c 5. c 6. a 7. b 8. c 9. b 10. a 11. c
12. (i) b (ii) a (iii) b 13. a r b r c q d p
144
COMPLEX NUMBERS AND QUADRATIC EQUATIONS - IVQuadratic Equations
Problem solving skills.If one root of ax2+bx+c=0 is n times the other, then (n+1)2 ac = nb2
If one root of ax2+bx+c=0 is square of the other, then (a2c)1/3+(ac2)1/3+b=0If , are roots of ax2+bx+c=0 then(i) – ,– are roots of ax2–bx+c=0
(ii)1
, 1
are roots of cx2+bx+a=0 ; ac 0
(iii) k , k are roots of ax2+kbx+k2c=0(iv) 2, 2 are roots of a2x2–(b2–2ac)x+c2=0If the sum of the coefficient of ƒ(x) = 0 is 0, then 1 is always a root of ƒ(x) = 0. Also x–1 is a facterof ƒ(x).In particular, for ax2+bx+c=0 if
a+b+c=0, then 1 is always a root and the other root = ac
( product of roots = ac
).
ƒ(x)=(x–a1)2+(x–a2)
2+.....................+(x–an)2, where ai R i.
ƒ(x) assumes its least value when x= na.......aa n21
While solving an equation, if you have to square, then additional roots will occur as the degree ofthe equation will change. In such cases, you have to check whether the roots satisfy the originalequation or not.
Solved Examples1 If , are roots of the equation x2–2x+3=0
Then the equation whose roots are3–3 2+5 –2 and 3– 2+ +5 is
(a) x2+3x+2=0 (b) x2–3x–2=0 (c) x2–3x+2=0 (d) NoneSolutions :
2–2 +3=0 and 2–2 +3=o3=2 2–3 and 3=2 2–3
P= 3–3 2+5 –2 = 2 2–3 –3 2+5 –2 = – 2+2 –2=3–2 = 1Similarly, we can show that Q = 3– 2+ +5 = 2
Sum = 1+2 = 3 and product = 1×2=2Hence x2–3x+2=0Ans (c)
2 If , are roots of the equatio x2+px– 2p21
=0, p R–{0}, then the minimum value of 4+ 4
145
is(a) 22 (b) 2–2 (c) 2 (d) 22
Solutions :4+ 4 = ( 2+ 2)2 –2 2 2 = (( + )2–2 )2 – 2( )2
= 2
22
p1p – 4p2
1 = p4+ 4p2
1+2
= 2
22
p21–p + 2+ 2
Min value is 2+ 2 .Ans (d)
3 Let p(x) be a polynomial of least possible degree with rational coefficients, having 31
7 + 31
49 as
one of its roots, then the product of all roots of p(x) = 0 is(a) 56 (b) 63 (c) 7 (d) 49
Solutions :
Let x = 31
7 + 31
49
Cubing x3=3
31
7 + 3
31
49 +3. 31
7 . 31
49 31
31
497
x3 = 7+49+3.7.32
31
497
x3 = 56+21xx3+0 x2–21x–56 = 0
Product of roots is 56Ans : (a)
4 If , , , are roots of x4+4x3–6x2+7x–9=0, then the value of (1+ 2)(1+ 2)(1+ 2)(1+ 2) is(a) 9 (b) 11 (c) 13 (d) 5
Solution :x4+4x3–6x2+7x–9 = (x– )(x– )(x– )(x– )Put x=i, i4+4i3–6i2+7i–9=(i– )(i– )(i– )(i– )
–2+3i = (i– )(i– )(i– )(i– )...............................(1)Put x = – i– 2–3i = (– i – )(– i – )(– i – )(– i – ).....................(2)Multiply (1) & (2)
146
4–9i2=( 2–i2)( 2–i2)( 2–i2)( 2–i2) 13 = (1+ 2)(1+ 2)(1+ 2)(1+ 2)
Ans : (c)5 If , , , are roots of 8x3+1001x+2008 =0, then the value of ( + )3+( + )3+( + )3 is
(a) 251 (b) 751 (c) 735 (d) 753Solution :
+ + = 0( + )3+( + )3+( + )3 = (– )3+(– )3+(– )3
= – 3 = – 3 82008–
= 753
Ans : (d)6 Total number of integral values of ‘n’ so that the equation x2+2x–n = 0 (n N) and n [5, 100] has
integral roots is(a) 2 (b) 4 (c) 6 (d) 8 and n [5,100]
Solution :x2+2x–n = 0
x2+2x+1 = n+1(x+1)2=n+1
x+1 = 1n n+1 should be perfect squaren [5,100]
n+1 [6,101]Perfect squares in the given interval are9, 16, 25, 36, 49, 64, 81, 1008 values Ans : (d)
7 If the equation p(q–r)x2+q(r–p)x+r(p–q) = 0
has equal roots, then q2
is equal to
(a) r1
p1
(b) p+r (c) rp1
(d) r1p
Solution :
Clearly x = 1 is one root and the other root is )r–q(p)q–p(r
. roots are equal, we have
)r–q(p)q–p(r
= 1 )r–q(p)q–p(rrootsofoductPr
rp–rq = pq–rq2rp=pq+rq
147
q2
= p1
+ r2
Ans : (a)
1 The largest interval for which x12–x9+x4–x+1>0 is(a) –4<x 0 (b) 0<x<1 (c) –100<x<100 (d) – <x<
2 Read the following passage and answer the questions.If a continuous function ƒ defined on the real line R, assumes positive and negative values in R, thenthe equation ƒ(x) = 0 has a root in R, for example, if it is known that a continous function ƒ on R ispositive at some point and its minimum value is negative , then the equations ƒ(x) = 0 has a root in R.Consider ƒ(x) = kex–x, x R where k R is a constant.(i) The line y=x meets y=kex for k 0 at
(a) no point (b) one point(c) two point (d) more than two points
(ii) The value of k for which kex–x=0 has only one root is
(a) e1
(b) e (c) 2loge (d) 1
(iii) For k>0, the set of all values of k for which kex–x=0 has two distinct roots is
(a) e1,0 (b) 1,
e1
(c) ,e1
(d) (0, 1)
3 If )3538(–25
315–26 = a2, then a is
(a) 31
(b) 31
(c) 3 (d) None of these
4* Solution of alog2 x + alogax + alog3 b = 0 , where a>0, b=a2x is(a) a–1/2 (b) a–4/3 (c) a1/2 (d) None of these
5 Solution of the system of equations
x+ 22 yxy–x3
= 3, y– 22 yxy3x = 0 is _________ or ________________
6 The number of ordered 4–tuple (x, y, z, w) where x, y, z, w [1, 10], which satisfies the inequalityxsin2
2 ycos2
3 zsin2
4 wcos2
5 120 is(a) 0 (b) 144 (c) 81 (d) infinite.
7 The number of solutions of the following inequality
22 xsin1
2 . 32 xsin1
3 . 42 xsin1
4 . .......... n2 xsin1
n n! where
xi (0,2 ) for i = 1,2,3.............n is(a) 1 (b) 2n–1 (c) nn (d) infinite
PRACTICE QUESTIONS
148
8 The number of solutions of |[x]–2x| = 4 is(a) infinite (b) 4 (c) 3 (d) 2
9 How many roots does the equation 3|x| |2–|x|| = 1 possess?(a) 1 (b) 2 (c) 3 (d) 4
10 Let S be the set of values of ‘a’ for which 2 lie between the roots of the quadratic equation x2+(a+2)x–(a+3)=0, then S is gives by(a) (– ,–5) (b) (5, ) (c) (– ,–5] (d) [5, )
11 Match the followingFor what values of m, the equation 2x2–2(m+1)x+m(m+1) = 0 has (m R)
ColumnI Columan II(a) both roots are smaller than 2 (p) {0,3}(b) both roots are grater than 2 (q) (0,3)(c) both roots lie in the interval (2, 3) (r) (– ,0) (3, )(d) exactly one root lie in the interval (2,3) (s)
(e) one root is smaller than 1, the other root (t) 326625–81,
32662581
is greater than 1
(f) both 2 & 3 lie between the roots (v) (– ,–1) [3, )12* The real roots of the equation
)signsradicaln(x32x2........x2x2x = x is
(a) 0 (b) 3 (c) 1 (d) None of these13 Solution of the equation 1+3x/2 = 2x is ___________14 The number of real solutions of the system of equations
x = 2
2
z1z2
, y = 2
2
x1x2
, z = 2
2
y1y2
is
(a) 1 (b) 2 (c) 3 (d) 415 If a, b, c > 0, a2 = bc and a+b+c = abc, then the least value of a4+a2+7 must be equal to
(a) 19 (b) 20 (c) 21 (d) 18
'Note : Questions with * have more than one correct option'Answers
1. d 2. (i) b (ii) a (iii) a 3. d 4. a,b5. (2,1),(+1,–1) 6. 7. b 8. b 9. d 10. a11. a q ; b p ; c r ; d v ; e s ; f t 12. a,b 13. 2 14.a 15.
c
149
SEQUENCES AND SERIES - IArithmetic Progression
SequenceA sequence is a function of natural numbers with codomain as the set of real numbers. It is said tobe finite or infinite according it has finite or infinite number of terms. Sequence a1, a2,........ an isusually denoted by {an} or <an>
SeriesBy adding or subtracting the terms of a sequence we get a series.
Arithmetic Progression (A.P.)A sequence is called an arithmetic progression, if the difference of two consecutive termis the same always.i.e. an–an–1 = d, (constant), nHere d is called the common difference (If d= 0 sequence is a constant sequence. if d>0 thesequence is increasing; if d<0,the sequence is decreasing)nth term of an A.P = a+(n–1)dnth term from end of an A.P = an+(n–1) (–d)where an is the last term and d, the common difference of the A.P.
Sum of n terms of an A.PThe sum Sn of n terms of an A.P is given by
Sn = )aa(
2n or
d)1–n(a22n
n1
Where a is the first term, an last term, d common differenceNote that a sequence is an A.P if and only if its nth term is a linear expression in n, and in thatcase its common difference is the coefficient of n.Also sum to n terms is of the form An2+Bn where A & B are constants, and the commondifference of the A.P is 2A.Note :(i) d = an– an–1
= (Sn–Sn–1)–(Sn–1–Sn–2)= Sn–2Sn–1+Sn–2
(ii) Sn–3Sn–1+3Sn–2–Sn–3= (Sn–Sn–1) –2 (Sn–1–Sn–2)+(Sn–2–Sn–3)= an–2an–1+an–2= 0 as an–2,an–1, an are in A.P
Selection of terms in an A.PIn case of an odd number of terms the middle term is a and common difference d while incase of even number of terms, middle terms are a–d, a+d and common difference is 2d.
No. of terms Terms Common difference.3 a–d, a, a+d d4 a–3d, a–d, a+d, a+3d 2d
N.
150
5 a–2d, a–d, a, a+d, a+2d dInsertion of Arithmeitc Means
If a, b, c are in A.P, then b= 2ca
is called the single arithmetic mean of a & c. Let a & b be two
given numbers and A1, A2,........... An are n A.M’s between them. Then a, A1,A2,...An, b are in A.P.
Common difference of this sequence d= 1na–b
.
A1= a+d, A2= a+2d etc. we can find all the arithmetic means.Properties of A.P.
1 If a1, a2, a3, ........ are in A.P; then a1 k, a2 k,a3 k,..................... are also in A.P..
2 If a1, a2, a3,................. are in A.P, then a1 , a2 ,a3 ,.................... and 1a,
2a, 3a
.............. are also in A.P ( ( 0)
3 If a1, a2,.......... an are in A.P, then an,an–1,...............a2,a1 is also an A.P with commondifference (–d)
4 If a1, a2, a3, ..................and b1, b2, b3, .......................... are two A.P.s thena1 b1,a2 b2,a3 b3,..... are also in A.P..
5 If a1, a2, a3,.............. and b1, b2, b3,................are two A.P.s then a1b1, a2b2, a3b3,...........and
1
1
b
a,
2
2ba
, 3
3
b
a,................. are NOT in A.P..
6 If 3 numbers are in A.P we may take them as a–d, a, a+d. If 4 numbers are in A.P, wecan take them as a–3d, a–d, a+d, a+3d.
7 In an arithmetic progression, sum of the terms equidistant form the beginning andend is a constant and equal to sum of first and last term.Iie for {an},a1+an = a2+an–1=a3+an–2=......
Also ar = 2aa krk–r , 0 k n–r..
8 Sum of n arithmetic means between two given numbers a & b is n times the singleA.M between them .
ie. A1+A2+...............+An = n 2ba
9 Also Sn = a1+a2+......+an= evenisnif);termsmiddletwoofsum(2n
.oddisnif);termmiddle(n
151
Solved Examples1 If x, y, z are real numbers satisfying the equation 25(9x2+y2)+9z2–15(5xy+yz+3zx) = 0, then
x,y,z are in(a) A.P (b) G .P (c) H.P (d) None of theseSolution:
We have (15x)2+(5y)2+(3z)2–(15x)(5y)–(5y)(3z)–(3z)15x) = 0or (15x–5y)2+(5y–3z)2+(3z–15x)2 = 0
a2+b2+c2–ab–bc–ca = 21
{(a–b)2+(b–c)2+(c–a)2}
(15x–5y)= 0,(5y–3z)= 0,(3z–15x) = 0
15x = 5y = 3z 5z
3y
1x
x = , y = 3 , z = 5So x, y, z are in A.P
Ans (a)2 The number of common terms of the two sequences 2,5,8,11..................299 and
3,5,7,9,11.....201.(a) 17 (b) 33 (c) 50 (d) 147Solution
Sequence of common terms is5,11,17..........whose nth term isan = 5+(n–1)6 = 6n–1an 201 6n–1 201
n 33 32
n = 33
Ans (b)
3 The value of x+y+z = 15. If a, x, y, z, b are in A.P, while the value of x1
+ y1
+ z1
is 35
. Ifa1
,
x1
, y1
, z1
, b1
are in A.P, then
(a) a = 1, b = 9 (b) a = 9 , b = 1 (c) can not find (d) None of theseSolution:
a+x+y+z+b = 25
(a+b)
x+y+z = 23
(a+b)
15 = 23
(a+b)
a+b = 10
152
Also a1
+ x1
+ y1
+ z1
+b1
= 25
b1
a1
x1
+ y1
+ z1
= 23
b1
a1
35
= 23
abba
35
= 23
. ab10
ab = 9
Solving we get a = 1, b=9 or a= 9, b = 1Ans = a,b
4 Let a1, a2, a3,............a11, be real numbers satisfying a1 = 15, 27–2a2>0 and ak = 2ak–1–ak–2 for
k = 3,4,..............11. If 11
a.......aa 211
22
21 = 90, then the value of 11
a.......aa 1121 is
equal toSolution
a1 = 15, ak=2ak–1–ak–2 a1, a2........a111 are in A.P.
11a.......aa 2
112
22
1 = 11
)d1015(.........)d15()15( 222
= 90
7d2+30d+27 = 0 d = –3, 79–
given a2< 227
d = –3 & d 79–
11a.......aa 1121 = 2
11 11
)103–30( = 0
Ans 05 Suppose A, B, C are defined as A = a2b+ab2–a2c–ac2, B = b2c+bc2–a2b–ab2 and C = a2c+c2a–
cb2–c2b, where a>b>c>0 and the equation Ax2+Bx+C = 0 has equal roots then a1
, b1
,c1
are in
(a) A.P (b) G.P (c) H.P (d) None of theseSolution :
A+B+C = 0, x = 1 is a root of Ax2+Bx+C = 0The other root = 1 ( roots are equal)
1×1 = AC
C = AA
a2c+c2a–cb2–c2b = a2b+ab2–a2c–ac2
c(a–b)(a+b+c) = a(b–c)(a+b+c) ac–cb = ab–ac ( a+b+c 0)2ac = ab+bc
11(15 11) d2
11(1+----+10 )
2 2
11 3od (1+2+10)
225+350 d +150 d =90
2
35d + 150 d +135 = 0 2
2
153
b2
= c1
+a1
a1
, b1
,c1
are in A.P.A.P.
Ans: (a)6 If a, b, c, d, are distinct integers in an increasing A.P such that d = a2+b2+c2, then a+b+c+d =
(a) –1 (b) 0 (c) 1 (d) 2Solution :
a, b, c, d ZLet b = a+ , c= a+2 , d = a+3 , Za+3 = a2+(a+ )2+(a+2 )2
5 2+3(2a–1) +3a2–a = 0 .................(1) is real
D 09(2a–1)2–4.5.(3a2–a) 0 24a2+16a–9 0
31–
– 1270 a
31–
+ 1270
a I a = –1,0If a = –1, = 1 and if a = 0 Z
a+b+c+d = 24
(a+d) = 2(2a+3 ) = 2(–2+3) = 2
Ans (d)7 Consider the sequence in the form of groups (1), (2,2), (3,3,3), (4,4,4,4), (5,5,5,5,5),............,
the 2000th term of the sequence is not divisible by(a) 3 (b) 9 (c) 7 (d) None of theseSolution
Let us write the terms in the groups as follows:(1), (2,2), (3,3,3).........consisting of 1,2, 3, 4, ............termsLet 2000th term fall in nth group. Then1+2+3+...........+(n–1)<2000 1+2+....................+n
2)1–n(n
<2002
)1n(n
n(n–1)<400 n(n+1)n(n–1)<400 and n(n+1) 400n2–n–400<0 and n2+n–400 0
n = 63i.e. 2000th term falls is 63 rd group.Also 2000th term is 63
Ans (d)8 If x18 = y21 =z28, then 3logyx,3logzy, 7logxz are in
(a) A.P (b) G.P (c) H.P (d) None of these
154
SolutionLet x18 = y21 =z28=18logx = 21log y = 28 logz = log
logyx = 1821
, logzy = 2128
, logxz =
2818
3logyx = 27
, 3logzy = 4, 7logxz = 29
3, 3logyx, 3logzy, 7logxz are in A.PAns (a)
1* If 1, logyx, logzy, –15logxz are in A.P, then
(a) z3 = x (b) x = y–1 (c) y = 3z1
(d) None of these
2 If 51+x+51–x, 2a
, 25x+25–x are three consecutive terms of an A.P, then a is
(a) 12 (b) 12 (c) = 12 (d) None of these3 If sin , sin2 , 1, sin4 and sin5 are in A.P, where – < < , then lies in the interval
(a) 2,
2–
(b) 3,
3–
(c) 6,
6–
(d) None of these
4 If the roots of x3–12x2+39x–28 = 0 are in A.P, then their common difference will be(a) 1 (b) 2 (c) 3 (d) 4
5 If the sides of a right triangle are in A.P, then the sum of the sines of two acute angles is
(a) 57
(b) 34
(c)2
1–5 +
215
(d) None of these
6 Read the passage and answer the following questions. Two consecutive numbers from 1, 2,
3, .........,n are removed. The arithmetic mean of the remaining numbers is 4105
.
(i) The value of n lies in(a) [45,55] (b) [52,60] (c) [41,49] (d) None of these(ii) The removed numbers(a) lie between 10 and 20 (b) are greater than 10(c) are less than 15 (d) none of these(iii) Sum of all numbers(a) exceeds 1600 (b) is less than 1500(c) lies between 1300 and 1500 (d) none of these
7 Concentric circles of radii 1, 2, 3, .....100 cm are drawn. The interior of the smallest circle is
PRACTICE QUESTIONS
155
coloured red and the angular regions are coloured alternately green and red, so that no twoadjacent regions are of same color. The total area of the given regions in sq.cm is equal to(a) 1000 (b) 5050 (c) 4950 (d) 515
8* If three identical fair unbiased dice are thrown together such that the numbers a, b and c, wherea,b, c {1,2,3,4,5,6} appear on each of them respectively. If r represents all possible distinctcases, represent the number of ways in which a+b+c = 9 and represents the number ofways of obtaining a+b+c = 8, then match the following.
Column I Column II(a) If represents the common difference of an A.P
such that the arithmetic mean of the squares of thesequantities exceeds the square of A.M by 9, then thenumber of terms of A.P are (p) 5C2
(b)),max(
),min(rr
),max( 5–C (q) 2
(c) If 6th term in the expansion of ( + )n is the greatestterm, then n is (r) 1
(d) ––
is equal to where [.] denotes the
greatest integer function (s) 109 A person is to count 4500 currency notes. Let an denotes the number of notes he counts in the
nth minute. If a1 = a2 = ............a10 = 150 and a10,a11...........are in A.P with common difference –2, then the time taken by him to count all notes is(a) 24 min (b) 34min (c) 125min (d) 135min
10 If a1, a2...............an are in A.P with common difference d 0 then (sind) (seca1 seca2+seca2seca3+.........secan–1secan) is equal to(a) cotan–cota1 (b) cota1–cotan (c) tanan–tana1 (d) tana1–tanan
Nnote:* Question with more than one option is correct.
Answers1. a,b,c 2. b 3. d 4. c 5. a 6. (i) a (ii) c (iii) b 7. b 8. a q ; b q ; c p,s ; d r9. b 10. c
156
SEQUENCES AND SERIES - IIGeometric Progression
A sequence of non-zero numbers is called a geometric progression if ratio of a term and theterm preceding to it is always a constant. This constant is called the common ratio of the G.P.
i.e. a1, a2, .........., an is in G.P. if ra
a
n
1n = const., Nn .
nth term of a G.P. = an = a rn–1.
nth term from end of a G.P = an. 1n
r1
where an is the last term and r the common ratio of the
G.P.
Sign of a +ve +ve –ve –ve
Range of r r>1 0<r<1 r>1 0<r<1
G.P. is increasing decreasing decreasing increasing
If r = 1, the sequence will be a constant sequence. If r is negative the terms of G.P. are alternatelypositive and negative and so the G.P is neither increasing nor decreasing.
Sum of n terms of a G.P
Let the first term of G.P be a common ratio r and last term an, then
i) Sum to n terms
Sn = 1rwhen ; na
1rwhen ;r1r1aor
1r1)a(r nn
Also Sn = 1raraor
r1raa nn ; when r 1
Sum of an Infinite G.P
Sum of an infinite G.P. with first term a and common ratio r (–1 < r < 1; r 0 or 0<|r|<1) is given
by S = r1a
If r>1, then the sum of an infinite G.P tends to infinity.
Selection of terms in a G.P
157
In case of an odd number of terms the middle term is a and common ratio is r while in case of
even number of terms, middle terms are ra
, ar and common ratio is r2.
No. of terms Terms Common ratio
3 ra
, a, ar r
4 3ra
, ra
, ar, ar3 r2
5 2ra
, ra
, a, ar, ar2 r
Note that it is convenient to take the terms as a, ar, ar2 ......... if the product of numbers is notgiven.
Insertion of geometric means
Note : If a & b are two numbers of opposite signs then geometric mean between them does not exist.
Solved Examples
1. Suppose a, b, c are in A.P and a2, b2, c2 are in G.P. If a<b<c and a+b+c = 23
, then the value of
a is
a. 221
b. 321
c. 31
211 d. 2
121
Solution : Let the numbers be A – d, A , A + d
Then A = 21
Numbers are 21
– d, 21
, 21
+d
a2, b2, c2 are in G.P.22
21
= 2
d21 2
d21
161
= 2
2d41
2d41
= 41
d2 = 21
d= + 21
158
Since a < b < c, a = 21
– 21
Ans. d
2. If an =n
1–n32
43)1(–.........
43
43–
43
and bn = 1–an, then the minimum natural number
no such that bn > an & n > no is
a. 4 b. 5 c. 6 d. 12
Solution : an =
431
431
43 n
i.e. an = n
431
73
bn > an 1 – an > an 2an < 1
n
431
76
< 1 1 – n
43
< 67
– n
43
< 61
(–3)n+1 < 22n–1
For n to be even, the inequality always holds. For n to be odd, it holds for n > 7.
Least natural number for which it holds is 6.
Ans. c
3. If a be the arithmetic mean of b and c and G1, G2 be the two geometric means between them,then G1
3 + G23 =
a. G1G2a b. 2G1G2a c. 3G1G2a d. none of these
Solution : b, a, c are in A.P
2a = b+c
Also b, G1, G2, c are in G.P
G12 = b G2 G1
3 = bG1G2
G22 = G1c G2
3 = cG1G2
G13 + G2
3 = G1G2 (b+c) = 2G1G2a
Ans. b
159
4. It is known that 8)1r2(1 2
21r
, then 21r r
1 is
a. 24
2
b. 3
2
c. 6
2
d. none of these
Solution:
Let ..........31
21
11
222 = x
21r r
1 = ..........
51
31
11
222 + ..........61
41
21
222
x = 8
2
+ 41
..........31
21
11
222
x = 8
2
+ 41
x 4x3
= 8
2
x = 6
2
Ans. c
5. If ai R, i = 1, 2, 3, .......n and all ai’s are distinct such thatn
2i
2i
1–n
1i1ii
21n
1i
2i 0axaa2xa , then a1, a2, ......... are in
a. G.P b. A.P c. H.P d. none of these
Solution :1n
1i(aix + ai+1)
2 < 0 1n
1i(aix + ai+1)
2 = 0
aix + ai+1 = 0 i = 1, 2, ......... n–1
i
1i
aa
= – x i = 1, 2, ...... n –1
a1, a2, ......... an are in G.P.
160
Ans . a
The 1025th term is the sequence
1, 22, 4444, 88888888, ........... is
a. 29 b. 210 c. 211 d. none of these
Solution : Number of digits in each term are in G.P.Let 1025th term =2n
then1+2+4+8+........+2n–1 < 1025 1+2+4+8+.....+2n
11–2
)1–2( n
< 1025 1.1–2
)1–2( 1n
2n–1 < 1025 2n+1–12n+1 1026 > 1024 = 210
n+1 > 10 n > 9n=10
Ans. bIf a1/x=b1/y=c1/z and a, b, c are in geometrical progression, then x,y, z are ina. A.P. b. G.P. c. H.P. d. None of these
Solution : a1/x=b1/y=c1/z =ka=kx, b=ky, c=kz
a, b, c are in G.P. b2 = ack2y=kx+z
2y=x+zAns. a
If the arithmetic mean of two numbers be A and geometric mean be G, then the numbers will be
a. A (A2–G2) b. 22 G–AA
c. )G–A)(GA( d. 2)G–A)(GA(A
6.
7.
8.
161
Solution : Let the number be a and b
A = 2
ba G2=ab
a and b are the roots of x2–2Ax+G2=0
x=2
G4–A4A2 22
(A2–G2)Ans. a
1. The sum of an infinite geometric series is 2 and the sum of the geometric series made from thecubes of this infinite series is 24. Then the series is
a. .......83
43–
233 b. ......
83
43
233
c. ......83–
43
23–3 d. none of these
2. Read the passage and answer the following questions.Let A1, A2,.....Am be arithmetic means between –2 and 1027 and G1, G2, .... Gn be Geometricmeans between 1 and 1024. Product of geometric means is 245 and sum of arithmetic means is1025×171.i. The value of n is
a. 7 b. 9 c. 11 d. none of theseii. The value of m is
a. 340 b. 342 c. 344 d. 346iii. The value of G1+G2+G3+....+Gn is
a. 1022 b. 2044 c. 512 d. none of theseiv. The common difference of the progression A1, A3, A5.....An is
a. 6 b. 3 c. 2 d. 1v. The numbers 2A171, 1G2
5 , 2A172 are ina. AP b. GP c. HP d. AGP
3. The difference between two numbers is 48 and the difference between their arithmetic meanand their geometric mean is 18. Then, the greater of two numbers isa. 96 b. 60 c. 54 d. 49
4. If ax=by=cz=dw, the value of x w1
z1
y1
is
a. loga(abc) b. loga(bcd) c. logb(cda) d. logc(dab)5. If three positive numbers x, y, z are in A.P. and also tan–1x, tan–1y, tan–1z are in A.P., then
a. x=y=z b. x y = z c. x = y z d. none of these6. If a1, a2, a3 are three consecutive terms of a G.P. with common ratio k. Then the values of k for
which the inequality a3 > 4a2 – 3a1, is satisfied is (if a1 > 0)
x= A
PRACTICE QUESTIONS
162
a. (1, 3) b. ,31,– c. R d. none of these7. The three successive terms of a G.P. will form the side of a triangle if the common ratio r lies in
a. 215,
21–5
b. ,2
51–
c. 21–5,– d. none of these
8. If for 0 < x < 2 , exp (sin2x+sin4x+sin6x+........ ) loge2) satisfies the quadratic equation
x2–9x+8=0, xcosxsinxcos–xsin
is
a. 2– 3 b. 2+ 3 c. 3 – 2 d. none of these9. Match the following
Column I Column IIa. If a, b, c are non zero real numbers such that
3(a2+b2+c2+1) = 2 (a+b+c+ab+bc+ca) then a, b, c are in (p) APb. If the square of difference of three numbers be in AP, then
their differences are in (q) GPc. If a–b, ax–by, ax2–by2(a, b 0) are in G.P. then
x, y, b–aby–ax
are in (r) HP
(s) equal10. If 1+p+p2+......+pn=(1+p) (1+p2)(1+p4)(1+p8) (1+p16) then the value of n (n N) is
a. 32 b. 16 c. 31 d. 1511. If sin , 2 (sin +1), 6sin +6 are in G.P., then the fifth term is
a. 81 b. 82 2 c. 162 d. none of these12. If x {1, 2, 3, ....9} and fn(x)=xxx......x(n digits) then fn
2(3)+fn(2) is equal toa. 2f2n(1) b. fn
2(1) c. f2n(1) d. 2f2n(4)13. The number of divisors of 1029, 1547 and 122 are in
a. AP b. GP c. HP d. none of these14. Let x1, x2,.....,xn be a sequence of integers such that
(i) –1 xi 2 for i=1, 2,....n(ii) xi+x2+....+xn=19(iii) x1
2+x22+....+xn
2=99Let m and M be the minimum and Maximum possible values of x1
3+x23+....xn
3 respectively, then
the value of mM
is _____
15. Let 16, 4, 1, .... be a geometric sequence. Define Pn as the product of the first n terms. Then the
163
value of 4
P1n
nn
is _____.
a. 64 b. 641
c. 32 d. none of these
Note:* Questions with more than one option is correct
ANSWERS
1. c 2. (i) b (ii) b (iii) a (iv) a (v) a 3. d 4. b
5. a 6. b 7. a 8. a
9. a p,q,s; b r; c p,q,s 10. c 11. c 12. c
13 a 14. 7 15. c
164
SEQUENCES AND SERIES - IIIHarmonic Progression
A sequence a1,a2......an of non-zero numbers is called a harmonic sequence if the sequence
,.......a1,........
a1,
a1
n21
is an A.P..
Eg: The sequence ......71,
51,
31
is an H.P. because the sequence 3,5,7...... is an A.P..
nth term of a H.P.The nth term of a H.P. is the reciprocal of nth term of the corresponding A.P. and the common
difference of the corresponding A.P is d. i.e. 12 a
1–a1d .
i. nth term of the H.P is given by
an =d1–n
a1
1
1 121 a1–
a11–n
a1
1212
21
a–a1–naaa
ii. nth term of the H.P. from end
d–1–na1
1a
n
1n
12n a1–
a11–n–
a1
1
=21n21
n21
a–aa1–n–aaaaa
Note: No term of H.P. can be zero and there is no general formula for finding out the sumof n terms of a H.P.
Harmonic MeanIf a,b,c are in H.P., then b is called the H.M. between a & c.Now a,b,c are in H.P.
c1,
b1,
a1
are is A.P..
c1
a1
b2
b = caac2
i.e. H.M. between a & c isca
ac2
165
Note: The single H.M. between a & b is baab2
.
The single H.M. (H) of n positive numbers a1,a2,.....an is given by
na1.......
a1
a1
a1
H1 n321 =
n321 a1.......
a1
a1
a1
n1
Insertion of Harmonic MeansLet a and b be two given numbers and H1,H2.....Hn be the H.M.’s between them. Then a,H1, H2.......Hn , b will be in H.P. Let d the common differece of the corresponding A.P.
th2nb term of H.P..
d1–2na1
1b
d1na1
b1
1na1–
b1
d
ab1nb–ad
da1
H1
1 ab1nb–a
a1
d2a1
H1
2 ab1nb–a2
a1
.
.
.
nda1
H1
n ab1nb–an
a1
E.g: Insert 4 H.M’s between 32
and 132
1423–
213
1na1–
b1
d
d = 1
251
23
H1
1
or52H1
166
272
23
H1
2
or72H2
293
23
H1
3
or92H3
2114
23
H1
4
or112H4
Note: The sum of reciprocals of n H.M’s between two numbers is n times the reciprocal ofsingle H.M. between them.i.e. if a,b are the numbers and H1,H2.....Hn be the H.M’s between them,
thenab2
banH1........
H1
H1
n21
= bandaof.M.H1n
Solved Examples1. If a1,a2,a3......a10 are in A.P. and h1,h2,h3,.....h10 are in H.P. If a1=h1 = 2 and a10 = h10 = 3,
then a4 h7 isa. 6 b. 7 c. 18 d. none of theseSolution:
a10 = 3 3 = 2+9d91d
h1,h2.....h10 are in H.P.
1110
d9h1
h1
1d921
31
541–d1
a4 = a1+3d = 2+3x91 =
37
312
and d6h1
h1
17 718h7
a4h7 = 67
18x37
Ans: a2. If a1,a2,a3.....an are in H.P., then a1a2 + a2a3+.........+an–1an will be equal
a. a1an b. na1an c. (n–1)a1an d. none of theseSolution:
n21 a1..,.........
a1,
a1 are in A.P..
12 a1–
a1 d
a1–
a1.......
a1–
a1
1–nn23
167
daa
a–a........aaa–a
aaa–a
1–nn
n1–n
32
32
21
21
a1 – a2 = da1a2
a2 – a3 = da2a3an–1–an = danan–1
Adding, we get a1– an = d(a1a2+a2a3+.......+anan–1)
i.e. a1a2+a2a3+......+anan–1= da–a n1
= 1–naaa–aa–a
n1n1
n1
= a1an(n–1)Ans: c
3. If A1,A2 ; G1,G2 and H1,H2 be two A.M.’s , G.M.’s and H.M.’s between two numbers a & b
repectively, then21
21
21
21
AAHHx
HHGG
a. 1 b. 0 c. 2 d. 3Solution:a, A1, A2, b are in A.P A1+A2 = a+ba, G1, G2, b are in G.P G1G2 = ab
a, H1, H2, b are in H.Pb1,
H1,
H1,
a1
21are is A.P
b1
a1
H1
H1
21
Given expression is 2121
21
H1
H1
AAGG
= 1b1
a1
baab
Ans: a
4. If nine A.M.’s and nine H.M.’s are inserted between 2&3, thenH6A ....(where A is any of
the A.M.’s and H the corresponding H.M.)a. 5 b. 3 c. 15 d. none of theseSolution:2,A1,A2.......A9,3 are in A.P.
d = 101
1na–b
Ai = 2 + id = 2+ 10i
; i = 1,2,......9
1an
=1a1
(n-1)d
1an
=1a (n-1)d–
–a n
d = (n-1)a an1 a1
1
168
31,
H1.,.........
H1,
H1,
21
921are in A.P..
D = 601–
ab)1n(b–a
; i = 1,2,......9
9,.....,2,1i;60
1–i21iD
21
H1
i
10i–3
H6
i; 9.,.........2,1i
Ai +iH
6= 5 ; i = 1,2,3.........9
Ans: a
5. If H1,H2.....H20 be 20 harmonic means between 2&3, then3–H3H
2–H2H
20
20
1
1
a. 20 b. 21 c. 40 d. 38Solution:
3–H3H
2–H2H
20
20
1
1
=
1
1
H1–
21
H1
21
+
20
20
H1–
31
H1
31
=d–
21–
21
d21
21
+d
31–
31
d–31
31
= d
d–32
d–d1 1–
d321–
d1–
= 2–1–32
d1
169
= 2–d42d1 d421–
32d1–22
21
31
Ans: 40
1. If x,y,z are in H.P., then the value of expression log(x+z) + log(x–2y+z) will bea. log(x–z) b. 2log(x–z) c. 3log(x–z) d. 4log(x–z)
2. If a,b,c,d are positive real numbers such that a+b+c+d = 2, then M = (a+b) (c+d) satisfies theequationa. 0 < M 1 b. 1 M 2 c. 2 M 3 d. 3 M 4
3. If a1,a2,a3.......,an are in H.P., then1–n21
n
n31
2
n32
1
a.....aaa...,,.........
a.....aaa,
a.....aaa
are in.a. A.P. b. G.P. c. H.P. d. none of these
4. If a,a1,a2.....a2n–1, b are in A.P. ; a,b1,b2.....b2n–1,b are in G.P. ; a,c1,c2.....c2n–1,b are in H.P.where a,b, are positive, then the equation anx
2 – bnx+cn = 0 has its rootsa. real and unequal b. real and equalc. imaginary d. none of these
5. If b–c, 2b–x, b–a are in H.P, then 2x–b,
2x–a and 2
x–c are in
a. A.P b. G.P c. H.P d. none of these
6. The first two terms of a H.P are 52
and 1312
respectively. Then the largest term is
a. 2nd term b. 6th term c. 4th term d. none of these
7. If a,b,c are in A.P, p,q,r are in H.P and ap, bq, cr are in G.P. then pr
rp
is equal to
a. ac–
ca
b. ac
ca
c. bq
qb
d. bq–
qb
8.* If a,b,c are in H.P, then the value of a1–
c1
b1
b1–
a1
c1
is
a. 2b1–
bc2
b. 22 c1–
ca2
c3
41
c. ab2–
b3
2 d. none of these
9. If H1,H2......Hn be n harmonic means between a and b, then b–HbH
a–HaH
n
n
1
1is equal to
a. 0 b. n c. 2n d. 1
PRACTICE QUESTIONS
170
10. If a,b,c be in G.P. and a+x,b+x,c+x in H.P., then the value of x is (a,b,c are district numbers)a. c. b. b c. a d. none of these
11.* The harmonic mean of two numbers is 4, their A.M.A and G.M.G satisfy the relation2A +G2=27. The numbers area. 6 and 3 b. 3 and 6 c. can’not find d. none of these
12. If n be a root of the equationx2(1–ab) –x(a2+b2) – (1+ab) = 0, then H1–Hn =
a. ab (a–b) b. ab(b–a) c. aba–b
d. none of these
13. If x,y,z are in A.P, ax,by,cz in G.P. and a,b,c in H.P., then xz
zx
a. 2 b. ac
ca
c. caac2
d. none of these
14. If a,b,c,d are in H.P., then the value of 2222 c1–
b1
d1–
a1
is
a. 1 b. 2 c. 3 d. 4
Note:*Questions with more than one option is correct
ANSWERS
1.b 2.a 3. c 4. c 5. b
6. a 7. b 8. a,c 9. c 10. b
11. a,b 12. a 13. b 14. c
171
SEQUENCES AND SERIES - IVInequalities based on A.M., G.M. and H.M.
Properties of A.M, G.M & H.M
Let A, G, H be the arithmetic, geometric and harmonic means of two positive numbers a & b.
Then, A = 2ba
, G = ab , H = baab2
i) A. H = 2ba
. baab2
= ab = G2
i.e. G2 = A.H
G is the geometric mean between A & H.
Again A – G = 2ba
– ab = 2
ba2
> 0
A > G
Also G2 = A.H
GA
HG
> 1
HG
> 1 or G > H
Combining, A > G > H
Note : If the numbers are equal, then A = G = H. Thus, A > G > H, equality holds when thenumbers are equal.
ii) The equation with a and b as its roots is x2 – 2Ax + G2 = 0.
or if A & G be the A.M and G.M between two positive numbers a & b then 22
22
GAAGAA
ba
iii) If A, G, H be the A.M, G.M, and H.M between three given numbers, a, b and c , then theequation having a,b,c as its roots is
x3 – 3Ax2 + HG3 3
x – G3 = 0
Proof : A = 3cba
a + b + c = 3A
G = (abc)1/3 G3 = abc
172
3c1
b1
a1
H1
H1
abc3cabcab
or ab + bc + ca = HG3
Habc3 3
Equation having a, b, c as roots is
x3 – (a + b + c)x + (ab + bc + ca)x2 – abc = 0
x3 – 3Ax + HG3 3
x – G3 = 0
Example : For distinct positive numbers x, y, z, prove that (x +y)(y + z)(z + x) > 8xyz
Solution : We have A > G
For positive numbers x and y, x + y > xy2
For positive numbers y and z, y + z > yz2
For positive numbers z and x, z + x > zx2
Multiplying, (x +y)(y + z)(z + x) > 222 zyx8
i.e., (x +y)(y + z)(z + x) > 8xyz
Solved Examples
1. If a, b, c are positive then prove that ((1+a)(1+b)(1+c))7 > 77a4b4c4.
Solution : (1+a)(1+b)(1+c) = 1 + a + b + c + ab + bc + ca + abc
> a + b + c + ab + bc + ca + abc
> 7(a . b . c . ab . bc . ca . abc)1/7 ( A > G )
i.e. (1+a)(1+b)(1+c) > 1 + 7(a4b4c4)1/7 > 7(a4b4c4)1/7
((1+a)(1+b)(1+c))7 > 77a4b4c4
2. Maximum value of xyz for positive values of x, y, z if yz + zx + xy = 12 is
a. 64 b. 43/2 c. 8 d. none of these
Solution : Apply A > G for yz, zx & xy
3xyzxyz
> (x2y2z2)1/3
173
312
> (xyz)2/3
(xyz) < 43/2
xyz < 8Ans: c
3. Maximum value of x2y3 where x & y lie in 1st quadrant in the line 3x + 4y = 5.
a. 165
b. 83
c. 85
d. 163
Solution : x2y3 = x . x . y . y . y
3x + 4y = 2x3
+ 2x3
+ 3y4
+ 3y4
+ 3y4
A > G
5/1
3y4.
3y4.
3y4.
2x3.
2x3
53y4
3y4
3y4
2x3
2x3
51
32
3yx16
55
x2y3 < 163
Ans. d4. If a2 + b2 + c2 = 1 = x2 + y2 + z2, then maximum value of ax + by + cz is (a, b, c, x, y, z are
positive real numbers)a. 4 b. 3 c. 2 d. 1
Solution : A > G
2222
xa2
xa a2 + x2 > 2ax
Similarly b2 + y2 > 2byc2 + z2 > 2cz
adding, (a2+b2+c2) + (x2+y2+z2) > 2(ax+by+cz)
ax+by+cz < 211
Ans. d
174
5. If a, b, c are positive then the minimum value of bac
acb
cba
is
a. 32
b. 23
c. 1 d. none of these
Solution : Apply A > H for ba1,
ac1,
cb1
accbba3
3ba
1ac
1cb
1
)cba(29
ba1
ac1
cb1
29
bacba
accba
cbcba
293
bac
acb
cba
23
bac
acb
cba
Ans. b
6. If x, y, z are three positive numbers, then (x+y+z) z1
y1
x1
> ______
a. 3 b. 9 c. 31
d. none of these
Solution : 3zyx
> (xyz)1/3 ( A > G)
Also3/1
z1.
y1.
x1
3z1
y1
x1
( A > G)
Multiplying, 9
z1
y1
x1)zyx(
> 1
175
Ans. b (or apply A > H for x, y, z to get the result)
7. Prove that : nC1 . (nC2)
2 (nC3)3 ....... (nCn)
n < 1n2n
Solution : Let S = nC1 + 2nC2 + 3nC3 + ....... n nCn
= n
1r r..nCr =
n
1rn. n–1Cr–1 = n 2n–1
Now A > G
n..........321C......CC......CCCCCC n
nn
nn
n3
n3
n3
n2
n2
n1
n
n.....211n
nn2
2n
1n C........CC
)1n(n2
nn
n22
n1
n1n
C........CC
2)1n(n
2.n
(nC1 . (nC2)
2 (nC3)3 ....... (nCn)
n) 2
)1n(nn
1n2
= 2
1n Cn
1n2
8. If a, b, c, d are in H.P. then
a. a+d > b+c b. ad > bc c. ad = bc d. none of these
Solution : a, b, c are in H.P b is the H.M of a & c.
A.M of a & c = 2ca
we have A > H 2ca
> b a+c > 2b
Similarly b + d > 2c
Adding, a+b+c+d > 2b+2c a+d > b+c
Also a, b, c are in H.P. b is the H.M of a & c
G.M of a & c = ca
G > H ca > b
176
Similarly db > c
Multiplying, abcd > bc ad > bc
Ans. a, b
1. If a+b+c = 1, then find k such that
1c11
b11
a1
abc27k
> k
a. 8 b. 7 c. 3 d. none of these
2. A rod of fixed length k slides along the coordinate axes. If it meets the axes at A(a,0) and
B(0,b), then the minimum value of 22
b1b
a1a is
a. 0 b. 8 c. 22
k44k d. 2
2
k44k
3. If positive numbers a, b, c be in H.P, the equation x2 – kx + 2b101 – a101 – c101 = 0 (k R) has
a. both roots imaginary b. one root is positive and other is negative
c. both roots positive d. both roots negative
4. If n N, n2
n
21nn >k where k is
a. (2n!)3 b. 2(n!)3 c. (n!)3 d. none of these
5. If x, y, z R+, then is yxxy
zxxz
zyyz
is always
a. < 21
(x+y+z) b. > xyz31
c. < 31
(x+y+z) d. > xyz21
6.0i 0j 0k
kji 3331
is
(i j k)
a. 271
b. 20881
c. 1 d. none of these
PRACTICE QUESTIONS
177
7. Minimum value of 3x)2x)(1x(
: x > 3 is
a. 223 b. 223 c. 323 d. 223
8. The least value of 6tan2 + 54cot2 + 18 is
i) 54 when A.M. > G.M is applied for 6tan2 54cot2 , 18
ii) 54 when A.M > G.M is applied for 6tan2 , 54cot2 and 18 is added further.
iii) 78 when tan2 = cot2 .
a. (iii) is correct b. (i) is correct (ii) is flase
c. (i) and (ii) are correct d. none of these
9. If A, G, H are A.M, G.M, H.M between the same two numbers, such that A–G=15 andA–H=27, then the numbers are
a. 100, 50 b. 120, 30 c. 90, 60 d. none of these
10. If a, b, c R, the square root of a2+b2+c2–ab–bc–ac is greater than or equal to
a.23
max {|b–c|, |c–a|, |a–b|} b. 23
max {|b–c|, |c–a|, |a–b|}
c. max {|b–c|, |c–a|, |a–b|} d.43
max {|b–c|, |c–a|, |a–b|}
11. If x1, x2, x3, x4 are four positive real numbers such that
x1 + 2x
1 = 4, x2 +
3x1
= 1, x3 + 4x
1 = 4, x4 +
1x1
= 1 then
a. x1 = x3 and x2 = x4 b. x2 = x4 but x1 x3
c. x1 x2 = 1, x3x4 = –1 d. x3x4 = 1, x, x2 1
12. If a, b, c > 0 and a (1–b) > 41
, b (1–c) > 41
, c (1–a) > 41
, then
a. never possible b. always true c. cannot be discussed d. none of these
13. If a, b, c are the sides of a triangle, then cb1
, ac1
, ba1
are also the sides of the triangle is
a. sometimes true b. always true c. cannot be discussed d. never true
14. Given n4 < 10n for a fixed positive integer n > 2, then
a. (n+1)4 < 10n+1 b. (n+1)4 > 10n+1 c. nothing can be said d. none of these
178
Answers
1. a 2. d 3. b 4. c 5. a 6. b
7. b 8. c 9. b 10. a 11. a 12. a
13. b 14. a
179
SEQUENCES AND SERIES - VArithmetico Geometric Series & Special Sequences
Arithmetico-geometric SeriesA series is said to be an arithmetico geometric series if its each term is formed by multiplyingthe corresponding terms of an A.P and a G.P.E.g. 1+2x+3x2+4x3+...........Here 1,2,3,4.....are in A.P and 1,x,x1,x3.......are in G.P
Sum to n termsLet Sn = a+(a+d)r + (a+2d)r2+......(a+(n–1)d)rn–1........(1)Multiply by r on both the sides
)2.......(rd1–na........rdaarrS n2n
(1) – (2) n1–n2n rd1–na–dr.....drdrar–1S
(n–1)terms
= n1–n
rd1–na–r–1
r–1dra
Sn = r–1rd1–na–
r–1r–1dr
r–1a n
2
1–n
Sum to infinity
If 1r and n , then 0rlim n
n
S = 2r–1dr
r–1a
Note: If we take the first term of a G.P to be b, then
Sn = r–1brd1–na–
r–1r–1dbr
r–1ab n
2
1–n
If 1r , then sum to infinity, 2r–1dbr
r–1abS
Use of Natural numbers1 Let Sr = 1r + 2r + 3r +.......................+nr, then
(i) S1 = 1 + 2 + 3 +..................+n = 2)1n(n
(ii) S2 = 12 + 22 + 33 +..................+n2 = 6)1n2)(1n(n
180
(iii) S3 = 13 + 23 + 33 +..................+n3 = 4
)1n(n 22
= S12
(iv) S4 = 14 + 24 + 34 +..................+n4 = 30
)1–n3n3)(1n2)(1n(n 2
= 5S2 (6S1–1)
(iv) S5 = 15 + 25 + 35 +..................+n5 = 12
)1–n2n2()1n(n 222
= 31
S12(4S1–1)
2 1 + 3 + 5 + ....................to n terms = n2
3 12 + 32 + 52 + .................to n terms = 3
)1–n4(n 2
4 13 + 33 + 53 + .................to n terms = n2 (2n2 –1)
5 1 – 1 + 1 –..................to n terms = 2
)1(––1 n
6 1 – 2 + 3 –.....................to n terms = 4
)1n2()1(––1 n
7 12 – 22 + 32 –...................to n terms = 2
)1n(n)1(– 1–n
= (–1)n–1S1
8 13 – 23 + 33 –.................to n terms = 8
1–)1–n6n4()1(– 231–n
ApplicationIf nth term of a sequence is given byTn = an3 + bn2 + cn + d, where a, b, c, d R, then
Sn = Tn = T1 + T2 +..............+Tn
= a 3n + b 2n + c n + d 1Note: If Tn is expressible as product of m consecutive numbers beginning with n,i.e. Tn = n(n+1) (n+2)....(n+m –1) then
Sn = n(n+1) (n+2)....(n+m –1) 1mmn
Eg. If Tn =n, then 21nnSn (Here m = 1)
E.g. Tn = n(n+1), then Sn = 32n1nn
(Here m = 2)
Method of differencesIf the differences of successive terms of a series are in A.P. or G.P., we can find Tn as follows(a) Denote nth term and the sum up to n terms by Tn &Sn respectively(b) Rewrite the given series with each term shifted by one place to the right
181
(c) Subtracting the above two forms of the series, find Tn.
(d) Apply Sn = nT .Note : Instead of determining the nth item of a series by the method of difference, we can use thefollowing steps to obtain the same(i) If the differences T2 –T1, T3 – T2,................etc are in A.P. Then take the nth term as
Tn = an2 + bn + c, a, b, c RDetermine a, b, c by putting n = 1, 2, 3 and equating them with the values of correspondingterms of the given series.
(ii) If the differences T2 –T1, T3 –T2, ...........etc are in G.P , with common ratio r, then take Tn=arn–1 + bn + c, a, b, c RDetermine a, b, c by putting n = 1, 2, 3 and equating them with the values of correspondingterms of the given series.
(iii) If the differences of the differences computed in step (i) are in A.P, then take Tn = an3 +bn2 + cn +dDetermine a, b, c by putting n = 1, 2, 3, 4 and equating them with the values of correspondingterms of the given series.
(iv) If the differences of differences computed in step (i) are in G.P with common ratio r, then takeTn = arn–1 + bn2 + cn + dDetermine a, b, c by putting n = 1, 2, 3, 4 and equating them with the values of correspondingterms of the given series.
Summation by "" (sigma) operator
i. n21
n
1rr T......TTT
ii. ntimesn......1111n
1r
iii.n
1rr
n
1rr ;TkkT k is a constant
iv.n
1r
1rr TT =
n
1r
1r
n
1rr TT
v.n
1jj
n
1ii
n
1iji
n
1j
TTTT
(Note that i & j are independent here)
vi Now considernji0
jfif Here three types of terms occur, for which i < j, i > j and
i = j. Also note that the sum of terms when i < j equal to the sum of the terms when i > jif f(i) and f(j) are symmetrical. In such case,
182
nji0nij0
n
0i
n
0j
)j(f)i(f)j(f)i(fjfif +ji
)j(f)i(f
=jinji0
)j(f)i(f)j(f)i(f2
nji0
)j(f)i(f
2
)j(f)i(f–)j(f)i(fn
0j i j
n
0i
When f(i) and f(j) are not symmetrical, the sum can be obtained by listing all the terms.
Example:1 i3i2i321iij4
1i
3
1j
(1+2+3) + (2+4+6) + (3+6+9) + (4+8+12) = 6+12+18+24 = 60
Also,4
1i
3
1j
4
1i
3
1j
602
4x3x2
5x4jiij (Since i & j are independent)
Note that n222
2n
1jin21ji a.....aa–a.....aaaa2
21
Example: 2 nji0
1
=2
1–1n
1i ji
n
1j
=2
1–11n
1j
n
1j
n
1i
= 2n C
21–nn
2n–n.n
Example:3Consider n
1i
n
1j
ij
There are 3 types of terms in this summation,i. Those terms when i < j (upper triangle)ii. Those terms when i > j (lower triangle)iii. Those terms when i = j (diagonal)It is shown in the diagram
ij 1 2 3 ...... n
1 1.1 1.2 1.3 ...... 1.n
2 2.1 2.2 2.3 ...... 2.n
3 3.1 3.2 3.3 ...... 3.n
.
n n.1 n.2 n.3 ...... n.n
.....
183
n
1i
n
1j
ij = sum of terms in upper triangle + sum of terms in lower triangle + sum of terms in
diagonal.n
1i
n
1j
ij =nji0 ji
ijij2 ( sum of terms in upper + lower tringles are same)
nji0
ij
2
ij–ijji
n
1i
n
1j
2
i–in
1i
n
1i
2n
1j
=2
61n21nn–
21nn.
21nn
Solved Examples1. Find sum of the series to n terms
Solution:1+3x+5x2+7x3+.........Let Sn = 1+3x+5x2+7x3+.........+(2n–3)xn–2 + (2n–1)xn–1........(1) xSn = x+3x2+5x3+............... +(2n–3)xn–1+(2n–1)xn ........(2)(1) – (2) givenSn (1–x) = 1+ (2x+2x2+2x3+.........+2xn–1) – (2n–1)xn
n–1 terms
=n
1–n
x1–n2–x–1
x–1x21
Sn = n2
1–n
xx–11–n2–
x–1x–1x2
x–11
2. to........x27x9x3 161
81
41
a. 3 b. 9 c. 31
d. none of these
Solution:
to........x27x9x3 161
81
41
to........x3x3x3 163
82
41
= to..........163
82
41
3= 3S..............(1)
Where S = )2......(..........to..........163
82
41
= 24nn - 1 ( 3 n + 2 )2
184
21
S = )3...(....................to..........162
81
(2) – (3) S 21–1 = to..........
161
81
41
=21
21–1
41
S = 1Substituting in (1)
to........27x9x3 161
81
41
= 331
Ans: a
3.1m 1n
nmm
2
3.m3.n3nm
a. 916
b. 169
c. 1 d. none of these
Solution: Let S =1m 1n
nmm
2
3.m3.n3nm
=1m 1n
nmm
n3
m3
m3
1
S =1m 1n nmm aaa
1............(i) n
3a&m3awhere
n
n
m
m
Intercharging m & n
S = 1m 1n nmn aaa
1...............(ii)
Adding (i) & (ii)
2S =1m 1n nmaa
1
=1m 1n
nm3mn
185
=1m
m3m
1nn3
n
= 43.
43
2S = 169
S = 329
Ans: d4. Sum to n terms of the series 12–22+32–42+..........is
Solution: Clearly nth term is negative or positive according n is even or odd.Case I when n is even.In this case the series is
(12–22)+(32–42)+..................+((n–1)2–n2)
= –{1+2+3+......+(n–1)+n} = 2)1n(n–
Case II when n is odd.In this case the series is
(12–22)+(32–42)+..................+{(n–2)2 – (n–1)2}+ n2
= 2n2
)1–n(n–
=2
)1n(n2
nn2
n2nn– 22
5. Find the sum of all possible products of first n natural numbers taken two by two
Solution: nji1
jixx =n
1i
n
1j ji
ij–ij21
= 61n21nn–
21nn
21 2
= 2n31–n1nn241
6. Find sum to n terms of ..........331
3221
2111
1424242
42n nn1nT
1n–n1nnn
22 1n–n1nn1n–n–1nn
21
22
22
=1m
m3m
n3n
1m=
13
+3
+22 3
33 +........ ) =S
13
+3
+22 3
33 +........ _ (i)S=
1
13 3
+233 +........ _ (ii)1
S =32
1+ 3 4
Subtratiy equation (ii) From (i)
3 3+
31
S =31
+2
2
1 13
........ +
31
S =2
1S =
34
Where= s s1 1
186
1nn1–
1n–n1
21T 22n
Putting n = 1,2,3......n and adding
1nn2nn
1nn1–1
21TS 2
2
2nn
7. is to.........4009
1447
365
43
a. 2 b. 1 c. 3 d. none of these
Solution: to.........4009
1447
365
43
= to..........5x4
94x3
73x2
52x13
2222
= to..........4x33–4
3x2–3
2x11–2
22
22
22
22
22
22
= to..........41–
31
31–
21
21–
11
222222
= 1112
Ans: b
8. Find the nth term of the following seriesi. 3+7+13+21+.......
Solution:1st consecutive differences 4,6,8,........are in A.P.
Tn = an2+bn+cPutting n = 1,2,3,a+b+c = 3, 4a+2b+c = 7, 9a+3b+c = 13
a = 1, b = 1, c = 1Tn = n2+n+1
ii. 5+7+13+31+85+..........Solution:1st consecutive differences 2,6,18,54,....... are in G.P. with common ratio 3
Tn = a.3n–1+bn+cPutting n = 1,2,3 we geta+b+c = 5, 3a+2b+c =7, 9a+3b+c = 13
a = 1, b = 0, c = 4Tn = 3n–1+4
2
1–11–11[2
13 3 7
.....12 n -n
112 n
1–+nS =n
187
Vn method
1. If 1r32r21
n a......aa1
a......aa1S
1–rnn1nn a......aaa1.........
1–rn1nnn a......aa
1T1–rn1n
n a......a1V (leave 1st term in denominator of Tn)
Vn–Vn–11–rn2n1n a......aa
12–rn1nn a......aa
1–
= 1–rnn1–rn1nn
a–aa......aa
1
Vn–Vn–1= Tn (1–r)d
Tn = )V–V()1–r(d
1–1–nn )V–V(
)1–r(d1–S 0nn
E.g 1nn
)1n(n1..........
3.11
2.11
2n1n1–
2.11
21
2n1nn1.................
4.3.21
3.2.11
3n2n1n1–
3.2.11
31
3n2n1nn1.................
5.4.3.21
4.3.2.11
2. If Sn = a1a2.....ar+a2a3......ar+1+.......+anan+1.....an+r–1
Tn = anan+1.......an+r–1 Vn = anan+1.......an+r–1an–rVn–Vn–1 = anan+1.....an+r – an–1an......an+r–1= anan+1.....an+r–1 (an+r – an–1)= Tn (r+1)d
Tn = )V–V()1r(d
11–nn )V–V(
)1r(d1S 0nn
1.2+2.3+.....+n(n+1) = n(n+1)3
2n
1.2.3.4+2.3.4.5+........+ n(n+1)(n+2)(n+3) = n(n+1)(n+2)(n+3)5
4n
Alternative method
1. )2n)(1n(n1..............
4.3.21
3.2.11S
,
188
= )2n)(1n(nn–)2n(.............
4.3.22–4
3.2.11–3
21
= )2n)(1n(1–
)1n(n1......
4.31–
3.21
3.21–
2.11
21
= )2n)(1n(1–
2.11
21
2. 1.2.3+2.3.4+......+n(n+1)(n+2) = )13(
)3n)(2n)(1n)(n(
= 13)3n(t n
i.e. Sum of the product of m consecutive natural 1m
)mn(TS nn
1. If R ,0 denote the set of values which satisfies the equation
42 ......x32 , then R =
a. 3–
b. 32,
3 c. 32,
3–
d. 32–,
3
2. Find the value of the expressionn
1i
i
1j
j
1k
1
3. If in a series tn = !1nn
, Then20
1nnt is equal to
a. !201!–20
b. !211!–21
c. )!1–n(21
d. none of these
4. The value of ......161
81
41log 52.0 is
a. 1 b. 2 c. 21
d. 4
5. nlim (1+3–1) (1+3–2) (1+3–4)(1+3–8)........ n–231 is equal to
a. 1 b. 21
c. 23
d. none of these
1 cosx cos cos x
PRACTICE QUESTIONS
189
6. nlim
termsnupto.........4.33.22.1ntermsnupto..........5.4.34.3.23.2.1 is equal to
a. 43
b. 41
c. 21
d. 45
7.* Sum to infinite terms of the series ......181tan
252tan
81tan
92tan
21tan 1–1–1–1–1–
a. 3tan 1– b. 31cot 1– c. 3
1tan 1– d. 3cot 1–
8. If f(x) = a0+a1x+a2x2+......+anx
n +.......and x–1)x(f
=b0+b1x+b2x2+.......+bnx
n+......If a0 = 1 and
b1 = 3 and b10 = k11–1, and a0, a1,a2,......are in G.P then k is
a. 2 b. 3 c. 21
d. none of these
9. The value of n for which 704+ 21
(704)+ 41
(704)+......up to n terms =
1984 –21 (1984)+ 4
1(1984).......upto n terms is
a. 5 b. 3 c. 4 d. 1010. If 12+22+32+......+20032 =(2003) (4007) (334) and (1) (2003) +2(2002) +(3) (2001).......+
(2003) (1) = (2003) (334) (x), then x equalsa. 2005 b. 2004 c. 2003 d. 2001
11. Read the passage and answer the following questionsLet T1,T2 .....Tn be the terms of a sequence and let (T2–T1) = T2
1, (T3–T2) = T21 ,......
(Tn–Tn–1) = T1n–1
Case I: If T11, T2
1...... T1n–1 are in A.P., then Tn is quardratic in ‘n’. If T1
1–T21,
T21–T3
1,....are in A.P., then Tn is cubic is n.Case II: If T1
1, T21...... 1
1–nT are not in A.P., but in G.P. , then TTn = arn+b where r is the
common ratio of the G.P. T11, T2
1,T31.....and a,b R. Again if TT1
1, T21...... 1
1–nT are not inG.P., but T2
1–T11,T3
1–T21,..... are in G.P., then Tn is of the form arn–1+bn+c and r is
the C.R. of the G.P. T21–T1
1,T31–T2
1, ...... and a,b,c R.i. The sum of 20 terms of the series 3+7+14+24+37+...... isa. 4010 b. 3860 c. 4240 d. none of theseii. The 100th term of the series 3+8+22+72+226+1036+.....is divisible by 2n,then maxi
mum value of n isa. 4 b. 2 c. 3 d. 5
iii. For the series 2+12+36+80+150+252+....., the value of 3n
n nTlim is (where Tn is the nth
term)
190
a. 2 b. 21
c. 1 d. none of these
12.* Match the followingColumn I Column II
a. If the sum of the series ......2113
32
75 p. 28
up to n terms is 5, then n =b. A term of the sequence 1,3,6,......is q. 10c. Sum of the series 1+2+3+......upto
7 terms is r. 36
d. If 1296n3 , then n is equal to s. 21
Note:* QuestionS with more than one option is correct
ANSWERS
1. b 2. 6
)2n)(1n(n 3. b 4. d
5. c 6. a 7. a,b 8. a
9. a 10. a 11. (i) c (ii) c (iii) c
12. a q,s; b p,q,r,s; c p; d p
191
SEQUENCES AND SERIES - VISequences and Series - Problem Solving
SequenceA sequence is a function of natural numbers with codomain as the set of real numbers. It is said tobe finite or infinite according it has finite or infinite number of terms. Sequence a1, a2,........ an isusually denoted by {an} or <an>
SeriesBy adding or subtracting the terms of a sequence we get a series.
Arithmetic Progression (A.P.)It is a sequence in which the difference between two consecutive terms is the same.For a sequence {an} which is in A.P, nth term an=a+(n–1)d= (last term) which is always a linearexpression in n)d=an–an–1 (If d= 0, then sequence is a constant sequence. if d>0 the sequence is increasing; ifd<0,the sequence is decreasing)nth term from the end an
1 = +(n–1)(–d) = –(n–1)d
Sum to n terms =
)a(2nor
d)1–n(a22n
(Sn is a quadratic expression in n; common difference = 21
coefficient of n2)
Also an = Sn–Sn–1Arithmetic mean
If a, b, c are in A.P, then b= 2ca
is called the single arithmetic mean of a & c. Let a & b be two
given numbers and A1, A2,........... An are n A.M’s between them. Then a, A1,A2,...An, b are in A.P.
Common difference of this sequence d= 1na–b
.
A1= a+d, A2= a+2d etc. we can find all the arithmetic means.Properties of A.P.1 If a1, a2, a3, ........ are in A.P; then a1 k, a2 k,a3 k,..................... are also in A.P..
2 If a1, a2, a3,................. are in A.P, then a1 , a2 ,a3 ,.................... and 1a, 2a
, 3a..............
are also in A.P ( 0)3 If a1, a2,.......... an are in A.P, then an,an–1,...............a2,a1 is also an A.P with common difference (–
d)4 If a1, a2, a3, ..................and b1, b2, b3, .......................... are two A.P.s then a1 b1,a2 b2,a3 b3,.....
are also in A.P.
192
5 If a1, a2, a3,.............. and b1, b2, b3,................are two A.P.s then a1b1, a2b2, a3b3,...........and
1
1
b
a,
2
2ba
, 3
3
b
a,................. are NOT in A.P..
6 If 3 numbers are in A.P we may take them as a–d, a, a+d. If 4 numbers are in A.P, we can takethem as a–3d, a–d, a+d, a+3d.
7 In an arithmetic progression, sum of the terms equidistant form the beginning and end is a constantand equal to sum of first and last term.ie for {an},a1+an = a2+an–1=a3+an–2=......
Also ar = 2aa krk–r , 0 k n–r..
8 Sum of n arithmetic means between two given numbers a & b is n times the single A.M betweenthem .
ie. A1+A2+...............+An = n 2ba
9 Also Sn = a1+a2+......+an= evenisnif);termsmiddletwoofsum(2n
.oddisnif);termmiddle(n
Geometric Progression (G.P.)It is a sequence in which the ratio of any two consecutive terms is the same. For a sequence {an}which is in G.P. nth term an = arn–1 (last term)
Common ratio r = 1–n
naa
(r 0 . If r>1, the sequence is an increasing sequence, if 0<r<1 then the
sequence is decreasing )
nth term from the end an1 = an
1–n
r1
(an1 = nth term from end)
Note : No term of G.P. can be zero
Sum to n terms Sn = 1r,na
1r,1–r
)1–r(a n
If |r|<1, the sum of the infinite G.P is given by S = r–1a
Geometric meanIf a, b, c are in G.P, then b2 = ac or b = ac is called the single geometric mean of a & c. Let a &b be two given numbers and G1, G2, .....Gn are n G.M.s between them. Then a, G1, G2,.........Gn,
193
b are in G.P. Common ratio of this sequence r = 1n1
ab
G1, = ar, G2 = ar2 etc. we can find all the geometric means.Properties of G.P.
1 If a1, a2, a3................... are in G.P., then a1k, a2k, a3k,.............. and ka1 , k
a 2 , ka3 , ...............are
also in G.P (k 0).
2 If a1, a2, a3,................ are in G.P., then 1a
1,
2a1
, 3a
1,....................and a1
n, a2n, a3
n,..........are also
in G.P.3 If a1, a2, a3,......... an are in G.P with common ratio r, then an, an–1.............a2, a1 is also in G.P. With
common ratio r1
.
4 If a1, a2, a3........ and b1, b2, b3,.......... are two G.P.s then a1 b1, a2 b2, a3 b3,........ are NOT in inG.P.
5 If a1, a2, a3, .............. and b1, b2, b3,................are two G.P.s then a1b1, a2b2, a3b3,...........and
1
1ba
, 2
2ba
, 3
3ba
,.............. are also in G.P..P.
6 If 3 numbers are in G.P., we may take them as ra
, a, ar. If 4 numbers are in G.P., we can take them
as 3ra
, ra
, a r, a r3.
7 In a geometric progression, product of the terms equidistant from the beginning and end is aconstant and equal to product of first and last term.ie For {an}
a1an = a2 an–1 = a3 an–2 =.......
Also ar = krk–r aa , 0 k n–r..8 Product of n geometric means between two given numbers a & b is nth power of the single G.M.
between them.
ie G1G2G3..........Gn = nab
9 If a1, a2, a3,............... are in G.P. (ai>0 i), then loga1,loga2,loga3,...... are in A.P. Its converse isalso true.
Harmonic Progression (H.P.)A sequence is said to be in H.P if the reciprocals of its terms are in A.P.
ie. if a1,a2,a3,....... an are in H.P., then 1a
1,
2a1
,......na
1 are in A.P..
194
For a sequence {an} which is in H.P.,
nth term an =
121 a1–
a1)1–n(
a1
1 = )a–a)(1–n(a
aa
212
21
nth term from end an1 =
12n a1–
a1)1–n(–
a1
1 =
)a–a)(1–n(a–aaaaa
21n21
n21
Note : No term of H.P. can be zero. There is no general formula for finding out the sum of n termsof H.P.
Harmonic mean
If a,b,c are in H.P; then b = caac2
is called the single H.M. between a & c. Let a & b be two given
numbers and H1, H2,..............,Hn are n H.M.s between them. then a, H1, H2,...... Hn, b are in H.P.The common difference d of the corresponding A.P is
d = ab)1n(b–a
1H1
= da1
, 2H
1 = d2
a1
etc. we can find all the harmonic means.
Note: The sum of reciprocals of n Harmonic means between two given numbers is n times thereciprocal of single H.M. between them.
ie1H
1+
2H1
+.......nH
1 = n
2b1
a1
Note : If a, b, c are three successive terms of a sequence. Then
c–bb–a
=
.P.Hinarec,b,aca
.P.Ginarec,b,aba
.P.Ainarec,b,aaa
Relation between A.M., G.M., and H.M.For positive numbers a1, a2, a3, .................an
A.M. = A = na.......aa n21
195
G.M. = G = n1
n21 a.......aa
H.M = H =
n21 a1......
a1
a1
n ,
A G H and G2 = AH.(equality holds if a1 = a2 =...............an)
Note : Also n
a......aa 2n
22
21
na........aa n1 2
(Root mean square inequality)
Note : The quadratic equation having a, b as its roots is x2–2Ax+G2 = 0 and a : b = A+ 22 G–A
: A – 22 G–A where A,G are respectively the A.M. and G.M. of a & bNote : Formation of progressionsTwo consecutive terms determine the required progression. If two numbers a & b are given, then(i) a, b, 2b–a is A.P.
(ii) a, b, a
b2 is G.P..
(ii) a, b, b–a2ab
is H.P..
Solved Examples.1 If the pth, qth and rth terms of an A.P are in GP, then the common ratio of the G.P is
(a) qrqp
(b) p–qq–r
(c) q–pr–p
(d) None of these
Solution : Tp, Tq, Tr are in G.P
p
q
TT
= q
r
TT
1–TT
p
q = 1–
TT
q
r
p
pq
TT–T
= q
qr
TT–T
p
q
TT
= pq
qr
T–TT–T
p
q
TT
= )D)1–p(A(–)D)1–q(A()D)1–q(A(–)D)1–r(A(
= p–qq–r
Ans : (b)2 If 4a2+9b2+16c2 = 2(3ab+6bc+4ca), where a, b, c are non–zero real numbers then a, b, c are in
(a) A.P. (b) G.P. (c) H.P. (d) None of theseSolution : Multiply by 2 on both sides
196
4a2+4a2+9b2+9b2+16c2+16c2–12ab–24bc–16ca = 0(2a–3b)2+(3b–4c)2+(4c–2a)2=02a=3b=4c=
2a , 3
b , 4c
2,3,4 are in AP 21
, 31
, 41
are in H.P..
2 , 3 , 4 are in HP gives
a, b, c are in HPAns (c)
3 If a, a1, a2, a3, ..........,a2n, b are in AP and a, g1, g2, g3, .....................g2n, b are in G..P. and h is thesingle harmonic mean of a & b, then
n21
n21
ggaa
+ 1–n22
1–n22ggaa
+...................+1nn
1nnggaa
is equal to
(a) hn2
(b) 2nh (c) nh (d) hn
Solution :a1+a2n = a2+a2n–1 =......................= an+an+1 = a+b andg1g2n = g2.g2n–1=.....................=gn.gn+1 = ab
Also h = baab2
Given expression = abba
+ abba
+............. abba
(n times)
= n abba
= h2.n
= hn2
Ans : (a)
4 If 0<x< 2 , then the minimum value of
(sinx+cosx+cosec2x)3 is
(a) 27 (b) 227
(c) 427
(d) None
Solution : Apply A.M GM
3cosec2xcosxsinx
31
cosec2xsinx.cosx.
3cosec2xcosxsinx
31
xcosxsin2xcosxsin
197
Cubing both sides
27x2eccosxcosxsin 3
21
Minimum of 3cosec2xcosxsinx = 227
Ans : (b)5 Sum of certain odd consecutive positive integers is 572 –132, then the integers are
(a) 25, 27, 29,..........111 (b) 27, 29, ..............113(c) 29, 31, 33,...........115 (d) None of theseSolution :(2m+1)+(2m+3)+..............n terms = 572–132
2n
{2.(2m+1)+(n–1)2} = 572–132
n(2m+n)=572–132
n2+2mn+m2–m2=572–132
(n+m)2–m2=572–132
n+m =57 and m = 13, Solve to get n = 44Hence, the series is27, 29, 31,.......................,113Ans : (b)
6 If x, y, z are three positive numbers in A.P, then the minimum value of x–y2yx
+ z–y2zy
is
(a) 2 (b) 4 (c) 41
(d) None of these
Solution :
put y= 2xz
in the given expression
= x–xz
2xzx
+ z–xz
z2
xz
= x2xz3
z2zx3
= 21
x2z3
21
z2x3
= 22
xz
zx
23
Now AM GM 2
xz
zx
zx.
xz
xz
zx
2
198
23
2+1 = 4
Ans : (b)7 If n arithmetic means are inserted between 50 and 200, and n harmonic means are inserted
between the same two numbers, then a2.hn–1 is equal to(a) 500 (b) 5000 (c) 10,000 (d) None of theseSolution :50, a1, a2,..................an, 200 are in AP ............................................... (1)Also, 50, h1, h2,................,hn 200 are in H.P
501
, 1h
1,
2h1
,.....................nh
1, 200
1 are in APAP
2001
, nh
1,
1–nh1
,.....................1h
1, 50
1 are in APAP
Multiply by 200×50 = 10,000
50,nh000,10
, 1–nh
000,10 , .........................
2h000,10
, 1h000,10
, 200 are in AP................. (2)
Now (1) and (2) are identical.
a2 = 1–nh
000,10 gives a2.hn–1 = 10,000
Ans : (c)
1 If a1, a2, .......... an are positive real numbers whose product is a fixed number c, then the minimumvalue of a1+a2+.........+an–1+2an is(a) n(2c)1/n (b) (n+1)c1/n (c) 2nc1/n (d) (n+1)(2c)1/n
2 If a, b, c are in A.P. and a2, b2, c2 are in G.P. If a<b<c and a+b+c = 23
, then the value of a is
(a) 221
(b) 321
(c) 31–
21
(d) 21–
21
3 Let ƒ(x) = ax2+bx+c, a 0 and =b2 – 4ac. If + , 2+ 2 & 3+ 3 are in G.P, then.P, then(a) 0 (b) b =0 (c) c = 0 (d) bc 0
4 If adbc
= dacb
= 3 d–ac–b
, then a, b, c, d are in
(a) A.P (b) G.P (c) H.P (d) A.G.P.5* The 4th term of the A.G.P. 6, 8, 8, ............. is
(a) 0 (b) 12 (c) 332
(d) 964
6 If x= 111......1(20digits), y=333.........3(10digits) and
PRACTICE QUESTIONS
199
z=222...........2(10digits) then zy–x 2
=
(a) 1 (b) 72 (c) 21
(d) 3
7 Read the passage and answer the questions that follow.An odd integer is the difference of two squares of integers.The cube of an integer is difference of two squares.The cube of an odd integer can be expressed as difference of two squares in two differentways.The difference of the cubes of two consecutive integers is difference of two squares.
(i) If 103= a2–b2, then a–b =(a) 5 (b) 0 (c) 10 (d) 15
(ii) If 93=a2–b2= c2–d2, a+b+c+d =(a) 720 (b) 750 (c) 800 (d) 810
(iii) 153–143=a2–b2, ab =(a) 90000 (b) 95940 (c) 99550 (d) 99540
8 Match the following :-
For the given number a and b, nn
1n1n
baba
is
Column I Column II(a) A.M. (p) for n=1(b) G.M (q) for n=1/2(c) H.M. (r) for n=0
(s) for n= –1/2(t) for n= –1
9 The sum of the products of the ten numbers 1, 2, 3, 4, 5 taking two at a time is(a) 165 (b) –55 (c) 55 (d) None of these
10 Let a1 = 0 and a1, a2, a3,...............an be real numbers such that |ai| = |ai–1+1| for all i, then the A.M.of the numbers a1, a2, a3................ an has the value A where
(a) A< 21–
(b) A<–1 (c) A 21–
(d) A = 21–
Note : Questions with more than one option is correct.
Answers1. a 2. d 3. c 4. c 5. c,d 6. a 7. (i) c (ii) d (iii) d8. a r, b s, c t 9. b 10. c
200
SEQUENCES AND SERIES - VIISequences and Series - Problem Solving
Some important Logarithmic and Exponential formulae1 If ax = y, then x = log ay2 log aa = 1 & log a1 = 03 na nloga
4 logamn = logam+logan
5 loga nm
= logam – logan
6 logamn = n logam
7 logba = blogalog
c
c
8 ma
alog n = nm
9 logba = blog1
a
10 loge (1+x) = x – 2x2
+ 3x3
– 4x4
+............
11 loge (1 – x) = – x – 2x2
– 3x3
– 4x4
............
12 loge(1+x) (1–x) = loge(1–x2) = –2 ............6x
4x
2x 642
13 loge x–1x1
= 2 ...................5x
3xx
53
14 loge 2 = 1 – 21
+ 31
– 41
+..................
15 ex = 1 + !1x
+!2
x2
+!3
x3
+ ...............
16 ex+e–x = 2 ............!4
x!2
x142
17 ex–e–x = 2 ............!5
x!3
x!1
x 53
201
18 e = 1+ !11
+ !21
+ !31
+...............
e is an irrational number & it lies between 2 & 3. e 2.7183
19 ay = alogy ee = 1+y(logea) +!2
y2
(logea)2+ !3
y3
(logea)3+..................
20 log2 log2 2............... = n where n is the number of square roots
21 baloga = alogbbVn - Method(i) To find the sum of series of the form
r21 a............aa1
+ 1r32 a............aa
1 + ....................+
1–rn1nn a............aa1
where a1, a2,
1–rn1nn
n ..a..........aa1T Here
........are in A.P. Let Vn = 1–rn2n1n a............aa
1 (avoiding first term for VVn ie an in Tn)
Tn = )1–r(d1–
(Vn–Vn–1) = )1–r(d1
(Vn–1–Vn)
put n = 1, 2, 3............. n and add to get Sn.
Sn = T1+T2+........Tn = )1–r(d1
(V0–Vn)
= )1–r(d1
1–rn2n1n1–r21 a......aa
1–a......aa
1
Eg: (a) 3.2.11
+ 4.3.21
+ ............... + )2n)(1n(n1
= )1–3.(11
)2n)(1n(1–
2.11
= 41
– )2n)(1n(21
(b) 4.3.2.11
+ 5.4.3.21
+ .......... )3n)(2n)(1n(n1
= )1–4.(11
)3n)(2n)(1n(1–
3.2.11
(ii) Summation of series of the form a1a2......ar+a2a3........ar+1+.....................+anan+1.........an+r–1 wherea1, a2.......are in A.P
202
Here Tn = anan+1.........an+r–1Let Vn = anan+1..............an–r+1an–r (Take one term extra in Tn for Vn)
Tn = d)1r(1
(Vn–Vn–1).
Put n = 1, 2, 3 ........and add to get Sn
Sn = T1+T2+.............+Tn = )1r(d1
(Vn–V0) = )1r(d1
(anan+1..........an+ r – a0a1a2...........ar) where
a0=a1 – d
Eg (a) 1.2+2.3 +....................+ n (n+1) = )12.(11
((n+1)(n+2) – 0.1.2) = 3)2n)(1n(n
(b) 1.2.3.4 + 2.3.4.5 +...........................+ n(n+1)(n+2)(n+3)
= )14.(11
(n(n+1)(n+2)(n+3)(n+4) – 0.1.2.3.4)
= 51
n(n+1)(n+2)(n+3)(n+4)
SOLVED EXAMPLES1 If the sides of a triangle are in A.P and the greatest angle of the triangle is double the smallest, then
the ratio of sides of the triangle is(a) 3 : 4 : 5 (b) 4 : 5 : 6 (c) 5 : 6 :7 (d) None of theseSolution :Applying sine rule, we have
sind–a
= )3–sin(a
= 2sinda
sind–a
= 3sin4–sin3a
= cossin2da
1d–a
= 2sin4–3a
= cos2da
gives cos = )d–a(2da
Also, 3–4sin2 = d–aa
3–4(1–cos2 ) = d–aa
gives –1+2
d–ada
= d–aa
d–aad4
= a 4ad = a2–ad gives a = 5d
sides a–d : a : a+d5d–d : 5d : 5d+d = 4 : 5: 6
Ans : (b)
a+d
A
B C
a–d
2
–
203
2 The sum of
21.
2.13
+ 3.24 2
21
+ 3
21.
4.35
+............. n terms is
(a) 1– n21n1
(b) 1– 1–n2.n1
(c) 1+ n21n1
(d) None of these
Solution :
Tn = n21.
)1n.(n2n
= n21.
1n1–
n2
Tn = n1–n 2).1n(1–
2.n1
Putting n = 1, 2, 3, ..............,n
T1 = 10 2.21–
2.11
T2 = 21 2.31–
2.21
T3 = 22 3.41–
2.31
.
.
.
Tn = n1–n 21n1–
2.n1
Adding , T1+T2+T3+.........+Tn = Sn
Sn = 1 – n21n1
Ans : (a)3 Coefficient of x49 in the expansion of (x–1)(x–3)(x–5).........................(x–99) is
(a) –992 (b) 1 (c) –2500 (d) None of theseSolution :(x–1)(x–3)(x–5) ................ (x–99)
= x50–S1x49+S2x
48.................Coefficient of x49 is –S1
2
Ans : (c)4 The coefficients of x15 in the product
(1–x)(1–2x)(1–22x)..............(1–215x) is(a) 2105–2121 (b) 2121–2105 (c) 2120–2104 (d) None of these
502 (1+99) =
204
Solution :(1–x)(1–2x)(1–22x)..............(1–215x)
= (–1)16(x–1)(2x–1)(22x–1).................. (215x–1)
= 21.22.23...................215(x–1) 21–x 22
1–x ........... 1521–x
= 2120.(x–1) 21–x 22
1–x ........... 1521–x
coeff of x15 is –2120 152 21...
21
211
= –2120.1
21–1
21–1 16
= –21211621–1 = 2105–2121
Ans : (a)5 The sum to 2n terms of the series
43
+ 47
+1615
+1631
+ 6463
+ 64127
+........................ is
(a) 3n– 32
n41–1 (b) 3n– 21
10n4
1–1 (c) 3n– 2113
n41
(d) None of these
Solution :Given expression
= 41–1 + 4
1–2 + 161–1 + 16
1–2 +...........................2n terms
= 3n –2 termsn..........161
41
= 3n – 2 41
41–1
41–1 n
= 3n – 32
n41–1
Ans : (a)
205
6 The sum to n terms of the series
41–1
1 +
41–31
1 +
41–531
1 + .........
(a) 1n2n2
(b) 1n2n4
(c) 1n22
(d) None
Solution :
Tn = termsn............5311
=
41–n
12
= 21–n
21n
1
= 21n
21–n
21–n–
21n
Tn = 1n21–
1–n212
T1 = 31–
112
T2 = 51–
312
.
.
.
Tn = 1n21–
1–n212
Adding, Sn = T1 +T2+.............+Tn=2 1n21–1 = 1n2
n4
Ans : (b)7 The sum of the series
3.11
+ 5.3.12
+ 7.5.3.13
+ ...................... is
(a) 1 (b) 21
(c) 23
(d) None
Solution :
Tn = 1n21–n2..............5.3.1n
206
= 21
)1n2)(1–n2...(..........5.3.11–1n2
= 21
1n2..........5.3.11–
1–n2.........5.3.11
T1 = 21
3.11–
11
T2 = 21
5.3.11–
3.11
.
.
.
)1n2.....(5.3.11–
)1–n2.....(5.3.11
21Tn
)1n2)(1–n2.....(5.3.11–1
21Sn
S = 21
(1–0) = 21
Ans : (b)
1* Let S1, S2,................be squares such that for each n 1 the length of a side of Sn equals thelength of a diagonal of Sn+1. If the length of a side of S1 is 10 cm, then for which of the followingvalues of n is the area of Sn less than 1 sq. cm.(a) 7 (b) 8 (c) 9 (d) 10
2 If a, b, c are is A.P and a2, b2, c2 are in H.P, then b2 =
(a) 2ca
(b) 2ca (c) 2ca–
(d) –2ca.
3 Let the HM & GM of two positive numbers a & b be in the ratio 4:5 then a : b is(a) 1 : 2 (b) 2 : 3 (c) 3 : 4 (d) 1 : 4
4 If cos(x–y), cos x, cos(x+y) are in H.P., then the value of cos x sec 2y
is
(a) 1 (b) 21
(c) 2 (d) 3
5 If x & y are positive real numbers and m, n are positive integers, then the minimum value of
)y1)(x1(yx
n2m2
nm
is
PRACTICE QUESTIONS
207
(a) 2 (b) 41
(c) 21
(d) 1
6* There are two numbers a & b whose product is 192 and the quotient of A.M. by H.M. of their
greatest common divisor and least common multiple is 48169
. The smaller of a & b is
(a) 2 (b) 4 (c) 6 (d) 127 Consider the sequence 1, 2, 2, 3, 3, 3, ...............where n occurs n times. The number that occurs
as 2007th term is(a) 61 (b) 62 (c) 63 (d) 64
8 Read the following paragraph and answer the questions.Let A1, G1, H1 denote the A.M., G.M., H.M of two distinct positive numbers.For n 2, let AAn–1 and Hn–1 have A.M., G.M., H.M as An, Gn, Hn respectively.(i) Which of the following statements is correct ?.
(a) G1>G2>G3 >....................(b) G1<G2<G3< ....................(c) G1=G2=G3 =....................(d) G1<G3<G5.................and G2>G4>G6>......
(ii) Which are of the following statement is correct ?.(a) A1>A2>A3>...................(b) A1<A2<A3<...................(c) A1>A3>A5>................... and A2<A4<A6<.........................(d) A1<A3<A5<................... and A2>A4>A6>...................
(iii) Which are of the following statement is correct ?.(a) H1>H2>H3>....................(b) H1<H2<H3<.................(c) H1>H3>H5>................... and H2<H4<H6<..................(d) H1<H3<H5<................... and H2>H4>H6>.................
9 If x,y, z>0 and x+y+z = 1, the )z–1)(y–1)(x–1(xyz
is necessarily
(a) 8 (b) 81
(c) 81
(d) None of these
10* Match the following :Column I Column II
(a) 3 numbers a,b, c between 2 & 18 such that (p) G–L = 4
(i) a+b+c = 25 (q) LG
= 4
(ii) 2, a, b are consecutive terms of an A.P. (r) G–L = 7
(iii) b, c, 18 are consecutive terms of a G.P. (s) LG
= 3
208
If G = max {a, b, c} & L = min {a, b, c}then (t) LG
+ LG
= 3
(b) 3 numbers a, b, c are in G..P. Such that(i) a+b+c = 70(ii) 4a, 5b, 4c are is A.P.
If G = Max{a, b, c} and L = Min{a, b, c}, then11 The coefficient of x203 is the expansion of (x–1)(x2–2)(x3–3) ....................(x20–20) is
(a) –35 (b) 21 (c) 13 (d) 2512 Read the following passage and answer the questions : -
Let ABCD be a unit square and 0< <1. Each side of the squareis divided in the ratio : 1– , as shown in the figure. Thesepoints are connected to obtain another square. The sides of newsquare are divided in the ratio : 1– and points are joined toobtain another square. The process is continued indefinitely.Let an denote the length of side and An the area of the nth square
(i) The value of for which 1n
AAn = 38
is
(a) 1/3, 2/3 (b) 1/4, 3/4 (c) 1/5, 4/5 (d) 1/2(2) The value of for which side of nth square equals the diagonals of (n+1)th square is
(a) 1/3 (b) 1/4 (c) 1/2 (d) 2/1
(iii) If a = 1/4 and Pn denotes the perimeter of the nth square then 1n
Pn equals
(a) 8/3 (b) 32/3 (c) 16/3 (d) 38
104
Answers1. b, c, d 2. c 3. d 4. c 5. b 6. b,d7. c 8. (i) c (ii) a (iii) b 9. b10. a r, s, b q 11. c 12. (i) b (ii) c (iii) d
D C
A –
–
–
–
B
209
SEQUENCES AND SERIES - VIIISpecial Series - Problem Solving
Arithmetico - Geometric Series (A.G.S.)If a1, a2, ............ an is an A.P. and b1, b2, ...........bn is a G.P, then the sequence a1b1, a1b2, .......anbnis said to be an A.G.S. The sequence is of the form ab, (a+d) br, (a+2d) br2,.......... ..............
Sum to n terms = Sn = r–1ab
+ 2
1–n
)r–1()r–1(dbr
– r–1
br)d)1–n(a( n
If –1 < r < 1, sum to infinite numbers is given by
S = r–1ab
+ 2)r–1(dbr
Important results1 Let Sr = 1r + 2r + 3r +.......................+nr, then
(i) S1 = 1 + 2 + 3 +..................+n = 2)1n(n
(ii) S2 = 12 + 22 + 33 +..................+n2 = 6)1n2)(1n(n
(iii) S3 = 13 + 23 + 33 +..................+n3 = 4
)1n(n 22
= S12
(iv) S4 = 14 + 24 + 34 +..................+n4 = 30
)1–n3n3)(1n2)(1n(n 2
= 5S2 (6S1–1)
(iv) S5 = 15 + 25 + 35 +..................+n5 = 12
)1–n2n2()1n(n 222
= 31
S12(4S1–1)
2 1 + 3 + 5 + ....................to n terms = n2
3 12 + 32 + 52 + .................to n terms = 3
)1–n4(n 2
4 13 + 33 + 53 + .................to n terms = n2 (2n2 –1)
5 1 – 1 + 1 –..................to n terms = 2
)1(––1 n
6 1 – 2 + 3 –.....................to n terms = 4
)1n2()1(––1 n
7 12 – 22 + 32 –...................to n terms = 2
)1n(n)1(– 1–n
= (–1)n–1S1
8 13 – 23 + 33 –.................to n terms = 8
1–)1–n6n4()1(– 231–n
Note 1 : (x+1)(x+2)(x+3)..................(x+n) = xn + A1 xn–1 + A2x
n–2 + A3xn–3+...................
210
Then A1 = 2)1n(n
A2 = 24)2n3)(1n(n)1–n(
A3 = 48
)1n(n)2–n)(1–n( 22
Note 2 : To obtain the sum ji
ai aj we use the identity
ji
2 ai aj = (a1+a2+.........+an)2 –(a1
2+a22+...+an
2)
More methods of summation of seriesIf nth term of a sequence is given byTn = an3 + bn2 + cn + d, where a, b, c, d R, then
Sn = Tn = T1 + T2 +..............+Tn
= a 3n + b 2n + c n + d 1I Method of differences
If the differences of successive terms of a series are in A.P. or G.P., we can find Tn as follows(a) Denote nth term and the sum up to n terms by Tn &Sn respectively(b) Rewrite the given series with each term shifted by one place to the right(c) Subtracting the above two forms of the series, find Tn.
(d) Apply Sn = nT .Note : Instead of determining the nth item of a series by the method of difference, we can use thefollowing steps to obtain the same(i) If the differences T2 –T1, T3 – T2,................etc are in A.P. Then take the nth term as
Tn = an2 + bn + c, a, b, c RDetermine a, b, c by putting n = 1, 2, 3 and equating them with the values of correspondingterms of the given series.
(ii) If the differences T2 –T1, T3 –T2, ...........etc are in G.P , with common ratio r, then takeTn= arn–1 + bn + c, a, b, c RDetermine a, b, c by putting n = 1, 2, 3 and equating them with the values of correspondingterms of the given series.
(iIi) If the differences of the differences computed in step (i) are in A.P, then take Tn = an3
+ bn2 + cn +dDetermine a, b, c by putting n = 1, 2, 3, 4 and equating them with the values of correspondingterms of the given series.
(iv) If the differences of differences computed in step (i) are in G.P with common ratio r, thentakeTn = arn–1 + bn2 + cn + dDetermine a, b, c by putting n = 1, 2, 3, 4 and equating them with the values of correspondingterms of the given series.
211
II Sum of series whose nth term is
Tn = )]nda][(d)1–n(a[1
Resolve Tn into partial fractions, (or express the Nr of Tn in terms of factors of Dr andsimplify), then find T1, T2, ............ Tn and add to get Sn.
III Sum of series in special form(a) Let the series consists of terms whose nth term
Tn = )d)1–n(a.....().........d2a)(da(a1
To find sum of such a series ( factors of Dr are in A.P.) as shown above, remove the leastfactor and multiply the denominator by the number of factors left out (here n–1), and alsoby the common difference (here d) change the sign and add a constant C.
Thus, Sn = )d)1–n(a.........().........d2a)(da(d)1–n(1
+ C
Find S1 and hence value of C. This gives the required sum.(b) Let the series consists of terms whose nth term Tn = a(a+d) (a+2d)..........(a+(n–1)d).
To find sum of such a series, as shown above, add one more factor and divide by the totalnumber of factors (here (n+1)) and also by the common difference (here d). Also add aconstant C.
Thus Sn = d)1n()nda)(d)1–n(a)........(d2a)(da(a
+ C
Find S1 and hence value of C. This gives the required sum.Note :
(i) for odd n, n = 2
21n
– 2
21–n
(ii) For any n, n3 = 2
21nn
– 3
2)1–n(n
(iii) For odd n, n3 = 33
21n
– 23
21–n
= 2
2)1n(n
– 2
2)1–n(n
SOLVED EXAMPLES
1 It is given that 411
+ 421
+ 431
+ ............. = 90
4
then, 411
+ 431
+ 451
+ ............. is equal to
(a)96
4
(b)45
4
(c)90
89 4
(d) None of these
Solution : Let 411
+ 431
+ 451
+ ............. = S
212
Now 411
+ 421
+ 431
+ ............. = 90
4
.........51
31
11
444 + .........61
41
21
444 = 90
4
S+ 421
.........31
21
11
444 = 90
4
S+ 161
. 90
4
= 90
4
S = 90
4
161–1 =
90
4
× 1615
= 96
4
Ans : (a)
2 If Sn = n
1rr
2
2r terms..........221
, then
Sn is equal to(a) 2n –(n+1) (b) 1 –2–n (c) n–1 + 2–n (d) 2n –1Solution :
Tr = n
2
2.termsr...................221
= r
r
2)1–2()1–2(1
= 1 – 2–r
Tr = r21–1
Sn = n – termsn................21
21
21
32
Sn = n –
21–1
21–1
21
n = n – (1– 2–n)
= n –1 + 2 –n
Ans : (c)3 The sum to n terms of the series
12 + 2.22 + 32 + 2.42 + 52 +.................... = 2
)1n(n 2
when n is even. When n is odd, the sum is
(a)2
)1n(n2
(b)2
)1–n(n 2
(c) n(n+1)2(2n+1) (d) None of these
213
Solution : Let n = 2k 12 + 2.22 + 32 + 2.42 + 52 +................+(2k–1)2 + 2(2k)2
= 2
)1k2(k2 2
Let n = 2k + 1 (odd) 12 + 2.22 + 32 + 2.42 +................+(2k–1)2 + 2(2k)2 + (2k+1)2
= 2
)1k2(k2 2
+ (2k + 1)2
= 2
)2k2()1k2( 2
= 2
)1n(n2
Ans : (a)4 If the sum to n terms of an A.P is cn(n–1); c 0, then the sum of squares of these terms is
(a) c2 n2 (n+1)2 (b)3c2 2
n(n–1)(2n–1)
(c)3c2 2
n(n+1)(2n+1) (d) None of these
Solution : Tn = Sn –Sn–1 = cn(n–1) –c(n–1)(n–2)= c(n–1){n–n+2}= 2c(n–1)
Tn2 = 4c2(n–1)2
Sn2 = 4c2{02+12+22+..................+(n–1)2}
= 4c2
6)1–n2)(1–n(n
= 3c2 2
n(n–1)(2n–1)
Ans (b)5 Let tr = 2r/2 + 2–r/2
The10
1r
2rt is equal to
(a) 202
1–210
21
(b) 192
1–210
21
(c) 19–2
1–220
21
(d) None of these
Solution : tr2 = 2r + 2–r + 2
S102 = (21+22+.........+210) + 102 2
1........21
21
+ 20
= 1–2
)1–2(2 10
+
21–1
21–1
21
10 + 20
214
= 211 –2 +1 – 1021
+ 20
= 211 – 1021
+ 19 = 10
21
21–2
+ 19
Ans : (b)6 Sum to n terms
1.(3n–1)+2. (3n–2) + 3.(3n–3) + .............n terms is
(a) 3)1–n5)(1n2(n
(b) 3)1n5)(1n2(n
(c) 6)1–n7)(1n(n
(d) None of these.
Solution :Tr = r(3n–r)
= 3nr–r2
Sn = n
1rrT =
n
1rrn3 –
n
1r
2r
= 3n. 2)1n.(n
– 6)1n2)(1n(n
= 2)1n(n
31n2–n3
= 2)1n(n
31–n2–n9
= 6)1–n7)(1n(n
Ans : (c)
7n
1r
2r – n
1m
m
1rr is equal to
(a)n
1r
n
1r
2 rr21
(b)n
1r
n
1r
2 r–r21
(c) 0 (d) None of these
Solution : n
1m
n
1r
2 –r 2)1m(m
= nn
1r
2 –r2
rr2
1r
215
= n
1r 2r–r
22
– n
1r 2r
= 21
n
1r
2r – 21 n
1rr =
n
1r
n
1r
2 r–r21
1*. For a positive integer n let
a(n) = 1 + 21
+ 31
+ 41
+................+ 1–)2(1
n then
(a) a(100) 100 (b) a(100)>100 (c) a(200) 100 (d) a(200)>1002 113–103+93–83+73–63+53–43+33–23+13 =
(a) 756 (b) 724 (c) 648 (d) 812
3 If a1, a2, ..........an+1 are in A.P with common difference d, then n
1r tan–1
1rraa1d
(a) tan–1
1n1aa1nd
(b) tan–1
1n1aa1d)1n(
(c) tan–1
1n1aa–1d)1–n(
(d) tan–1
1n1aa–1d)1–n(
4 The sum to 50 terms of
213
+ 22 215
+ 222 3217
+ ............. is
(a) 1750
(b) 17100
(c) 17150
(d) 17200
5 The sum of the first 10 common terms of the series 17, 21, 25, .................... and 16, 21, 25, ... is(a) 1100 (b) 1010 (c) 1110 (d) 1200
6 Match the following :Column I Column II
(a) 12–22+32 –................. to 21 terms (p) 680(b) 13–23+33–43 + ........... to 15 terms (q) 2556(c) 12+32+52 + ................ to 8 terms (r) 1856(d) 13+33+53 + ............... to 6 terms (s) 231
7 The sum of the series 1r
1r4eccos 41– is
(a) (b) /2 (c) /4 (d) None of these
8 Let n
1r
4r = ƒ(n), then n
1r (2r–1)4 is equal to
PRACTICE QUESTIONS
216
(a) ƒ(2n) – 16ƒ(n), n N (b) ƒ(n)–16ƒ 21–n
when n is odd
(c) ƒ(n)–16ƒ 2n
when n is odd (d) None of these
9 Match the following : -Column I Column II
(a) If n = 210, then 2n is divisible bythe greatest prime number which is greater than (p) 16
(b) Between 4 & 2916 is inserted odd number (2n+1)G.M’S. Then the (n+1)th G.M. is divisible by greatest (q) 10odd integer which is less than
(c) In a certain progression, three consecutive termsare 40, 30, 24, 20. Then the integral part of the (r) 34next term of the progression is more then
(d) 1+ 32 510
57
54
+............. to = ba
, where (s) 30
HCF(a,b) = 1, then a–b is less then
10 If S= 13
12 +
241
2 + 35
12 +
461
2 +............... then the value of S1
is ..............
11 The value of the ratio ............41
31
211 222 .................
41–
31
21–1 222 is .......
12 nlim n
1r )1r2.........(..........9.7.5.3.1
r is equal to
(a) 31
(b) 23
(c) 21
(d) None of these
13 If (12–t1) + (22–t2) + ............. + (n2–tn) = 3n
(n2–1), then tn is equal to
(a) n2 (b) 2n (c) n2–2n (d) None of these14 If (1+3+5+.........+p) + (1+3+5+.............+q) = (1+3+5+.................+ r) where each set of
parantheses contains the sum of consecutive odd integers as shown, the smallest possible value ofp+q+r (where p>6) is..........................(a) 12 (b) 21 (c) 45 (d) 54
Answers1. a,d 2. a 3. a 4. b 5.c 6. a s, b r, c p,d q7. c 8. a 9. a p,q,r,s; b r,s; c p,q; d r,s,s 10. 2 11.212.c 13.d 14. b
217
PERMUTATION AND COMBINATIONS -IProperties of nPr and nCr
Fundametal principle of counting
(i) Addition principle : If an operation can be performed in ‘m’ different ways and another operationwhich is independent of the first operation, can be performed in ‘n’ different ways then either ofthem can be performed in (m+n) ways.
(ii) Multiplication Principle. If an operation can be performed is ‘m’ different ways; following witha second operation can be performed in ‘n’ different ways, then the two operations in successioncan be performed in ‘mn’ ways.
Factorial
The continued product of first n natural numbers is n!.
n! = 1 . 2 . 3 . ............. n
value of 0! is 1
Exponent of prime p is n!
Let n be a positive integer and p, a prime number. Then the exponent of p is n! is given by
where s in the largest positive integer such that
ps < n < ps+1
PermutationNumber of permutations of n distinct things taking r(1 r n) at a time is denoted by nPr.
rn P = )!r–n(
!n
Number of ways of filling r places using n things if repetition is allowed = nr
Circular PermutationNumber of circular permutations of n things = (n–1)!
Number of circular permutations of n different things taken r at a time = rPr
n
Number of circular permutations of n different things when clockwise and anticlockwise circular
permutations are considered as same is 2)!1–n(
Note : When position are marked, circular arrangement is assumed to be linear.Combination
Number of combinations of n distinct things taking r at a time is denoted by nCr.
nCr = )!r–n(!r!n
Note: nPr = r! (nCr)
218
Properties of nPr and nCr
i) nPn = n! ; nCo = nCn = 1
ii) nPr = n–1Pr–1+ r n–1Pr–1
iii) n
1rr . rPr = n+1Pn+1 –1
iv) nCr = nCs either r=s or r+s = n
v) nCr is greatest , odd. isn if ;
21n r
even isn if ; 2nr
vi) nCr = nCn–r = rn
n–1Cr–1
vii) r1rn
CC
1–rn
rn
viii) nCr–1 + nCr = n+1Cr
Solved Examples
1. How many 5-digit numbers divisible by 3 can be formed using digits 0, 2, 4, 6, 8, 9, if repetition notallowed?
Soluton:Sum of digits =0+2+4+6+8+9=29so 5-digit numbrs can be formed using the digits 0, 4, 6, 8, 9 or 0, 2, 4, 5, 9 as sum of digits isdivisible by 3.
total number of numbers formed is2(4.4.!)=192
Ans. 1922. How many 5-digit numbers divisible by 4 can be formed using digits 0, 2, 4, 7, 8, 9, if repetition not
allowed?Soluton:
Number is divisible by 4, if the last two digits of the number are a multiple of 4, so we can have 04,08, 20, 24, 28, 40, 48, 72, 80, 84, 92 at the last two places. Hence total number of numbers canbe 5.(4P3)+6.3.3P2 as the first place cannot have zero.
the answer is 228.Ans. 228
219
3. How many 5-digit numbers divisible by 6 can be formed using digits 0, 2, 4, 5, 6, 8, if repetition notallowed.
Soluton:Sum of digits is0+2+4+5+6+8=25
the numbers can be formed with the digits 0, 2, 5, 6, 8 as their sum is a multiple of 3 providedunit’s place has an even number. So it can be done in 4!+3.3.3!=78 ways.
4. The total number of odd natural numbers that can be formed with the digits 1, 3, 1, 5,4, 1, 4 andare greater than 2 million are(a) 120 (b) 160 (c) 180 (d) none
Soluton:Odd digits Even digits1, 3, 5 41 41Number of arrangements can be
!2!3!5
5
3
!2!2!5
1
!2!5
1
4
!3!5
3
!3!5
5
!2!2!5
1
5
!2!3!5
3
Ans. c5. The total number of 4 digit numbers greater than 4000, whose sum of digits is odd is
a. 2800 b. 3000 c. 3600 d. none of theseSoluton:
Thousands place can be filled in 6 ways. Hundred’s and Ten’s place can be filled in 10 ways each.First 3 places given the sum either odd or even. In either case last place can be filled in 5 ways.
Number of 4 digit numbers is= 6×10×10×5=3000
Toal no. = 10+30+60+20+20+30+10=180
220
Ans. b6. Mohan writes a letter to five of his friends and addresses them. The number of ways in which the
letters can be placed in the envelopes so that that three of them are in the wrong envelopes isa. 44 b. 119 c. 21 d. 20
Soluton:
5C2.3! !31–
!21
!11–1 =20 (dearrangement of n pairs)
Ans. d
7. In how many ways the squares of the figure given below be filled up with letters of the word‘ROHINI’, so that each row contains atleast one letter.
Soluton:
!2!62–C6
8 =26×360=9360 ways
as six squares can be slected in such a way that no row is empty is (8C6–2)=26 ways.Ans. 9360
1. Number of polynomials of the form x3+ax2+bx+c that are divisible by x2+1, where a, b, c {1,2, 3, .....9, 10}
a. 30 b. 1000 c. 10 d. none of these
2. The number of ways in which we can select four numbers from 1 to 30 so as to exclude everyselection of four consecutive number is
a. 27378 b. 27405 c. 27399 d. none of these
3. The number of ordered pairs of integers (x, y) satisfying the equation x2+6x+y2 = 4 is
a. 2 b. 8 c. 6 d. none of these
4. How many six digit numbers are there in which sum of the digits is divisible by 5
a. 180000 b. 540000 c. 5x105 d. none of these
5. Ten IIT and 2 DCE students sit in a row. The number of ways in which exactly 3 IIT students sitbetween 2 DCE students is
a. 10C3 2! 3! 8! b. 10! 2! 3! 8! c. 5! 2! 9! 8! d. none of these
6. Let there are n>3 circles. The value of n for which the number of radical centres is equal to thenumber of radical axis is (assume that all radical axis and radical centres exist and are different)
a. 7 b. 6 c. 5 d. none of these
7. Total number of integers ‘n’ such that 2 < n < 2000 and HCF of n and 36 is one, is equal to
a. 666 b. 667 c. 665 d. none of these
PRACTICE QUESTIONS
221
8. Match the following :-
Column I Column II
a) The number of five digit numbers having p) 77 the product of digits 20 is
b) A man took 5 space plays out of an engine q) 31to clean them. The number ofways in whichhe can place atleast two plays in the enginefrom where they came out is
c) The number of integers between 1 and 1000 r) 50inclusive in which atleast two consecutivedigits are equal is
d) Value of 9ji1
j i 151
s) 181
9. The number of ordered triplets (a, b, c) such that LCM (a, b) = 1000, LCM (b, c) 2000 andLCM (c, a) = 2000 is _______
10. Read the paragraph and answer the questions that follow :-
Number of ways of distributing n different things into r different groups is rn when blank groupsare taken into account and is rn –rC1 (r–1)n + rC2(r–2)2 ........ + (–1)r–1 rCr–1 when blank groupsare permitted.
i) 4 candidates are competing for two managerial posts. In how many ways can the candidates beselected?
a. 42 b. 4C2 c. 24 d. none of these
ii) 8 different balls can be distributed among 3 children so that every child receives at least oneball is
a. 38 b. 8C3 c. 83 d. none of these
iii) 5 letters can be posted into 3 letter boxes in
a. 35 ways b. 53 ways c. 5C3 ways d. none of these
Answers
1. c 2. 3. b 4. a 5. a
6. c 7. a 8. a-r, b-q, c-s, d-p 9. 70
10. (i)-b, (ii)-d, (iii)-a
222
PERMUTATION AND COMBINATIONS -IISimple Applications on nPr and nCr
Important Results1. Sum of digits in the unit place of all numbers formed by a1, a2.........an taken all at a time is given by
(n–1)! (a1+a2+......an) if repetition of digits is not allowed.2. Sum of all the numbers which can be formed using the digits a1, a2.........an (repetition not allowed)
= (n–1)! (a1+ a2+........+an)9
1–10n
= (n–1)! (sum of digits) timesn1.......11
3. Number of integral solutions of linear equations and unequations (Multinomial Theorem)i. Total number of non negative integral solutions of x1+x2+......xr = n is n + r–1Cr–1
Total number of positive integral solutions of x1+x2+......xr = n is n–1Cr–1ii. In order to solve inequations of the form x1+x2+......+xr n, we introduce artificial (dummy)
variable xr+1 such that x1+x2+.......+xr+xr+1 = n where xr+1 0.Number of solutions of this equation are same as the number of solutions of inequationx1+x2+.....xr n.
iii. Number of solutions of nq.....32 is
includednotiszeroifx–1.......x–1x–1xin xoft coefficien
includediszeroifx–1.......x–1x–1x–1in xoft coefficien1–q1–21–q......21n
1–q1–31–21–n
4. Number and sum of divisorsLet N = ap bq cr where a, b, c are primes & p, q, r Z .
i. Number of divisors of N = (p+1) (q+1) (r+1)Sum of divisors of N = (1+a+a2+....ap) (1+b+b2+....bq) (1+c+c2+....cr)
ii. Number of ways in which N can be resolved as a product of two factors is
squareperfectaisNif11c1b1a21
squareperfectnotisNif1c1b1a21
5. Number of ways in which a composite number N can be resolved into two factors which arerelatively prime (or co prime) to each other is 2n–1 where n is the number of different primefactors is N.
6. DivisibilityCondition for divisibility of a numberA number abcde will be divisible1. by 4 if 2d+e is divisible by 4
223
2. by 8 if 4c+2d+e is divisible by 83. by 3 if a+b+c+d+e is divisible by 34. by 9 if a+b+c+d+e is divisible by 95. by 5 if e = 0 or 5
6. by 11 if placesevenatdigitofsum
placesoddatdigitsofSum
db–eca is divisible by 111
7. by 6 if e = even and a+b+c+d+e is divisible by 38. by 18 if e = even and a+b+c+d+e is divisible by 9
Solved Examples1. Ten different letters are printed round a circle. The number of different ways in which we can select
three letters so that no two of them are consecutive isa. 26 b. 50 c. 56 d. 72
Soluton:Number of selections is10C3–10 – 10.6C2= 50 ways as total ways of selection is 10C3 (number of selections isnC3–n – n.n–4C2)Number of ways when three are consecutive is 10Number of ways when two are consecutive is 10.6C2. Subtract these two cases from the totalnumber of ways.Ans. b
2. The numberof triangles whose vertices are the vertices of an octagon but none of whose sideshappen to come from the sides of octagon isa. 24 b. 52 c. 48 d. 16
Soluton:Proceding in a similar way as in solved example 1, we have number of selection as8C2 – 8 –8.4C2 =16 (number of selections is nC3–n – n.n–4C2)Ans. d
3. The maximum number of points in which 4 circles and4 straight lines intersect isa. 26 b. 50 c. 56 d. 72
Soluton:Maximum number of points of line - line intersection = 4C2=6Maximum number of points circle - circle intersction is 4P2=12Maximum number of points of line circle intersection is (2×4)×4=32
total number of points of intersection is 6+12+32=50.Ans. b
4. In a plane two families of lines are given by y = x+r and (y = –x+p) where r {0, 1, 2, 3, 4}andp {0, 1, 2, ......, 9}. The number of squares of diagonals of length 3 units formed by theselines isa. 36 b. 24 c. 20 d. none of these
Soluton:It means we have to select two lines from each family in such a way that there is a gap of 2lines between the selected lines. First pair can be selected in two ways and second pair canbe selected in seven ways. Hence, number of squares selected is 7×2 = 14
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Ans: d5. The total number of words that can be made using letters of the word CALCULATE so that
each word starts and ends with a consonant is
a. 2!7.5
b. 2!7.3
c. !7.2 d. none of these
Solution:Consonants VowelsCLT AUECL A
Arrangements can be as follows.
C !2!2!7
C
C !2!7
L
L !2!7
C
C !2!2!7
T
T !2!2!7
C
C !2!2!7
T
L !2!2!7
T
T !2!2!7
T
Total number of arrangements can be 21
21
21
21
2111
21
!2!7
= 2!7.5
6. The number of 5 digit numbers of different digints in which middle digits is the largest is
a.9
4n4
n P b. 33(3!) c. 30(3!) d. none of these
Solution: Fix the middle digit. Number of arrangements is(4P4 – 3P3) + (5P4 – 4P3) +........+ (9P4 – 8P3)Ans: d
225
1. Last digit of (1!+2!+.....+2005!)500 isa. 9 b. 2 c. 7 d. 1
2. Number of integral solutions of x+y+z = 0 with ,5–x ,5–y 5–z isa. 134 b. 136 c. 138 d. 140
3. Let x1,x2.....,xk be divisors of positive integer n (excluding 1 & n). If x1+ x2+.....+xk = 75, thenk
1i ix1
is equal to
a. 2n75
b. n75
c. k75
d. none of these
4. The total number of ways in which n2 number of identical balls can be put in n numbered boxes(1,2,3,....,n) such that ith box contains at least i number of balls is
a. 1–nn C
2
b.1–n
1–n C2 c.
1–n2
2–nn
C2
d. none of these
5. Total number of positive integral solutions of 15 < x1+x2+x3 20 isa. 685 b. 785 c. 1125 d. none of these
6. If n is selected from the set {1,2,3....10} and the number 2n+3n+5n is formed. Total number ofways of selecting n so that the formed number is divisible by 4 is equal toa. 50 b. 49 c. 48 d. none of these
7. A is a set containing n different elements. A subset P of A is chosen. The set A is reconstructedby replacing the elements of P. A subset Q of A is again chosen. The number of ways of choos-ing P and Q so that P Q contains exactly two elements isa. nC3×2n b. nC2×3n–2 c. 3n–2 d. none of these
8. The number of three digit numbers of the form xyz such that x < y and yz isa. 276 b. 285 c. 240 d. 244
9. Number of ordered triplets (x,y,z) such that x,y,z are primes and xy +1 = z isa. 0 b. 1 c. 3 d. none of these
10. Read the passage and answer the following questions:-Suppose a lot of n objects having n1 objects one kind, n2 objects of second kind, n3 objects ofthird kind,....,nk objects of kth kind satisfying the condition n1+n2+….+nk = n, then the numberof possible arrangements / permutations of m objects out of this lot is the coefficient of xm in the
expansion of m!ni
0 ix
i. The number of permutations of the letters of the word AGAIN taken three at a time isa. 48 b. 24 c. 36 d. 33
ii. The number of permutations of the letters of the word EXAMINATION taken 4 at a time isa. 136 b. 2454 c. 2266 d. none of these
iii. The number of permutations of the letters of the word EXERCISES taken 5 at a time is
PRACTICE QUESTIONS
226
a. 2250 b. 30240 c. 226960 d. none of theseiv. The number of ways in which an arrangement of 4 letters of the word PROPORTION can be
made isa. 700 b. 750 c. 758 d. none of these
v. The number of permutations of the letters of the word SURITI taken 4 at a time isa. 360 b. 240 c. 216 d. none of these
ANSWERS
1. d 2. b 3. b 4. c
5. a 6. b 7. b 8. a
9. b 10. (i) d, (ii) b, (iii) a, (iv) c, (v) d
227
PERMUTATIONS & COMBINATIONS - IIISimple Applications on nPr and nCr
Important Results1 Total number of selections of one or more objects from n different objects
= nC1+nC2+....+nCn=2n–1
2 Total number of selections of any number of things from n identical things
= .selectedbetoisthingoneleastatwhen;nallowedisthingszeroofselectionwhen;)1n(
3 Total number of selections from p like things, q like things of another type and r distinct things
= )selectedbecannotallornoneif(2–2)1q)(1p()selectedbetothingoneleastatif(1–2)1q)(1p(
r
r
4 Total number of selections of r things from n different things when each thing can be repeated unlimitednumber of times = n+r–1Cr–1
5 Total number of ways to divide n identical things among r persons = n+r–1Cr–16 Results on distribution
Distribution of n things to r boxes
Given Condition Number of ways
n distinct things Empty boxes are allowed rn
r distinct boxes Empty boxes are not allowed coefficient of xn in n!(ex–1)r
n identical things Empty boxes are allowed n+r–1Cr–1
r distinct boxes Empty boxes are not allowed n–1Cr–1
7 Division of items into groups(i) Groups of unequal size.
Number of ways in which (m+n+p) items can be divided into unequal groups containing
m, n, p items is !p!n!m
)!pnm(
Number of ways to distribute (m+n+p) items among 3 persons in the group containing
m,n & p items is !p!n!m
)!pnm( 3!
(ii) Groups of equal sizeNumber of ways in which (mn) different items can be divided equally into m groupseach containing n objects
= ttanimporisgroupsoforderif;
)!n()!mn(
ttanimpornotisgorupsoforderif;!m)!n(
)!mn(
m
m
Note(i) If there are m items of one kind, n items of another kind, then the number of ways of
228
choosing r items out of these = coefficient of xr in (1+x+x2+...........+xm)(1+x+x2+.......+xn)(ii) If there are m items of one kind n items of another kind, then the number of ways of choosing
r items such that at least one item of each kind is included= coefficient of xr in (x+x2+ ......+xm) (x+x2+....+xn)
8 Results related with points, Lines, Rectangle, Polygon, Circle, etc.(i) If these are n points in the plane, number of line segments nC2(ii) Number of points n, then the number of triangles nC3(iii) Number of diagonals in a regular polygon having n sides = nC2–n(iv) Number of parallelograms when a parallelogram is cut by two sets of m lines parallel to
its sides = m+2C2 m+2C2
(v)n
1r
2
n
1r
3
rsofNumber
rglestanrecofNumber
(vi) n
1r
21n
21m
)1r–n)(1r–m(sofNumber
)1n)(1m(4
mnCCglestanrecofNumber
(vii) Maximum number of parts in which a plane can be divided by n straight lines = 1+n
1rr
(viii) Maximum number of points of intersection of n straight Lines = 1×nC2(ix) Maximum number of points of intersection of n circles = 2×nC2(x) Maximum number of points of intersection of n parabolas = 4×nC2
De-arrangementNumber of arrangement of m things in a row so that none of them occupies its original place is
m! !m1)1(–.......
!31–
!21
!11–1 m
Exponent of prime p in n!
Ep(n!) = pn
+ 2pn
+ 3pn
+..........+ kpn
where k is the largest positive integer such that
pk n<pk+1
Solved Examples1 The number of zeroes at the end of 100! is
(a) 23 (b) 24 (c) 25 (d) None of theseSolution :
5100
+ 25100
+ 35100
+....................
n
n
n
m
229
= 20+4+0= 24Ans (b)
2 The total number of integral solutions of the triplet (x,y,z) for the equation xyz=24 is(a) 30 (b) 60 (c) 120 (d) None of theseSolution :
24.1.1 !2!3
= 3
12.2.1 3! = 66.4.1 3! = 68.3.1 3! = 6
6.2.2 !2!3
= 3
4.3.2 3! = 6Total = 30
30 positive integral solutionsTotal number of integral solutions with negative integers included is 30×4 = 120Ans (c)
3 The total number of squares in a chess board is(a) 64 (b) 65 (c) 204 (d) None of theseSolution :
12+22+32+................+82 = 6)116)(18(8
= 204
Ans (c)4 20 lines pass through a given plane. The maximum number of parts in which the plane can be
divided is(a) 210 (b) 211 (c) 212 (d) None of theseSolution :
Use 1+ n
= 1+ 20 = 1+ 221.20
= 2111
Ans (b)5 The number of quadrilaterals that can be formed using 10 points in a plane out of which 4 are
collinear is(a) 210 (b) 209 (c) 185 (d) None of theseSolution:
10C4–4C4–
4C3. 6C1 = 185
Ans (c)
6 The total number of distinct rational numbers x such that 0<x<1 and x = qp
where
p,q {1,2,3,4,5,6} is
230
(a) 15 (b) 13 (c) 11 (d) None of theseSolution :
Values of p Possible rational numbers
1 21
, 31
, 41
, 51
, 61
2 32
, 42
, 52
, 62
3 43
, 53
, 63
4 54
, 64
5 65
Out of 15 possible rational numbers, only 11 are distinct.Ans (c)
7 The sum of 5 digit number in which only odd digits occur without repetition is(a) 277775 (b) 555550 (c) 1111100 (d) None of theseSolution :Sum of n digit numbers
= (Sum of digits) 1–10
)1–10( n
(n–1)!
= (1+3+5+7+9) 1–10
)1–10( 5
(5–1)!
= 25 × 11111 × 24= 6666600
Ans (d)
1 An n digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbersare to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this ispossible is(a) 6 (b) 7 (c) 8 (d) 9
2 Match the following :Consider all possible permutations of the letters of the word ENDEANOEL
Column I Column II(a) The number of permutations containing the word ENDEA is (p) 5!(b) The number of permutations in which the letter E occurs in
the first and last position is (q) 2×5!(c) The number of permutations in which none of the letters
D, L, N occurs in the last five positions is (r) 7×5!(d) The number of permutations in which the letters A, E, O
occur only in odd positions is (S) 21×5!
PRACTICE QUESTIONS
231
3 Five balls of different colors are to be placed in 3 boxes of different sizes. Each box can hold all 5balls. The number of ways we can place the balls so that no box is empty, is(a) 116 (b) 126 (c) 144 (d) 150
4 A student is allowed to select atmost n books from a collection of (2n+1) books. If number ofways in which he can select atleast one book is 63, then n =(a) 3 (b) 4 (c) 6 (d) 5
5 A rectangle with sides 2m–1, 2n–1 is divided into squares of unit length by drawing lines parallel tosides of a rectangle. The number of rectangles with odd side length is(a) (m+n+1)2 (b) mn(m+1)(n+1) (c) m2n2 (d) 4m+n–1
6 Out of 5 apples, 10 mangoes and 15 oranges, the number of ways of distributing 15 fruits each totwo persons, is(a) 56 (b) 64 (c) 66 (d) 72
7* Match the followingColumn I column II
(a) The number of positive integral solutions of theequation x1x2x3x4x5 = 1050 is , then is divisible by (p) 3
(b) Let y be the element of the set A = {1,2,3,5,6,10,15,30} (q) 4and x1, x2, x3 be integers such that x1x2x3 = y.If be (r) 5the number of integral solutions of x1x2x3 = y, then is divisible by (s) 8
(c) Let a be a factor of 120. If be the number of positiveintegral solutions of x1x2x3=a, then is divisible by (t) 16
8 The maximum number of points into which 4 circles & 4 straight lines intersect is(a) 26 (b) 50 (c) 56 (d) 72
9 A is a set containing n elements. A subset P1 is chosen and A is reconstructed by replacing theelements of P1. The same process is repeated for subsets P2, .......Pm with m>1. The number of waysof choosing P1, P2, .......,Pm, so that P1 P2 ...... Pm = A is is(a) (2m–1)mn (b) (2n–1)m (c) m+nCm (d) None of these
10 Number of points having position vector kcjbia where a,b, c {1,2,3,4,5,} such that2a+3b+5c is divisible by 4 is(a) 70 (b) 140 (c) 210 (d) 280
11 Read the passage and answer the following questions.A is a set containing n elements. A subset P of A is chosen and the set A is reconstructed byreplacing the elements of P. A subset Q of A is chosen again. Find the number of ways ofchoosing P & Q when(i) Q is subset of P is(a) 3n (b) 2n (c) n.3n–1 (d) None of these(ii) P & Q contain just one element is(a) 2n (b) 3n (c) n.3n–1 (d) None of these(iii) P = Q is(a) 2n (b) 3n (c) n.3n–1 (d) None of these
Note:* Questions with more than one option is correct.
232
Answers1. b 2. a p; b s; c q; d q 3. d 4. a 5. c6. c 7. a p,r; b q,s,t; c q,s,r,t,t 8. b 9. d10. a 11. (i) a (ii) c (iii) a
233
PERMUTATIONS AND COMBINATIONS - IVSimple Applications on nPr and nCr
Important Results1. Sum of digits in the unit place of all numbers formed by a1, a2.........an taken all at a time is given by
(n–1)! (a1+a2+......an) if repetition of digits is not allowed.2. Sum of all the numbers which can be formed using the digits a1, a2.........an (repetition not allowed)
= (n–1)! (a1+ a2+........+an)9
1–10n
= (n–1)! (sum of digits) timesn1.......11
3. Number of whole number solutions i0xi (non-negative) of x1+x2+..... xr = n is n–1Cr–1
Number of solutions of nq.....32 is
includednotiszeroifx–1.......x–1x–1xin xoft coefficien
includediszeroifx–1.......x–1x–1x–1in xoft coefficien1–q1–21–q......21n
1–q1–31–21–n
4. Number and sum of divisorsLet N = ap bq cr where a, b, c are primes & p, q, r Z .
i. Number of divisors of N = (p+1) (q+1) (r+1)Sum of divisors of N = (1+a+a2+....ap) (1+b+b2+....bq) (1+c+c2+....cr)
ii. Number of ways in which N can be resolved as a product of two factors is
squareperfectaisNif11c1b1a21
squareperfectnotisNif1c1b1a21
5. Number of ways in which a composite number N can be resolved into two factors which arerelatively prime (or co prime) to each other is 2n–1 where n is the number of different primefactors is N.Multinomial Theorem:Coefficient of xr in (1–x)–n = n+r–1Cr. Number of ways of making a selection from m+n+p = Nthings where p are alike of one kind, m alike of second kind and n alike of third kind taken r ata time is given by coefficient of xr is expansion of(1+x+x2+…….xm) (1+x+x2+…….xn) (1+x+x2+…….xp)
Example: Number of selection of 4 letter words from the letters for the ward PROPROTIONis
N,I,T,OOO,RR,PPCoefficient of x4 is (1+x+x2) (1+x+x2) (1+x+x2+ x3) (1+x) (1+x) (1+x)Condition for divisibility of a number
234
A number abcde will be divisible1. by 4 if 2d+e is divisible by 42. by 8 if 4c+2d+e is divisible by 83. by 3 if a+b+c+d+e is divisible by 34. by 9 if a+b+c+d+e is divisible by 95. by 5 if e = 0 or 5
6. by 11 if placesevenatdigitofsum
placesoddatdigitsofSum
db–eca is divisible by 111
7. by 6 if e = even and a+b+c+d+e is divisible by 38. by 18 if e = even and a+b+c+d+e is divisible by 9
Solved examples1. The number of divisors of (6!)3! is
a. 364 b. 9100 c. 2275 d. 75Solution: (6!)3! = (24.32.51)6
= 224.312.56
= 25x13x7 = 2275Ans: c
2. The number of ways in which three district number in AP can be selected from1,2,3--------,24 isa. 132 b. 572 c. 264 d. 150
Ans: 12C2+12C2 = 1322.
1.211.12
(First and last number should either be both even or both odd
and the middle number is average of the two)3. If x,y,z are integers and 0x , 1y , 2z and x+y+z = 15, then the number of ordered
triplets (x,y,z) isa. 91 b. 455 c. 17C2 d. none of theseSolution: 0x , 1y , 2z
Apply n+r–1Cr–1 when n = 12, r = 312+3–1C3–1 = 14C2 = 2
13x14= 91
Ans: a4. a,b,c,d are odd natural numbers such that a+b+c+d = 20, then number of quadrapulets (a,b,c,d)
isa. 165 b. 455 c. 310 d. 255Solution: Let a = 2p+1, b = 2q+1, c = 2r+1, d = 2s+1
p+q+r+s = 88+4–1C4–1 = 11C3 = 165 (distribution of alike objects)
Ans: a5. The number of positive integral solutions of x+y+z 10 is ________________.
12 objects ( alike) are to be distriuted among 3 persons. (distribution of alike objects)
no.of divisors
Put y-1=Y,z-2=z , x+y+z=12. so, we can say1
235
Solution: Let x+y+z+a = 10 0a&za,1Z,1y,1xwhere )Required number = n+r–1Cr–1= 7+4–1C4–1 = 10C3
= 1203.2.18.9.10
1. Number of divisors of the form (4n+2); 0n of the integer 240 isa. 4 b. 8 c. 10 d. 13
2. If r,s,t are prime numbers and p,q are the positive integers such that LCM of p,q is r2s4t2, thenthe number of ordered pairs (p,q) isa. 252 b. 254 c. 225 d. 224
3. The number of seven digit integers, with sum of the digits equal to 10 and formed by using thedigits 1,2 and 3 only, isa. 55 b. 66 c. 77 d. 88
4. Let n and k be positive integers such that 21k Cn . The number of solutions (x1,x2,....xk);
kx,.......2x1x k21 all integers satisfying x1+x2+......+xk = n is
a.k
2k–nC b.
k2k–1–nC c.
1–k2k–1–nC d.
1–k2k–1nC
5. The number of divisors of the form 4n+1, 0n of the number 1010111111313 isa. 750 b. 840 c. 924 d. 1024
6. The number of positive integer solution of the equation 101x
99x
is
a. 2500 b. 2499 c. 1729 d. 1440
7. Let N be natural number. If its first digit (form the left) is deleted , it gets reduced to 57IV
. The
sum of all the digits of N isa. 15 b. 18 c. 24 d. 30
8. The number of positive integral pairs (x,y) such that yx,2007
1y1
x1
is
a. 5 b. 6 c. 7 d. 89. The number of ordered triplets of positive integers which satisfy the inequality 45zyx15
isa. 45C2–
14C2 b. 45C3–14C3 c. 46C3–
15C3 d. none of these
PRACTICE QUESTIONS
236
10.* Match the following:Column I Column II
a. Total number of functions f{1,2,3,4,5} p. divisible by 11{1,2,3,4,5} that are into and f(i) i is q. divisible by 7
b. If x, x2 x3 = 27.5.2 then the number of r. divisible by 3solution sets for (x1, x2, x3) where
1x,Nx ii is s. divisible by 4c. Number of factors of 3780 are divisible
by either 3 or 2 or both isd. Total number of divisors of n = 25.34.510
that are of the form 1,24 is11. Read the passage and answer the following questions
Five balls are to be placed in 3 boxes. Each can hold all the five balls. In how many ways canwe place the balls so that no box remains empty, wheni. Balls and boxes are all different
a. 150 b. 6 c. 50 d. 2ii. balls are identical but boxes are different
a. 150 b. 6 c. 50 d. 2iii. balls are different but boxes are identical
a. 150 b. 6 c. 50 d. 2iv. balls as well as boxes are identical.
a. 150 b. 6 c. 50 d. 2
Note: * Questions with more than one option is correct
ANSWERS
1. a 2. c 3. c 4. c 5. c 6. b 7. a 8. c
9. b 10. a p,s; b q,r; c p,s; d r 11. (i) a (ii) b (iii) c (iv) d
237
BINOMIAL THEOREM - IPrinciple and simple applications
Binomial Theorem for Positive Integral Index
If x and y are real, then for all n N
(x+y)n = nC0xn + nC1x
n–1y+nC2xn–2y2 + ....... + nCrx
n–ryr + ....+nCn–1x1yn–1+nCny
n
=n
0r nCrx
n–ryr
(x–y)n = nC0xn – nC1x
n–1y1 + nC2xn–2y2......nCr(–1)rxn–ryr+.... 1–n)1(–
1–nnC x1yn–1+nCn(–1)nyn
(x+y)n + (x–y)n=2{nC0xn + nC2x
n–2y2 + nC4xn–4y4+.......]
(x+y)n – (x–y)n=2{nC1xn–1 a1+nC3x
n–3a3+nC5xn–5y5+........]
(1+x)n = nC0+nC1x+nC2x
2 +........nCrxr+......nCnx
n
(1–x)n=nC0– nC1x+nC2x2 –........nCr(–1)rxr+......+nCn(–1)nxn
Properties of Binomial Expansion
(i) The number of terms in the expansion of (x+y)n where n N is (n+1).
(ii) The sum of exponents of x & y in (x+y)n is equal to n, the index of the expansion.
(iii) Since nCr=nCn–r; r=0, 1, 2, ......n, the binomial coefficients equidistant from the begin
ning and the end are equal.
i.e. nC0=nCn,
nC1 = nCn–1 and so on.
(iv) The general term is the expansion of (x+y)n is given by
Tr+1 = nCr xn–r yr
(v) Coefficient of (r+1)th term in the expansion of (1+x)n is nCr=coefficient of xr.
(vi) If n is odd, then {(x+y)n+(x–y)n} and {(x+y)n–(x–y)n} have same number of terms equal to
21n
.
If n is even, then
{(x+y)n+(x–y)n} has 12n
terms and
{(x+y)n–(x–y)n} has 2n
terms.
(vii) Middle term
If n is even then in the expansion of (x+y)n, 12n
th terms is the middle term.
238
If n is odd natural number, then 21n
th and 23n
th are the middle terms in the
expansion of (x+y)n.
(viii) Let S = (x+y)n = nC0xn+nC1x
n–1y+.......+nCn–1xyn–1+nCnyn where Nn
=n
0rnCrx
n–ryr
Replacing r by n–r we get,
S = n
0r nCn–r x
ryn–r
= nCnyn+nCn–1y
n–1x+......+nCn–1yxn–1+nCnxn.
i.e. By replacing r by n–r, we are writing the binomial expansion in the reverse order
Properties of Binomial Coefficient
(i) Sum of two binomial coefficients, nCr+nCr–1=
n+1Cr
(ii) nCr = rn
n–1Cr–1
(iii)1–r
nr
n
CC
= r1r–n
(iv) nCr = nCs either r = s or r+s = n
Multinomial Theorem (For a positive integral index)
(x1+x2+........+xk)n = !nk!.....n!n!n
21 nk
k2n
21n
1 x......xx ,
Where n1+n2+......+nk=n and 0 < n1, n2, .......nk< n
• The greatest coefficient in this expansion is rr–k ))!1q(()!q(!n
where q is the quotient
and r is the remainder when n is divided by k.Eg. Find the greatest coefficient in (x+y+z+w)15
n=15, k=4 we have 15=4×3+3 i.e. q=3, r=3 greatest coefficient = 31 )!4()!3(!15
• Number of distinct terms in the expansion is n+k–1Ck–1 (Total number of terms).• Number of positive integer solutions of x1+x2+....+xk=n is n–1Ck–1.• Number of non negative integer solutions of x1+x2+....+xk=n is n+k–1Ck–1.• Sum of all the coefficients is obtained by setting x1=x2=.....xk=1.
239
Greatest Coefficient and Greatest term
Consider the binomial expansion of (x+y)n. where n W. For a given value of n,
Maximum value of nCr is nCn/2 if n is even
Maximum value of nCr is 2
1–nn C =
21n
n C if n is odd.
To find the greatest term in the expansion of (x+y)n,
(i) Find m =
yx1
1n
(ii) If m is an integer we have mth and (m+1)th terms as greatest terms.
(iii) If m is not an integer, then ([m]+1)th term is the greatest term where [.] denote the greatestinteger < m.
Divisibility Problems
From the expansion
(1+a)n = 1+nC1a+nC2a2+.....+nCna
n, we can see that
(i) (1+a)n–1 is a mutliple of a = M(a)
(ii) (1+a)n–1–na is a mutliple of a2 = M(a2)
(iii) (1+a)n–1–na– 2a2
)1–n(n is a mutliple of a3 and so on.
For example
• (1+8)50–1 = 9n–1 is M(8)
• (1+8)50–1 –50×8= 9n–399 is M(82) = M(64)
• (1+8)50–1 –50×8– 282
4950= M(83) = M(512) and soon
Binomial Theorem for any index (for negative or fractional index)
If n Q, then (1+x)n=1+nx+!2
)1–n(n x2+!3
)2–n)(1–n(n x3+..... provided x < 1.
• For any index n, the general term in the expansion of
i. (1+x)n is Tr+1= !r)1r–n().........1–n(n xr
ii. (1+x)–n is Tr+1= !r)1–rn().........1n(n)1(– r
xr
240
iii. (1–x)n is Tr+1= !r)1r–n().........1–n(n)1(– r
xr
iv. (1–x)–n is Tr+1= !r)1–rn().........1n(n xr
• The following expansions should be remembered (for x < 1).i. (1+x)–1=1–x+x2–x3+ ........ ii. (1–x)–1=1+x+x2+x3 + ........ iii. (1+x)–2=1–2x+3x2–4x3+ ........ iv. (1–x)–2=1+2x+3x2+4x3+ ........
• Note : The expansion in ascending powers of x is valid if x is small. If x is large (i.e.
x > 1), then we may find it convenient to expand in powers of x1
, which then will be
small.Exponential series
• ex = 1+ .....!3
x!2
x!1
x 32
• e = 1+ .....!3
1!2
1!1
1(e 2.72 )
• e +e–1= 2 .....!6
1!4
1!2
11
• e –e–1= 2 .....!7
1!5
1!3
1!1
1
Logarithmic seriesFor –1 < x 1
loge(1+x)=x– .....4x–
3x
2x 432
• loge(1–x)= –x– .....4x–
3x–
2x 432
, –1 x < 1
• log x–1x1
=2 .....5x
3xx
53
, –1 < x < 1
• loge 2=1– .....41–
31
21
0.693
Solved Examples
1. If C0, C1, C2,......Cn denote the coefficients in the binomial expansion of (1+x)n, prove that :
C1+2C2+3C3+.....+nCn=n.2n–1
241
i.e.n
1rr.Cr=n.2n–1
Solution :
We have
C1+2C2+3.C3+.....+nCn
= n
1rr.Cr
= n
1rr..nCr [ Cr =
nCr]
= n
1rr.. r
n n–1Cr–1 1–r
1–nr
n C.rnC
= n
1rn–1Cr–1
= n(n–1C0+n–1C1+....+n–1Cn–1)=n(1+1)n–1 1x
= n.2n–1
2. If C0, C1, C2,......Cn denote the coefficients in the binomial expansion of (1+x)n, prove that:C0+3C1+5C2+.....+(2n+1)Cn=(n+1).2n.
Solution
We have,
C0+3C1+5C2+.....+(2n+1)Cn
= n
0r(2r+1)Cr
= n
0r(2r+1)nCr [ Cr=
nCr]
= n
0r(2r..nCr+
nCr)
= n
0r2r..nCr+
n
0rnCr
= 2n
1rr.. 1–r
1–n C.rn +
n
0rnCr 1–r
1–nr
n C.rnC
242
= 2nn
1rn–1Cr–1
+n
0rnCr
= 2n.2n–1+2n
= n.2n+2n=(n+1)2n
3. If C0, C1, C2,......Cn denote the coefficients in the binomial expansion of (1+x)n, prove that:
12.C1+22.C2+32.C3+.....+n2.Cn=n(n+1)2n–2
SolutionWe have,12.C1+22.C2+32.C3+.....+n2.Cn
= n
1rr2.Cr
= n
1rr2.nCr
= n
1r[r(r–1)+r] nCr
= n
1rr(r–1). r
n.rC1–r1–n.
rn n
1r2–r
2–n1–r
1–n C
1–r1–n
n
1r2–r
2–nn
2rCnC)1–n(n
= n(n–1)(n–2C0+n–2C1+....+n–2Cn–2)
+n(n–1C0+n–1C1+....+n–1Cn–1)
= n(n–1).2n–2+n.2n–1
= n(n–1+2)2n–2
= n(n+1)2n–2
4. If C0, C1, C2,......Cn denote the coefficients in the binomial expansion of (1+x)n, provethat:13.C1+23.C2+33.C3+.....+n3.Cn=n2(n+3)2n–3
SolutionWe have,13.C1+23.C2+33.C3+.....+n3.Cn
= n
1rr3.Cr
= n
1rr3.nCr
243
= n
1r[r(r–1)(r–2)+3r(r–1)+r] nCr
= n
1rr(r–1)(r–2)nCr+
n
1r3r (r–1)nCr+
n
1rr..nCr
= n
3rr(r–1)(r–2).
3–r3–n C.
2–r2–n.
1–r1–n.
rn
+ n
2r3r(r–1) 1–r
1–nn
1r2–r
2–n Crn.rC
1–r1–n.
rn
1–r1–n
n
1r2–r
2–nn
2r3–r
3–nn
3rCnC)1–n(n3C)2–n)(1–n(n
= n (n–1) (n–2) {n–3C0+n–3C1+.......n–3Cn–3
+3n(n–1) (n–2C0+n–2C1+......n–2Cn–2
+ n {n–1C0 +n–1C1 +......+n–1Cn–1}
= n (n–1) (n–2). 2n–3+3n(n–1).2n–2 +n.2n–1
= {(n–1) (n–2) +6(n–1) +4} n2n–3
= n (n2+3n) 2n–3
= n2 (n+3) 2n–3
5. If C0,C1,C2.....,Cn–1,Cn denote the binomial coefficients in the expansion of (1+x)n, prove
that: 2
)1n(nCC.n......
CC.3
CC.2
CC
1–n
n
2
3
1
2
0
1
Solution: We have,1–n
n
2
3
1
2
0
1
CC.n......
CC.3
CC.2
CC
=n
1r 1–r
r
CCr
=n
1r 1–rn
rn
CC.r
=n
1r r1r–n.r
r1r–n
CC
1–rn
rn
=n
1r
1r–n
=n
1r
r–1n
244
=n
1r
r–1nn
= 2)1n(n–1nn
= 2)1n(n
6. If C0,C1,C2.....,Cn–1,Cn denote the binomial coefficients in the expansion of (1+x)n, prove
that: (C0+C1) (C1+C2) (C2+C3) (C3+C4)......(Cn–1+Cn) = !n
)1n(C.....CCC n1–n210
Solution:We, have (C0+C1) (C1+C2) (C2+C3)......(Cn–1+Cn)
= C0C1C2.....Cn–11–n
n
1
2
0
1
CC1.....
CC1
CC1
= (C0C1....Cn–1)n
1r 1–rn
rn
CC1
= (C0C1....Cn–1)n
1r r1r–n1
= (C0C1....Cn–1)n
1r r1n
= (C0C1....Cn–1)n
1r
n
!n1n
7. If C0,C1,C2,.....Cn denote the binomial cofficients in the expansion of (1+x)n, prove that
1n1–2
1nC......
3C
2CC
1nn21
0
Solution: we have, 1n
C......3
C2
CC n210
=n
0r
r
1rC
=n
0rr
n C.1r
1
=n
0rr
n C.1r1n.
1n1
245
=n
0rr
n C.1r1n
1n1
=n
0r1r
1n C1n
1r
n1r
1n C.1r1nC
= 1n1n
31n
21n
11n C....CCC
1n1
= 01n
1n1n
11n
01n C–C....CC
1n1
= 1–21n
1 1n
1. The sum ________..........)!5–n(!5
1)!3–n(!3
1)!1–n(!1
1
2. If the coefficient of xn in (1+x)101 (1–x+x2)100 is non-zero, then n cannot be of the forma, 3r+1 b. 3r c. 3r+2 d. none of these
3. The coefficient of xr; 1–nr0 , is the expansion of (x+3)n–1+(x+3)n–2(x+2) +(x+3)n–2
(x+2)2+......(x+2)n–1 area. nCr (3
r – 2n) b. nCr (3n–r – 2n–r) c. nCr (3
r – 2n–r) d. none of these4. The number of real negative terms in the binomial expension of (1+ix)4n–2, ,Nn x > 0 is
a. n b. n+1 c. n–1 d. 2n5. (n+2) nC0 2
n+1 – (n+1) nC1 2n+n.nC2 2
n–1......is equal toa. 4 b. 4n c. 4(n+1) d. 2(n+2)
6.1k
1–k
n11k
a. n(n–1) b. n(n+1) c. n2 d. (n+1)2
7. The sum of rational term in 1063 532 is equal toa. 12632 b. 1260 c. 126 d. none of these
8. Last two digit of (23)14 area. 01 b. 03 c. 09 d. none of these
9.* If n154 = I+f, where n is an odd natural number, I is an interger and 0<f<1, then
a. I is a natural number b. I an even integer
c. (I+f) (1–F) = 1 d. 1–f =n
5410. The number of rational numbers lying in the interval (2002,2003) all whose digits after the
decimal point are non-zero and are in decreasing order is
a.9
1ii
9 P b.10
1ii
9 P c. 29–1 d. 210–1
PRACTICE QUESTIONS
246
11. Match the following:Column I Column II
a. mC1 nCm – mC2
2nCm + mC3 3nCm..... p. The coefficent of xm in the expansion of
+(–1)m–1mCmmnCmis ((1+x)n–1)m
b. nCm + n–1Cm + n–2Cm +......+mCm is q. The coefficent of xm inxx1 1n
c. C0Cn+ C1Cn–1+......CnC0 is r. The coefficent of xn in (1+x)2n
d. 2k nC0 – 2k–1 nC1 n–1Ck–1+(–1)k nCk s. The coefficent of xk in the expansion
n–kC0 is (1+x)n
Note:* Question with more than one option is correct
ANSWERS
1.!n
2 1–n
2. c 3. b 4. a
5. c 6. c 7. d 8.
9. a,c,d 10. c 11. a p; b q; c r; d s
247
BINOMIAL THEOREM - IIApplications of Binomial Coefficients
1 Bino-geometric seriesn
nn2
2n
1n
on xC............xCxCC =(1+x)n
eg.nC0+nC1.3+nC2.3
2+.............+nCn3n = (1+3)n = 4n
2 Bino-arithmetic series
nn
2n
1n
on C)nda(.................C)d2a(C)da(Ca
This series is the sum of the products of corresponding terms of
nn
2n
1n
on C....,.........C,C,C (binomial coefficients) and a, a+d, a+2d, ............,a+nd(arithmetic progression)Such series can be solved either by(i) eliminating r in the multiplier of binomial coefficient from the (r+1)th terms of the series
(i.e. using r rn C = n n–1 Cr–1) or
(ii) Differentiating the expansion of xa(1+xd)n or (If product of two or more numericals occur,then differentiate again and again till we get the desired result)
eg. Prove that nC0+2 nC1+3nC2+..............+(n+1)nCn = (n+2). 2n–1
nC0+2.nC1+.....+(n+1) nCn
= n
0r)1r( nCn
= n
0rr
nCr + n
0rr
n C
= n
1r rn.r n–1Cr–1 +
n
1rr
n C
= n.2n–1+2n = (n+2)2n–1
ORConsider the expansionnC0+
nC1 x+nC2 x2+..........+nCn x
n = (1+x)n
Multiply by xnC0x+nC1 x
2+nC2 x3+.........+nCn x
n+1 = x(1+x)n
Differentiate w.r.t.xnC0+
nC1 2x+nC23x2+..........+nCn(n+1)xn = x n(1+x)n–1+(1+x)n
Put x = 1nC0+2nC1+3nC2+............+(n+1)nCn = n2n–1+2n = (n+2)2n–1
3. Bino-harmonic series
aC0
n+
daC1
n+
d2aC2
n+.........+
ndaCn
n
This series is the sum of the products of corresponding terms of
nn
2n
1n
on C....,.........C,C,C (binomial coefficients) and
248
a1
, da1
, d2a1
,................ nda1
(harmonic progression)
Such seris can be solved either by(i) eliminating r in the multiplier of binomial coefficient from the (r+1)th term of the series
(ie using 1r
1 r
n C =
1n1 n+1Cr+1) or
(iii) integrating suitable expansionNote(i) If the sum contains C0, C1, C2............Cn are all positive signs, integrate between limits 0
to 1(ii) If the sum contains alternate signs (i.e.+ & –) then integrate between limits –1 to 0(iii) If the sum contains odd coefficients (i.e. Co, C2, C4,.......) then integrate between –1 to
+1.(iv) If the sum contains even coefficient (i.e. C1, C3, C5,........) the find the difference between
(i) & (iii) and then divide by 2(v) If in denominator of binomial coefficient is product of two numericals then integrate
two times first time take limits between 0 to x and second time take suitable limits
eg: prove that 1C0
n
+ 2C1
n +
3C2
n +..................+
1nCn
n =
1n1–2 1n
nC0 + 2C1
n +
3C2
n+..............+
1nCn
n =
n
0r
rn
1rC
= 1n
1
n
0r
rn
1rC
nCr
= 1n
1 n
0r1r
1n C
= 1n
1 (2n+1–n+1C0)
= 1n
1 (2n+1–1)
ORConsider the expansion
(1+x)n = nC0+nC1x+nC2x
2+...........+nCnxn
Integrate between limit to 0 to 11
0
1n
1n)x1(
= 1
0
1nn
n3
2n
2
1n
0n
1nxC
........3
xC2
xCxC
1n1–
1n2 1n
= nC0+ 2C1
n +
3C2
n+...........+
1nCn
n
249
nC0+ 2C1
n +
3C2
n+...........+
1nCn
n =
1n1
(2n+1–1)
4. Bino-binomial series
nn
r–nn
2rn
2n
1rn
1n
rn
0n CC...........CCCCCC or
0n
rm
2–rn
2m
1–rn
1m
rn
0m CC...........CCCCCCSuch series can be solved by multiplying two expansions, one involving the first factors ascoefficient and the other involving the second factors as coefficients and finally equatingcoefficients of a suitable power of x on both sides.
Prove Thatn–1C0
nC1+n–1C1
nC2+n–1C2
nC3+..........+n–1Cn–1nCn = 2n–1Cn–1
We have(nC0x
n+nC1xn–1+nC2x
n–2+.......+nCn–1x+nCn) (n–1C0+n–1C1x+n–1C2x
2+.........+n–1Cn–1xn–1) = (1+x)n
(1+x)n–1
(nC0xn+nC1x
n–1+nC2xn–2+....+nCn–1x+nCn) (n–1C0+
n–1C1x+n–1C2x2+......+n–1Cn–1x
n–1) = (1+x)2n–1
Equate the coefficients of xn–1 on both sides, n–1C0nC1+
n–1C1nC2+..............+n–1Cn–1
nCn = 2n–1Cn–1Note : For the sake of convenience, the coefficients nC0,
nC1,.........nCr,......
nCn are usually denotedby C0, C1,..........Cr,.......Cn respectively
Use of complex numbers in Binomial TheoremWe know (cos +sin )n = cosn +sinn .Expand and the binomial and then equating the real and imaginary parts, we getcosn = cosn –nC2cosn–2 sin2 +nC4cosn–4 sin4 +................sinn = nC1cosn–1 sin –nC3cosn–3 sin3 +nC5cosn–5 sin5 +...............
tann = .....tanC–tanCtanC–1...............tanCtanC–tanC
66
n44
n22
n
55
n31
n1
n
Solved Examples1. If C0, C1, C2, C3,...........Cn–1, Cn denote the binomial coefficients in the expansion of (1+x)n, prove
that
C0C1+C1C2+C2C3+.......+Cn–1Cn
= )!1n()!1–n()!n2(
= n2.n.
)!1n()1–n2(5.3.1
Solution :Using binomial expansion, we have
(1+x)n = C0+C1x+C2x2+........+Crx
r+.....+Cnxn.......(A) and
250
(1+x)n = C0xn+C1x
n–1+C2xn–2+.........+Crx
n–r+.............+.........+Cn–1x+Cn.....(B)Multiplying (A) and (B), we get
(1+x)2n
= (C0+C1x+C2x2+.....+Crx
r+.....+Cnxn)×(C0x
n+C1xn–1+C2x
n–2+.....+Crxn–r+......+Cn–1x+Cn)
or (C0+C1x+C2x2+.......+Crx
r+.......+Cnxn)(C0x
n+C1xn–1+C2n–2+......+Crx
n–r+......+Cn–
1x+Cn)=(1+x)2n..................(C)Equating the coefficients of xn–1 on both sides of (C), we get
C0C1+C1C2+.......+Cn–1Cn=2nCn–1
C0C1+C1C2+.......+Cn–1Cn=2nCn–1 =
)!1–n()!1n()!n2(
Now,
)!1–n()!1n()!n2(
= )!1–n()!1n()n2)(1–n2)(2–n2(..........6.5.4.3.2.1
= )!1–n()!1n(n2........6.4.2)1–n2.......(5.3.1
= )!1–n()!1n()1–n2.......(5.3.1
= )!1n()1–n2.......(5.3.1
)!1–n(2
= n2.n
)!1n(1
Hence,
C0C1+C1C2+C2C3+.........+Cn–1Cn = )!1n()!1–n()!n2(
= )!1n(n1 n
2. If C0, C1, C2, C3,...........Cn–1, Cn denote the binomial coefficients in the expansion of (1+x)n, provethat
C02+C1
2+C22+..........Cn
2 = 2)!n()!n2(
= n2!n
)1–n2....(5.3.1
Solution :Using binomial expansion, we have
.2( )–
( )–
( )–
T + 1= xGeneral Term
c2n1
251
(1+x)n = C0+C1x+C2x2+........+Crx
r+.....+Cnxn.......(A) and
(1+x)n = C0xn+C1x
n–1+C2xn–2+.........+Crx
n–r+.............+.........+Cn–1x+Cn.....(B)Multiplying (A) and (B), we get
(1+x)2n
= (C0+C1x+C2x2+.....+Crx
r+.....+Cnxn)×(C0x
n+C1xn–1+C2x
n–2+.....+Crxn–r+......+Cn–1x+Cn)
or (C0+C1x+C2x2+.......+Crx
r+.......+Cnxn)(C0x
n+C1xn–1+C2n–2+......+Crx
n–r+......+Cn–
1x+Cn)=(1+x)2n..................(C)Equating the coefficients of xn on both sides of (C), we get
C02+C1
2+C22+.......Cn
2=2nCn
C02+C1
2+C22+.......Cn
2 = !n!n)!n2(
Now
!n!n)!n2(
= !n!n
)n2)(1–n2)(2–n2(..........5.4.3.2.1
= !n!n
)n2)(2–n2........(6.4.2)1–n2.......(5.3.1
= !n!n
n).1–n........(3.2.12)1–n2.......(5.3.1 n
= !n!n
!n2)1–n2.......(5.3.1 n
= n2!n
)1–n2.......(5.3.1
Hence, C02+C1
2+C22+.......Cn
2 = !n!n)!n2(
= n2!n
)1–n2.......(5.3.1
3. If C0,C1,C2,.......Cn denote the binomial fulfillments in the expansion of (1+x)n, prove that ;C0
2–C12+C2
2–C32+......+(–1)n Cn
2
= evenisnif,C.)1(–oddisnif,0
2/nn2/n
Solution : We have,(1+x)n = (C0+C1x+C2x
2+....+Cnxn) .........(i)
Also,(1+x)n = (C0x
n+C1xn–1+..........+Cn–1x+Cn)...........(ii)
Replacing x by –x in (i), we get(1–x)n = C0–C1x+C2x
2–C3x3+.......+(–1)nCnx
n........(iii)Multiplying (ii) and (iii), we get
(C0–C1x+C2x2–C3x
3+.....+(–1)nCnxn)×(C0x
n+C1xn–1+C2
n–2+........+Cn–1x+Cn) = (1+x)n(1–x)n
or (C0–C1x+C2x2–C3x
3+.........+(–1)nCnxn)×(C0x
n+C1xn–1+C2x
n–2+........+Cn–) = (1–x2)n........(iv)
T = xGeneral Term
c2n+1
252
Equating coefficients of xn on both sides of (iv), we getC0
2–C12+C2
2–C32+......+(–1)nCn
2 = Coeffecient of xn in (1–x2)n.............(v)Clearly, RHS of (v) contains only even powers of x when it is expanded with the help of binomialtheorem. Therefore,
Coefficient of xn in (1–x2)n = 0, if n is an odd natural number..If n is even, suppose (r+1) the term in the binomial expansion of (1–x )n contains xn. We have,
Tr+1 = nCr(–1)r(x2)r = nCr(–1)rx2r
For this term to contain xn, we must have,2r = n r = n/2
Coeff. of xn = nCn/2(–1)n/2
Hence,C0
2–C12–C2
2–C32+.....+(–1)nCn
2
= evenisnif,C.)1(–oddisnif,0
2/nn2/n
4. If (1+x)n = C0+C1x+C2x2+....+Cnx
n prove that n
0ssr
n
0r)CC( = (n+1)
Sloution: We have,n
0ssr
n
0r)CC(
= n
0ss
n
0s
n
0rr
n
0rCC
= n
0ss
n
0r
n
0sr
n
0r
CC
= n
0s
n2 +n
0r
n2
= (n+1)2n+(n+1)2n
= 2(n+1)2n
= (n+1)2n+1
5. If (1+x)n = C0+C1x+C2x2+....+Cnx
n prove that n
0ssr
n
0rCC = 22n
Solution: we have,n
0ssr
n
0rCC
= n
0ssr
n
0rCC
= n
0rr
n C.2
2
253
= 2n
0rrC
= 2n.2n = (2n)2 = 22n
ALITER n
0ssr
n
0rCC =
n
0rrC
n
0ssC = 2n.2n = 22n
6. If (1+x)n = C0+C1x+C2x2+....+Cnx
n prove that
nsr0(Cr+Cs) = n.2n
Solution: We have,n
0ssr
n
0r)CC( =
n
0rsr CC +2
nsr0sr CC
(n+1)2n+1 = 2n
0rrC +2
nsr0sr CC
(n+1)2n+1 = 2 n2 +2 nsr0
sr CC
n.2n+1 = 2 nsr0
sr CC
nsr0sr CC = n.2n
7. If (1+x)n = C0+C1x+C2x2+....+Cnx
n prove that nsr0
CrCs = 21
(22n–2nCn)
Solution: We have,n
0ssr
n
0rCC =
n
0r
2rC + 2
nsr0srCC
22n = 2nCn+2nsr0
srCC
nsr0srCC =
21
[22n–2nCn]
ALITER We have,2n
0rrC =
n
0r
2rC +2
nsr0srCC
(2n)2 = 2nCn+2nsr0
srCC
nsr0srCC =
21
[22n–2nCn]
.....
Cr
CsC0 C1 C2 ...... Cn
C0 2C0 C0+C1 C0+C2 ...... C0+Cn
C1 C1+C0 2C1 C1+C2 ...... C1+Cn
C2 C2+C0 C2+C1 2C2 ...... C2+Cn
.
Cn Cn+C0 Cn+C1 Cn+C2 ...... 2Cn
.....
Cr
CsC0 C1 C2 ...... Cn
C0 C02 C0C1 C0C2 ...... C0Cn
C1 C1C0 C12 C1C2 ...... C1Cn
C2 C2C0 C2C1 C22 ...... C2Cn
.
Cn CnC0 CnC1 CnC2 ...... Cn2
n
.
254
1 If n is an integer between 0 and 21, then the minimum value of n!(21–n)! is attained for n =(a) 1 (b) 10 (c) 12 (d) 20
2 404C4–4C1
303C4+4C2
202C4–4C3
101C4 is equal to(a) (401)4 (b) (101)4 (c) 0 (d) (201)4
3 If (3+x2008+x2009)2010 = a0+a1x+a2x2+.......+anx
n, then the value of a0– 21
a1– 21
a2+a3– 21
a4–
21
a5+a6.......is
(a) 32010 (b) 1 (c) 22010 (d) None of these4
2
5n
4n
2n
1n
3n
0n .........CCCC
21–...........CC +
43 2
5n
4n
2n
1n .......C–CC–C =
(a) 3 (b) 4 (c) 2 (d) 1
5 Value of 20
0r)r–20(r (20Cr)
2 is equal to
(a) 400 39C20 (b) 400 40C19 (c) 400 39C19 (d) 400 38C206 If for z as real or complex , (1+z2+z4)8 = C0+C1z
2+C2z4+............+C16z
32, then(a) C0–C1+C2–C3+............+C16 = 1 (b) C0+C3+C6+C9+C15 = 37
(c) C2+C5+C8+C11+C14 = 36 (d)C1+C4+C7+C10+C13+C16 = 37
7 Read the passage and answer the following questionsAny complex number in polar form can be an unpleasing in Euler’s form as cos +sin = ei which
is useful is finding the sum of series n
0r
rr
n sinicosCn
0rr
n rsinircosC
= rin
0rr
n eC = (1+ei )n
Also we know that the sum of binomial series does not change if r is replaced by n–r.
(i) Value of 100
0rr
100 C sin(rx0) is .....
(a) 2100 2x
sin (50x) (b) 2100sin50x cos2x
(c) 21012x
(d) 2101sin100 50xcos50x
ii. In triangle ABC, the value of 50
0rr
50 C ar bn–r cos(rb–(50–r)A) is equal to (a,b,c are sides
opposite to A, B, C & S in semi perimeter)(a) c49 (b) (a+b)50 (c) (25–a–b)50 (d) None of these
cos
(50x) sin
)(cos )(
PRACTICE QUESTIONS
255
(ii) If f(x) = rx2cosC
rx2sinC
50
0rr
50
50
0rr
50
, then f ( /8) is
(a) 1 (b) –1 (c) irrational value (d) None of these8 Match the following
Column I Column II
(a) jij
10i
10 CC (p)2
C–2 102020
(b) nji0j
10i
10 CC (q) 220 – 20C10
(c)nji0
j10
i10 CC (r) 220
(d)10
0jj
10i
1010
0iCC (s)
2C2 10
2020
9 The coefficient of n n is the expansion of (1+ )n (1+ )n ( + )n is
(a)n
0r
2r
n C (b)n
0r
22–r
n C (c)n
0r
23r
n C (d)n
0r
3r
n C
10 If C1,C2..........Cn are binomial Coefficients, then the value of C12–2C2
2+3C32–..........–2nC2
2n is(a) n2 (b) (–1)n–1n (c) 2(–1)n–1 n 2n–1Cn (d) –n2
'Note : Questions with * have more than one correct option'
ANSWERS
1. b 2. b 3. 4. d
5. d 6. a,b,d 7. (i) a (ii) c (iii) a
8. a q; b s; c p; d r 9. d 10. c
256
BINOMIAL THEOREM - IIIPrinciple and simple applications - Problem Solving
1. Binomial Theorem for positive integral index.If x, y R and n N
(x+y)n = nCoxn+nC1x
n–1y+nC2xn–2y2+.........+ nCrx
n–ryr+....+nCnyn=
n
0r
nCrxn–ryr
Properties• Number of terms of the above expansion is (n+1)• The binomial coefficients equidistant from the beginning and the end in a binomial expansion
are equal.• General term = Tr+1=
nCr xn–ryr
n is even : only one middle term th
12n
term .
Middle term
n is odd : Two middle termsth
21n
and th
23n
term .
n is even 2n
n C .
Greatest coefficient
n is odd 2
1–nn C and
21n
n C .
Greatest TermTo find numerically greatest term in the expansion of (1+x)n
i. Calculate m = 1x1nx
ii. If m is an integer, then Tm and Tm+1 are equal and both are greatest term.iii. If m is not an integer then T[m]+1 is the greatest term.
Note : To find the greatest term in the expansion of (x+y)n, find the greatest term in n
xy1 and
then multiply by xn (since (x+y)n=xn n
xy1
2. Multinomial Theorem (for a positive integral index)
(x1+x2+x3+.....xk)n =
!n!.....n!n!n!n
k321
knk
3n3
2n2
1n1 x.....xxx
where ni {0, 1, 2,....n}, n1+n2+.....+nk=n
257
• The greatest coefficient in this expansion is rr–k ))!1q(()!q(!n
where q is the quotient and
r is the remainder when n is divided by k.Eg. Find the greatest coefficient in (x+y+z+w)15
n=15, k=4 we have 15=4×3+3 i.e. q=3, r=3 greatest coefficient = 31 )!4()!3(!15
• Number of distinct terms in the expansion is n+k–1Ck–1 (Total number of terms).• Number of positive integer solutions of x1+x2+....+xk=n is n–1Ck–1.• Number of non negative integer solutions of x1+x2+....+xk=n is n+k–1Ck–1.• Sum of all the coefficients is obtained by setting x1=x2=.....xk=1.3. Binomial Theorem for Negative or Fractional IndicesIf n Q, then
(1+x)n=1+nx+!2
)1–n(n x2+!3
)2–n)(1–n(n x2+..... provided x < 1.
• For any index n, the general term in the expansion of
i. (1+x)n is Tr+1= !r)1r–n().........1–n(n xr
ii. (1+x)–n is Tr+1= !r)1–rn().........1n(n)1(– r
xr
iii. (1–x)n is Tr+1= !r)1r–n().........1–n(n)1(– r
xr
iv. (1–x)–n is Tr+1= !r)1–rn().........1n(n xr
• The following expansions should be remembered (for x < 1).i. (1+x)–1=1–x+x2–x3+ ........ ii. (1–x)–1=1+x+x2+x3 + ........ iii. (1+x)–2=1–2x+3x2–4x3+ ........ iv. (1–x)–2=1+2x+3x2+4x3+ ........
• Note : The expansion in ascending powers of x is valid if x is small. If x is large (i.e. x > 1),
then we may find it convenient to expand in powers of x1
, which then will be small.
4. Exponential series
• ex = 1+ .....!3
x!2
x!1
x 32
• e = 1+ .....!3
1!2
1!1
1(e 2.72 )
• e +e–1= 2 .....!6
1!4
1!2
11
258
• e –e–1= 2 .....!7
1!5
1!3
1!1
1
5. Logarithmic seriesFor –1 < x 1
loge(1+x)=x– .....4x–
3x
2x 432
• loge(1–x)= –x– .....4x–
3x–
2x 432
, –1 x < 1 or 1x
• log x–1x1
=2 .....5x
3xx
53
, –1 < x < 1
• loge 2=1– .....41–
31
21
0.693
Solved Examples
1. Let (1+x)n = n
0r
rrxa ,
then 1–n
n
2
3
1
2
0
1
aa1.......
aa1
aa1
aa1 is equal to
a.!n1n 1n
b.!n1n n
c. )!1–n(n 1–n
d. None of these
Solution :(1+x)n = nC0+
nC1x+nC2x2+.........+ nCnx
n
= a0+a1x +a2x
2 +...........+anxn (given)
Comparing a0=nC0, a1=
nC1, a2=nC2,.......... and so on.
1–nn
nn
2n
3n
1n
2n
0n
1n
CC1.......
CC1
CC1
CC1
= n1n–n1.......
32–n1
21–n1
10–n1
=!n1n
nn1........
3n1.
2n1.
1n1 n
Ans. b
2. If an=n
0r rn C1
, then n
0r rn Cr
equals
a. (n–1).an b. nan c. 21
nan d. None of these
259
Solution :
Let S = n
0r rn Cr
nn
3n
2n
1n
0n C
n..........C3
C2
C1
C0S ....(1)
Also,n
n3
n2
n1
n0
n C0..........
C3–n
C2–n
C1–n
CnS ....(2)
Adding (1) and 2
nn
2n
1n
0n C
n.........Cn
Cn
CnS2
nn
2n
1n
0n C
1.........C1
C1
C1nS2
2S=nan S= 21
nan
Ans. c3. If (1+x)10=a0+a1x+a2x
2+a3x3+.......+a10x
10, then (a0–a2+a4–a6+a8–a10)2 + (a1–a3+a5–a7+a9)
2 isequal toa. 310 b. 210 c. 29 d. none of theseSolution :Put x=i and x = –i
(1+i)10= (a0–a2+a4–a6+a8–a10) + i(a1–a3+a5+a5–a7+a9) ....(1)Also, (1–i)10= (a0–a2+a4–a6+a8–a10) – i(a1–a3+a5–a7+a9) ....(2)Multiply (1) and (2)((1+i)(1–i)10=(a0–a2+a4–a6+a8–a10)
2 + (a1–a3+a5–a7+a9)2
210=(a0–a2+a4–a6+a8–a10)2 + (a1–a3+a5–a7+a9)
2
Ans. b4. If (1+x+2x2)20=a0+a1x+a2x
2+a3x3+.......+a40x
40, then a0+a2+a4+........+a38 is equal toa. 219(220–1) b. 220(219–1) c. 219(220+1) d. none of theseSolution :Put x=1 and x = –1 and adding we get 420+220=2(a0+a2+a4+....+a38+a40)
239+219=a0+a2+.....+a38+220 2040 2a
a0+a2+a4+.....+a38=239+219–220
= 219(220+1–2)= 219(220–1)Ans. a
5. The coefficient of x13 in the expansion of (1–x)5 (1+x+x2+x3)4 isa. 4 b. –4 c. 0 d. none of theseSolution :Coefficient of x13 in = (1–x)5 (1+x+x2+x3)4 = (1–x)5 ((1+x)(1+x2))4
260
= (1–x) {(1–x)(1+x)(1+x2)}4
= (1–x) {(1–x4)4
= (1–x) (1–4x4+6x8–4x12+x16)coefficient of x13 is –1×–4=4
Ans. a6. The sum 20C0+
20C1+20C2+.............. + 20C10 is equal to
a. 220+ 2)!10(!20
b. 219– 2)!10(!20
21
c. 219+ 2)!10(!20
d. none of these
Solution :In the expansion of (1+x)20, put x=1220 = 20C0+
20C1+20C2+.............. + 20C9+
20C10+....... 20C20= 2(20C0+
20C1+.............. + 20C10)– 20C10 (
nCr=nCn–r)
2C2 10
2020
=20C0+20C1+.............. + 20C10
= 219 + 2)!10(!20
21
=20C0+20C1+.............. + 20C10
Ans. d
1. The coefficient of x4 in 10
2x3–
2x
is
a. 256405
b. 259405
c. 263450
d. None of these
2. Let Tn denotes the number of triangles which can be formed using the vertices of a regularpolygon of n sides. If Tn+1–Tn=21, then n equalsa. 5 b. 7 c. 6 d. 4
3. The sum m
1i i–m20
i10
, where qp
=0 if p>q, is maximum when m is
a. 5 b. 10 c. 15 d. 204. Coefficient of t24 in (1+t2)12 (1+t12) (1+t24) is
a. 12C6+3 b. 12C6+1 c. 12C6 d. 12C6+25. If n–1Cr=(k2–3)nCr+1, then k belongs to
a. 2,–– b. ,2 c. ]3,3[– d. 2,36. If (1+ax)n=1+8x+24x2+......., then a = __ and n = ____.
7. The greatest term in the expansion of 20
3113 is
a. 271
720
b. 811
620
c. 920
91
d. none of these
PRACTICE QUESTIONS
261
8. If x= 612 , then the integral part of [x] is
a. 98 b. 197 c. 196 d. 1989. The greatest integer m such that 5m divides 72n+23n–3.3n–1 for n N is
a. 0 b. 1 c. 2 d. 3
10. If )bx–1)(ax–1(1
=a0+a1+a2x2....., then an =
a.a–bb–a 1n1n
b.a–ba–b 1n1n
c.a–ba–b nn
d.a–bb–a nn
11. Read the passage and answer the following questions.If n is a positive integer and a1, a2, a3 .....am C, then
(a1+a2+a3+.....+am)n =mn
m3n
32n
21n
1m321
a.......a.a.a.!n!......n!n!n
!nwhere n1, n2, n3..... nm are all
non-negative integers subject to the condition n1+n2+n3+....+nm=n.i. The number of distinct terms in the expansion of (x1+x2+x3+.....+xn)
4 is
a. n+1C4 b. n+2C4 c. n+3C4 d. n+4C4ii. The coefficient of x3y4z in the expansion of (1+x–y+z)9 is
a. 2320 b. 2420 c. 2520 d. 2620iii. The coefficient of a3b4c5 in the expansion of (bc+ca+ab)6 is
a. 40 b. 60 c. 80 d. 100iv. The coefficient of x39 in the expansion of (1+x+2x2)20 is
a. 5.219 b. 5.220 c. 5.221 d. 5.223
v. The coefficient of x20 in (1–x+x2)20 and in (1+x–x2)20 are respectively a and b, thena. a=b b. a>b c. a<b d. a+b=0
12*. Match the following :Column I Column II
a. If n be the degree of the polynomial (p) 2
)1x3( 27
27
2 )1x3(–x–)1x3(x ,
then n is divisible by (q) 4b. In the expression of (x+a)n there is only one middle term (r) 8
for x=3, a=2 and seventh term is numerically (s) 16greatest term, then n is divisible by (t) 32
c. The sum of the binomial coefficients in the expansion of(x–3/4+nx5/4)m, where m is positive integer lies between200 and 400 and the term independent of x is equals 448.Then n5 is divisible by
'Note : Questions with * have more than one correct option'ANSWERS
1. a 2. b 3. c 4. d 5. d 6. a=2, n=4 7. a8. b 9. c 10. b 11. (i) c, (ii) c (iii) b (iv) c (v) b12. a p, q, r; b p, q, r, s; c p, q, r, s, t
262
BINOMIAL THEOREMFor Positive Integral Index - Problem Solving (Lecture-04)
Summation of Series (involving binomial coefficients)1 Bino-geometric series
nn
n22
n1
no
n xC............xCxCC =(1+x)n
2 Bino-arithmetic series
nn
2n
1n
on C)nda(.................C)d2a(C)da(Ca
This series is the sum of the products of corresponding terms of
nn
2n
1n
on C....,.........C,C,C (binomial coefficients) and a, a+d, a+2d, ............,a+nd(arithmetic progression)Such series can be solved either by(i) eliminating r in the multiplier of binomial coefficient from the (r+1)th terms of the
series (i.e. using r rn C = n n–1 Cr–1)
or(ii) Differentiating the expansion of xa(1+xd)n or (If product of two or more numericals occur,
then differentiate again and again till we get the desired result)3 Bino-harmonic series
aC0
n+
daC1
n+
d2aC2
n+.........+
ndaCn
n
This series is the sum of the products of corresponding terms of
nn
2n
1n
on C....,.........C,C,C (binomial coefficients) and
a1
, da1
, d2a1
,................ nda1
(harmonic progression)
Such seris can be solved either by(i) eliminating r in the multiplier of binomial coefficient from the (r+1)th term of theseries
(ie using 1r
1 r
n C =
1n1 n+1Cr+1) or
(iii) integrating suitable expansionNote(i) If the sum contains C0, C1, C2............Cn are all positive signs, integrate between
limits 0 to 1(ii) If the sum contains alternate signs (i.e.+ & –) then integrate between limits –1 to 0(iii) If the sum contains odd coefficients (i.e. Co, C2, C4,.......) then integrate between –1
to +1.(iv) If the sum contains even coefficient (i.e. C1, C3, C5,........) then find the difference between
(i) & (iii) and then divide by 2
263
(v) If in denominator of binomial coefficient is product of two numericals then integrate twotimes first time take limits between 0 to x and second time take suitable limits
4 Bino-binomial series.
nn
r–nn
2rn
2n
1rn
1n
rn
0n CC...........CCCCCC or
0n
rm
2–rn
2m
1–rn
1m
rn
0m CC...........CCCCCCSuch series can be solved by multiplying two expansions, one involving the first factors ascoefficient and the other involving the second factors as coefficients and finally equatingcoefficients of a suitable power of x on both sides.
Binomial coefficients1 C0+C1+C2+C3+............2n
2 C0–C1+C2–C3+............=03 C0–C1+C2–C3+...........+Cr(–1)r= n–1Cr(–1)r ;r<n4 Co+C2+C4+C6+............=2n–1
5 C1+C3+C5+C7+............=2n–1
6 C0–C2+C4–C6+..........= n2 cos 4
n
7 C1–C3+C5–C7+.......= n2 sin 4
n
8 C0+C4+C8+C12+.......= 21
4ncos22
n1–n
9 C1+C5+C9+C13+.......= 21
4nsin2 1–n
10 C0+C3+C6+C9+.......= 31
3ncos22n
11 C1+2C2+................= 1–nr 2.nrC
12 C1–2C2+3C3................= 0rC)1(– r1–r
13 12C1+22C2+..................=n(n+1)22n–2
14 12C1–22C2..............= 015 C0
2+C12+C2
2+.............= 2nCn
16 C02–C1
2+C22–C3
2.............= evenisnifC)1(–oddisnif0
2/nn2/n
17 nji0 CiCj = 22n–1 – 2n–1Cn
18 nji0 (Ci–Cj)2 = (n+1) 2nCn–22n
2n
264
Note : Consider the equation x1+x2+...........+xr = n,. n N.Number of positive integral solutions = n–1Cr–1Number of non negative integral solutions = n+r–1Cr–1
Solved Examples
1 The value of n
1rr
n2 rC is
(a) n.22n–1 (b) 22n–1 (c) 2n–1+1 (d) None of theseSolution:
n
1r rn2r 2n–1Cr–1
= 2nn
1r1–r
1–n2 C
= 2n(2n–1C0+2n–1C1+
2n–1C2+......2n–1Cn–1)
= 2n. 2
2 1–n2 = n.22n–1 [ 22n–1 = C0+C1+.........+C2n–1 22n–1= 2(C0+C1+C2+...... Cn–1)]
Ans (a)2 The coefficient of x5 in the expansion of (1+x)21+(1+x)22+.........+(1+x)30 is
(a) 31C5–21C5 (b) 31C6–
21C6 (c) 30C6–20C6 (d) None of these
Solution:Co-efficient of x5 in (1+x)21+(1+x)22+............+(1+x)30
= Co-efficient of x5 in 1–)x1(1–)x1()x1( 1021
coefficient of x6 in (1+x)31 – (1+x)21 is 31C6 – 21C6Ans (b)
3 The number of distinct terms in the expansion of (x+y–z)16 is(a) 136 (b) 153 (c) 16 (d) 17Solution :Apply n+r–1Cr–1 to get number of terms 16+3–1C3–1 = 18C2 = 153Ans (b)
4 If I is the integral part of (2+ 3 )n and f is the fractional part. Then (I+f) (1–f) is equal to(a) 0 (b) 1 (c) n (d) None of theseSolution :Let (2+ 3 )n = I+f..........................(1) (0 f<1)
and (2– 3 )n = F.............................(2) (0<F<1)(1) + (2) gives (adding 0 <f+F<2)2(nC0.2
n+nC2.2n–2( 3 )2+........) = I+f+F
I+f+F is an even integerI+F is an integer
+
265
f+F = 1 ( 0 <f+F<2)F = 1–f
(I+F)(1–f) = (2+ 3 )n (2– 3 )n = (4–3)n = 1Ans (b)
5 If the middle term of (1+x)2n (n N) is the greatest term of the expansion, then the interval inwhich x lies is
(a) n2n,
n1n
(b) n1n,
n1–n
(c) n1n,
1nn
(d) None of these
Solution :Tn Tn+1 & Tn+1 Tn+22nCn–1 x
n–1 2nCn.xn & 2nCn.x
n 2nCn+1xn+1
nn2
1–nn2
CC
x &1n
n2n
n2
CC
x.
x 1n–n2n
& x n–n21n
x n1n,
1nn
Ans (c)6 If C0, C1, C2, ...................Cn are the binomial coefficients in expansion of (1+x)n, n being even, then
C0+(C0+C1)+(C0+C1+C2)+..........+(Co+C1+......+Cn–1) is equal to(a) n.2n (b) n.2n–1 (c) n.2n–2 (d) n.2n–3
SolutionC0+(C0+C1)+.........+(C0+C1+.......Cn–2)+(C0+C1+..........Cn–1)
= (Cn)+(Cn+Cn–1)+.......+(C0+C1+.........+Cn–2)+(C0+C1+.....+Cn–1)
= 2n+2n+2n+... 2n
times (Adding the terms equidistant from the begining and the end)
= 2n
. 2n = n.2n–1
Ans (b)
7 The number of terms in the expansion of 100
33 1
x1x is
(a) 201 (b) 200 (c) 300 (d) 100cSolution
100
33 1
x1x = C0+C1 3
3
x1x +C2
2
33
x1x +............+C100
100
33
x1x
gives terms of x3, x6, ..............x300, 3x1
, 6x1
,.......... 300x1
and a constant term
3
266
201 termsAns (a)
Exercise1 If Cr stands for nCr, then the sum of the series
2!n
!2n!
2n
(C02–2C1
2+3C22–..........+(–1)n (n+1)Cn
2) when n is an even positive integer, is equal
to(a) (–1)n/2 (n+2) (b) (–1)n (n+1) (c) (–1)n/2 (n+1) (d) None of these
2 030
1030
– 130
1130
+.........+ 2030
3030
is equal to
(a) 30C11 (b) 60C10 (c) 30C10 (d) 65C55
3 If r = 0, 1, 2, ........10, let Ar, Br and Cr denote, respectively, the coefficent of xr in the expansions
of (1+x)10, (1+x)20 and (1+x)30. Then 10
1rrA (B10 Br – C10Ar) is equal to
(a) B10 – C10 (b) A10 (210B –C10A10)
(c) 0 (d) C10–B10
4 Value of 2k 0n
kn
–2k–11n
1–k1–n
+ 2k–22n
2–k2–n
– .......+(–1)k kn
0k–n
is
(a) kn
(b) 1–k1–n
(c) 1 (d) None of these
5n
0r
r)1(– nCr )temsmtoup.........
215
27
23
21
r4r3
r
r2
r
r
r
=
(a) )1–2(21–2
nmn
mn
(b)n–m2–2 nm
(c) 1 (d) None of these
6 If n2
0rra (x–2)r =
n2
0rrb (x–3)r and ak = 1 for all k n, then bn =
(a) nCn (b) 2n+1Cn+1 (c) 2n+1Cn (d) None of these7 If (1+x)n = C0+C1x+C2x
2+..........+Cnxn, then the sum of the products of the Ci’s taken two at a
time represented by jiCC (0 i<j n) is equal to
(a) 22n–1 (b) 2n – 2)!n(2)!n2(
(c) 22n–1 – 2)!n(2)!n2(
(d) None of these
267
8 Given sn = 1+q+q2+..........+qn and Sn = 1+ 21q
+ 2
21q
+ .........+n
21q
, q 1 then
n+1C1+n+1C2s1+
n+1C3s2+.........+n+1Cn+1sn =
(a) 2n Sn (b) Sn (c) nn
2S
(d) None of these
9 nlim
n
0r rn
rn)3r(1
=
(a) e (b) e–1 (c) e+1 (d) e–2
10 The coefficient of x8 is the expansion of 28642
!8x
!6x
!4x
!2x1 is
(a) 3151
(b) 3152
(c) 1051
(d) 2101
11 Match the following :Column I Column II
(a) The sum of binomial coefficients of terms containingpower of x more than x20 in (1+x)41 is divisibile by (p) 239
(b) The sum of binomial coefficients of rational terms inthe expansion of (1+ 2 )42 is divisible by (q) 240
(c) If21
22
x1x
x1x = a0x
–42+a1x–41+a2x
–40+........+a82x40, (r) 241
then a0+a2+.....+a82 is divisible by(d) The sum of binomial coefficients of positive real terms
in the expansion of (1+ix)42 (x>0) is divisible by (s) 238
12 Read the passage and answer the questions that follow:An equation a0+a1x+a2x
2+.......+a99x99+x100 = 0 has roots 99C0,
99C1 ,99C2,........
99C99.(i) The value of a99 is(a) 298 (b) 299 (c) –299 (d) None of these(ii) The value of a98 is
(a)2
C–2 99198198
(b)2
C2 99198198
(c) 499999 C–2 (d) None of these
(iii) The value of (99C0)2+(99C1)
2+............+(99C99)2 is
(a) 98a2 – 299a (b) 2
98a – 299a (c) 2
99a –2a98 (d) None of theseAnswers
1. a 2. c 3. d 4. a 5. a 6. b7. c 8. a 9. d 10. a 11. a p, b r, c r, d, d q12. (i) c (ii) a (iii) c
TRIGONOMETRY - ITrigonometric Functions
1. a. Measurement of angles. There are three systems of measurement of angles.i. Sexagesimal system
Here 1 right angle = 90° (degrees)1° = 60' (minutes)1' = 60" (seconds)
ii. Centrimal systemHere 1 right angle = 100g (grades)1g = 100' (minutes)1' = 100" (seconds)
iii. Circular system. Here an angle is measured in radians. One radian correspondsto the angle subtended by arc of length ‘r’ at the centre of the circle of radiusr. It is a constant quantity and does not depend upon the radius of the circle.
b. Relation between the three systems:c = 180° = 200g = 2rt s .
c. If is the angle subtended at the centre of a circle of radius ‘r’, by an arc of length
then r .
Note that here , r are in the same units and is always in radians.
CONVENTION FOR PERPENDICULAR AND BASE IN A RIGHT TRIANGLESide opposite to 90° is called hypotenuse and side opposite to angle considered for T--ratios is known as perpendicular and third remaining side is base.
2. T-RATIOS (or Trigonometrical functions)
,bptan,
hbcos,
hpsin
,pheccos
bhsec and
pbcot
‘p’ perpendicular; ‘b’ base and ‘h’ stands for hypotenuse.
3. DOMAINS AND RANGES OF TRIGONAL METRICAL FUNCTIONSFunction Domain Rangesinx R [–1,1]cosx R [–1,1]tanx R– Zn:2/1n2 R
268
cotx R– Zn:n Rsecx R– Zn:2/1n2 ,11,––cosecx R– Zn:n ,11,––
4. Signs of trigonometrical functions in different quadrants:i. I quadrant: All t-ratios are positive tii. II quadrant : sin and cosec are positive and all others are negative.iii. III quadrant: tan and cot are positive and all others are negative.iv. IV quadrant: cos and sec are positive and all others are negative.
5. Values of t-ratios of some standard angles:
6. a. Trigonometric functions of 2 n + , Zn will be same as ofi.e., sin (2 n + ) = sin , cos(2 n + ) = cos , etc.
b. Trigonometrical functions of – , for all values ofsin(– ) = –sin , cos(– ) = cos ,tan(– ) = –tancot(– ) = –cot , sec(– ) = sec ,cosec(– ) = –cosec
c. The values of t-ratios of any angle can be expressed in terms of an angle in firstquadrant.
Let A = 2.n where .
20,Zn Then
i. sin sin2
n if n is even
= +cos
ii. cos cos2
n if n is even
= -sin
iii. tan tan2
n if n is even
= cot , if n is odd
, if n is odd
, if n is odd
269
iv. cot cot2
n if n is even
= tan , if n is odd
v. sec sec2
n if n is even
= cosec , if n is odd
vi. cosec eccos2
n if n is even
= sec , if n is oddThe sign R.H.S. is decided from the quardrant in which A lies.
7. IDENTITIES
1. sin .cosec =1 or cosec = sin1
2. cos .sec =1 or sec = cos1
3. tan .cot =1 or cot = tan1
4. tan = cossin
5. cot = sincos
6. sin2 +cos2 =1 or sin2 =1 – cos2
or cos2 =1 – sin2
7. sec2 –tan2 =1 or sec2 =1+ tan2
or tan2 = sec2 –18. cosec2 – cot2 =1 or cosec2 =1+ cot2
or cot2 = cosec2 –19. sin(A+B) = sinA cosB + cosA sinB10. sin(A–B) = sinA cosB – cosA sinB11. cos(A+B) = cosA cosB – sinA sinB12. cos(A–B) = cosA cosB + sinA sinB
13. tan(A+B) = BtanAtan–1BtanAtan
14. cot(A+B) = BcotAcot1–BcotAcot
270
15. tan(A–B) = BtanAtan1Btan–Atan
16. cot(A–B) = Acot–Bcot1BcotAcot
17. 2sinA cosB = sin(A+B) + sin(A–B)18. 2cosA sinB = sin(A+B) – sin(A–B)19. 2cosA cosB = cos(A+B) + cos(A–B)20. 2sinA sinB = cos(A–B) – cos(A+B)
21. sinC + sinD = 2sin 2D–Ccos
2DC
22. sinC – sinD = 2cos 2D–Csin
2DC
23. cosC + cosD = 2cos 2D–Ccos
2DC
24. cosC – cosD 2sin 2D–Csin
2DC
25. sin 2 = 2sin cos = tan1tan2
2
26. cos 2 = cos2 – sin2 = 2cos2 –1
= 1–2sin2 =tan1tan–1
2
2
27. 1+ cos 2 = 2cos2 or cos = 2cos1
28. 1– cos 2 = 2sin2 or sin = 22cos–1
29. tan = 2cos12sin
2sin2cos–1
= 2cos12cos–1
30. tan 2 = tan–1tan2
2
31. sin3 = 3sin – 4sin3
32. cos 3 = 4cos3 – 3cos
271
, –
33. tan3 =tan3–1
tan–tan32
3
34. sin2A – sin2B = sin(A+B).sin(A–B)= cos2B – cos2A
35. cos2A– sin2B = cos(A+B).cos(A–B)= cos B- sin A36. sin(A+B+C) = sinA cosB cosC + sinB cosA cosC
+ sinC cosA cosB – sinA sinB sinC= cosA cosB cosC [tanA + tanB +tanC – tanA tanB tanC]
37. cos(A+B+C) = cosA cosB cosC – sinA sinB cosC– sinA cosB sinC – cosA sinB sinC
= cosA cosB cosC [1–tanA tanB – tanB tanC – tanC tanA]
38. tan(A+B+C) = AtanCtan–CtanBtan–BtanAtan–1CtanBtanAtan–CtanBtanAtan
39. sin + sin + sin 2 +.......sin 1–n
=
2sin
2nsin
21–nsin
40. cos + cos +cos 2 +.......cos 1–n
=
2sin
2nsin
21–ncos
8. Some t-ration of 180,720,360,540,150, ,217,
2167,
2122
000
90,810,270,630 etc.
1. sin180 = 072cos4
1–5
2. cos360 = 054sin4
15
3. cos180 = 072sin4
5210
4. sin360 = 054cos4
52–10
272
2 2
5. sin750 =015cos
2213
6. tan750 = 015cot1–313
7. sin150 =075cos
221–3
8. tan150 = 2– 3 = 075cot131–3
9. tan00
2167cot1–2
2122
10. tan00
2122cot12
2167
11. cot 2346217
0
=0
2188tan1223
12. tan 23–4–6217
0
=0
2182cot1–22–3
13. sin90 = 081cos4
5–553
14. cos90 = 081cos4
5–553
15. sin270 = 063cos4
5–3–55
16. sin630 = 027cos4
5–355
17. sinbcos will always lie in the interval },ba,ba{– 2222 i.e. the maximum
and minimum value of a sinbcos is 2222 ba–ba repectively..
18. For 0 < < , minimum value of a eccosbsin is ab2
19. For 2/2/– , minumum value of a cos +b sec is ab2
273
20. For2
3or /20 , Minimum value of a tan + b cot is ab2
21. Periods of sinx, cosx, secx, cosecx, is 2 and period of tanx and cotx is .22. If a function is periodic with period , i.e. f(x+ ) = f(x), then period of thefunction
pqisx
qpf
23. If f(x) and g(x) are periodic functions with periods and repectively, then period of
the function f(x) g(x), f(x).g(x) or )x(g)x(f
is L.C.M of and
24. LCM of rational numbers is orsdenominati of HCFnumerators of LCM
25. Expression of sin(A/2) in terms of sinA
Asin12Acos
2Asin
2
and Asin–12Acos–
2Asin
2
so that Asin12Acos
2Asin ......(1)
and Asin–12Acos–
2Asin ......(2)
By addition and subtraction, we have
Asin–1Asin12Asin2 ......(3)
and Asin–1Asin12Acos2 ......(4)
The ambiguities of sign in relation (1) and (2) is determined by the followingdiagram.
274
S
P
X
V
R
X
Q
Y
sin(A/2) + cos(A/2) +sin(A/2) – cos(A/2) +
/4< A/2<3 /4
sin(A/2) + cos(A/2)+sin(A/2) – cos(A/2) –
5 /4<A/2 <7 /4
sin(A/2)+ cos(A/2) –
sin(A/2) – cos(A/2) + sin(A/2) – cos(A/2) –
sin(A/2) + cos(A/2) –
3 /4<A/2<5 /4 – /4<A/2< /4
1. i. If 1sin + 2sin +............ nsin = n, then
1sin = 2sin =............= nsin =1
ii. If 1cos + 2cos +........... ncos = n, then
1cos = 2cos =........... ncos = 1
2. i. sin + eccos = 2 sin =1ii. cos + sec = 1 cos =1
3. i. sin150 + cos150 = sin750 + cos750 = 23
ii. cos15 – sin15 = sin75 – cos75 = 21
4. i. tan 150 + cot 150 = tan 750 + cot 750 = 21
ii. cot 150 – tan 150 = tan 750 – cot 750 = 32
5. i. cos – 60cos 0 – –60cos 0 = 0
ii. cos + 120cos 0 + –120cos 0 = 0
iii. cos + 240cos 0 + –240cos 0 = 0
iv. sin – 60sin 0 + –60sin 0 = 0
v. sin + 120sin 0 – –120sin 0 = 0
vi. sin + 240sin 0 – –240sin 0 = 0
275
6. i. 45tan 0 –45tan 0 =1
ii. 45cot 0 –45cot 0 =17. i. If A+B = 450 then (1+ tanA) (1+ tanB) = 2
ii. If A+B = 1350 then (1– tanA) (1– tanB) = 2iii. If A+B = 450 then (1– cotA) (1– cotB) = 2iv. If A+B = 1350 then (1+ cotA) (1+ cotB) = 2
8. i. If cos x + cos y = a, sin x + sin y = b, thenab
2yxtan
ii. If cos x – cos y = a, sin x – sin y = b, thenba–
2yxtan
iii. If cos x – cos y = a, sin x + sin y = b, thenba
2y–xtan
iv. If cos x + cos y = a, sin x – sin y = b, thenab
2y–xtan
9. i. sin 60sin 0 –60sin 0 = 3sin41
ii. sin 120sin 0 –120sin 0 = 3sin41
10. i. cos 60cos 0 –60cos 0 = 3cos41
ii. cos 120cos 0 –120cos 0 = 3cos41
11. i. tan 60tan 0 –60tan 0 = 3tan
ii. tan 120tan 0 –120tan 0 = 3tan
12. i. cot 60cot 0 –60cot 0 = 3cot
ii. cot 120cot 0 –120cot 0 = 3cot
1 If cot +tan = m and cos1
– cos = n, then
(a) m(mn2)1/3–n(nm2)1/3 = 1 (b) m(m2n)1/3–n(mn2)1/3 = 1(c) n(mn2)1/3–m(nm2)1/3 = 1 (d) n(m2n)1/3–m(mn2)1/3 = 1SolutionGiven that
cot +tan = m1+tan2 = mtansec2 = mtan
276
PRACTICE QUESTIONS
alsocoscos–1 2
= n
cossin2
= n
tan2 = n secsquaringtan4 = n2 sec2
= n2m tantan3 = n2mtan = (n2m)1/3
sec2 = m(n2m)1/3
sec2 –tan2 = 1m(n2m)1/3–(n2m)2/3 = 1m(n2m)1/3–(n4m2)1/3 = 1m(n2m)1/3–n(nm2)1/3 = 1
Correct option is ‘a’
2 If cos2 = 2cos–31–2cos3
, then tantan
equals
(a) 1 (b) –1 (c) 2 (d) – 2Solution
cos2 = 2cos–31–2cos3
2
2
tan1tan–1
= 2
2
2
2
tan1tan–1–3
1–tan1
)tan–1(3
= tan1–tan33tan–tan3–3
22
22
= 2
2
tan42tan4–2
2
2
tan1tan–1
= 2
2
tan21tan2–1
By componendo and dividendo we get,
2tan2–2
= 2tan4–2
277
1–
2 = 2
2
tantan
tantan
= 2
correct options are ‘c’ & ‘d’3 cot15°+cot75°+cot135°–cose30° is equal to
(a) –1 (b) 0 (c) 1 (d) None of theseSolution
cot15°+cot75°+cot135°–cosec30°= cot15°+tan15°–tan45°–cosec30°
= 30sin2
– 1 – 2
= 4 – 3= 1
correct option is c4 If sin(y+z–x), sin(z+x–y), and sin(x+y–z) are in A.P then tanx, tany and tanz are in
(a) A.P (b) GP (c) HP (d) None of theseSolution
sin(y+z–x), sin(z+x–y) and sin(x+y–z) are in APsin(z+x–y)–sin(y+z–x) = sin(x+y–z)–sin(z+x–y)
2coszsin(x–y) = 2cosxsin(y–z)sinx cosz cosy–cosx siny cosz = cosx siny cosz–cosx cosysinzDivide by cosx cosy cosz we gettanx–tany = tany–tanztanx+tany = 2tany
tanx,tany,tanz are in A.Poption a is correct
5 If + = 90, then the maximum value of sin sin is(a) 1 (b) 1/2 (c) 3/2 (d) None of theseSolution
+ = 90sin sin = sin sin(90– )= sin cos
= 22
sin cos
= 22sin
We knew that – 1 sin2 1 or 21–
22sin
21
278
sin +15° = 15°
cos15° sin15°cos15°
= sin15° cos15°
2
30sin2
=
[cot15°+tan
]
maximum value of sin sin = 21
correct option is ‘b’EXERCISE
1. The least value of secA+secB+secC in an acute angle triangle is
a. 3 b. 6 c. 2 d. none of these
2. The sumof maximum and minimum values of cos2 –6sin cos +3sin2 +2 is
a. 102 b. c. d. none of these
3. Let f( )=sin (sin +sin3 ). Then f( ) is
a. 0 only when 0 b. 0 only real
c. 0 for all real d. 0 only when 0
4. Let (0, 4 ) and t1=(tan )tan t2=(tan )cot t3=(cot )tan and t4=(cot )cot then
a. t1> t2 > t3 > t4 b. t4> t3 > t1 > t2 c. t3> t1 > t2 > t4 d. t2> t3 > t1 > t2
5. For a positive integer n, let
fn( )=tan 2 (1+sec ) (1+sec2 ) (1+sec4 )......(1+sec2n ). Then
a. f2 16 =1 b. f3 32 =1 c. f4 64 =1 d. f5 128 =1
6. The maximum value of (cos 1)(cos 2) (cos 3)........(cos n) under the restrictions
0 1, 2 ... n 2 and (cot 1), (cot 2)......(cot n)=1 is
a. 2/n2
1b. n
2
1c. n2
1d. 1
7. If cos4 + , sin4 + are the roots of the equaton x2+b(2x+1)=0 and cos2 + ,sin2 are theroots of the equation x2+4x+2=0, then b is equal toa. 1 b. –1 c. 2 d. –2
8. If in ABC, tanA+tanB+tanC=6 and tanAtanB=2, then sin2A: sin2B:sin2C isa. 8:9:5 b. 8:5:9 c. 5:9:8 d. 5:8:9
9. If )1k(kAcosA3tan
279
a. k21–k
A3cosAcos 2
b. 1–kk2
AsinA3sin
c. 31k d. k>3
10. If (x–a) cos + ysin = (x–a) cos + ysin = a and tan 2 – tan 2 = 2b, then
a. y2 = 2ax – (1– b2)x2 b. tan 2 = x1
(y + bx)
c. y2 = 2ax – (1– a2)x2 d. tan 2 = x1
(y – bx)
11. PassageIncreasing product with angles are in GP
cos cos2 cos22 .......cos2n =
1–2 if,
21–
In12
if,21
n ifsin2
sin2
nn
nn
n
n
On the basis of above infromation, answer the following questions.
i. The value of 76cos
74cos
72cos is
a. 21–
b. 21
c. 41
d. 81
ii. If = 13 then the value of 6
1r
rcos is
a. 641
b. 641–
c. 321
d. – 81
iii. The value of sin14 sin 143
sin 145
sin 147
sin 149
sin 1411
sin 1413
is
a. 1 b. 81
c. 321
d. 641
iv. The value of sin18 sin 185
sin 187
is
280
a. 161
b. 81
c. 81
d. –1
v. The value of 64 3 sin 48 cos 48 cos 24 cos12 cos 6 is
a. 8 b. 6 c. 4 d. –112. Matrix Match Type
Column I Column II(a) In triangle ABC, 3sinA+4cosB=6 and
3cosA+4sinB = 1 then c can be (p) 60°(b) In any triangle if(sinA+sinB+sinC)(sinA+sinB–sinC)
= 3sinAsinB then the angle c (q) 30°(c) If 8sinxcos5x–8sin5xcosx = 1 then x = (r) 165°(d) ‘O’ is the centre of the inscribed circle in a 30°–60°–90°
triangle ABC with right angled at c. If the circle is tangents to A B at D then the angle COD is (s) 75°
a q ; b p ; c s ; d r13. Assertion and Reason type questions
A : Both A and R individually true and R is the correct explanation of AB : Both A and R individually true and R is not the correct explanation of AC : A is true but R is falseD : A is false but R is true
If A+B+C = , then(i) Assertion (A) : cos2A+cos2B+cos2C has its minimum value 3/4
Reason (R) : Maximum value of cosAcosBcosC is 1/8(a) A (b) B (c) C (d) D(ii) Assertion (A) : sin /18 is a roots of 8x3–6x+1 = 0
Reason (R) : For any R, sin3 = 3sin –4sin3
(a) A (b) B (c) C (d) D(iii) In any ABC
Assertion (A) : n 2Ccot
2Bcot
2Acot = 2
Ccotn2Bcotn
2Acotn
Reason (R) : n 3231 = n1+ n 3 + n 32
Answers1. b 2. c 3. c 4. b 5. a, b, c, d6. a 7. b, c 8. b, d 9. a, b, c, d10. a, b, d 11. (i) d(ii) a (iii). d(iv). b (v). b 12.a q ; b p ; c s ; d r 13. (i) a (ii) a (iii) b
281
TRIGONOMETRY - IITrigonometric Functions
1 sin47° + sin61°–sin11°–sin25° is equal to(a) sin36° (b) cos36° (c) sin7° (d) cos7°Solution
sin47+sin61°–sin11°–sin25°(2sin54°cos7°) – (2sin18°cos7°)= 2cos7°(sin54°–sin18°)= 2cos7°(cos36°–sin18°)
= 2cos7° 41–5–
415
= 2cos7° 21
= cos 7°correct option is ‘d’
2 If cosx
=
32–cos
y =
32cos
z , then x+y+z is equal to
(a) 1 (b) 0 (c) –1 (d) none of theseSolution
We have
cosx
=
32–cos
y =
32cos
z
we know that each ratio is equal to rdenominato of sumNumerator of sum
cosx
=
32–cos
y =
32cos
z =
32cos
32–coscos
zyx
32cos
32–coscos
=cos +2cos cos 32
(cos(A+B)+cos(A–B)=2cosAcosB)
= cos +2cos 21–
282
now,
= 0 x+y+z = 0
correct option is ‘b’
3(a) The value of 8cos1 8
3cos18
5cos18
7cos1 is
(a) 21
(b) 8cos (c) 8
1(d)
2221
Solution
87cos = cos 8
– = – cos 8
83cos = cos 8
–2 = sin 8
85cos = cos 82 = – sin 8
= 8cos1 8
sin18
sin–18
cos–1
= 8cos–1 2
8sin–1 2
= sin2
8 cos2
8
= 4
8cos
8sin2
2
= 44
sin2
= 81
correct option is ‘c’3 If x1,x2...xn are in AP whose common difference is , then the value of
sin (secx1secx2+secx2secx3+.......+secxn–1secxn) is
(a)n1 xcosxcos
)1–nsin((b)
n1 xcosxcosnsin
(c) sin(n–1) cosx1cosxn (d) sinn cosx1cosxnSolutionWe havesin secx1secx2+sin secx2secx3+............+sin secxn–1secxn
= 21
12
xcosxcos)x–xsin(
+ 32
23
xcosxcos)x–xsin(
+............+ n1–n
1–nn
xcosxcos)x–xsin(
283
=21
1212
xcosxcosxsinxcos–xcosxsin
+32
2323
xcosxcosxsinxcos–xcosxsin
+.............
+n1–n
1–nn1–nn
xcosxcosxsinxcos–xcosxsin
= tanx2–tanx1+tanx3–tanx2+........+tanxn–tanxn–1= tanxn–tanx1
= 1n
1n1n
xcosxcosxsinxcos–xcosxsin
= 1n
1n
xcosxcos)x–xsin(
= 1n xcosxcos
)1–nsin(( xn = x1+(n–1) xn–x1 = (n–1) )
correct option is ‘a’
4 Let ƒ(n) = 2cosnx n N, then ƒ(1) ƒ(n+1) –ƒ(n) is equal to(a) f(n+3) (b) f(n+2) (c) f(n+1)f(2) (d) f(n+2)f(2)Solution
We have f(n) = 2cosnx n Nf(1) = 2cosxf(n+1) = 2cos(n+1)xf(1) f(n+1) = 4cosx cos(n+1)xf(1)f(n+1) – f(n) = 4cosx cos(n+1)x–2cosnx= 2[2cosxcos(n+1)x–cosnx]= 2[cos(n+2)x+cosnx–cosnx]= 2cos(n+2)x= f(n+2)
option ‘b’ is correct5 The ratio of the greatest value of 2–cosx+sin2x to its least value is
(a) 1/4 (b) 9/4 (c) 13/4 (d) None of these
Solution2–cosx+sin2x= 2–cosx+1–cos2x= –(cos2x+cosx)+3
= – 2
21xcos + 4
1 + 3
= – 2
21xcos + 4
13
284
= 413
–2
21xcos
Maximum value 413
occurs at cosx = – 21
Minimum value occurs at cosx = 1
413
– 2
211
413
– 49
= 44
= 1
The ratio of greatest to the least is 413
option ‘c’ is correct6 If sin(120– ) = sin(120– ), 0< , < , then find the relation between and .
SolutionIf sinA = sinBThen A = B or A = –BHere, sin(120– ) = sin(120– )
120– = 120– or 120– = –(120– ) = or 120– = 60+ + = 60
7 If x,y,z are in AP, then xcos–zcoszsin–xsin
is equal to
(a) tany (b) coty (c) siny (d) cosySolution
We have x,y,z are in APx+z = 2y
Here, xcos–zcoszsin–xsin
=
2z–xsin
2zxsin2
2z–xsin
2zxcos2
285
° °°° ° °
° ° °
= ysinycos
= cotycorrect option is ‘b’
1. It cosbabcosacos , then 2tan equals
a. 2
tanbaba
b. 2
cosbaba
c. 2
sinbaba
d. none
2. If acos2 + bsin2 = c has and as its solution, then tan + tan equals
a. cba2
b. acb2
c. bac2
d. none
3. The value of 00 10cos3
10sin1
equals
a. 1 b. 4 c. 2 d. 0
4. 87cos
85cos
83cos
8cos 4444 equals
a. 21
b. 1 c. 23
d. 2
5. n5sin
n3sin
nsin ............ n terms equals
a. 1 b. 0 c. 2n
d. none
6. If 14sin114sin1x then one of the values of x is
a. – tan b. cot c. 4tan d. 4cot
7. The value of cos120 + cos840 + cos1560 + cos1320 is
a. 21
b. 1 c. – 21
d. 81
286
PRACTICE QUESTIONS
8. sin60 – sin660 + sin780 – sin 420 is
a. –1 b. – 21
c. 21
d. 1
9. 6532cos
6516cos
658cos
654cos
652cos
65cos equals
a. 81
b. 161
c. 321
d. 641
10. sin360 sin720 sin1080 sin1440 equals
a. 165
b. 163
c. 161
d. none
11. Passage
If A, B, C be the angles of a triangle, then
a) sin2A = 4sinA sinB sinC
b) cos2A = 1 – 4cosA cosB cosC
c) sinA = 2Ccos2
Bcos2Acos4
d) cosA = –1 + 2Csin2
Bsin2Asin4
e) tanA = tanA tanB tanC ie., S1 = S3
f ) 2Btan2
Atan = 1 or 2Acot = 2
Ccot2Bcot2
Acot
Answer the following questions based upon above passage.
i) In a triangle ABC CsinBsinAsinCsinBsinAsin
equals
a. 2Bcot2
Atan b. 2Btan2
Acot c. 2Bcot2
Acot d. 2Btan2
Atan
ii) sin2A + sin2B + sin2C – 2cosAcosBcosC equals
a. 1 b. 2 c. 3 d. 4
iii) I n a ABC, whose angles are acute and positive such that A+B+C= and
2Ccot2
Bcot2Acot = K then
a. K < 3 b. K < 33 c. K > 33 d. none
287
iv) tanA, tanB, tanC are the roots of the cubic equation x3 – 7x2+11x –7 = 0 then A + B + C equals
a. 2 b. c. 0 d. none
12. Matchning type question :-Column I Column II
a) sin(B+C–A) + sin(C+A–B) + p. 1– 2Ccos
2Bcos
2Acos2
sin(A+B–C)=
b) 2Csin
2Bsin
2Asin 222 = q. 1
c) If 2Ctan
2Btan
2Atan 222 = K r. 4sinA sinB sinC
then K >Assertion and reason type questionsA : Both A and R are individually true and R is the correct explaination of AB : Both A and R are individually true and R is not the correct explaination of AC : A is true but R is falseD : A is false but R is true
13. Assertion (A) : tan + 2tan2 + 4tan4 + 8tan8 + 16cot16 = cotReason (R) : cot – tan = 2cot2a. A b. B c. C d. D
14. Assertion (A) : 22
)yx(xy4sec x = y
Reason (R) : sec > 1a. A b. B c. C d. D
15. Assertion (A) : If A, B, C, D be the angles of a cyclic quadrilateral thencosA + cosB + cosC + cosD = 0Reason (R) : sinA + sinB + sinC + sinD = 0a. A b. B c. C d. D
Answers :-1. a 2. b, c 3. b 4. c 5. b6. a,b,c,d 7. c 8. b 9. d 10. a11. i-c, ii-b, iii-c, iv-b 12. a-r, b-p, c-q 13. a14. a 15. c
288
TRIGONOMETRY - IIITrigonometric Functions - Problem Solving
Trigonometrical ratios and Identities
1 An angle is positive, if it is measured in anti clockwise direction and is negative if it is measuredin clock wise direction
2 Angle in radian = circletheofRadiusarctheoflength
i.e; = r3 System of Measurement of Angles
Sexagesimal system (D) Centesimal system (G) Circular system (C)
1 right angle = 90° (90 degrees) 1 right angle = 100g (100 grades) 1 right angle = 2 radians
1° = 601 (60 minutes) 1g = 1001 (100 minutes) 180° = 11 = 6011(60seconds) 11 = 10011(100 seconds)
Note : 90D
= 100G
=C2
OR (see the graph below)
4 Basic trigonometrical identities(i) sin2 + cos2 = 1
|sin | 1 and |cos | 1– 1 sin 1 and – 1 cos 1
(ii) sec2 – tan2 = 1|sec | 1 sec –1 or sec 1
tan may take any real value(iii) cosec2 –cot2 = 1
|cosec | 1 cosec –1 or cosec 1
cot may take any real value.5 Sign of Trigonometrical Ratio : To find the sign of a trigonometrical ratio, remember the sentence
Add Sugar To CoffeeA S T C1stquadrant 2nd quadrant 3rd quadrant 4th quadrant.Where A stands for all ratios are positive in 1st quadrant
S stands for sin & its reciprocal are positive in 2nd quadrantT stands for tan & its reciprocal are positive in 3rd quadrantC stands for cos & its reciprocal are positive in 4rd quadrant
Two rightangles
200g180°
c
289
6 Domain & Range
Function Domain Range
sin R [–1,1]cos R [–1,1]
tan R– Zn,2
)1n2( R
cot R– Zn,n R
cosec R– Zn,n R–(–1,1)
sec R– Zn,2
)1n2( R–(–1,1)
7 Trigonometric ratios in terms of each of the other
sin cos tan
sin sin 2sin–1 2sin–1sin
cos 2sin–1 coscos
cos–1 2
tan 2tan1tan
2tan11
tan
cosec eccos1
eccos1–eccos 2
1–eccos1
2
secsec
1–sec2
sec1
1–sec2
cot 2cot11
2cot1cot
cot1
8 Sum and Difference formula.(i) sin (A B) = sin A cos B cosA sin B(ii) cos (A B) = cos A cos B sin A sin B
(iii) tan (A B) = BtanAtan1BtanAtan
(iv) tan ( 4 A) = Atan1Atan1
sd
290
(v) cot (A B) = AcotBcot
1BcotAcot
(vi) sin(A+B)sin(A–B) = sin2A – sin2 B = cos2B–cos2A(vii) cos(A+B)cos(A–B) = cos2A – sin2 B = cos2B–sin2A
9 Multiple and half angles
(i) sin2 = 2tan1
tan2cossin2
Also, 2sin1 = |cos sin |
(ii) cos2 =
2
2
2
2
22
tan1tan–1
sin2–11–cos2
sin–cos
Also 1+cos2 = 2cos2
1–cos2 = 2sin2
(iii) tan2 = 2tan–1tan2
Also 2sin2cos–1
= tan
2cos12cos–1
= tan2
(iv) sin3 = 3
sinsin–3
sin4
sin4–sin3 3
or sin3 = 43sin–sin3
(v) cos3 = 3
coscos–3
cos4
cos3–cos4 3
or cos3 = 4cos33cos
291
(vi) tan3 =
3tantan–
3tan
tan3–1tan–tan3
2
3
10 Transformation formulae
(i) sin C + sin D = 2sin 2DC
cos 2D–C
sin C – sin D = 2cos 2DC
sin 2D–C
cos C + cos D = 2cos 2DC
cos 2D–C
cos C – cos D = –2sin 2DC
sin 2D–C
(ii) 2 sin A cos B = sin(A+B) + sin (A–B)2 cos A sin B = sin(A+B) – sin (A–B)2 cos A cos B = cos(A+B) + cos (A–B)2 sin A sin B = cos(A–B) – cos (A+B)
(ii) tan A tan B = BcosAcos)BAsin(
cot A cot B = BsinAsin)ABsin(
(iv) cot A – tan A = 2cot2A
cotA+tan A = 2cosec2A = AcosAsin1
(v) 1 tan A tan B = )BAcos(
(vi) cosA sinA = 2 sin A4 = 2 cos A
411 Three angles
sin(A+B+C) = cosA cosB cosC (tanA+tanB+tanC–tanAtanBtanC)cos(A+B+C) = cosAcosBcosC(1–tanAtanB–tanBtanC–tanCtanA)
tan(A+B+C) = AtanCtan–CtanBtan–BtanAtan–1CtanBtanAtan–CtanBtanAtan
292
12 Trigonometrical series
(i) sinA+sin(A+d)+sin(A+2d)+...............+sin(A+(n–1)d) =
2dsin
2ndsin
2d)1–n(Asin
(ii) cosA+cos(A+d)+cos(A+2d)+..........+cos(A+(n–1)d) =
2dsin
2ndsin
2d)1–n(Acos
(iii) cosA. cos2A. cos22A...........................cos 2n–1 A = Asin2
)A2sin(n
n
(iv) (2cos –1) (2cos2 –1) (2cos22 –1)..........(2cos2n–1 –1) = 1cos212cos2 n
; n N
(v) tan +2tan2 +22tan22 +23tan23 +............+2n tan2n +2n+1cot2n+1 = cot ; n N
(vi) n2cos22........222 = 2 cos , n N where there are n square root signs
on left hand side.
13 Greatest and least value of asin bcos is 22 ba and – 22 ba respectively
i.e. – 22 ba asin bcos 22 baAlso sin2 + cosec2 2
cos2 + sec2 2tan2 + cot2 2
14 Method of componendo and dividendo.
If qp
= ba
then by componendo and dividendo we can write qpq–p
= bab–a
or q–pqp
= b–aba
PeriodicityAll the six trigonometric functions are periodic sin , cos , cosec , sec are periodic withperiod 2 where tan and cot are period c with period
293
Solved Examples1 If ƒ( )= sin4 +cos4 +1, then the range of ƒ( ) is
(a) 2,23
(b) 23,1 (c) [1 ,2] (d) None of these
Solution :ƒ( ) = (sin2 +cos2 )2 –2sin2 cos2 +1
= 2 – 2
2sin2
ƒ( ) min = 2 – 21
(1) = 23
and ƒ( ) max = 2– 21
(0) = 2
Ans : (a)2 If 4n = , then the value of
cot .cot2 .cot3 ..............cot(2n–1) is(a) 1 (b) –1 (c) (d) None of theseSolution :
cot .cot(2n–1) =cot .cot(2n – )=cot .cot –2 =cot .tan =1
Product of terms equidistant from the beginning and end is 1
The middle term is cotn = cot 4 =1
Given expression = 1.1.1......................n times = 1Ans : (a)
3 If tan 2 and tan 2 are roots of the equation 8x2–26x+15 = 0 then cos( + ) =
(a) – 725627
(b) 725627
(c) 1 (d) None of these
Solution :
tan 22 =
2tan
2tan–1
2tan
2tan
=
815–1
826
= 726–
cos( + ) =
2tan1
2tan–1
2
2
=
496761
49676–1
= 725627–
Ans (a)
294
4 If cos(x–y), cosx, cos(x+y) are in H.P, then 2ysec.xcos equals
(a) 1 (b) 2 (c) 2 (d) None of theseSolution :
cosx = )y–xcos()yxcos()yxcos()y–xcos(2
cosx = ycosxcos2)ysin–x(cos2 22
2cos2x cosy = 2cos2x–2sin2y2sin2y = 2cos2x(1–cosy)
2(4sin2
2y
cos2
2y
) =2cos2x. (2sin2
2y
)
2ycos
xcos2
2
= 2 cosx.sec 2y
= 2
Ans (c)
5 If tan 9 , x, tan 1815
are in A.P and tan 9 , y, tan 187
are in A.P, then
(a) 2x = y (b) x = y (c) x = 2y (d) None of theseSolution :2x = tan20° +tan50° 2y = tan20°+tan70°
2x = 50cos.20cos50sin.20cos50cos.20sin
2y = 70cos.20cos70sin.20cos70cos.20sin
2x = 20cos.50cos70sin
2y = 20cos.20sin90sin
2x = 20cos.50cos20cos
2x = 50cos1
= 40sin1
.......................(1) 2y = 40sin2
.....................(2)
From (1) and (2)2x = yAns : (a)
6 The value of sin n + sin n3
+ sin n5
+..............n terms is
(a) 1 (b) 0 (c) 2 (d) None of these
295
Solution :
Given Series =
n2sin
2n2.)1–n(
n.2
sinn2
nsin
=
n2sin
n–
nsin
nsin
=
n2sin
sin.n
sin = 0
Ans : (b)
71–n
1r
2
nrcos =
(a) 2n
(b) 21–
2n
(c) 1–2n
(d) None of these
Solution :1–n
1r
2
nrcos =
1—n
1r 2r2cos1
= 21 1–n
1r1 + 2
1 1–n
1r
rcos
= 21
21–n
n)1–n(2cos..........
n4cos
n2cos
= 21
21–n
.
nsin
2n
)1–n(2n
2
cos2n2sin
= 21
21–n
.
nsin
cos.n
sin = 2
1–2
1–n = 2
n – 1
Ans : (c)
1 Let 4,0 and t1 = (tan )tan , t2 = (tan )cot , t3 = (cot )tan and t4 = (cot )cot , then
(a) t1>t2>t3>t4 (b) t4>t3>t1>t2 (c) t3>t1>t2>t4 (d) t2>t3>t1>t4
2 n
296
(c)
)()( /n
PRACTICE QUESTIONS
2* For a positive integer n, let ƒn( ) = 2
tan
(1+sec )(1+sec2 )(1+sec22 )...........(1+sec2n ), then
(a) ƒ2 16 =1 (b) ƒ3 32 =1 (c) ƒ4 64 =1 (d) ƒ5 128 =1
3 Two rays are drawn through a point A at an angle 30°. A point B is taken on one of them at a distance‘a’ from the point A. A perpendicular is drawn from the point B to the other ray, and anotherperpendicular is drawn from its foot to A B to meet AB at another point from where the similarprocess is repeated indefinitely. The length of the resulting infinite polygonal line is
30°
a
A B1B3
B
B4
B2
(a) a(2+ 3 ) (b) a(2– 3 ) (c) a (d) None of these4 If cos6 +sin6 +ksin22 =1(0< < /2), then k is
(a) 3/4 (b) 1/4 (c) 1/3 (d) 1/85 In an acute angled triangle ABC, the least value of secA+secB+secC is
(a) 6 (b) 8 (c) 3 (d) none of these
6 The sum to infinite tems of the series cos 3 + 21
cos 32
+ 31
cos 33
+................ is
(a) 0 (b) 2 (c) 1 (d) 6
7 If = 72
, then the value of tan tan2 +tan2 tan4 +tan4 tan is
(a) 0 (b) –7 (c) 413
(d) 2
8 If sin(y+z–x), sin (z+x–y) sin(x+y–z) be is A.P., then tanx, tany, tanz are in(a) A.P. (b) G.P (c) H.P. (d) None of these
9* 16cot76cot16cot76cot3
=
(a) tan16° (b) cot76° (c) tan46° (d) cot44°
10 If x cos +ysin = 2a, xcos +ysin = 2a and 2sin 2 sin 2 =1, then
(a) cos +cos = 22 yxax2
(b) cos cos = 22
22
yxy–a2
297
(c) y2=4a(a–x) (d) cos +cos =2cos cos11 Match the following :-
Column I Column II(a) The maximum value of y = cos(2A+ )+cos(2B+ ) (p) 2sin(A+B)
where A&B are constants is(b) The maximum value of y = cos2 A+cos2 B where
A+B is a constant & A,B 2,0 is (q) 2sec(A+B)
(c) The minimum value of y = sec2A+sec2B where
A+Bis a constant & A,B 4,0 is (r) 2cos(A+B)
(d) The minimum value of (s) 2cos(A–B)
y = )BA(2cos2–cottan where A, B are
constants and 2,0 is
12 Let A0 A1 A2 A3 A4 A5 be a regular hexagon inscribed in a circle of unit radius. Then the productof the lengths of the line segments A0 A1, A0A2 and A0 A4 is
(a) 43
(b) 33 (c) 3 (d)2
33
13 If A, B, C, D are the smallest positive angles in ascending order of magnitude which have theirsines equal to the positive quantity k, then the value of
4sin 2A
+3sin 2B
+2sin 2C
+sin 2D
is
(a) k–12 (b) k12 (c) k2 (d) None of these14 Read the paragraph and answer the following questions .
If , , , are the solutions of the equation tan 4 = 3tan3 , no two of which have
equal tangents, then(i) The value of tan +tan +tan +tan is
(a) 1/3 (b) 8/3 (c) –8/3 (d) 0(ii) The value of tan tan tan tan is
(a) –1/3 (b) –2 (c) 0 (d) None of these
(iii) The value of tan1
+ tan1
+ tan1
+ tan1
is
(a) –8 (b) 8 (c) 2/3 (d) 1/3
298
15* Which of the following is / are correct ?
(a) )x(sinloge)x(tan > )x(sinloge)x(cot , x (0, /4)
(b) )ecx(cosloge4 < )ecx(cosloge5 , x (0, /2)
(c))x(cosloge
21
< )x(cosloge
31
, x (0, /2)
(d) )x(tanelog2 < )x(sinloge2 , x (0, /2)
'Note : Questions with * have more than one correct option'
Answers1. b 2. a, b, c 3. a 4. a 5. a 6. a 7. b8. a 9. c, d 10. c 11. a s ; b r ; c q ; d p 12. c13. b 14. (i) d (ii) a (iii) 8 15. a, b, c, d
299
TRIGONOMETRY - ITrigonometric Functions - Problem Solving
1 Values of trigonometrical ratios of some particular angles
(i) sin 217 =
226–2–4
cos 217 =
22624
tan 217 = 1–22–3
cot 217 = 23 12
(ii) sin15° = cos75° = 221–3
cos15° = sin75° = 2213
tan15° = cot75° = 3–2cot15° = tan75° = 32
(iii) sin22 21
= 21
2–2
cos22 21
= 21
22
tan22 21
= 1–2
cot22 21
= 12
(iv) sin18° = cos72° = 4
1–5
cos18° = sin72° = 4
5210
sin36° = cos54° = 4
52–10
300
cos36° = sin54° = 4
15
(v) cos9° = 21
18sin–118sin1
(vi) cos27° = 21
36cos–136cos1
2 Conditional identitiesIf A, B, C are angles of a triangle (i.e. A+B+C= ) then
tanA+tanB+tanC= tanAtanBtanCcotAcotB+cotBcotC+cotCcotA = 1
tan 2A
tan 2B
+tan 2B
tan 2C
+tan 2C
tan 2A
=1
cot 2A
+cot 2B
+cot 2C
= cot 2A
cot 2B
cot 2C
sin2A+sin2B+sin2C = 4sinAsinBsinCcos2A+cos2B+cos2C = – 1 – 4cosAcosBcosC
sinA+sinB+sinC = 4cos 2A
cos 2B
cos 2C
cosA+cosB+cosC = 1 + 4sin 2A
sin 2B
sin 2C
3 Trigonometric ratios of sum of more than three angles.sin(A1+A2.................+An) = cosA1cosA2.............cosAn(S1–S3+S5–...............)cos(A1+A2.................+An) = cosA1cosA2.............cosAn(1–S2+S4–S6+..............)
tan(A1+A2.................+An) = .....S–SS–1.......–SS–S
642
531
where S1 = tanA1 = sum of tangents of angles
S2 = tanA1tanA2 = sum of tangents taken two at a time etc.In particular, if A1=A2=..................An= A, then
S1 = n tanA ; S2 = nC2tan2A ; S3=nC3tan3A etc.
sin nA = cosnA (nC1tanA–nC3tan3A+nC5tan5A–.....................)cos nA = cosnA (1–nC2tan2A+nC4tan4A–.....................)
tan nA = ......................–AtanCAtanC–1...........–AtanCAtanC–AtanC
44
n22
n
55
n33
n1
n
301
Solved Examples
1 If ƒ(x) = xcot1xcot
and + = 45
, then the value of ƒ( ).ƒ( ) is
(a) 2 (b) – 21
(c) 21
(d) None of these
Solution :
ƒ( ).ƒ( ) = cot1cot
. cot1cot
= tan11
. tan11
= tan11
.–
4tan1
1 = tan1
1 ×
tan1tan–11
1
= tan11
2tan1
= 21
Ans : (c)2 The value of tan81°–tan63°–tan27°+tan9° equals
(a) 1 (b) 2 (c) 3 (d) 4Solution :(tan81°+tan9°)–(tan63°+tan27°)=(cot9°+tan9°)–(cot27°+tan27°)
= 9cos9sin1
– 27cos27sin1
= 18sin2
– 54sin2
= 1–542
– 1542
= 1–5
15–158 = 4
28= 4
Ans : (d).3 The number of integral values of k for which the equation 7cosx+5sinx = 2k+1 has a unique
solution is(a) 4 (b) 8 (c) 10 (d) 12Solution :
747
. cosx+ 745
sinx = 741k2
sin(x+ ) = 74
1k2
Now –174
1k21
302
2
1–74– k
21–74
–4.8 k 3.8 k = –4, –3, –2, –1, –0, 1, 2, 3
i.e. 8 values.Ans : (b)
4 If 21
ysinxsin
and 23
ycosxcos
where x,y 2,0 then tan (x+y) =
(a) 13 (b) 14 (c) 17 (d) 15Solution :
sin2x+cos2x=1
41
sin2y+ 49
cos2y= 1
cosy=22
3 and tany=
35
Also sinx = 24
5 and tanx =
335
tan(x+y)= ytan.xtan–1ytanxtan
=
35.
335–1
35
335
= 35–9
535
= 4
54× 3 = 15
Ans : (d)
5 If + = 2 and + = , then tan is equal to
(a) 2(tan +tan ) (b) tan +tan (c) tan +2tan (d) 2tan +tan Solution :
= –
tan = tan( – ) = tan.tan1tan–tan
303
tan = –
2tan.tan1
tan–tan
tan = 11tan–tan
2tan = tan –tantan = tan +2tan
Ans : (c)
67
1r
2
16rtan =
(a) 34 (b) 35 (c) 37 (d) None of theseSolution : Given series can be simplified to
16cot
16tan 22
+ 162cot
162tan 22
+ 163cot
163tan 22
+1
General pattern is tan2 +cot2
= 22
44
cossincossin
= 22
22
cossincossin2–1
= 2sin42 – 2
= 4cos–124
–2 = 4cos–18
– 2
2–
4cos–1
8 + 2–
2cos–1
8+ 2–
43cos–1
8 + 1
= 1–2
28 –2+8–2+
1228
–2+1
= 1–2
28 +
1228
–6+8+1
= 1–2
28–162816 +3 = 32+3 = 35
Ans : (b)
304
Exercise
1 If pn+1= )p1(21
n , then cos .......pppp–1
321
20
is equal to
(a) 1 (b) –1 (c) p0 (d)0p
1
2 If A, B, C are acute positive angles such that A+B+C= and cotA cotBcotC= k, then
(a) k 331
(b) k 331
(c) k< 91
(d) k> 31
3 If xy = 1, then xy–1yx
=
(a) xyz1
(b) xyz4
(c) xyz (d) None of these4 The value of cot16°cot44°+cot44°cot76°–cot76°cot16° is
(a) 3 (b) 31
(c) 31–
(d) –3
5 The number of solutions of tan(5 cos ) = cot(5 sin ) for in (0,2 ) is(a) 28 (b) 14 (c) 4 (d) 2
6 If cos x = tan y, cos y = tan z and cos z = tan x, then a value of sin x is equal to(a) 2cos18° (b) cos18°(c) sin18° (d) 2sin18°
7 Let n be an odd integer. If sinn = n
0rbr sinr , , then
(a) b0 = 1, b1=3 (b) b0 =0, b1=n(c) b0= –1 b1= n (d) b0=0, b1=n2–3n+3
8 If 2/–e < < 2 , which is larger, cos elog or )(cosloge
(a) cos elog (b) )(cosloge
(c) both are equal (d) None of these
91–n
1r nr2cos)r–n( for n 3 is ____________
(a) 2n
(b) n
(c) (n–3) (d) None of these
305
10* Match the following :-Column I Column II
(a) In an acute angled ABC, the least values (p) – = 2
of Asec & Atan2 are and respectively, then
(b) In ABC, the least values of )2/A(eccos (q) – = 3
& )2/A(sec2 and & respectively then (r) – = 4
(c) In ABC, the least values of cosec 2A
cosec 2B
cosec 2C
(s) 3 –2 =0
& Aeccos 2 are & respectively, then (t) 2 –3 = 0
11 In any ABC, the minimum value of Asin–CsinBsin
Asin is
(a) 3 (b) 0 (c) 4 (d) None ofthese
12 If cos 7 , cos 73
, cos 75
, are the roots of the equation 8 x3–4x2–4x+1 = 0.
On the basis of above information, answer the following questions :-
(i) The value of sec 7 +sec 73
+sec 75
is
(a) 2 (b) 4 (c) 8 (d) None ofthese
(ii) The value of sin14 sin 143
sin 145
is
(a) 41
(b) 81
(c)47
(d)87
(iii) The value of cos14 cos 143
cos 145
is
(a) 41
(b) 81
(c)47
(d)87
(iv) The equation whose roots are tan2
7 , tan2
73
, &tan2
75
, is
(a) x3–35x2+7x–21=0 (b) x3–35x2+21x–7=0(c) x3–21x2+35x–7=0 (d) x3–21x2+7x–35=0
(v) the value of 3
1r
2tan7
1–r2 3
1r
2cot7
1–r2 is
(a) 15 (b) 105 (c) 21 (d) 147
306
13 If a = sin18 sin 185
sin 187
, and x is the solution of the equation y = 2[x]+2 and y = 3[x–2], then
a =
(a) [x] (b) x1
(c) 2[x] (d) [x]2
14 If tan , tan , tan are the roots of x3–px2–r =0, then the value of (1+tan2 )(1+tan2 ) (1+tan2 )is equal to(a) (p–r)2 (b) 1+(p–r)2 (c) 1–(p–r)2 (d) None ofthese
15 If tan is an integral solution of 4x2–16x+15<0 and cos is the slope of the bisector of theangle in the first quadrant between the x & y axes, then the value of sin( + ): sin( – ) isequal to(a) –1 (b) 0 (c) 1 (d) 2
'Note : Questions with * have more than one correct option'
Answers1. c 2. a 3. a 4. a 5. a 6. d 7. b8.a 9. a 10. a q,s ; b p,t ; c r 11. a12. (i) b (ii) b (iii) d (iv) c (v) b 13. b 14. b 15. c
307
TRIGONOMETRY - IITrigonometric Equations
1. TRIGONOMETRIC EQUATION. An equation involving one or more trigonometrical ratiosof unknown angles is called a trigonometrical equation.
2. SOLUTION OF TRIGONOMETRIC EQUATION. A value of the unknown angle whichsatisfies the given equation is called a solution of the equation.
(a) Principal solution. The smallest numerical value, positive or negative for the angleof a trigonometrical equation is called its principal solution. If a positive angle as well as anegative angle of smallest value (and equal) are available, then we take the positive valueas the principal values.
(i) The principal solution of sin 1k,k is the value of in the interval 2,
2–
which satisfy the equation.
(ii) The principal solution of cos 1k,k is the value of in the interval [0, ]which satisfy the equation.
(iii) The principal solution of tan k,k the value of in the interval 2,
2– ,
which satisfy the equation.
(b) General solution. Since all the trigonometric functions are many one onto, therefore, thereare infinite values of for which trigonometric functions have the same value. Therefore,all such possible values of for which trigonometric ratios are same is known as generalsolutions of the variable angle .
3. GENERAL SOLUTIONS OF TRIGONOMETRIC RATIOS(i) If sin = 0, then = n ,n Z (set of integers)(ii) If cos = 0, then = (2n+1) ,n Z(iii) If tan = 0, then = n ,n Z
(iv) If sin = sin then = n + (–1)n , n Z, 2,
2–
(v) If cos = cos , then = 2n , n Z, [0, ]
(vi) If tan = tan , then = n + , n Z 2,
2–
(vii) If sin = 1, then = 2n + 2 , n Z
(viii) If cos = 1, then = 2n n Z(ix) If sin2 = sin2 or cos2 = cos2 or tan2 = tan2 then = n ± , n Z
308
(x) If sin (a + b) = sin , then n +(–1)n ab–
. First of all we have to write the
general solution as such for a + b and then, find as a + b = n + (–1)n.
= a1
[n +(–1)n –b]
(xi) For n Z, sin n = 0 and cos n = (–l)n
sin(n + )=(–1)n sin
cos(n + ) = (–l)n cos
(xii) If n is an odd integer, then
sin 2n
(–l) 21–n cos 2
n
sin 2n
(–l) 21–n cos
sin)1(–2
ncos 21n
4. IMPORTANT POINTS TO REMEMBER
(i) For equations of the type sin = k or cos = k, one must check that |k| 1.
(ii) Avoid squaring the equation, if possible, because it may lead to extraneous solutions.
(iii) Do not cancel the common variable factor from the two sides of the equations which arein a product because we may loose some solutions.
(iv) The answer should not contain such values of which make any of the terms undefined orinfinite.
(v) Check that denominator is not zero at any stage while solving equations.
(vi) (a) If tan or sec is involved in the equation, should not be odd multiple of 2 .
(b) If cot or cosec is involved in the equation, should not be a multiple of or 0.
(vii) If two different trigonometric ratios, such as, tan and sec are involved then aftersolving we cannot apply the usual formulae for general solution, because periodicityof the functions are not same.
(viii) If L.H.S. of the given trigonometric equation is always less than or equal to k and RHS isalways greater than k, then no solution exists. If both the sides are equal to k for samevalue of , then solution exists and if they are equal for different value of , then solutiondoes not exist.
309
EXAMPLES
1 The general solution of the equation )rxsin()xrcos( 2 = 21
is
(a) 2m + 6 , m I (b) )1n(n)1m4(
2 m I
(c) )1n(n)1–m4(
2 ,m I (d) None of these
Solutionn
1r
2 )rxsin()xrcos( = 21
= n
1r
2 )rxsin()xrcos(2 = 1
= n
1r
22 ]x)r–rsin(–x)rr[sin( = 1
= n
1r]x)1–r(rsin–x)1r(r[sin = 1
= sin2x–sin0+sin6x–sin2x+sin12x–sin6x+.............+sin(n(n+1)x)–sin(n(n–1)x) = 1= sin[n(n+1)x]–sin0 = 1= sin(n(n+1)x) = 1
= sinn(n+1)x = sin 2
n(n+1)x = 2m2 , m I
x = )1n(n2m4
, m I
= )1n(n)1m4(
2 , m I
option ‘b’ is correct2 If [sinx] + [ 2 cosx] = –3, x [0,2 ] [ [.] denotes the greatest integer function], then x belongs
to
(a) 45, (b) 4
5, (c) 2,4
5(d) 2,
45
Solution[sinx] + [ 2 cosx] = –3
310
sin(n + ) = sin
[sinx] = –1 [ 2 cosx] = –2
– 1 sinx<0 and –2 2 cosx<–1
– 2 cosx< 21–
–1 sinx<0 and –1 cosx<– 21
( –1 cosx 1)
since both sinx and cosx are negative
i.e. x ( ,2 ) and x 45,
43
x 45,
option ‘a’ is correct3 The number of solutions of the equation sinx+2sin2x = 3+sin3x in the interval [0, ] is
(a) 0 (b) 1 (c) 2 (d) 3Solution
We havesinx+2sin2x = 3+sin3xsin3x–sinx–2sin2x+3 = 02cos2xsinx–4sinxcosx+3 = 02sinx(cos2x–2cosx)+3 = 02sinx(2cos2x–1–2cosx)+3 = 0sinx(4cos2x–4cosx–2)+3 = 0sinx{(2cosx–1)2–3}+3 = 0sinx(2cosx+1)2+3(1–sinx) = 0
0 x 0 sinx 1
1–sinx 0sinx(2cosx–1)2 0
each term is equal to zero1–sinx = 0 sinx = 1
cosx = 0sinx(2cosx–1)2 0
no solution
4 The number of values of in the interval 2,
2– suggesting the equation
2sec3 = tan4 +2tan2 is
(a) 1 (b) 2 (c) 3 (d) None of thesesolution
311
We have2sec
3 = tan4 +2tan2
= (1+tan2 )2 –1= (sec2 )2 –1
let sec2 = x x 1x
3 = x2–1
graph of y = x3 & y = x2–1
intersect at one pointWhen x = 2 ; y = 3
sec2 = 2sec = 2
takes two values in 2,
2–
option ‘b’ is correct
5 If 0 x and xsin2
81 + xcos2
81 = 30 then x is equal to
(a) 6 (b) 2 (c) (d) 4Solution
we havexsin2
81 + xcos2
81 = 30xsin2
81 + xsin–1 2
81 = 30
xsin2
81 + xsin 2
8181
= 30
let xsin2
81 = yy2–30y+81 = 0y2–27y–3y+81 = 0(y–27)(y–3) = 0
xsin2
81 = 27 or 3xsin4 2
3 = 33 or 31
xsin4 2 = 3 or 1
sin2x = 43
or 41
sinx = 23
or 21
y
(0,1)
O
(0,–1)
y=x –12
X
x)3(y
312
0 x sinx = 23
or 21
3 , 32
, 6 , 65
6 The equation a sinx+bcosx = c
where |c| > 22 ba has(a) one solution (b) two(c) no solution (d) infinite number of solutionSolution
We haveasinx+bcosx = c
22 baa
sinx + 22 bab
cosx = 22 bac
sinx cos +cosx sin = 22 bac
sin(x+ ) = sin >1 |c| > 22 ba i.e. 22 ba|c|
> 1
not possiblecorrect option is c
1. If 2cos = 12 21cos ,then =
a. n2 b. 4n2 c. 3
n2 d. none
2. It 3cos2 – 32 sin cos – 3sin2 = 0 then equals
a. 62n
b. 6–
2n
c. 32n
d. 3–
2n
3. The general solution ofsinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x is
a. 8n b. 82
n
c. 82n)1(– n
d. 23cotn2 1–
313
PRACTICE QUESTIONS
4. The number of points of intersection of 2y = 1 and y = cosx in 2x
2–
is
a. 1 b. 2 c. 3 d. 45. The number of values of x in the internal 5,0 satisfying the equation 3sin2x – 7sinx + 2
= 0 isa. 0 b. 5 c. 6 d. 10
6. The sum of all the solutions of the equation cos3
cos –3
cos =41 , 6,0 is
a. 15 b. 30 c. 3100
d. none
7. The number of values of 4,0 satisfying the equation 2xsin–xcos3 is
a. 0 b. 2 c. 4 d. 8
8. If tan,cos,sin61
are in G.P., then equals
a. 3n2 b. 6
n2 c. 3)1(–n n d. 3
n
9. The number of solutions of the equation tanx + secx = 2cosx lying in the internal2,0 is
a. 0 b. 1 c. 2 d. 310. The general solution of equation sin2 sec + 3 tan = 0 is
a. = 3)1(–n 1n b. n
c. 6)1(–n 1n d. 2
n
11. PassageSolution of equations asinx bcosx = c.General value satisfying two equations.acos bsin = c, satisfying two equations.a. The equation acos bsin = c be first converted to r cos = c where
a = rcos ,b = rsinb. satisfying two equations
Find the common value of lying between 0 and 2 and then add n2Answer the following questions based upon above passage(i) The number of intergral values of k for which the equation 7cosx + 5sinx = 2k+1 has a
solution isa. 4 b. 8 c. 10 d. 12
314
(ii) If cos3x + sin6
7–x2 = – 2 then x equals
a. 1–m63 b. 1m6
3 c. 1m23 d. none
(iii) The values of x such that x– and satisfying the equation are given byto.....xcosxcosxcos1 32
8 = 43 x equals
a. 3 b. 32
c. 3–
d. 32–
(iv) The number of solution of the equaiton esinx – e–sinx –4 = 0 isa. 1 b. 2 c. 4 d. 0
Answers1. b, c 2. a 3. b 4. b 5. c 6. b 7. c 8. a9. c 10. b 11. (i). b (ii). b (iii), a, b, c, d (iv). b
315
TRIGONOMETRY - IIITrigonometric Equations
1. The number of values of x in the internal [0, 3 ] satisfying the equation 2sin2x+5sinx–3=0 is
a. 6 b. 1 c. 2 d. 4.
Solution :
We have 2sin2x+5sinx–3=0
2sin2x+6sinx–sinx–3=0
(2sinx–1) (sinx+3)=0
sinx= 21
( sin x 3) ( –1 sin x 1)
y=sinx and y= 21
intersect in 4 points
number of values of x in the internal [0, 3 ] is 4.
2. The number of values of in the internal 2,
2– such that 5
nfor n=0, 2,1 and tan
= cot5 as well as sin 2 =cos4 is _______.
Solution :
Given, tan = cot
tan =tan 5–2
2 –5 =n +
6 = 2 –n
=12 – 6n
Also cos4 =sin2
cos4 =cos 2–2
4 =2n 2–2
2o
1
–1
Y
X2 2
3
25
21y
21
316
= 2–
82n
Taking positive Taking negative
82n
23
8–
2n
2
3 =n + 4 =n – 4
= 12n4
=(4n–1) 4
=(4n+1)12
These values of gives only 3 common solution lies in the interval 2,
2– .
3. The number of all possible values of , where 0 < < , for which the system ofequaitons(y+z)cos3 =(xyz)sin3
xsin3 = z3sin2
y3cos2
and (xyz) sin3 =(y+2z)cos3 +ysin3
have a solution (x0, y0, z0) with y0z0 0 is
Solution :
Given equations are xsin3 = yzzy
cos3
xsin3 – y3cos
–z3cos = 0 .....(1)
xsin3 – y
3cos2 –
z3sin2 = 0 ....(2)
xsin3 – z1 cos3 –
y2
cos3 – z1 sin3 ....(3)
from (2) and (3) we get
2sin3 = cos3 +sin3
317
sin3 =cos3
tan3 =1=tan 4
3 = 4 , 45
, 49
or = 12 , 125
, 129
4. The solution of the equation 4sin4x+cos4x=1 is
a. x=2n b. x=n +1 c. x=(n+2) d. none of these
Solution:
Given that 4sin4x+cos4x=1
4sin4x+(cos2x–1) (cos2x+1)=0
4sin4x–sin2x(cos2x+1)=0
sin2x[4sin2x–cos2x–1]=0
sin2x[5sin2x–2]=0
sin2x=0 or sin2x= 52
sinx=0 or sinx= 52
x=n x=n where sin52
option d is correct.
5. The solution of the equation [sinx+cosx]1+sin2x=2, – x is
a. 2 b. c. 4 d. 43
Solution : [sinx+cosx]1+sin2x
x2sin1
x4
sin2
at x= 4 , 4
2sin1
44sin2 = 2
2 =2
Option c is correct.
318
or
6. The most general solutions 2sinx+2cosx=21–2
1are
a. n – 4 b. n + 4 c. n +(–1)n4 d. 2n 4
Solution : We have 2sinx+2cosx=21–2
1
AM GM
222 xcosxsin
xcosxsin 22
xcosxsin 222 (Equality holds when sinx=cosx)
xxsin22
But the minimum value of sinx+cosx is 2–
sinx+cosx= x4
sin2
and–1 x4
sin2 1
22– 2x4
sin
sinx=cosx
tanx=1=tan4
x=n4
option b is correct.
7. The set of values of satisfying the inequation 2sin2 –5sin +2>0, where 0 < < 2 is
a. 2,6
56
,0 b. 2,6
56
,0
c. 2,3
23
,0 d. none of these
cos
319
2
Solution : Given inequation 2sin2 –5sin +2>0
2sin2 –4sin –sin +2>0
2sin (sin –2)–1(sin –2)>0
(2sin –1) (sin –2) > 0
sin < 21
or sin > 2
but sin > 2 not possible
sin < 21
If x ,6
56
,0 for x
sin < 21
,6
56
0 for .
correct option is ‘d’
8. Passage
Suppose equation is f(x)–g(x)=0 or f(x)=g(x)=y say. Then draw the graphs of y=f(x) and y=g(x).If graphs of y=f(x) and y=g(x) cuts at one, two, three ...., no points, then number of solutionsare one, two, three, ... zero respectively.
On the basis of above information, answer the following questions.
1. The number of solution of sinx=10x
is
a. 4 b 6 c. 8 d. none of these
2. Total number of solutions of the equation 3x+2tanx= 25
in x [0, 2 ] is equal to
a. 1 b 2 c. 3 d. 4
3. Total number of solutions of sin{x}=cos{x}, where {.} denotes the fractional part, in [0,2 ] is
a. 3 b. 5 c. 7 d. none of these
4. If 1–sinx= 2–x
23
+a has no solution when a R+ then
a. a R+ b. a > 23
+ 3
o
1
–1
Y
x
y = sinx
6 65
21 2
1y
320
c. a 323,0 d. a 32
3,23
Solution :
1. Graphs of y=sinx and y = 10x
meet exactly six times. Hence no of solutions =6
o
1
Y
X– – –
y=sinx
1
– –
10x
y
2. We have 3x+2tanx= 25
in x [0, 2 ]
tanx = 45
– 2x3
2 23 2o
y
x
y=tanx
2x3
45y
Graphs of y = 45
– 23
and y = tanx meet exactly three times in [0, 2 ]
Thus number of solution = 3
321
3. sin{x}=cos{x}
Graphs of y=sin{x} and y=cos{x} meet excatly 7 times in [0, 2 ]
o o o o o o o
o
1
x
y
4. slope of y=1–sinx is –cosx
slope of y= 2–x
23
+a for x > 2 is 23
P2
Y
XO2
cosx=– 23
x = 67
P = 23,
67
If y= 2–x
23
+a passes throgh ‘P’ then
a= 23
– 3
a > 3–
23
322
1. Sin + 3 cos =6x–x2–11, 0 , x R, holds for
a. no values of x and b. one value of x and two values of
c. two values of x and two values of d. two points of values of (x, )
2. For 0 x 2 , then 21y–y212 2xeccos 2
a. is satisfied by exactly one value of y b. is satisfied by exactly two value of x
c. is satisfied by x for which cos x=0 d. is satisfied by x for which sin x=0
3. Let tanx–tan2x > 0 and 2sinx < 1. Then the intersection of which of the following two setssatisfies both the inequalities?
a. x > n , n Z b. x > n –6 , n Z
c. x < n 4– , n Z d. x < n 6 , n Z
4. The equation (cosp–1)x2+(cosp)x+sinp=0 in the varibale x has real roots. Then p can take anyvalue in the interval
a. (0, 2 ) b. (– , 0) c. 2,
2– d. (0, )
5. Let 2sin2x + 3sinx–2>0 and x2–x–2<0 (x is measured in radians). Then x lies in the inerval
a. 65,
6 b. 65,1– c. (–1, 2) d. 2,
6
6. If (cosec2 –4)x2+(cot + 3 )x + cos22
3=0 holds true for all real x, then the most general
values of can be given by (n Z)
a. 2n + 611
b. 2n + 65
c. 2n 67
d. n 611
7. If sin4x+cos4y+2=4 sinx cosy and 0 x, y 2 then sinx+cosy is equal to
a. –2 b. 0 c. 2 d. 23
323
PRACTICE QUESTIONS
8. The solution of the equation sin10x+cos10x= 1629
cos42x is
a. x= 4n
+ 8 , n I b. x=n 4 , n I
c. x=2n 2 , n I d. none of these
9. The most general values of x for which sinx+cosx=min {1, a2–4a+6},a R are given by
a. 2n , n N b. 2n 2 , n N
c. n +(–1)n4 – 4 ,n N d. none of these
10. Number of solutions of the equations y= 31
[sinx+[sinx]] and [y+[y]]=2cosx where [.] denotes
the greatest integer function is
a. 0 b. 1 c. 2 d. infinite
11. If 21xsin
23–x2sinxcos =1, then possible values of x are
a. n or n + (–1)n6 , n I b. n or 2n + 2 or n +(–1)n
6 , n Z
c. n + (–1)n6 , n I d. n , n I
12. If xcos1xcos 2
2(1+tan22y) (3+sin3z)=4, then
a. x may be multiple of b. x cannot be an even multiple of
c. z can be a multiple of d. y can be a multiple of 2
13. Matrix match type
Column I Column II
a. If 3cos2 –2 3 sin cos –3sin2 =0 than = p 6 , 65
324
b. If rsin =3, r=4(1+sin ) q. 2n + 67
where 0 then =
c. If sin = 21– and tan = 3
1, then r. 6 , 3 , 6
5, 3
2
the general value of which satisfies both the equations is
d. If 0 x and xcosxsin 22
8181 =30 then x is s. 62n
14. If x2cos2x2sin3 + xcos2x2sin–1 2
3 =28, then the value of x are given by
a. tanx=1 b. tanx=–1 c. cosx=0 d. none of these
15. If }2log)....xsinxsinx{(sin e642
e satisfies the equation x2–9x+8=0, then value of xsinxcosxcos
,
0 < x< 2 is
a. 1321
b. 1–321
c. 21
d. 0
Integer type questions .16. If , satisfy the equaiton
12sin +5cos =2 –8 +21, then –2 , is ___________.17. The value of x and y satisfy the equaiton tan2(x+y)+cot2(x+y)=1–2x–x2, then the value of x2–
3x+2 is ___________.
18. The integral value of p for which p cosx – 2sinx = p–22 has a solution, is _______
Answers
1. b, d 2. a,b,c 3. a,d 4. d 5. d 6. a,b 7. c 8. a
9. c 10. a 11. b 12. a,d 13. a s ; b p ; c q ; d r ;14. b,c 15. b 16. 0 17. 6 18. 2
325
TRIGONOMETRY - IVTrigonometric Equations - Problem Solving
Equations involving trigonometric functions of unknown angles are called trigonometric equations.Numerically least angle is called the principal value (solution).Since, trigonometric functions are periodic, a solution is generalised by means of periodicity of the function.The solution consisting of all possible solutions of a trigonometric equation is called its general solution .
Trigonometric Equations General solutions (n Z)
sin = 0 = ncos = 0 = (2n+1) /2tan = 0 = ncot = 0 = (2n+1) /2sin = sin = n +(–1)n
cos = cos = 2ntan = tan = n +sin2 = sin2 = ncos2 = cos2 = ntan2 = tan2 = n
Method of an auxiliary angleEquations of the form a cos b sin = c are equivalent to the elementary trigonometric equations
cos( )= 22 bac
where sin = 22 bab
, cos = 22 baa
If |c| > 22 ba , then a cos b sin = c has no solution.
If |c| 22 ba , then put
22 ba|c|
= cos so that
cos( ) = cos = 2n
= 2n ; n ZNote :(i) While solving, if you are squaring a trigonometric equation (which should be avoided as for as
possible), check the solution for extraneous roots.(ii) Do not cancel terms containing unknown terms which are in product (it may cause loss of a valid
solution)(iii) Remove all those values (angles) which make any of the terms undefined or infinite.Solving trigonometric equations graphicallySketch the graph of each side of the equation. Now look for all the points of intersection within the giveninterval.
326
Solved Examples1 The number of distinct solutions of
sin5 cos3 = sin9 cos7 in 2,0 is
(a) 4 (b) 5 (c) 8 (d) 9Solution :
2sin5 cos3 = 2sin9 cos7sin8 +sin2 =sin16 +sin2sin16 –sin8 = 02cos12 .sin4 = 0
sin4 =0 or cos12 = 0
gives = 4n
or gives = (2n+1) 24
= 0, 4 , 2 or = 24 , 243
, 245
, 247
, 249
, 2411
9 possible valuesAns : (d)
2 The number of values of x in [0,5 ] satisfying 3cos2x–10cosx+7=0 is(a) 5 (b) 6 (c) 8 (d) 10Solution :3(2cos2x–1) –10cosx+7 = 03cos2x–5cosx+2= 0
cosx = 32
, 1.
Clearly the lines y = 32
, 1 intersect the graph at 8 points
Hence 8 valuesAns : (c)
3 If sin2 –2sin –1=0 is to be satisfied for exactly 4 distinct values of [0,n ], n N ; thenthe least value of n is(a) 2 (b) 6 (c) 4 (d) 8Solution :
sin2 –2sin –1=0sin = 1 2sin = 1– 2 (as 1+ 2 is rejected)
clearly four solutions lie in [0,4 ] least value of n is 4
o
1
–1
2
Y
X3
2
2
o
1
–1
Y
X
327
3cos2x–3cosx+2= 0(3cox–2) (cox–1) = 0
Also 5th solution lies in [0,5 ] greatest value can be 5.
Ans : (c)4 The most general solution of secx–1 = 1–2 tanx is
(a) n + 8 (b) 2n , 2n + 4 (c) 2n (d) None of these
Solution :Given equation can be simplified to(1–cosx) = 1–2 sinx
2sin2
2x
– 1–2 2sin 2x
cos 2x
= 0
2sin 2x
2xcos)1–2(–
2xsin = 0
sin 2x
= 0 or tan 2x
= 1–2
2x
= n tan 2x
= tan 8
x = 2n 2x
= n + 8
x = 2n + 4Ans : (b)
5 Sum of all the solutions of
cosx.cos x3 . cos x–
3 = 41
, x [0,6 ] is
(a) 15 (b) 30 (c) 3110
(d) None
of theseSolution :
41
cos3x = 41
cos3x=1 cos3x=cos0 3x=2n x= 3n2
x=0, 32
, 34
, 36
, 38
,..............., 318
Adding we get the sum as 30
328
,n z
,n z
Ans : (b)6 The number of solutions of cosx= |1+sinx| for x [0,3 ], is
(a) 3 (b) 2 (c) 4 (d) None of theseSolution :It is evident from the figure that, the two graphs intersect at 3 points.
o
1
–1
Y
X
Ans (a)7 The number of solutions of xsin2
16 + xcos2
16 = 10 in x [0,2 ] is(a) 8 (b) 6 (c) 4 (d) 2Solution :
xsin2
16 + xsin–1 2
16 = 10
put xsin2
16 = y y+ y16
= 10
y2–10y+16 = 0 gives y = 8,2
xsin2
16 = 2, 8 = 41
16 & 43
16
sin2x = 41
, 43
Clearly lines y = 43,
41
intersect the graph in 8 points
Hence 8 possible solutions.Ans : (a)
1 Let P = cos2cos–sin: and
Q = sin2cossin: be two sets. Then
(a) P Q and Q–P (b) Q P (c) P Q (d) P = Q
2 The number of values of in the interval 2,
2–
such that 5n
for n= 0, 1, 2 and
tan =cot5 as well as sin2 = cos4 is ___________(a) 0 (b) 1 (c) 2 (d) 3
2o
1
–1
X
329
PRACTICE QUESTIONS
3 The positive integer value of n>3 satisfying the equation
nsin
1 =
n2sin
1 +
n3sin
1 is
(a) 5 (b) 6 (c) 7 (d) 8
4* All values of in the interval 2,
2–
satisfying (1–tan )(1+tan )sec2 +2tan2 = 0
(a) 3 (b) 3–
(c) 0 (d) None of these.
5 Match the following :-Column I Column II
(a) Number of roots of cos7x+sin4x= 1 (p) 1in the interval [0,2 ]
(b) Value of ‘a’ for which a2– 2a+sec2 (a+x) = 0 (q) 0has solution
(c) Number of solutions of |cosx| = 2[x] (r) 46 The solution set of |4sinx–1|< 5 , |x|< is
(a) 54,––
10,
5– ,
109
(b) 10–,
109–
107,
103
(c) 109–,–
103,
10– ,
107
(d) 109,
107–
7 The number of points (x,y) inside the circle x2+y2=4 satisfying the equation tan4x+cot4x+1=3sin2y is(a) 1 (b) 2 (c) 4 (d) 8
8 If sinx = 2sin , cosx = tany, tanx=cosz and cosy = tanz, then is(a) 18° (b) 36° (c) 54° (d) 72°
9 Read the following and then answer the questions.Consider the system of equationssinxcos2y = (a2–1)2+1, cosxsin2y = a+1(i) the number of values of a for which the system has a solution is
(a) 1 (b) 2 (c) 3 (d) infinite.(ii) The number of values of x [0,2 ] when the system has solution for permissible values of a
is(a) 1 (b) 2 (c) 3 (d) 4
(iii) The number of values of y [0,2 ] when the system has solution for permissible values of a(a) 2 (b) 3 (c) 4 (d) 5
10 The total number of solutions of sin{x} = cos{x} in[0,2 ] is(a) 5 (b) 6 (c) 8 (d) None of these
330
11 If sinx+cosx= y1y , x [0, ] then
(a) 4x , y=1 (b) y = 2 (c) 4
3x (d) None of these.
12 If the inequatlity sin2x+acosx+a2 > 1+cosx hods for any x R then the largest negative integralvalue of ‘a’ is(a) –4 (b) –3 (c) –2 (d) –1
13 The number of solutions of x [0,2 ] for which [sinx+cosx] = 3+[–sinx]+[–cosx] is(a) 0 (b) 4 (c) infinite (d) 1
14 The arithmetic mean of the roots of the equation 4cos3x–4cos2x–cos(315 +x) =1 in the interval(0,315) is(a) 50 (b) 51 (c) 100 (d) 315
15 Values of x & y satisfying the equationsin7y = |x3–x2–9x+9| + |x3–x2–4x+4| + sec22y+cos4y are
(a) x=1, y=n , n Z (b) x=1, y=2n + 2 , n Z
(c) x=1, y=2n , n Z(d) None of these
'Note : Questions with * have more than one correct option'
Answers1. d 2. d 3. c 4. a, b 5. a r ; b p ; c q6. c 7. c 8. a 9. (i) a (ii) b (iii) d 10. b11. a 12. b 13. a 14. b 15. b
331
Properties of Triangles
1. SINE RULE
kc
Csinb
Bsina
Asin
or Csinc
Bsinb
Asina
2. COSINE RULE
i.bc2
a–cbAcos222
ii.ca2
b–acBcos222
iii.ab2
c–baCcos222
3. PROJECTION FORMULAEi. b cos C + c cosB = aii. c cos A + a cosC = biii. a cos B + b cosA = c
4. NAPIER’S ANALOGY
i. 2Acot
cbc–b
2C–Btan
ii. 2Bcot
aca–c
2A–Ctan
iii. 2Ccot
bab–a
2B–Atan
332
TRIGONOMETRY - I
5. a. i. bcc–sb–s
2Asin
ii. caa–sc–s
2Bsin
iii. abb–sa–s
2Csin
b. i. bca–ss
2Acos
ii. cab–ss
2Bcos
iii. abc–ss
2Ccos
c. i. a–ssc–sb–s
2Atan
ii. b–ssa–sc–s
2Btan
iii. c–ssb–sa–s
2Ctan
d. i. c–sb–sa–ss
2Acot
ii. a–sc–sb–ss
2Bcot
iii. b–sa–sc–ss
2Ccot
e. i. c–sb–sa–ssbc2
bc2Asin
ii. c–sb–sa–ssca2
ca2Bsin
iii. c–sb–sa–ssab2
ab2Csin
333
6. RADIUS OF THE CIRCUMCIRCLE ‘R’
4abc
Csin2c
Bsin2b
Asin2aR
7. RADIUS OF THE INCIRCLE ‘r’
2Btanb–s
2Atana–s
sr = 2
Csin2Bsin
2AsinR4
2Ctanc–s
8. RADII OF THE EXCIRCLES r1,r2 and r3
i. 2Ccos
2Bcos
2AsinR4
2Atans
a–sr1
ii. 2Ccos
2Bsin
2AcosR4
2Btans
b–sr2
iii. 2Csin
2Bcos
2AcosR4
2Ctans
c–sr3
334
iv. etc
2Acos
2Ccos
2Bcosa
r1
v. r1+r2+r3 = r + 4r
vi. r1
r1
r1
r1
321
vii. r1r2+r2r3+r3r1 = s2
viii. a cosA + b cosB + c cosC = 4R sinA sinB sinCix. a cotA + b cotB + c cotC = 2(R+r)
Also,
2Acos
2Csin
2Bsina
r
2Bcos
2Asin
2Csinb
=
2Ccos
2Bsin
2Asinc
9. If length of the median AD, BE and CF are given, then sides can be determined by usingthe formula
2222 AD–CF2BE234a
Similarly 2222 BE–AD2CF234b
and 2222 CF–BE2AD234c
10. Distance of the orthocentre from the vertex A is 2R cosA.11. PEDAL TRIANGLE
Triangle formed by joining the foot of perpendiculars drawn from vertices to opposite sides ofa given tri-angle is called pedal triangle.Length of the sides of pedal triangle of given ABCare a cos A, b cos B and c cos C respectively..Angles of the pedal triangle are 180° – 2A, 180° – 2B and 180° – 2C respectively.
12. EXCENTRIC TRIANGLEIf I1, I2, and I3 are the centres of excircles, then 321 III is called excentric triangle. I2, A, I3 ;
I3, B, I1 and I2, C, 12 are collinear. ABCis the pedal of 321 III . Incentre ‘I’ of of
the ABCwill be orthocentre of 321 III .
13. If median AD inclined at angles ,, with BC, CA and AB respectively then
222 a–c2b2Csinbsin
222 a–c2b2Csinasin and
222 a–c2b2Bsinasin
335
14. Distance between orthocentre and circumcentre= CcosBcosAcos8–1R
15. Distance between circumcentre and incentre is
r2–RR2Csin
2Bsin
2Asin8–1R
16. Distance between circumcentre and centre of excircles are
1r2RR2Ccos
2Bcos
2Asin81ROI1
2r2RR2Ccos
2Bsin
2Acos81ROI 2
33 r2RR2Csin
2Bcos
2Acos81ROI
17. Length of the angle bisectors
2Asincb
Asinbc2Acos
cbbc2AD
2Bcos
acca2BE and 2
Ccosba
ab2CF
18. Area of quadrilateral ABCD, if sum of a pair opposite angle is 2 is
cosabcd–d–sc–sb–sa–s 2
where 2s = a+b+c+d19. m-n THEOREM
(m+n) cot = m cot – n cot(m+n) cot = n cotB – m cotC
336
SOME EXTRA TIPS
1. i. 0C–Bsina3
ii. abc3C–Bcosa3
2. i. If a cosB = b cosA, then the triangle is isoscelesii. If a cos A = b cos B, then the triangle is isosceles or right anglediii. If a2+b2+c2 = 8R2, then the triangle is right anglediv. If cos2A + cos2B + cos2C = 1, then the triangle is right angled
v. If cosA = Csin2Bsin
, then the triangle is isosceles
vi. If Ccosc
Bcosb
Acosa
, then the triangle is equilateral
vii. If cosA+cosB+cosC = 23
,then the triangle is equilateral
viii. If sinA+sinB+sinC =2
33, then the triangle is equilateral
ix. If tanA+tanB+tanC = 33 , then the triangle is equilateral
x. If cotA+cotB+cotC = 3 , then the triangle is equilateral3. a. The circumcentre lies (i) inside an acute angled triangle (ii) outside an obtuse angled
triangle.b. The circumcircle of a right angled triangle is the mid point of the hypotenuse.c. The orthocentre of a right angled triangle is the vertex at the right angle.
4. Triangle is equilateral if any one of the following holds:a. R = 2r b. r1 = r2 = r3 c. r:R:r1 = 1:2:3
5. Triangle is right angled if r : R : r1 = 2:5:126. If r1,r2,r3 are in H.P. iff a,b,c are in A.P.7. In an equilateral triangle:
a. area =4a3 2
b. 3aR c. 2
Rr
d. 2R3rrr 321 e. 3:2:1rRr 1
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EXAMPLES
1 In triangle ABC, 2ac sin CB–A(21
is equal to
(a) a2+b2–c2 (b) c2+a2–b2 (c) b2–c2–a2 (d) c2–a2–b2
Solution
Given that 2ac sin CB–A(21
= 2ac sin )B2–(21
= 2ac sin B–2
= 2ac cos B
= 2ac ac2b–ca 222
= a2+c2–b2
Correct option is ‘b’
2 Let ABC be a triangle such that ACB = 6 and let a,b and c denote the lengths of the side
opposite to A,B and C, respectively. The values of x for which a = x2+x+1, b= x2–1 andc = 2x+1 is (are)(a) – 32 (b) 1+ 3 (c) 2+ 3 (d) 4 3Solution
Given that ACB = 6
cos( ACB) = cos C = ab2
c–ba 222
cos 6 = )1–x)(1xx(2)1x2(–)1–x()1xx(
22
22222
23 =
)1–x)(1xx(2
)1–x
(
)x–x)(2x3x(22
2222
3 = 1xx
1–x)2x3x(x2
22
3 = 1xx
1–x2x22
2
338
23 =
)1–x)(1xx(2
1––22
22x4 2x3
3x 2x
23 =
)1–x)(1xx(2
1––22
22x x =
2- 2x– 1)( )
2x2–3 + 2–3 x+ 13 = 0
x = 2–32
33–2
x = – 32 , 1+ 3
as x>0 x = 1+ 3Correct option is ‘b’
3 In ABC, interval angle bisector of A meets side BC in D. DE AD meets AC in E and ABABin F. Then
(a) AE is HM of b and c (b) AD = cbbc2
cos 2A
(c) EF = cbbc4
sin 2A
(d) AEF is isosceles
SolutionIn AFE, we get AF = AE
AFE is an isosceles ar( ABC) = ar ( ABD)+ar( ADC)
= 21
bc sinA = 21
c ADsin 2A
+ 21
bADsin 2A
2bcsin 2A
cos 2A
= ADsin 2A
(b+c)
AD = cb
2Acosbc2
Also AD = AE cos 2A
AEcos 2A
= cb
2Acosbc2
AE is HM of b and c
Again EF = 2DE = 2AD tan 2A
= cb
2A
tan.2A
cosbc2.2
= cb
2A
sinbc4
Hence option a, b, c, d are correct
2A
A
b
E
CDB
F
c
a
2A
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4 In a triangle ABC with fixed base BC, the vertex A moves such that cosB+cosC = 4 sin22A
If a, b and c denote the lengths of the sides of the triangle opposite to the angles A,B and Crespectively, Then(a) b+c = 4a (b) b+c = 2a(c) locus of point A is an ellipse (d) locus of point A is a pair of straight lines.Solution
Given that cosB+cosC = 4sin22A
2cos2
CB cos
2C–B
= 4sin22A
2CBcos
2C–Bcos
= 12
Componendo and Dividendo
2CB
cos–2
C–Bcos
2CB
cos2
C–Bcos
= 1–212
2C
sin2B
sin2
2C
cos2B
cos2
= 3
tan2C
tan2B
= 31
sa–s
= 31
3s–3a=s2s=3aa+b+c = 3ab+c = 2a AB+AC>BC
locus of A is an ellipse.Hence options ‘b’ & ‘c’ are correct
1 The roots of the equation 6x2–5x+1 = 0 are tan 2A
and tan 2B
where A, B, C are the angles of a
triangle, then(a) a2+b2>c2 (b) a2+b2 = c2 (c) a2–b2= c2 (d) None of these
340
sin2A
= sin2 2
CBcos
2CB
--
- =
PRACTICE QUESTIONS
2 If aAcos
= bBcos
= cCcos
and the side a = 2 then area of triangle is
(a) 1 (b) 2 (c)23
(d) 3
3 If in a triangle PQR, sinP, sinQ, sinR are in A.P then(a) The altitudes are in A.P (b) The altitudes are in H.P(c) The medians are in G.P (d) The medians are in A.P
4 If the sides a, b, c of ABC are in A.P then cot 21
A, cot 21
B, cot 21
C are in
(a) A.P (b) G.P (c) H.P (d) None of these5 If radius of the incircle of a triangle with sides 5p, 6p and 5p is 6, then p is equal to
(a) 4 (b) 6 (c) 8 (d) 10
6 In any ABC, Asin1AsinAsin2
is always greater than
(a) 9 (b) 3 (c) 27 (d) 107 If c2 = a2+b2, then 4s(s–a)(s–b)(s–c) =
(a) s4 (b) b2c2 (c) c2a2 (d) a2b2
8 If a, b, c, d be the sides of a quadrilateral, then the minimum value of 2
222
dcba
is equal to
(a) 21
(b) 31
(c) 41
(d) 1
9 bcr1 + ca
r2 = abr3 =
(a) r1–
R21
(b) 2R–r (c) r–2R (d) R21–
r1
10 Passage
(a) Asina
= 2R, R = 4abc
(b) r = 3 = (s–a) tan 2A
= 4R sin 2A
sin 2B
sin 2C
(c) r1 = a–s = s tan 2A
= 4Rsin 2A
cos 2B
cos 2C
(d) If P be the orthocenter of a ABC and its distances PA from the vertex A and PL fromthe sides BC are PA = 2RcosA, PL = 2RcosBcosC
Answer the following questions based upon above passage(i) If r1 = 2r2 = 3r3, then
(a) 54
ba
(b) 45
ba
(c) 53
ca
(d) 35
ca
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(ii) 21r1
+ 22r1
+ 23r1
+ 2r1
=
(a) 2
222
scba
(b) 2
2a(c) 4R (d) 4r
(iii) ar–r1 + b
r–r2 =
(a)1ra
(b)2rb
(c)3rc
(d) None of these
(iv) In a triangle ABC, let c = 2 . If r is the inradius and R is the circumradius of the
triangle then 2(r+R) is equal to(a) a+b (b) b+c (c) c+a (d) a+b+c
Answers1. b 2. d 3. b 4. a 5. a 6. c 7. d 8. b9. d 10. (i).b,d (ii). b (iii). c (iv). a
342
TRIGONOMETRY - IIProperties of Triangles
EXAMPLES1 In a ABC the inradius and three exradii are r, r1, r2 and r3 respectively. In usual notations the
value of r.r1.r2.r3 is equal to
(a) 2 (b) 2 (c) R4abc
(d) None of these
Solution
r.r1.r2 r3 = s
.a–s
.b–s c–s
= 2
4
= 2
correct option is ‘b’2 The distance between the circumcentre and the orthocenter of a triangle ABC is
(a) R osC8cosAcosBc–1 (b) R osC8cosAcosBc1
(c) R osC4cosAcosBc1 – (d) None of theseSolution
Let O & P be circumcenter and orthocentre respectively of ABCOF AB.We have OAF = 90°–C ( AOF = C)
= PAE (In ADC)OAP = A– OAF– PAEAE
= A–(90– C)–(90– C)= A+2 C–180= A+2 C–( A+ B+ C)= C– B
OA = R (circum radius) and PA = 2RcosAIn OAPOP2 = OA2+PA2–2(OA) (PA)COS(C–B)
= R2+4R2cos2A–4R2cosAcos(C–B)= R2+4R2cosA(cosA–cos(C–B)= R2+4R2cosA(–cos(B+C)–cos(C–B))= R2–4R2cosA(cos(B+C)+cos( –B))= R2–4R2cosA(2cosBcosC)= R2–8R2 cos AcosBcosC
OP2 = R2 (1–8cosAcosBcosC)OP = R osC8cosAcosBc–1Hence ‘a’ is the correct option
3 The radii r1, r2, r3, of escribed circles of the triangle ABC are in HP. If its area is 24sq.cm and itsperimeter is 24cm, then the lengths of its sides are
A
E
P
CDB
F
cc
O
C
343
° °°
(a) 4, 6, 8 (b) 3, 9, 11 (c) 6, 8, 10 (d) None of theseSolution
Given that r1, r2 r3 are in HP
i.e. a–s
, b–s
, c–s
are in A.PA.P
or s–a, s–b, s–c are in A.Por a, b, c are in APa+c = 2ba+b+c = 243b = 24b = 8
s = 224
= 12
a+c = 16c = 16–a
2 = s(s–a)(s–b)(s–c)24×24 = 12(12–a)(12–8)(12–(16–a))24×24 = 12×4(12–a)(a–4)12 = (12–a)(a–4)a2–16a+60 = 0a2–10a–6a+60 = 0(a–10)(a–6) = 0a = 10 or a = 6a = 10,c = 6, b=8 or a= 6, b=8, c= 10correct option is ‘c’
4 In ABC the value of ccos1rr 21 is always equal to
(a) 2r (b) 2R (c)R
22r (d)r
2R 2
SolutionWe know
r1+r2 = 4R sin2A
cos2B
cos2C
+4Rsin2B
cos2C
cos2A
= 4Rcos2C
2Bsin
2Acos
2Bcos
2Asin
= 4Rcos2C
2BAsin
=4Rcos2C
2Ccos
= 4Rcos22C
= 2R(1+cosC) (1+cos2 =2cos2 )
344
Ccos1rr 21 = 2R
5 Let ABCD be a quadrilateral with area 18 with side AB parallel to the side CD and AB = 2CD.Let AD be perpendicular to AB and CD.If a circle is drawn inside the quadrilaterals ABCDtouching all the sides, then its radius is
(a) 3 (b) 2 (c)23
(d) 1
SolutionGiven that AB = 2CDLet CD = a AB = 2a
B(2a,0)and c(a,2r)
Let the centre of in circle be (r,r)Where r is the radius of the circleSince AB || to CD
ABCD is a trapezium
ar(ABCD) =21
h(sum of parallel sides)
= 21
×2r(a+2a)
18 = 3arar = 6
Now BC is a tangent to the circle(x–r)2+(y–r)2 = r2
Equation of tangent is y = – 2r2
(x–2a)
2rx+ay–4ar = 0
r = 22
2
ar4ar4–arr2
r = 22
2
ar4
ar3–r2
squaring4r4+a2r2 = 4r4+9a2r2–12ar3
12ar3 = 8a2r2
3r = 2a [ ar = 6]3r2 = 2ar3r2 = 12r2 = 4r = 2
6 In a triangle ABC, a : b : c = 4 : 5 : 6. The ratio of the radius of the circumcircle to that of theincircle is
345
(a)4
15(b)
511
(c)7
16(d)
316
Solution
R = 4abc
, r = s
rR
= 4abc
×s
= 24sabc
= ))c–s)(b–s)(a–s(s(4sabc
= )c2–s2)(b2–s2)(a2–s2(abc2
= )a–cb)(b–ac)(c–ba(abc2
= 7.5.36.5.4.2
= 7
16
7 The value of 21r1
+ 22r1
+ 23r1
+ 2r1
is
(a) 0 (b) 2
222 cba(c) 222
2
cba(d)
222 cba
Solution
21r1
= 2
2)a–s(; 2
2r1
= 2
2)b–s(; 2
3r1
= 2
2)c–s(; 2r
1 = 2
2s
21r1
+ 22r1
+ 23r1
+ 2r1
= 2
2222 )c–s()b–s()a–s(s
= 2
2222 )cba(s2–cbas4
= 2
22222 s4–cbas4(a+b+c = 2s)
= 2
222 cba
346
1 In ABC, a b c, if csinBsinAsin
cba323
333
= 8, then the maximum value of a is
(a)21
(b) 2 (c) 8 (d) 64
2 Sides of triangle ABC are in AP. If a <min{b,c}, then cosA may be equal to
(a)b2
b4–c3(b)
c2b4–c3
(c)b2
b3–c4(d)
c2b3–c4
3 In a ABC, angles A,B,C are in AP. Then cxlin
|C–A|CsinAsin4–3
is
(a) 1 (b) 2 (c) 3 (d) 44 In a triangle ABC, 2a2+4b2+c2 = 4ab+2ac, then the numerical value of cosB is equal to
(a) 0 (b)83
(c)85
(d)87
5 If a, b, c be the sides of a ABC and if roots of the equation a(b–c)x2+b(c–a)x+c(a–b) = 0 are
equal, then sin22A
, sin22B
, sin22C
are in
(a) AP (b) GP (c) HP (d) AGP
6 In a ABC sides a, b, c are in AP and !9!1
2 +
!7!32
+ !5!51
= )!b2(
8a
then the maximum value
of tanA tanB is equal to
(a)21
(b)31
(c)41
(d)51
7 If A, B, C, D are the angles of quadrilateral, then AcotAtan
is equal to
(a) tanA (b) cotA (c) Atan2 (d) Acot2
8 With usual notations, if in a ABC, 11
cb =
12ac
= 13
ba, then cosA: cosB: cosC is equal
to(a) 7 : 19 : 25 (b) 19 : 7 : 25 (c) 12 : 14 : 20 (d) 19 : 25 :20
9 In and ABC, Asin1AsinAsin2
is always greater than
(a) 9 (b) 3 (c) 27 (d) None of these10 The radius of the circle passing through the centre of incircle of ABC, and through the end
points of BC is given
(a) 2a
cosA (b) 2a
sec 2A
(c) 2a
sinA (d) asec 2A
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PRACTICE QUESTIONS
11 If in a ABC, a, b, c are in AP and p1, p2, p3 are the altitude from the vertices A, B, C respectivelythen(a) p1, p2, p3 are in AP (b) p1, p2, p3 are in HP
(c) p1+ p2 +p3 R3
(d)1p
1 +
2p1
+ 3p1
R3
12 If tanA, tanB are the roots of the quadratic abx2–c2x+ab = 0, where a, b, c are the sides of atriangle, then
(a) tanA = ba
(b) tanB = ab
(c) cosC = 0 (d) tanA+tanB = abc2
13 If sin is the GM between sin and cos , then cos2 is equal to
(a) 2sin2 –4 (b) 2cos2 –
4 (c) 2cos24 (d) 2sin2
4
14 Passage 1If p1, p2, p3 are altitudes of a triangle ABC from the vertices A, B, C respectively and is thearea of the triangle and s is the semipermanent of the triangle. On the basis of above information,answer the following questions :
(i) If 1p
1 +
2p1
+ 3p1
= 21
then the least value of p1, p2, p3 is
(a) 8 (b) 27 (c) 125 (c) 216
(ii) The value of 1pAcos
+ 2pBcos
+ 3pCcos
is
(a)r1
(b)R1
(c)R2
Cba 222
(d)1
(iii) The minimum value of c
pb 12
+ a
pc 22
+ b
pa 32
is
(a) (b) 2 (c) 3 (d) 6(iv) The value of p1
–2+p2–2+p3
–2 is
(a) 2
2
4
a(b) 3
3
8
a(c) 2
2
4
a(d) 2
2
8
a
(v) In the triangle ABC, the altitudes are in AP, then(a) a, b, c are in AP (b) a, b, c are in HP(c) a,b,c are in GP (d) angles A, B, C are in AP
15 Passage IIIn a triangle if the sum of two sides is x and this product is y (x 2 y ) such that(x+z) (x–z) = y where z is the third side of the triangle.
348
On the passes of above information, answer the following questions.(i) Greatest angle of the triangle is(a) 105 (b) 120 (c) 135 (d) 150(ii) Cerium radius of the triangle is(a) x (b) y (c) z (d) None of these(iii) In radius of the triangle is
(a) )xz(2y
(b) )yx(2z
(c)xz3y (d)
yx3z
(iv) Area of the triangle is
(a)4
3y (b)4
3x (c)43z (d) None of these
(v) The sides of the triangle are
(a)2
y4–xx 2
, z (b)2
z4–yy 2
,z (c)2
x4–zz 2
,z (d) None of these
Assertion and Reason
16 Assertion (A) : In any ABC, the minimum value of r
rrr 321 is
Reason (R) : AM GM(a) A (b) B (c) C (d) D
17 Assertion (A) : If A, B, C D are angle of a cyclic quadrilateral then Asin = 0Reason (R) : If A, B, C, D are angles of cyclic quadrilateral then Acos = 0(a) A (b) B (c) C (d) D
18 Assertion (A) : In any ABC, the square of the length of the bisector AD is bc 2
2
)cb(a–1
Reason (R) : In any ABC, length of bisector AD is cb
bc2 cos
2A
(a) A (b) B (c) C (d) DInteger Type Questions
19 If the radius of the circumcircle of a triangle is 12 and that of the incircle is 4, then the squareof the sum of radii of the escribed cirle must be
20 In a ABC, the maximum value of 1000 cba2A
cosa 2
must be
21 Matrix Match TypeColumn I Column II
(a) In a ABC, if 2a2+b2+c2 = 2ac+2ab, then (p) ABC is equilateral (b) In a ABC, if a2+b2+c2 = 2 b(c+a), then (q) ABC is right angled
(c) In a ABC,if a2+b2+c2 = bc+ca 3 , then (r) ABC is scalene
349
(s) ABC is scalene rightangled
(t) Angles B, C, A are in AP
Answers1. b 2. d 3. a 4. d 5. c 6. b 7. a 8. a9. c 10. b 11. b,d 12. a, b, c, d 13. a,c 14. (i). b (ii). b(iii). d (iv). c (v). b 15. (i). b (ii). d (iii).c (iv). a (v). a16. b 17. d 18. a 19. 2704 20. 0750 21. a p,t b q,s c q, r, t.
350
TRIGONOMETRY - IIIProperties of Triangles - Problem Solving
Properties and Solutions of TrianglesNotation :Vertices A,B,CSides a, b, cCentroid G (Point of intersection of medians )Orthocentre O (Point of intersection of altitudes)Circumcentre S (Point of intersection of perpendicular bisectors of the sides)Incentre I (Point of intersection of internal bisectors of the angles)Excentres I1,I2,I3 (Point of intersection of internal bisector of an angle & external bisectorsof the other two angles).Circumradius R (radius of circle with centre S and passing through the vertices)Inradius r (radius of circle with centre I and touching the sides)Exradii r1, r2, r3 ( radii of circles with centres I1,I2,I3 respectively and touching the sides)
Semiperimeter s = 2cba
Area of triangle Concepts and Formula1 Sine law
In any triangle ABC, Asina
= Bsinb
= Csinc
2 Cosine lawIn any triangle ABC
cosA = bc2
a–cb 222 or a2 = b2+c2–2bc cosA
cosB = ac2
b–ac 222
or b2 = c2+a2–2ca cosB
cosC = ab2
c–ba 222
or c2 = a2+b2–2ab cosC
3 Projection formulaa = bcosC+ccosBb = ccosA+acosCc = acosB+bcosA
4 Napier’s analogy (Tangent rule)
tan 2C–B
= cbc–b
cot 2A
351
tan 2A–C
= aca–c
cot 2B
tan 2B–A
= bab–a
cot 2C
5 Auxiliary formulae(Trigonometrical ratios of half angles of a triangle)
sin 2A
= bc
)c–s)(b–s(; sin 2
B=
ca)a–s)(c–s(
; sin 2C
= ab
)b–s)(a–s(;
cos 2A
= bc
)a–s(s; cos 2
B=
ca)b–s(s
; cos 2C
= ab
)c–s(s;
tan 2A
= )a–s(s)c–s)(b–s(
; tan 2B
= )b–s(s)a–s)(c–s(
; tan 2C
= )c–s(s)b–s)(a–s(
;
)a–s(s )b–s(s )c–s(s6 Area of Triangle
Area of triangle ABC = )c–s)(b–s)(a–s(s (Hero’s formula)
= 21
absinC = 21
bcsinA = 21
casinB
= 21
)CBsin(CsinBsina2
= 21
)ACsin(AsinCsinb2
= 21
)BAsin(BsinAsinc2
7 sinA = bc2
= bc2
)c–s)(b–s)(a–s(s
sinB = ca2
= ca2
)c–s)(b–s)(a–s(s
sinC = ab2
= ab2
)c–s)(b–s)(a–s(s
Also aAsin
= bBsin
= cCsin
= abc2
8 m – n RuleIn any triangle,(m+n) cot = m cot – n cot
= n cot B – m cot C9 Circumcircle of a triangle.
Circle passing through the angular point of a ABC is called its circumcircle. Its radius isdenoted by R. The circum centre may lie within, outside or upon one of the sides of the
A
B Cnm
352
triangle. In a right angled triangle the circum centre is the mid-point of hypotenuse.
R = Asin2a
= Bsin2b
= Csin2c
R = 4abc
10 Incircle of a triangleThe circle which touches the sides is called inscribed circle or incircle. Its radius is denoted byr.
r = s
r = (s–a) tan 2A
= (s–b)tan 2B
= (s–c)tan 2C
r =
2Acos
2Csin
2Bsina
=
2Bcos
2Csin
2Asinb
=
2Ccos
2Asin
2Bsinc
r = 4Rsin 2A
sin 2B
sin 2C
11 Escribed circles of a triangle.The circle which touches side BC and two sides AB & AC produced of ABC is calledescribed circle opposite to the angle A. Its radius is denoted by r1 .Similarly r2 & r3 denote theradii of escribed circles opposite to angles B & C.
r1 = a–s ; r2 = b–s ; r3 = c–s
r1 = s tan 2A
; r2 = s tan 2B
; r3 = s tan 2C
r1 =
2Acos
2Ccos
2Bcosa
; r2 =
2Bcos
2Ccos
2Acosb
r3 =
2Ccos
2Acos
2Bcosc
r1 = 4Rsin 2A
cos 2B
cos 2C
; r2 = 4Rcos 2A
sin 2B
cos 2C
; r3 = 4Rcos 2A
cos 2B
sin 2C
12 Orthocentre and Pedal triangleThe triangle MNP formed by joining the feet of the altitudes is called the pedal triangleThe distance of orthocentre O from vertices A,B,C are 2RcosA, 2RcosB and 2RcosCDistance of O from sides are 2RcosBcosC, 2RcosCcosA and 2RcosAcosB.
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In MNP,,M = –2C; MN = acosAN = –2B; NP = ccosCP = –2A; MP = bcosB
i.e. sides of pedal triangles are a cosA(Rsin2A), bcosB(Rsin2B) and ccosC(Rsin2C). Circumradii of the triangles OBC, OCA & OAB are equal.
13 Note :Orthocentre of ABC is the incentre of pedal triangle MNP..Incentre I of ABC is the orthocentre of I1,I2I3.Centroid of ABC lies on the line joining the circumcentre to the orthocentre and divides it inthe ratio 1:2Circumcentre of the pedal triangle bisects the line joining the circumcentre of the triangleto the orthocentre.
14 Excentral Triangle.The triangle formed by joining the three excentres I1, I2, & I3 of
ABC is called the excentral or excentric triangle.Here AI1 I2 I3
BI2 I1 I3& CI3 I1 I2
So clearly ABC is pedal triangle of excentral triangle I1I2I3
Sides of excentral trangle are 4R cos 2A
, 4R cos 2B
and 4Rcos 2C
and its angles are
2 – 2A
, 2 – 2B
, 2 – 2C
II1 = 4Rsin 2A
; II2 = 4Rsin 2B
; II3 = 4Rsin 2C
15 Nine point circleCircle circumscribing the pedal triangle of a given trianglebisects the sides of the given triangle and also the linejoining the vertices of the given triangle to the orthocentreof the given triangle. This circle is known as nine pointcircle.i.e. Nine point circle passes through the mid point ofthe sides, feet of the perpendiculars and the mid pointsof the line joining the orthocentre to the angular points.
16 Distance between special pointsDistance between excentre and circumcentre
OI1 = R 2Ccos
2Bcos
2Asin81
A
M N
P
O
B C
AI3 I2
I1
I
B C2B
A
B CL D
EFO
354
OI2 = R 2Ccos
2Bsin
2Acos81
OI3 = R 2Csin
2Bcos
2Acos81
Distance between cirumcentre & orthocentre is CcosBcosAcos8–1
Distance between cirumcentre & incentre is Rr2–R 2
Distance between incentre & orthocentre is CcosBcosAcosR4–r2 22
Perimeter (P) and area (A) of a regular polygon of n sides inscribed in a circle of radius r are
given by P = 2nrsin n and A = 21
nr2sin n2
.
Perimeter (P) and area (A) of a regular polygon of n sides circumscribed about a given circle
of radius r are given by P = 2nrtan n and A = 21
nr2tan n2
17 Length of medians (Apollonious rule)
m12 =
2cb 22
– 4a 2
m22 =
2ac 22
– 4b2
m32 =
2ba 22
– 4c2
Also m12+m2
2+m32 = 4
3(a2+b2+c2)
If A = 90°, then m22+m3
2 = 5m12
Area = 34
)m–m)(m–m)(m–m(m 321 where 2m = m1+m2+m3.
18 Altitudes
h1 = a2
, h2 = b2
, h3 = c2
,
Area is given by 1
= 4 321 h
1–h1
h1–
h1
h1–
h1
h1 where h
2 =
1h1
+2h
1+
3h1
19 Length of internal bisectors of angles A,B,C are given by cbbc2
cos 2A
; acca2
cos 2B
;
baab2
cos 2C
respectively
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20 Some useful results.
r1+r2 = 4Rcos2
2C
r3–r = 4Rsin2
2C
.
r1+r2+r3–r = 4Rrr1r2r3 = 2
1r1
+ 2r1
+ 3r1
= r1
2r1
+ 21r1
+ 22r1
+ 23r1
= 2
222 cba
s = 133221 rrrrrr = 21rr
= srrr 321 =
21
321
rrrrr
r = 21
321
rrrrr
R = 21
133221
rr4)rr)(rr)(rr(
a = 21
321
rr)rr(r
; b = 21
132
rr)rr(r
; c = 21
233
rr)rr(r
21 Solution of triangles(a) Solution of a general triangle.
Given To find Formulae used
a, b, c A, B, C Find = )c–s)(b–s)(a–s(s
sinA = bc2
, sinB = ca2
, sinC = ab2
,
or tan 2A
= )a–s(s etc.
or cosA = bc2
a–cb 222
etc.
(if a, b, c are sides)
356
a, b, C c, A, B 2BA
= 90° – 2C
& tan 2B–A
= bab–a
cot 2C
c = AsinCsina
(sine rule)
or c2= a2+b2–2abcos C (cosine rule)
a, A, B C, a, b C = 180° –(A+B)
(or c, A, B) b = AsinBsina
, c = AsinCsina
(sine rule)
(or b = CsinBsinc
, a = CsinAsinc
as the case
may be)
a, b, A c, B,C sinB = ab
sinA ........................(i)
C = 180° – (A+B)....................(ii)
c = AsinCsina
..............................(iii)
Now, the following cases arise.
A<90° A<90° A<90° A>90°& a<bsinA & a= bsinA & a>bsinA (ambiguous case)
Here sinB = aAsinb
sinB = 1 sinB = aAsinb
If a b, then B is
also obtuse whichis not possible
sin B>1not possible B = 90° gives 2 angles If a>b, then A>B & C No triangle is only one triangle such that will be an acute angle
possible is possible B1+B2 = 180° So solution existswhich is right two trianglesanlged at B are possible
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(b) Solution of a right angled triangleLet C = 90°
Given To find Formulae used
a,b A, B, c c2 = a2+b2
(two sides) tanA= ba
(or tanB = ab
)
B = 90°–A(or A = 90°–B)c,a A, B, b b2 = c2–a2
(hypotenuse & SinA = ca
one side) b = ccosA or b=a cot AB = 90° – A
a, A B, b, c B = 90° – A(Side and angle) b = a cot A
c = Asina
c,A B, a, b B = 90° – A(hypotenuse & angle) a = c sin A
b = c cos A
Solved Examples
1 If in ABC, a = 6, b = 3 and cos(A–B) = 54
then its area is
(a) 8 (b) 9 (c) 6 (d) None of theseSolution : Using Napier’s analogy,
tan 2B–A
= bab–a
cot 2C
and tan 2B–A
= )B–Acos(1)B–Acos(–1
31
= 363–6
cot 2C
tan 2B–A
= 54154–1
= 31
cot 2C
= 1
C = 90°
Area of = 21
absinC = 21
.6.3.sin90° = 9 square units
Ans (b)
358
2 If the angles A, B, C are the solutions of the equation tan3x–3k.tan2x–3tanx+k = 0, then theABC is
(a) isosceles (b) equilateral (c) acute angled (d) None of theseSolution: tanA, tanB, tanC are roots of the given equation
tanA+tanB+tanC = 3k, tanAtanB+tanBtanC+tanCtanA= –3 and tanA.tanB.tanC = – kBut tanA+tanB+tanC = tanAtanBtanC
3k = – k 4k = 0 gives k= 0tanAtanBtanC = 0 tanA = 0 or tanB = 0 or tanC = 0 which is not possible
Ans : (d)3 In ABC
sin4A+sin4B+sin4C = sin2Bsin2C+2sin2Csin2A+2sin2Asin2B, then A =
(a) 6 , 65
(b) 6 , 35
(c) 65
, 3 (d) None of these
Solution : Given equation is,a4+b4+c4 = b2c2+2c2a2+2a2b2 (using sine rule)
a4+b4+c4+2b2c2–2c2a2–2a2b2 = 3b2c2.(b2+c2–a2)2 = 3b2c2
22
2222
cb4)a–cb(
= 43
2222
bc2a–cb
= 2
23
cos2A = 43
cosA = 23
A = 6 , 65
Ans (a)4 If the median of a triangle through A is perpendicular to AB, then
(a) 2tanA+tanB = 0 (b) 2tanA–tanB = 0 (c) tanA–2tanB = 0 (d) tanA+2tanB=0Solution :tanA = – tan (180–A)
= – 2 cx
= – 2 cx
= – 2tanBtanA+2tanB = 0Ans (d)
C
D
E A
2x
Bc
b
x
–Ac
2a
2a
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5 In ABC, medians AD and BE are drawn. If AD = 4, DAB = 6 and ABE = 3 , then the area
of ABC is (in square units)
(a) 3364
(b) 338
(c) 3316
(d) 3332
Solution :In ABG
cos 6 = ABAG
23
= 32
ABAD
( AD = 4)
23
= AB342
AB = 3316
Area of ABD = 21
AB.AD. sin 6
= 21
. 3316
.4. 21
= 3316
units2
Area of ABC = 2 area of ABD = 3332
units2
Ans (d)6 In ABC, a2+b2+c2 = ac+ 3 ab, then the triangle is
(a) equilateral (b) isosceles (c) right angled (d) None of theseSolution :a2+b2+c2–ac– 3 ab = 0
ac–c2+ ab3–b
4a3 2
2
= 0
2
c–2a
+
2
b–2
a3 = 0
2a
= c & a23
= b
A
B C
E
G
6
3
D
360
a = 2c & a = 3b2
a = 2c = 3b2
=
a = , c = 2 , b = 23
a2 = b2 + c2
right angledAns (c)
7 In ABC, if cotA+cotB+cotC = 3 , then show that the triangle is equilateralSolution :In ABC, tanA+tanB+tanC = tanAtanBtanC
cotBcotC+cotCcotA+cotBcotA = 1 ..........................(1)Let cotA=x, cotB=y, cotC=zgiven that x+y+z = 3Squaring,
x2+y2+z2+2xy+2yz+2zx =3×1x2+y2+z2+2xy+2yz+2zx = 3 (xy+yz+zx) using (1)
i.e. x2+y2+z2–xy–yz–zx = 02x2+2y2+2z2–2xy–2yz–2zx = 0(x–y)2+(y–z)2+(z–x)2 = 0x=y & y = z & z = xx = y = zcotA = cotB = cotCA = B = CThe triange is equilateral.
1 In ABC, B = 3 , C = 4 , Let D divides BC internally in the ratio 1:3, then CADsinBADsin
is equal to
(a) 61
(b) 31
(c) 31
(d)32
2 If in PQR, sinP, sinQ, sinR are in A.P., then(a) The altitudes are in A.P (b) The medians are in G.P(c) The altitudes are in H.P (b) The medians are in A.P
3 In PQR, R = 2 , if tan 2P
and tan 2Q
are the roots of ax2+bx+c = 0 (a 0), then
(a) a+b = c (b) b+c = a (c) c+a = b (d) b = c
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PRACTICE QUESTIONS
4 In ABC, 2ac sin 2)CB–A(
=
(a) a2+b2–c2 (b) c2+a2–b2 (c) b2–c2–a2 (d) c2–a2–b2
5 In ABC, let C = 2 . If r is the inradius and R is the circumradius of the triangle, then
2(r+R) is equal to(a) a+b (b) b+c (c) c+a (d) a+b+c
6 In radius of a circle which is inscribed in an isosceles triangle one of whose angle is 32
is 3 , then
area of triangle is(a) 4 3 (b) 12–7 3 (c) 12+7 3 (d) None of these
7 If the angles A,B&C of a triangle are in A.P and if a, b and c denote the lengths of the sides
opposite to A,B&C respectively, then the value of the expression ca
sin2C+ ac
sin2A is
(a) 21
(b)23
(c) 1 (d) 3
8 Read the following passage and answer the questions.Consider the circle x2+y2 = 9 and the parabola y2 = 8x. They intersect at P & Q in the first andfourth quadrants, respectively. Tangents to the circle at P & Q intersect the x–axis at R andtangents to the parabola at P&Q intersect the x axis at S.(i) The ratio of the areas of PQS and PQR is(a) 1: 2 (b) 1:2 (c) 1:40 (d) 1:8(ii) The radius of the circumcircle of the PRS is(a) 5 (b) 3 3 (c) 3 2 (d) 2 3(iii) The radius of the incircle of the triangle PQR is
(a) 4 (b) 3 (c) 38
(d) 2
9* Internal bisector A of ABC meets side BC at D.A line drawn through D perpendicular to ADintersects the side AC at E & side AB at F. If a,b,c represent sides of ABC, then
(a) AE is the HM of b & c (b) AD = cbbc2
cos 2A
(c) EF = cbbc4
sin 2A
(d) AEF is isosceles
10* A straight line through the vertex P of a PQR intersects the side QR at the point S and thecircumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then
(a) PS1
+ ST1
< SRQS2
(b) PS1
+ ST1
> SRQS2
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(c) PS1
+ ST1
< QR4
(d) PS1
+ ST1
> QR4
11* In a ABC with fixed base BC, the vertex A moves such that cosB+cosC = 4sin2
2A
. If a, b and c
denote the lengths of the sides of the triangle opposite to the angles A, B and C respectively, then(a) b+c = 4a (b) b+c = 2a (c) locus of point A is an ellipse(d) locus of point A is a pair of straight line
12 Match the following :Column I Column II
(a) In ABC , if aAcos
= bBcos
= cCcos
(p) 2 63
and the side a = 2, then area of the triangle is (q) 61
(b) In ABC, a b c, if CsinBsinAsin
cba333
333
= 7,
then the maximum possible value of a is (r) 3 7(c) Two sides of a triangle are given by the roots of the
equation x2–2 3 x+2 = 0. the angle between the
sides is 3 .The perimeter of the triangle is (s) 3
13 In ABC if cosA+CosB+cosC = 47
, then rR
=
(a) 34
(b) 43
(c) 32
(d) 23
14 In ABC, sides a,b,c are in A.P and !9!12
+ !7!32
+ !5!51
= )!b2(8a
then the maximum value of
tanAtanB is
(a) 21
(b) 31
(c) 41
(d) 51
15 If A1A2A3......An be a regular polygon of n sides and 21AA
1 =
31AA1
+ 41AA
1, then
(a) n = 5 (b) n = 6 (c) n = 7 (d) None of these
'Note : Questions with * have more than one correct option'Answers
1. a 2. c 3. a 4. b 5. a 6. c7. d 8 (i) c (ii) b (ii) d 9. a,b,c,d 10. b,d11. b,c 12. a s; b r; c p 13.a 14. b 15. c
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TRIGONOMETRY - IInverse Trigonometric Functions
A ƒunction ƒ: A B is invertible iff it is a bijection. The inverse of ƒ is denoted by ƒ–1 and is definedas ƒ–1(y) = x ƒ(x)=y. Trigonometric functions are periodic and hence they are not bijective. But ifwe restrict their domains and codomains they can be made bijective and we can obtain their inverses.
1 Sin–1 x:The symbol sin–1x or arcsinx denote the angle so that sin = x. As a direct meaning, sin–1x isnot a function, as it does not satisfy the requirements for a rule to become a function. But by asuitable choice [–1,1] as its domain and standardized set [– /2, /2] as its range, then rule sin–
1x is a single valued functionThus sin–1x is considered as a function with domain [–1,1] and range [– /2, /2]The graph of y = sin–1x is as shown below which is obtained by taking the mirror image, of theportion of the graph of y = sin x from x = – /2 to x = /2, on the line y = x
cos–1x:By following the discussions, similar to above, we have cos–1 x or arccos x as a function with domain[–1,1] and range [0, ]The graph of y = cos–1 x is similarly obtained as the mirror image of the portion of the graph of y =cos x from x = 0 to x =
tan–1x:We get tan–1x or arctanx as a function with domain R and range (– /2, /2). Graph of y = tan–1x
364
365
cosec–1x:cosec–1x or arccosec x is a function with domain R–(–1,1) and range [– /2, /2] – {0}. Graph ofy = cosec–1x
sec–1x:sec–1x or arcsec x is a function with domain R–(–1,1) and range [0, ] – { /2}.Graph of y = sec–
1x is
cot–1x:cot–1x or arccot x is a function with domain R and range (o, ). Graph of y = cot–1x is
Property : “–x”The graphs of sin–1x, tan–1x, cosec–1x are symmetric about origin.Hence we get sin–1(–x) = –sin–1x
tan–1(–x) = –tan–1xcosec–1(–x) = –cosec–1x
Also the graphs of cos–1x, sec–1x, cot–1x are symmetric about the point (0, /2). From this,we get
cos–1(–x) = –cos–1xsec–1(–x) = –sec–1xcot–1(–x) = –cot–1x
366
Notes :
(i) x2+y2 1 & x,y 0 sin–1x + sin–1 y 2 and x2+y2 1 & x,y 0
2 sin–1x + sin–1y
(ii) xy<1and x,y 0 0 tan–1x +tan–1y< 2 ; xy>1 and x,y 0 2 <tan–1x +tan–1y<
(iii) For x<0 or y<0 these identities can be used with the help of property “–x”i.e. change x or y to –x or –y which are positive
Domain & range of inverse trigonometric FunctionsFunction Domain Range (Principal value branch)sin–1x [–1,1] 2/–,2/–cos–1x [–1,1] [0, ]tan–1x (– (– /2, /2)cot–1x (– (0, )cosec–1x ( – –1] [1, ) [– /2,0) (0, /2]sec–1x ( – –1] [1, ) [0, /2) ( /2, ]
Note : If no branch of an inverse trigonometric function is mentioned, then it means the principal valuebranch of the function.
Properties of Inverse Trigonometric Functions1
(i) sin–1(sinx) = Zn,
21n2x
2–1n2,x–1n2
Zn,2
n2x2
–n2,xn2–
Y
X0 2
y=–x y=
x–2
y=x
Period = 2 & it is an odd function.
(ii) cos–1(cosx) = Zn,n2x)1–n2(,x–n2Zn,1n2xn2,xn2–
367
Y
X
y=2
+xy=2
–x
2–2 –
y=–x y=x
0
Period=2 and it is an even function
(iii) tan–1(tanx) = Zn,2
nx2
–n,xn–
Y
y=+x
y=–x
y=x
/2/2 0
Period = (iv) cot–1(cotx) = Zn,)1n(xn,xn–
Y
X
y=+x
y=x–y=
x
0
Period =
(v) sec–1(secx) = Zn
2–n2(x,n2x)1n2(,n2x–
22x,)1n2(xn2,n2–x
368
Y
y=2
+x
y=2–x
y=–x y=x
X0
Period = 2
(vi) cosec–1(cosecx) = Zn,n
2)1n2(x
2–)1n2(,x–)1n2(
2n2x
2–n2n2–
Y
X
y=–x y=
x–2y=–x
y=x 0
Period = 2 2. (i) sin(sin–1x) = x, –1 x 1 (ii) cos(cos–1x) = x, –1 x 1
Y
X0 1
1
–1–1
Y
X0 1
1
–1–1
(iii) tan(tan–1x) = x, x R (iv) cot(cot–1x) = x, x R
369
Y
X0 1
1
–1–1
Y
X0 1
1
–1–1
(v) sec(sec–1x) = x, x R (– –1] [1,
Y
X0 1
1
–1–1
(vi) cosce(cosce–1x)= x, x R (– –1] [1,
Y
X0 1
1
–1–1
3. (i) sin–1x+cos–1x= 2 , –1 x 1
(ii) tan–1x+cot–1x = 2 , x R
(iii) sec–1x+cosce–1x = 2 , x R (– –1] [1,
4. (i) sin–1x=cosec–1
x1
, –1 x 1
370
(ii) cos–1x=sec–1x1
, –1 x 1
(iii) tan–1x= 0x,x1cot–
0x,x1cot
1–
1–
5 (i) sin– 1(–x) = –sin–1x, –1 x 1(ii) cos–1(–x) = –cos–1x, –1 x 1(iii) tan–1(–x) = –tan–1 x, – x<(iv) cot–1(–x) = –cot–1x, – x<(v) cosec–1(–x) = –cosec–1x, x –1or x 1(vi) sec–1(–x ) = –sec–1x, x –1or x 1
6. Conversions of inverse trigonometric functions
(i)x 1
2x–1
sin–1x = 2
1–21–
x–1xtanx–1cos
= 2
1–1–2
1–
x–11sec
x1
xx–1cot
(ii)
x
12x–1
cos–1x =x
x–1tanx–1sin2
1–21–
= x1sec
x–11
x–1xcot 1–
2
1–
2
1–
(iii)x
1
2x+1
Cosec
Cosec
371
tan–1x =
21–2
1–
1–
2
1–
2
1–
x1x
x1eccos
x1cot
x11
x1sin
7. (i)
1yx&0y,x1–ifx–1yy–1xsin––
1yx&1y,x0ifx–1yy–1xsin–
1yx&0xyifor
1yx&1y,x1–ifx–1yy–1xsin
ysinxsin
22221–
22221–
22
22221–
1–1–
(ii)
1yx&1y0,0x1–ifx–1y–y–1xsin––
1yx&0y1–,1x0ifx–1y–y–1xsin–
1yx&0xyifor
1yx&1y,x1–ifx–1y–y–1xsin
ysin–xsin
22221–
22221–
22
22221–
1–1–
8. (i)0yx&1y,x1–ify–1x–1–xycos–2
0yx&1y,x1–ify–1x–1–xycosycosxcos
221–
221–1–1–
(ii)yx&1x0,0y1–ify–1x–1xycos–
yx&1y,x1–ify–1x–1xycosycos–xcos
221–
221–1–1–
9. (i)
ifxy–1yxtan–
ifxy–1yxtan
xifxy–1yxtan
ytanxtan
1–
1–
1–
1–1–
(ii)
–ifxy1y–xtan–
–ifxy1y–xtan
–xifxy1y–xtan
ytan–xtan
1–
1–
1–
1–1–
Cos
Sec
372
Remark : If x1, x2,..........xn R, then tan–1x1+tan–1x2+.........+tan–1xn
= tan–1......s–ss–1
........s–ss–s
642
7531
Where sk is the sum of the product of x1,x2,............ xn taken k at a time.ie. s1 = x1 + x2+..........+xn = xi
s2 = x1x2+x2x3+.......+xn–1xn= x1x2.s3 = x1x2x3.................etc.
10. (i)
21–x1–if ,x–12xsin––
1x2
1if ,x–12xsin–
21x
21–if ,x–12xsin
xsin2
21–
21–
21–
1–
(ii)
21–x1–if ,4x–3xsin––
1x21if ,4x–3xsin–
21x
21–if ,4x–3xsin
xsin3
31–
31–
31–
1–
11. (i) 0x1–if ,1–2xcos–21x0 if ,1–2xcos
xcos221–
2–11–
(ii)
21–x1–if ,3x–4xcos2
21x
21–if ,3x–4xcos–2
1x21if ,3x–4xcos
xcos3
31–
31–
31–
1–
12. (i)
–1 xif ,x–1
2xtan–
1 xif ,x–1
2xtan
1x1– if ,x–1
2xtan
xtan2
21–
21–
21–
1–
373
(ii)
31 xif ,
3x–1x–3xtan–
31 xif ,
3x–1x–3xtan
31x
31– if ,
3x–1x–3xtan
xtan3
2
31–
2
31–
2
31–
1–
Note : If |x| 1 then 21–
2
21–
21–1–
x–1x2tan
x1x–1cos
x1x2sinxtan2 .
If |x|>1, change x to x1
in the above.
Note : In cases of identities in inverse trigonometric functions, principal values are to be taken. As suchsigns of x,y etc., will determine the quadrant in which the angles will fall. In order to bring the anglesof both sides in the same quadrant, adjustment by is to be made.
1 Evaluate tan–1 tan(–6)Solution:
We know that tan–1 (tan ) = if – 2 < 2
(2 –6) 2,
2–
tan(2 –6) = – tan6 = tan(–6)Therefore tan–1tan(–6) = tan–1 tan (2 –6) = 2 –6
2 If 2tan–1x+sin–12x1
x2 is independent of x, then
(a) x [1, ) (b) x [–1,1] (c) x (– ,–1] (d) None of theseSolution:
Sin–1 2x = 1|x,xtan2
1–x)xtan2(–1x,xtan2–
1–
1–
1–
2tan–1x+ –2tan–1x = when x [1, )and 2tan–1x– –2tan–1x = – when x (– ,–1]Ans a,c.
3 If 0<x<1, then 2x1 [{xcos(cot–1x)+sin(cot–1x)}2–1]1/2 =
(a) 2x1x
(b) x (c) x 2x1 (d) 2x1
374
Solution:
2x1
2/12
2
1–
2
1– 1–x1
1sinsinx1
xcoscosx
2x1
2/12
221–
x11
x1xx
2x1 2/12
2 1–x1 = x.x1 2
Ans (c)
4 If x 2,
2– , then the value of tan–1
4xtan
+ tan–1x2cos35
x2sin3 is
(a) 2x
(b) 2x (c) 3x (d) x
Solution:
tan–14
xtan + tan–1
xtan1)xtan–1(35
xtan1xtan6
2
2
2
tan–14
xtan + tan–1 xtan28
xtan62
tan–1
)xtan4(4xtan3–1
xtan4xtan3
4xtan
2
2
2
as xtan4xtan3.
4xtan
2 < 1
tan–1 xtan16xtanxtan16
2
3
tan–1 (tanx)x
Ans d.5 An integral solution of the equation
tan–1x+tan–1 y1
= tan–13 is
(a)(2,7) (b) (4,–13) (c) (5,–8) (d) (1,2)
375
Solution:
tan–1x+tan–1 y1
= tan–1 3
tan–1
yx–1
y1x
= tan–1 3 if yx
<1
yx–1
y1x
= 3
x+ y1
= 3– yx3
y = x–31x3
which satisfied by options a, b, c and dAns a, b, c, d
6 Sum to the n terms of the series
cosec–1 10 +cosec–1 50 +cosec–1 170 +............+cosec–1 )2n2n)(1n( 22 is
(a) 0 (b) (c) tan–1(n+1) – 4 (d) cot–1(n+1) – 4Solution
Let = cosec–1 )2n2n)(1n( 22
cosec2 = (n2+1) (n2+2n+2)= (n2+1)2 + 2n(n2 +1) +n2+1
1+cot2 = (n2+n+1)2+1cot = n2+n+1
tan = 1nn1
2 = n)1n(1n–)1n(
= tan–1(n+1) –tan–1nThus, sum to n terms of the series
= (tan–12–tan–11) + (tan–13–tan–12)+...........+(tan–1(n+1)–tan–1n)= tan–1(n+1)–tan–11
= tan–1(n+1) – 4
376
Ans. c7 In a ABC, if A = tan–12, B = tan–13 then C is equal to
(a) 3 (b) 4 (c) 6 (d) None of these
Solution:A+B+C =C = –(A+B)= –(tan–12+tan–13)
= – 6–15tan 1–
xy>1
= – +tan–11
= 4Ans. b
1 If x satisfies the inequation x2–x–2>0, then a value exits for(a) sin–1x (b) sec–1x (c) cos–1x (d) None of these
2 If [sin–1x] +[cos–1x] = 0, where x is a non negative real number and [.] denotes the greatestinteger function, then complete set of values of x is(a) (cos1,1) (b) (–1,cos1) (c) (sin1,1) (d) (cos1,sin1)
3 If cos–1x+cos–1y+cos–1z = 3 , then xy+yz+zx is(a) –3 (b) 0 (c) 3 (d)
4 If sin–1x+sin–1y = 32
then cos–1x+cos–1y =
(a) 32
(b) 3 (c) 6 (d)
5 The greatest of tan1, tan–11, sin–11, sin1, cos1 is(a) sin1 (b) tan1 (c) tan–11 (d) None of these
6 The value of cos–1(cos12)–sin–1(sin12) is(a) 0 (b) (c) 8 –24 (d) None of these
7 cot–1 cos –tan–1 cos = x, then sin x =
(a) tan2
2(b) cot2
2 (c) tan (d) cot 28 If sin–1x = 2sin–1 a has a solution for
(a) |a|2
1(b) |a|
21
(c) all real values of a (d) |a|< 21
PRACTICE QUESTIONS
377
9 If cos–1x–cos–1
2y
= , then 4x2–4xycos +y2 is equal to
(a) 2sin2 (b) 4 (c) 4sin2 (d) –4sin2
10 If sin–15x
+cosec–145
= 2 , then the value of x is
(a) 4 (b) 5 (c) 1 (d) 3
11 The value of cot 32tan
35eccos 1–1–
is
(a) 176
(b) 173
(c) 174
(d) 175
Assertion | Reasoning
12 Let f(x) = sin–12x1
x2
Statement-1 : f (2) = – 52
and
Statement-2 : sin–1 2x1x2
= – 2tan–1x, x>1.
(a) Statement-1 is True, statement-2 is True ;statement-2 is a correct explanationfor statement-1
(b) Statement-1 is True, statement -2 is True ; statement - 2 is NOT a correct explanation forstatement - 1
(c) Statement -1 is True, statement - 2 is False(d) Statement -1 is False, statement - 2 is True.Comprehension (Q.No. 13 to 15)
Given that tan–12x–1
x2 =
1–x,xtan21x,xtan2–
1|x|,xtan2
1–
1–
1–
sin–12x–1
x2 =
1–x),xtan2(–1x,xtan2–
1|x|,xtan2
1–
1–
1–
and sin–1x+cos–1x = 2 for –1 x 1.
378
13 sin–14x
x42 +2tan–1
2x– is independent at x then :
(a) x [1, ) (b) x [1,1] (c) [–2,2] (d) x [–3,4]
14 If (x–1) (x2+1)>0, then sin xtan–x–1x2tan
21 1–
21–
=
(a) –1 (b) 1 (c) 21
(d) None of these
15 If cos–12x91
x6 = – 2 + 2 tan–1 3x then x
(a) (– ,–1) (b) 31,
31– (c) ,
31
(d) None of these
16 Match the followingColumn I Column II
(a) If cos–1a+cos–1b+cos–1c = 3 then ab+bc+ca is (p) 2n
(b)10
1ii
1– xcos = 0, then 10
1iix (q) sin–1x – 6
(c)n2
1ii
1– xsin = n , thenn2
1iix is (r) 3
(d) f(x) = sin–12x–1
21–x
23
, – 21
x 1 is (s) 10
Answers1. b 2. d 3. c 4. b 5. b 6. c 7. a 8. b9. c 10. d 11. a 12. a 13. c 14. a 15. c16. a r, b s, c p, d q
379
TRIGONOMETRY IIInverse Trigonometric Functions
Hyperbolic functions(i) sinh(–x) = – sinhx odd function
cosh(–x) = coshx even functiontanh(–x) = –tanhx odd function
(ii) Function Domain Rangesinh–1x R Rcosh–1x (0, ) (1, )tanh–1x R (–1,1)coth–1x R–{0} R–[–1,1]sech–1x (0, ) (0,1)cosech–1x R–{0} R–{0}
(iii) sinh (sinh–1x) = x sinh–1(sinhx) = xcosh (cosh–1x) = x cosh–1(coshx) = xtanh (tanh–1x) = x tanh–1(tanhx) = xsinh (sin –1x)=xn sinh (sinh–1xn) = xn
(iv) sinh–1x = loge 1xx 2
cosh–1x = loge 1–xx 2
tanh–1x = 21
loge 1–x1x
x > 1, x< –1
coth–1x = 21
loge 1x1–x
x > 1, x< –1
sech–1x = loge xx–11 2
0 < x 1
cosech–1x = 0xif
xx1–1log
0xifx
x11log
2
e
2
e
(v) sinh–1x = cosech–1x1
sinh–1x = cosh–1 1x2
cosh–1x = sinh–1 1–x2
sinh (cosh–1x) = 1–x2
380
1 Total number of positive integral values of n sothat the equation cos–1x+(sin–1y)2 = 4
n 2
and
(sin–1y)2–cos–1x = 16
2
are consistent, is equal to
(a) 1 (b) 4 (c) 3 (d) 2Solution
we have 2(sin–1y)2 = 16
)1n4( 2
016
)1n4( 2
4
2
– 41
n 47
Also 2cos–1x = 16
)1–n4( 2
016
)1–n4( 2
41
n8
+ 41
n = 1Ans (a)
2 The minimum value of (sin–1x)3+(cos–1x)3 is equal to
(a)32
3
(b)325 3
(c)329 3
(d)32
11 3
SolutionLet y = (sin–1x)3+(cos–1x)3
= (sin–1x+cos–1x) {(sin–1x)2+(cos–1x)2–sin–1x.cos–1x}
= 2 {(sin–1x+cos–1x)2–3sin–1x.cosx}
= 2xsin–
2xsin3–
41–1–
2
= 2 4xsin
23–)x(sin3
21–21–
= 2 4163–
16.3xsin
23–)x(sin3
2221–21–
= 2 164–xsin3
221–
2
2
On adding the two given equations
On subtracting the two given equations
381
Minimum value of y = 2 .16
2
= 32
3
since 2
1–
4–xsin 0
Ans: (a)
3 If A = 2tan–1 1–22 and B = 3sin–131
+sin–153
. then
(a) A = B (b) A<B (c) A>B (d) None of theseSolution
A = 2tan–1 1–22 = 2tan–1(1.828)>2tan–1 3
A> 32
3sin–1
31
= sin–1
3
314–
31.3 = sin–1
274–1 = sin–1
2723
= sin–1(0.852)
3sin–1
31
<sin–123
sin–153
= sin–1(0.6)< sin–123
B< 3 + 3 = 32
Hence A>BAns (c)
4 The complete solution set of sin–1(sin5)>x2–4x is(a) |x–2| < 2–9 (b) |x–2| > 2–9 (c) |x| < 2–9 (d) |x| > 2–9Solution
sin–1sin5>x2–4xsin–1sin(5–2 )>x2–4x
5–2 >x2–4x (5–2 ) 2,
2–
x2–4x+2 –5<0(x–2)2<9–2|x–2|< 2–9
Ans (a)
5 Let (x,y) be such that sin–1(ax)+cos–1y+cos–1bxy = 2 .
Match the statements in column I with statements in column IIColumn I Column II
(a) If a = 1 and b = 0, then (x,y) (p) Lies on the circle x2+y2 = 1(b) If a = 1 and b = 1, then (x,y) (q) Lies on the (x2–1) (y2–1) = 0
382
(c) If a = 1 and b = 2, then (x,y) (r) Lies on y = x(d) If a = 2 and b = 2, then (x,y) (s) Lies on the (4x2–1) (y2–1) = 0Solution
cos–1y+cos–1bxy = 2 – sin–1 ax
cos–1y+cos–1bxy = cos–1axLet cos–1y = A, cos–1bxy = B and cos–1ax = C
A+B = CB = A–C
cos(A–C) = cosBcosAcosC+sinAsinC = cosB
y ax+sinA sinC = bxysin A sin C = bxy–axysin2A sin2C = (b–a)2x2y2
(1–a2x2)(1–y2) = x2y2(b–a)2
(a) Put a = 1 and b = 0(1–x2)(1–y2) = x2y2
x2+y2 = 1(b) Put a = 1 and b = 1
(1–x2)(1–y2) = 0(x2–1)(y2–1) = 0
(c) Put a = 1 and b = 2(1–x2) (1–y2) = x2y2
x2+y2 = 1(d) Put a = 2 and b = 2
(1–4x2)(1–y2) = 0(4x2–1)(y2–1) = 0
(a) (p), (b) (q), (c) (p), (d) (s)
6 If sin–1 ..........4x
2x–x
32
+ cos–1 ..........4x
2x–x
642
= 2 for 0<|x|< 2 , then x equals
(a) 21
(b) 1 (c) – 21
(d) –1
Solution
sin–1 ..........4x
2x–x
32
= 2 – cos–1 ..........4x
2x–x
642
sin–1 ..........4x
2x–x
32
= sin–1 ..........4x
2x–x
642
..........4x
2x–x
642 = ..........
4x
2x–x
32
or sin + cos = 1/2–1 –1
so,
383
2x––1
x2
2
= 2x––1
x1r
r–1aS
2
2
x2x2
= x2x2
2x2+x3 = 2x+x3
2x(x–1) = 0x = 0, 1x = 1 (0<|x|< 2 )
Ans (b)7 Let a, b, c be positive real numbers. Let
= tan–1
bc)cba(a
+ tan–1
ca)cba(b
+ tan–1
ab)cba(c
, then tan =
(a) 4 (b) 2 (c) (d) None of these
SolutionLet a+b+c = u
= tan–1
bcau
+tan–1
cabu
+tan–1
abcu
zx–yz–xy–1xyz–zyxtan
ztanytanx tanusecan you Also
1–
–1–1–1
= + tan–1
acbc
bcau–1
cabu
bcau
+ tan–1
abcu
, bcau
cabu
= cu
= ccba
= cba
+1>1
= + tan–1
cu–1
abcu)ba(
+ tan–1
abcu
= + tan–1
)u–c(c.
abcu)c–u( + tan–1
abcu
= – tan–1
abuc + tan–1
abcu
= Ans (c)
384
1 If tan–1x+tan–1y+tan–1z = or 2 then
(a) x+y+z = 3xyz (b) x+y+z = 2xyz (c) xy+yz+zx = 1 (d) None of these2 If [cos–1x] + [cot–1x] = 0 , where x is a nonnegative real number and [.] denotes the greatest integer
function then complete set of values of x is(a) (cos1,1) (b) (cot1,1) (c) (cos1,cot1) (d) None of there
3 Range of the function f(x) = cos–1(–{x}), where {.} is fractional part function is
(a) ,2 (b) ,
2 (c) ,2 (d) 2
,0
4 The sum of solutions of the equation 2sin–1 1xx2 +cos–1 xx2 = 23
is
(a) 0 (b) –1 (c) 1 (d) 25* Which of the following is a rational number
(a) sin 31tan3tan 1–1–
(b) cos 43sin–
21–
(c) log2 863sin
41sin 1–
(d) 35cos
21tan 1–
Assertion | Reasoning
6 Statement 1 : sin–1e
1>tan–1
1
Statement 2 : sin–1x > tan–1y for x>y, x,y (0,1)(a) Statement -1 is True, Statement -2 is True, Statement -2 is a correct explanation for
Statement- 1.(b) Statement -1 is True, Statement-2 is True, statement-2 is NOT correct explanation for
Statement - 1(c) Statement-1 is True, Statement -2 is False(d) Statement 1 is False, Statement -2 is True.
Comprehension (Q.No. 7 to 9)
It is given that A = (tan–1x)3+(cot–1x)3 where x>0 and B = (cos–1t)2+(sin–1t)2 where t 21,0
and sin–1x+cos–1x = 2 for –1 x 1 and tan–1x+cot–1x = 2 for all x R.
7 The interval in which A lies is
(a) 2,
7
33
(b) 10,
40
33
(c) 8,
32
33
(d) None of these
8 The maximum value of B is
PRACTICE QUESTIONS
385
(a)8
2
(b)16
2
(c)4
2
(d) None of these
9 If least value of A is m and maximum value of B is M then cot–1cot MM–m
=
(a) – 87
(b) 87
(c) – 8 (d) 8Single Integer Answer type Question
10 The number of all positive integral solutions of the equation tan–1x+cos–1 2y1y
= sin–1103
,
are .....................
11 If cos–1(4x3–3x) = a+b
cos–1x, for –1<x<– 21
then [a+b+2] is .............
12 Match the statement of column I with values of column II
Column I Column II
(a) Absolute difference of greatest and least value
of 2 (sin2x–cos2x) (p) 4(b) Absolute difference of greatest and least value
of x2–4x+3, x [1,3] is (q) 6(c) Greatest value of tan–1
x1x–1
, x [0,1], is (r) 4
(d) Absolute difference of greatest and least
value of cos–1x2, x 2
1,2
1– , is (s) 1
13* If a tan–1x+cot–1x+sin–1x b, then
(a) a = 4 (b) a = 0 (c) b = 2 (d) b =
14 cot–1(2.12)+cot–1(2.22)+cot–1(2.32)+..............upto is equal to
(a) 4 (b) 3 (c) 2 (d)
15 Number of solutions of the equation tan–11x2
1+tan–1
1x41
= tan–12x
2 is
(a) 1 (b) 2 (c) 3 (d) 4
'Note : Questions with * have more than one correct option'
Answers
1. c 2. c 3. c 4. b 5. a, b, c 6. a 7. c 8. c 9.d
10. 2 11. –2 12. a r, b s, c 0, d q 13. b,d 14. a 15. a
386
TRIGONOMETRY - IIIInverse Trigonometric Functions - Problem Solving
A ƒunction ƒ: A B is invertible iff it is a bijection. The inverse of ƒ is denoted by ƒ–1 and isdefined as ƒ–1(y) = x ƒ(x)=y. Trigonometric functions are periodic and hence they are notbijective. But if we restrict their domains and codomains they can be made bijective and wecan obtain their inverses.
Domain & range of inverse trigonometric FunctionsFunction Domain Range (Principal value branch)sin–1x [–1,1] 2/–,2/–cos–1x [–1,1] [0, ]tan–1x (– (– /2, /2)cot–1x (– (0, )cosec–1x ( – –1] [1, ) [– /2,0) (0, /2]sec–1x ( – –1] [1, ) [0, /2) ( /2, ]
Note : If no branch of an inverse trigonometric function is mentioned, then it means the principalvalue branch of the function.
Properties of Inverse Trigonometric Functions1
(i) sin–1(sinx) = Zn,
21n2x
2–1n2,x–1n2
Zn,2
n2x2
–n2,xn2–
Y
X0 2
y=–x y=
x–2
y=x
Period = 2 & it is an odd function.
(ii) cos–1(cosx) = Zn,n2x)1–n2(,x–n2Zn,1n2xn2,xn2–
Y
X
y=2
+xy=2
–x
2–2 –
y=–x y=x
0
Period=2 and it is an even function
387
(iii) tan–1(tanx) = Zn,2
nx2
–n,xn–
Y
y=+x
y=–x
y=x
/2/2 0
Period = (iv) cot–1(cotx) = Zn,)1n(xn,xn–
Y
X
y=+x
y=x–y=
x
0
Period =
(v) sec–1(secx) = Zn
2–n2(x,n2x)1n2(,n2x–
22x,)1n2(xn2,n2–x
Y
y=2
+x
y=2–x
y=–x y=x
X0
Period = 2
388
(vi) cosec–1(cosecx) = Zn,n
2)1n2(x
2–)1n2(,x–)1n2(
2n2x
2–n2n2–
Y
X
y=–x y=
x–2y=–x
y=x 0
Period = 2 2. (i) sin(sin–1x) = x, –1 x 1 (ii) cos(cos–1x) = x, –1 x 1
Y
X0 1
1
–1–1
Y
X0 1
1
–1–1
(iii) tan(tan–1x) = x, x R (iv) cot(cot–1x) = x, x R
Y
X0 1
1
–1–1
Y
X0 1
1
–1–1
389
(v) sec(sec–1x) = x, x R (– –1] [1,
Y
X0 1
1
–1–1
(vi) cosce(cosce–1x)= x, x R (– –1] [1,
Y
X0 1
1
–1–1
3. (i) sin–1x+cos–1x= 2 , –1 x 1
(ii) tan–1x+cot–1x = 2 , x R
(iii) sec–1x+cosce–1x = 2 , x R (– –1] [1,
4. (i) sin–1x=cosec–1
x1
, –1 x 1
(ii) cos–1x=sec–1x1
, –1 x 1
(iii) tan–1x= 0x,x1cot–
0x,x1cot
1–
1–
5 (i) sin– 1(–x) = –sin–1x, –1 x 1(ii) cos–1(–x) = –cos–1x, –1 x 1(iii) tan–1(–x) = –tan–1 x, – x<(iv) cot–1(–x) = –cot–1x, – x<
390
(v) cosec–1(–x) = –cosec–1x, x –1or x 1(vi) sec–1(–x ) = –sec–1x, x –1or x 1
6. Conversions of inverse trigonometric functions
(i)x 1
2x–1
sin–1x = 2
1–21–
x–1xtanx–1cos
= 2
1–1–2
1–
x–11sec
x1
cosecx
x–1cot
(ii)
x
12x–1
cos–1x =x
x–1tanx–1sin2
1–21–
= x1sec
x–11
x–1xcot 1–
2
1–
2
1–
(iii)x
1
2x+1
tan–1x =
21–2
1–
1–
2
1–
2
1–
x1x
x1eccos
x1cot
x11
x1sin
7. (i)
1yx&0y,x1–ifx–1yy–1xsin––
1yx&1y,x0ifx–1yy–1xsin–
1yx&0xyifor
1yx&1y,x1–ifx–1yy–1xsin
ysinxsin
22221–
22221–
22
22221–
1–1–
cosec
cos
sec
391
(ii)
1yx&1y0,0x1–ifx–1y–y–1xsin––
1yx&0y1–,1x0ifx–1y–y–1xsin–
1yx&0xyifor
1yx&1y,x1–ifx–1y–y–1xsin
ysin–xsin
22221–
22221–
22
22221–
1–1–
8. (i)0yx&1y,x1–ify–1x–1–xycos–2
0yx&1y,x1–ify–1x–1–xycosycosxcos
221–
221–1–1–
(ii)yx&1x0,0y1–ify–1x–1xycos–
yx&1y,x1–ify–1x–1xycosycos–xcos
221–
221–1–1–
9. (i)
ifxy–1yxtan–
ifxy–1yxtan
xifxy–1yxtan
ytanxtan
1–
1–
1–
1–1–
(ii)
–ifxy1y–xtan–
–ifxy1y–xtan
–xifxy1y–xtan
ytan–xtan
1–
1–
1–
1–1–
Remark : If x1, x2,..........xn R, then tan–1x1+tan–1x2+.........+tan–1xn
= tan–1......s–ss–1
........s–ss–s
642
7531
Where sk is the sum of the product of x1,x2,............ xn taken k at a time.ie. s1 = x1 + x2+..........+xn = xi
s2 = x1x2+x2x3+.......+xn–1xn= x1x2.s3 = x1x2x3.................etc.
392
10. (i)
21–x1–if ,x–12xsin––
1x2
1if ,x–12xsin–
21x
21–if ,x–12xsin
xsin2
21–
21–
21–
1–
(ii)
21–x1–if ,4x–3xsin––
1x21if ,4x–3xsin–
21x
21–if ,4x–3xsin
xsin3
31–
31–
31–
1–
11. (i) 0x1–if ,1–2xcos–21x0 if ,1–2xcos
xcos221–
2–11–
(ii)
21–x1–if ,3x–4xcos2
21x
21–if ,3x–4xcos–2
1x21if ,3x–4xcos
xcos3
31–
31–
31–
1–
12. (i)
–1 xif ,x–1
2xtan–
1 xif ,x–1
2xtan
1x1– if ,x–1
2xtan
xtan2
21–
21–
21–
1–
(ii)
31 xif ,
3x–1x–3xtan–
31 xif ,
3x–1x–3xtan
31x
31– if ,
3x–1x–3xtan
xtan3
2
31–
2
31–
2
31–
1–
393
Note : If |x| 1 then 21–
2
21–
21–1–
x–1x2tan
x1x–1cos
x1x2sinxtan2 .
If |x|>1, change x to x1
in the above.
Note : In cases of identities in inverse trigonometric functions, principal values are to be taken. Assuch signs of x,y etc., will determine the quadrant in which the angles will fall. In order tobring the angles of both sides in the same quadrant, adjustment by is to be made.
13. Hyperbolic functions(i) sinh(–x) = – sinhx odd function
cosh(–x) = – coshx even functiontanh(–x) = tanhx odd function
(ii) Function Domain Rangesinh–1x R Rcosh–1x (0, ) (1, )tanh–1x R (–1,1)coth–1x R–{0} R–[–1,1]sech–1x (0, ) (0,1)cosech–1x R–{0} R–{0}
(iii) sinh (sinh–1x) = x sinh–1(sinhx) = xcosh (cosh–1x) = x cosh–1(coshx) = xtanh (tanh–1x) = x tanh–1(tanhx) = xsinhn (sinh–1x)=xn sinh (sinh–1xn) = xn
(iv) sinh–1x = loge 1xx 2
cosh–1x = loge 1–xx 2
tanh–1x = 21
loge 1–x1x
x > 1, x< –1
coth–1x = 21
loge 1x1–x
x > 1, x< –1
sech–1x = loge xx–11 2
0 < x 1
cosech–1x = 0xif
xx1–1log
0xifx
x11log
2
e
2
e
394
(v) sinh–1x = cosech–1x1
sinh–1x = cosh–1 1x2
cosh–1x = sinh–1 1–x2
sinh (cosh–1x) = 1–x2
Solved Examples1. The sum to infinite terms of the series
tan–1
31
+ tan–1
71
+ tan–1 131
+............................ is
(a) 6 (b) 4 (c) 3 (d) None of these
Solution : By method of difference
Tn = tan–12nn1
1
Tn = tan–1 )1n(n11
= tan–1 )1n(nn–1n
= tan–1 (n+1) – tan–1n
Tn = tan–1 (n+1) – tan–1nT1 = tan–1 2– tan–11T2 = tan–13 – tan–12T3 = tan–14 – tan–13...Tn = tan–1 (n+1) – tan–1nAdding, Sn = T1 + T2 +............ + Tn
= )1n(tan 1– – tan–11
S = tan–1 –tan–1 1= 44–
2Ans : b
2 The sum to infinite terms of the series
tan–1 21.21
+ tan–122.2
1 + tan–1
23.21
+.................. is
(a) 4 (b) 3 (c) 2 (d) None
395
Solution :
Tn = tan–1 2n21
= tan–12n4
2 = tan–1 1–n21n21
1–n2–1n2
Tn = tan–1 (2n+1) – tan–1 (2n–1)T1 = tan–1 3– tan–11T2 = tan–1 5– tan–13T3 = tan–1 7– tan–13...
)1–n2(tan–)1n2(tanTn 1–1–
Adding, Sn = T1+T2+.......Tn = Tan–1(2n+1)–tan–11
S = tan–1 –tan–1 = 44–
2Ans : a
3 The value of
tan–1 xycy–xc
1
1 +tan–1
21
12
cc1c–c
+ tan–1
32
23
cc1c–c
+......................+ tan–1
nc1
is
(a) tan–1yx
(b) tan–1
xy
(c) tan–1 x–tan–1y (d) None
Solution Write the series as
tan–1
1
1
c1
yx1
c1–
yx
+ tan–1
21
21
c1
c11
c1–
c1
+ tan–1
32
32
c1
c11
c1–
c1
+ ........+ tan–1
n1–n
n1–n
c1
c11
c1–
c1
+ tan–1 c1
1
1–1–
c1tan–
yxtan +
2
1–
1
1–
c1tan–
c1tan +
3
1–
2
1–
c1tan–
c1tan +
.................. + n
1–
1–n
1–
c1tan–
c1tan + tan–1
nc1
= tan–1 yx
Ans : a
4. The number of positive integral solutions of the equation tan–1x + cos–12y1
y
= sin–1103
is
(a) 1 (b) 2 (c) 3 (d) None
1
n
396
Solution : tan–1 x + tan–1y1
= tan–1 3
tan–1 y1
= tan–1 3 – tan–1x
y1
= x31x–3
y = x–3x31
Put x = 1, then y = 2 Put x = 2, then y = 7 (1,2) & (2,7) are two sets.
Ans : b
5. If cot–16
n; n N, then the maximum value of n can be
(a) 4 (b) 5 (c) 6 (d) None
Solution :n
< cot 6 ( cot–1x is a decreasing function)
n < 3n < 5.43 n = 5 (max)
Ans : b6. The value of
2sec412cos
43–2sincotsin 1–1–1–1–
(a) 0 (b) 4 (c) 2 (d) None
Solution : 4
3–2sin 1–
832–4sin 1–
= 21–
22
32–13sin
=
2
1–
221–3sin
= 221–3sin 1–
397
= 1212sinsin 1–
4612cotsin 1–
= sin–1 cot 2= sin–1 0 = 0
Ans : a7. The greatest value of (tan–1x)2 + (cot–1x)2 is .......................
Solution :(tan–1x)2 + (cot–1x)2 = (tan–1x+cot–1x)2 – 2tan–1xcot–1x
= 4
2
– 2 tan–1x xtan–2
1–
Let tan–1 x= y, then LHS= 4
2
– 2y y–2
= 4
2
– y +2y2 = 2 416.2–
162y–y
2222
= 2 84–xtan
221–
Minimum value is 8
2
Ans :8
2
398
* More than one options questions
1. If (tan–1x)2 + (cot–1 x)2 = 8
5 2
, then x equals
(a) –1 (b) 1 (c) 0 (d) None of these
2. If sin–1 .........–4x
2x–x
32
+ cos–1 .........–4x
2x–x
642
= 2 for 0<|x|< 2 , then
x equals
(a) 21
(b) 1 (c) 21–
(d) –1
3. Match the conditions / expressions in column I with statement in column II.
Let (x,y) be such that sin–1(ax)+cos–1y + cos–1(bxy)= 2Column I Column II
(a) If a=1 & b = 0, then(x,y) (p) lies on the circle x2+y2=1(b) If a=1 & b = 1, then(x,y) (q) lies on (x2–1) (y2–1) = 0(c) If a=1 & b = 2, then(x,y) (r) lies on y = x(b) If a=2 & b = 2, then(x,y) (s) lies on (4x2–1) (y2–1) = 0
4 Sum to n terms of the series
cosec–1 10 +cosec–1 50 + cosec–1 170 + ..........+cosec–1 2n2n1n 22 is
(a) 0 (b) (c) tan–1
4–1n (d) cot–1
4–1n
5 Match the following.Let t1 = (sin–1x)Sin–1x ,t2=(sin–1x)Cos–1x ,t3 =(cos–1x)Sin–1x ,t4 = (cos–1 x)cos–1x
Column I Column II(a) x (0,cos1) (p) t1>t2>t4>t3
(b) x 21,1cos (q) t4>t3>t1>t2
(c) x 1sin,2
1(r) t1>t2>t4>t3
(c) x 1,1sin (s) t3>t4>t1>t2
6. Read the passage & answer the following questions
If tan–1x : tan–1y = 1:4 (where |x|< tan 6 ) then
(i) The value of y as an algebraic function of x will be
PRACTICE QUESTIONS
399
(a) 1x6–x)x1(x4
24
2
(b) 1x6–x)x–1(x4
24
2
(c) 1x6x)x1(x4
24
2
(d) None of these.
(ii) The root of the equation x4 – 6x2+1=0 is
(a) tan12 (b) tan 4
(c) tan 8 (d) tan167 If a sin–1 x – b cos–1x=c, then a sin–1x + b cos–1x is
(a) 0 (b) ba)a–b(cab
(c) 2 (d) ba)a–b(cab
8. )1r(r1–r–rSin 1–
n
1r is
(a) tan–1
4–n (b) tan–1
4–1n
(c) tan–1 n (d) tan–1 1n9 If [cot–1x] + [cos–1x] = 0, then complete set of values of x is
(a) (cos1, 1] (b) (cot1,cos1) (c) (cot1, 1)(d) None of these
10*. If (sin–1x+sin–1w) (sin–1y+sin–1z) = 2 , then
4N3N
2N1N
wzyx
D wher N1,N2,N3,N4 W
(a) has a maximum value of 2 (b) has a minimum value of 0(c) 16 different D are possible (d) has a minimum value of –2.
11. The value of k (k>0) such that the length of the longest interval in which the function
ƒ(x)=sin–1 |sinkx|+cos–1(coskx) is constant is 4 is / are
(a) 8 (b) 4 (c) 12 (d) 1612*. Match the following
Column I Column II
(a) (sin–1x)2 + (sin–1y)2 = 2
2
(p) 1
x3+y3=
400
(b) (cos–1x)2 + (cos–1y)2 = 2 2 (q) –2x5+y5=
(c) (sin–1x)2 + (sin–1y)2 = 4
4
(r) 0
|x–y| =(d) |sin–1x–sin–1y| = (s) 2
xy =Answers
1 a 2 b3 a p, b q, c p, d s 4 c5 a q, b s, c r, d s 6 (i) b (ii) a7 d 8 c9 c 10 a, c, d11 b 12 a q,r, s , b q , c r,s d p
__________________________________
_________
401
Mathematical ReasoningLogical Statements, Tautology and Contradiction
Mathematical LogicStatement (Proposition): A sentence which is either true or false but not both is called a statement.They are denoted by p, q, r, s,...............A true statement is called a valid statement. If a statementis false we call it invalid statement.
Truth TableIt is a tabular device to obtain the truth value of a compound statement or to check thevalidity of a simple or compound statementNumber of horizontal lines in a truth table depends on the number of substatements presentin it.If the problem involves n simple statements then number of rows is 2n.i.e.No. of statements No.of rows
1 21 = 22 22 = 43 23 = 84 24 = 16
Logical Connectives(i) Conjunction ( )
A compound statement joining two statements by “and” is called a conjunction and isdenoted by .i.e. the conjunction of two statements p and q is denoted by p q. p q is true only ifboth the components p and q are true otherwise it is false.
(ii) Disjunction ( )A compound statement joining two statements by “or” is called a disjunction and isdenoted by .i.e. the disjunction of two statements p & q is true or both p and q is true otherwise p qis false.We can express both p q and p q is tabular form as under. T stands for true and Fstands for false.
p q p q p q
T T T TT F F TF T F TF F F F
Quantifiers : phrases like “there exists” ( ) and “for all” ( ) are called quantifiers.(iii) Implications ( )
p q. One way implicationHere p is called antecedent or hypothesis or premise and q is called consequence orconclusion.p q is the same for each of the following(i) If p then q
402
(ii) P is sufficient for q(iii) q is necessary condition for p(iv) p only if q(v) q if p(vi) q follows from p(vii) q is consequence of pSince a true statement cannot imply a false statement, p q is always true except whenp is true and q is false . It may also be useful to note that p q is equivalent to ~p qNote : The contrapositive of a conditional statement is formed by negating both thehypothesis and the conclusion and then interchanging the resulting negations.In otherwards, the contrapositivse negates and switches the parts of the sentence. Itdoes both the jobs of the INVERSE and the CONVERSE.IMPORTANT : Contrapositive has the same truth value as the original conditionalstatement.Note :
Statement Converse Inverse Contrapositive Negation p q q p ~p ~q ~q ~p ~(p q)
The inverse and the converse of a conditional statement are logically equivalent to each other,just as the conditional and its contrapositive are logically equivalent to each other.
(iv) Two way implication ( )p q, “p implies and implied by q” or “p if and only if q”p q is true if both p & q are true or both false and false when one of the statements is trueand the other is false.
p q p q q p p qT T T T TT F F T FF T T F FF F T T T
(v) NegationIf p is a statement then negation p is written as ~ p.If p is true then ~ p is false. If p is false then ~ p is true
p ~pT FF T
Note :~ (~p) = p~ (p q) = ~ p ~q~ (p q) = ~ p ~q~ (p q) = (p ~q)~ (p q) = (p ~q) (~p q)
(vi) NOR ( )Let p & q be two statements. Then “p q” is called Joint Denial or “NOR” statement
403
(combination of NOT and OR) and read as “Neither p nor q”. “p q” can also be written as (p q) or ~ (p q).
Joint Denial is true only when p and q both are false.
p q p q = ~ (p q)
T T FT F FF T FF F T
(vii) NAND ( )Let p & q be two statements. Then “p q” is called NAND statement (combination ofNOT and AND) and is written as “p q”“p q” is also written as (p q) or ~ (p q) This statement is false only if both p & qare true.
p q p q = ~ (p q)T T FT F TF T TF F T
Here is a table that shows a commonly used Precedence of logical operators
Operator Precedence
~ 12345
Use of brackets(i) If negation (i.e or ~) is repeated in the same statement then there is no need of
bracket.(ii) If in a statement, the connectives of same type are present, then brackets are applied
from left.(iii) If different connectives are used in a statement, then we remove the bracket of lower
order connective. But we cannot remove the bracket of higher order connective.For example:(i) p (q r) = p q r (order of is less than order of )(ii) p (q r) p q r
404
Table of Symbols
(i) If p then q p q(ii) p if q q p(iii) p only if q q p(iv) p unless q ~q p(v) p is a sufficient condition for q p q(vi) p is a necessary condition for q q p(vii) A sufficient condition for p is q q p(viii) A necessary condition for p is q p q(ix) In order that p is sufficient that q q p(x) In order that p is necessary that q p q(xi) p if and only if q P q(xii) p is a necessary and sufficient condition for q p q
Tautology and Contradiction (Fallacy)A statement whose truth value is always T (i.e. True) is called a tautology and the statementwhose truth value is always F(i.e.False) is called a contradiction. Negation of tautology is acontradiction and vice versa.
Logical equivalenceTwo compound statements are said be logically equivalent if both have same truth values for allpossible assignments given to the variables.
DualityTwo compounds are said to be dual of each other if either can be obtained from the other byinterchanging and provided both remain valid.For e.g. the dual of (p q) r is (p q) r..
Algebra of StatementsCommutative laws (i) (p q) (q p)
(ii) (p q) (q p)Associative laws (i) p (q r) (p q) r
(ii) p (q r) (p q) rDistributive laws (i) p (q r) (p q) (p r)
(ii) p (q r) (p q) (p r)Idempotent laws (i) (p p) p
(ii) (p p) pAbosorption laws (i) p (p q) p
(ii) p (p q) pDe Morgan’s laws (i) ~ (p q) (~ p ~q)
(ii) ~ (p q) (~ p ~q)Detachment law ((p q) p) qChain law ((p q) (q p)) (p r)Identity laws (t is tautology and (i) p t = t p = pf is contradiction) (ii) p f = f p = pCompliment laws (i) p ~p = t
(ii) p ~p = t
405
Solved Examples1 If p (q r) is false, then the truth values of p,q,r are respectively
(i) T, F, F (b) F, T, T (c) T, T, F (d) F, F, FSolution
p q is false only when p is true and q is falsep (q r) is false when p is true and q r is false and q r is false when both q and r are false
Hence, truth values of p, q, r are respectively T, F, F.Ans: (a)
2 Test the validity of the argument (S1, S2; S), whereS1 ; p q, S2 : ~ p and S : q.
SolutionIn order to test the validity of the argument (S1, S2; S), we first construct the truth tablefor the conditional statement.S1 S2 S i.e [(p q) ~p] qThe truth table is as given below:
p q S1 = p q S2= ~ p S1 S2 S=q S1 S2 Si.e. S1 S2 q
T T T F F T TT F T F F F TF T T T T T TF F F T F F T
We observe that the last comumn of the truth table for S1 S2 S contains T only. Thus, S1 S2 S is a tautology.Hence, the given argument is valid.
3 If (p ~q) (~q q) is(a) a contradiction(b) a tautology(c) neither a tautology nor a contradiction(d) both a tautology and a contradiction
Solution(p ~q) (~p q) (p ~p) (~q q)
(f f) f
(f false)(By using associative laws and commutative laws)
(p ~q) (~p q) is a contradiction.Ans : (a)
4 Which of the following is logically equivalent to ~ (~ p q) ?(a) ~ p q (b) p q (c) ~ p ~ q (d) p ~ qSolution
Since ~ (p q) = p ~ q~ (~ p q) = p ~ q
Ans : (c)
406
5 If p He is a carpenter and q = He is making a table.Then write down the following statement into symbols:(i) He is a carpenter and making a table.(ii) He is a carpenter but is not making a table.(iii) It is false that he is a carpenter or making a table.(iv) Neither he is a carpenter nor he is making a table.(v) He is not a carpenter and he is making a table.(vi) It is false that he is not a carpenter or is not making a table.(vii) He is a carpenter or making a table.SolutionThe solution of above compound statements in terms of p and q are given below :(i) p q (ii) p q (iii) (p q) (iv) p q(v) p q (vi) ( p q) (vii) p q
6 Write in words the converse, inverse, contrapositive and negation of the implication “If 2 isless then 3, than 1/3 is less than 1/2.Solution
Let p 2 is less than 3,q 1/3 is less than 1/2.Then implication is p q :(i) Converse of p q is q p. In words q p means “If 1/3 is less than 1/2, then 2
is less than 3”.(ii) Inverse of p q is p q. Thus in words, p q means “ If 2 is not less than 3,
then 1/3 is not less than 1/2”.(iii) Contrapositive of p q is q p. Thus in words q p means “If 1/3 is not less
than 1/2, then 2 is not less than 3”.(iv) Negation of p q is (p q). Thus in words (p q means “It is false than p implies
q”7 The statement p (q p) is equivalent to
(a) p (p q) (b) p (p q) (c) p (p q) (d) p (p q)Solution
p (q p) = ~ p (q p)= ~ p (~q p) since p ~ p is always true= ~ p p q = p (p q)
Ans : (b)8 Statement - 1:
~ (p ~ q) is equivalent to p q.Statement - 2:(~ p ~q) is a tautology..a. Statement - 1 is True, Statement - 2 is True ; Statement - 2 is a correct explanation forStatement - 1b. Statement - 1 is True, Statement - 2 is True ; Statement - 2 is NOT a correct explanationfor Statement - 1c. Statement - 1 is True, Statement - 2 is Falsed. Statement - 1 is False, Statement - 2 is True
407
Solution :
p q p q ~ p p ~ q ~(p ~ q) ~q ~p ~ q
T T T F F T F TT F F T F TF T F T F FF F T T F T T T
Ans (c)Exercise
1 Are the following statements equivalent:“If the traders do not reduce the price then the government will take action against them”. “Itis not true that the traders do not reduce the prices and government does not take actionagainst them”
2 Which of the following is false?(a) (p q) (~q ~ p) is a contradiction(b) (p ~ p) is a tautology(c) ~(~ p) p is a tautology(d) (p ~ p) is a contradiction
3 If each of the statement p ~ q ; q r ; ~ r is true, then(a) p is false (b) p is true (c) q is true (d) None of these
4 Which of the following is true?(a) ~(p (~ q)) (~ p) q (b) (p q) (~q) is a tautology(c) ~(p (~ p)) is a contradiction (d) None of these
5 Which of the following is the contrapositive of ‘If two triangles are identical, then these are similar’?(a) If two triangles are not similar, then these are not identical(b) If two triangles are not identical, then these are not similar(c) If two triangles are not identical, then these are similar(d) If two triangles are not similar, then these are identical
6 The contrapositive of the converse of p ~ q is(a) ~ q p (b) p q (c) ~ q ~ p (d) ~ p ~ q
7 ~(p q) (~p q) is equivalent to(a) q (b) p (c) ~ p (d) ~ q
8 Negation of the compound proposition.If the examination is difficult, then I shall pass if I study hard.(a) The examination is difficult and I study hard but I shall not pass(b) The examination is difficult and I study hard and I shall pass(c) The examination is not difficult and I study hard and I shall pass(d) None of these
9 If p is true, q is false and r is false, then which of the following is true?(a) (p q) r (b) p ~ (q r) (c) (p q) r (d) p ~(q r)
10 If p “she goes to market” and q “She buys some fruits”.Then choose the correct symbol for the given statements :(i) Either she goes to market or she buys some fruits:(a) p q (b) p q (c) ~ p q (d) p ~ q
FT
FF
408
(ii) If she goes to market, then she buys some fruits:(a) ~ p q (b)p q (c) p q (d) ~ p q(iii) Neither she go to market nor she buy some fruits:(a) ~ p ~ q (b) p q (c) p q (d) ~ p ~ q(iv) She does not go to market and she buys some fruits:(a) ~ p q (b) ~ (p q) (c) p ~ q (d) p q(v) She does not go to market unless she buys some fruits :(a) p q (b) q p (c) ~ q ~ p (d) p q
Answers1. Yes 2. a 3. a 4. a 5. a6. a 7. c 8. a 9. b 10. (i)a (ii) b (iii) d (iv) a (v) c
409
STATISTICSMeasures of Central Tendency (Averages)
i. Arithmetic Unclassified datan
1iix
n1x
Mean or ungrouped frequency
Mean x distribution1i
n
1iiii fN;fx
N1x
Step deviation methodn
1iiiuf
N1hAx
where A is assumed mean, h class interval
and ui = hA–xi
Combined meank21
kk2211
n......nnxn......xnxnx
where k21 x.....x,x are means of kgroups having n1,n2......nk members.
Weighted Arithmetic (i) n
1ii
n
1iii
w
w
xwx if wi be the weight of
Mean variable xi
(ii) Short cut method n
1ii
n
1iii
ww
w
dwAx
Aw = assumed mean, di = deviationswi = weights
ii. Geometric Unclassified data G = n/1n21 x....xx or
Mean G = antilogn
1iixlog
n1
Frequency distribution N/1fff n
n
2
2
1
1x......xxG where
n
1iifN or
G = antilogn
1iii xlogf
N1
iii. Harmonic Unclassified datan
1i ix1
nH
Mean Freqency distribution n
1i i
i
xf
NH ; n
1iifN
Mat
ham
etic
al A
vera
ges
410
Median Unclassified data a. n is oddth
21n
value.
b. n is even
A.M. ofth
2n
and th
12n
value
Ungrouped frequency The value of x for which the cumulative
distribution frequency is just greater than 2N
.
Grouped frequency hf
cf–2N
M
distribution = lower limit of median classf = frequency of median classh = width of median classcf = cumulative frequency of the class justbefore the median class.
Mode Unclassified data Value which appears most frequently in thedistribution.
Ungrouped frequency Value of x which has greatest frequency.distribution
Grouped frequency hf–f–f2
f–fM11–0
1–00
distribution = lower limit of modal classf0 = frequency of modal classf–1 = frequency of pre-modal classf1 = frequency of post modal classh = length of modal classIf 2f0 – f–1 – f1 = 0,
Then hf–ff–f
f–fM101–0
1–00
Note: Like median, the other partition values quartiles, deciles, percentiles etc can be deter-mined.
ith quartile Qi is given by Qi = hf
cf–4
iN
; i = 1, 2, 3
Where symbols have same meanings as in median. Clearly, the second quartile Q2 is median.First quartile is called lower quartile and the third quartile is upper quartile.
Posi
tion
Ave
rage
411
Measures of DispersionThe degree to which numerical data tend to spread about an average value is called variation ordispersion of the data. It measures the scatterendness of various observation about some centralvalue.
Range Difference of the largest and smallestvalues
Quartile Deviation 2Q–Q 13
Coefficient of quartile13
13
QQQ–Q
where Q1, Q3 are respectivelydeviation the first & third quartiles.
Mean Devialion unclassified datan
1i
a–xn1
where a = A.M,Median or Mode as the case may be
ungrouped frequency M.D =n
1iii a–xf
n1
where a = A.M,distributon Median or Mode as the case may be &
(M.D is least when
measured from then
1iifN
median)
Grouped frequency M.D = n
1iii a–xf
n1
where a = A.M,distribution Median or Mode as the case may be &
n
1iifN
Standard Unclassified datan
1i
2n
1ii
2i
n
1i
2i n
x–x
n1x–x
n1
deviation ( ) Ungrouped frequencyn
1i
2ii x–xf
N1
distributionn
1i
2n
1iii
2ii N
xf–xf
N1
where
n
1iifN
412
Grouped frequency
n
1i
2n
1iii
2ii N
uf–uf
N1h
distribution where,n
1iifN & h
A–xu ii
A assumed meanh class interval
Root mean square Unclassified datan
1i
2i a–x
n1s ; a is asumed mean
deviation (s) Ungrouped frequencyn
1i
2ii a–xf
N1s ;
n
1iifN
distribution
Grouped frequencyn
1i
2ii a–xf
N1s ;
n
1iifN
distribution
Note that s2= 2 + d2 where d = a–xClearly s is least when d = 0 i.e. a–xi.e. root mean square devian is least when deviations are taken from x .Note: median can be determined from graph also. It is the abscissa of the point ofintersection of “less than” ogive and “more than” ogive.Note: i) The algebraic sum of the deviations of all the values of the variable from their mean
is zero.
i.e. 0x–xi , for ungrouped distribution and 0x–xf ii , for groupeddistribution.
ii. The sum of the squares of the deviations of the variable is minimum when taken aboutarithmetic mean.
iii. Let x & y be two variables, b,c two constants and u = bx+cy. Then ycxbu wherex & y are A.M.S of xi’s &yi’s, when b = c = 1, u = x+y & yxui.e. The mean of sum of two variables is equal to sum of their means (it is true for morethan two variables also)
Note:i. Standard deviation is independent of shift of origin but depends upon change of scale.
i.e. if hxy , then y h
x
ii. Square of S.D., ie. 2 is called the variance.
413
iii. Coefficient of variation c.v. = x × 100.
The distribution for which the coefficent of variance is less is more consistent.iv. S.D. of the combined group of two groups having means 21 x,x ; standard
deviations 1 , 2 and number of elements n1, n2 is given by
22
222
21
211
21
dndnnn
12 where d1 = x–x1 and d2 = x–x2
21
2211
nnxnxnx (combined mean)
Also 2Range2
Symmetric and skew DistributionIn a symmetrical distribution, mean, median and mode coincide. Here frequencies aresymmetrically distributed on both sides of the central value. (ie. same number of frequenciesare distributed at the same linear distance on either side of mode).The frequency curve is bell-shaped and mean = median = mode
In a skew distribution, the variation doesnot have symmetry.
Note: In a moderately skewed distribution, Mean – Mode = 3(Mean – Median)
Normal Distributioon
Positive Skewness
NegativeSkewness
414
Solved Examples
1. In any disrete series when all values are not same, the relation between M.D about mean andS.D is
a. M.D = S.D b. D.SD.M c. M.D < S.D d. D.SD.M
Solution:
For a distribution (xi,fi) i = 1,2,.....n
x–xfN1
ii2x
2 & x–xfN1D.M ii
2
ii2
ii22
x dfN1–df
N1D.M– where di = x–xi
= 02d
22x .D.M
S.D. M.D.
Ans: c
2. The A.M. of n observation is x . If the sum of n–5 observations is a, then the mean of remaining5 observation is
a. 5axn
b. 5a–xn
c. axn d. none of these
Sloution:
If m is the mean of 5 observations, then
55–n
m55–n
a5–nx
21
2211
nnxnxnx
xn = a+5m
5a–xnm
Ans: b
415
3. If 1x and 2x are means of two distributions such that 1x < 2x and x is the combined mean, then
a. 1xx b. 2xx c. 2xxx 21 d. 21 xxx
Solution:
If n1 & n2 are the number of items in two distributions having means 21 x&x .
21
2211
nnxnxnx
121
22111 x–
nnxnxnx–x = 0
nnx–xn
21
12212 xx
1xx
Similary, 0nnx–xnx–x21
2112
2xx
21 xxx
Ans: d
4. The mean of 51C......
5C,
3C,
1C 50
504
502
500
50
is
a.51250
b.51249
c.17x39
249
d. none of these
Solution:
(1+x)50 + (1–x)50 = 5050
5022
500
50 xC......xCC2
Integrating with limits 0 to 1
1
0
51
5050
3
250
050
51xC....
3xCxC2 =
1
0
5151
51x1–
51x1
416
512
512.
21
51C........
3CC
505150
502
50
050
Mean = 26x51
250
=17x39
249
Ans: c
1. Mean of n terms is x . If these x items are successively increased by 2,22,23,........2n, then thenew mean is
a.n
2x1n
b.n2
n2x
1n
c.n2x
n
d. none of these
2. The weighted A.M. of first n natural number whose weights are equal is
a. 21n
b. 21n2
c. 31n2
d. 61n1n2
3. If G is the G.M. of the product of k sets of observations, with G.M.’s G1,G2......Gk respectively,then G is equal to
a. logG1+logG2+........+logGk b. logG1+logG2........+logGk
c. G1G2......Gk d. none of these
4. The mean square deviation of n observations x1,x2,......xn about –2 and 2 are 18 and 10respectively. Then, S.D of the given set is
a. 1 b. 2 c. 3 d. 4
5. A car owner buys petrol at `7.50, `8.00 and `8.50 per litre for the 3 successive years. If hespends 4,000 each year, then the average cost per litre of petrol is
a. `8 b. `8.25 c. `7.98 d. none of these
6. The mean of the values 0,1,2,.....,n with the corresponding weights nC0,nC1,.....
nCn repectivelyis
a.1n
2n
b. 1nn2 1n
c. 21n
d. 2n
PRACTICE QUESTIONS
417
7. The quartile deviation of daily wages (in Rs.) of 7 persons is given below:
12,7,15,10,17,17,25 is
a. 14.5 b. 3.5 c. 9 d. 4.5
8. If a variable x takes values xi such that bxa i , for i = 1,2,....,n, Then
a. bxvara b. 22 bxvara
c. xvar4a2
d. xvara–b 2
9. For (2n+1) observations x1,–x1,x2,–x2,.....,xn,–xn and 0, where xi’s are different, let S.D. andM.D. denote standard deviation and mean deviaiton about median, then which is true
a. S.D.< M.D. b. S.D > M.D.
c. S.D. = M.D. d. nothing can be said in general
10. The AM and variance of 10 observations are 10 and 4 respectively. Later it is discovered thatone observation was incorrectly read as 8 instead of 18. Then, the correct value of mean andvariance are
a. 20,9 b. 20,14 c. 11,9 d. 11,5
11. In a frequency distribution, the mean and median are 21 and 22 respectively, then its mode isapproximately
a. 25.5 b. 24.0 c. 22.0 d. 20.5
12. For a symmetrical distribution Q1 = 20 and Q3 = 40, the median of the data is
a. 20 b. 30 c. 40 d. 10
13. If the mean deviation of 1,1+d,1+2d,......,1+100d from their mean is 255, then d =
a. 20.0 b. 10.1 c. 20.2 d. 10.0
14. The mean value of 1C0
30
,3C2
30
......,21C20
30
,22C21
30
,.......31C30
30
is
a.31231
b.31431
c.31213
d. none of these
418
15. Read the paragraph and answer the questions that follow:
If x1, x2 ,x3 are n values of the variable x, then mathematical averages. Arithmetic Mean (A.M.),Geometric Mean (G.M.) and Harmonic Mean (H.M.) are count by the following formula’s.
A.M. = nx.....xxx n321 0xx
n1
i
n
1ii
G.M. = ,x....x.x n/1n21 if each xi (i=1,2,.....,n) is positive and
H.M. = n
1i in21 x1
n1
1
x1.......
x1
x1
n
* In case of frequency distribution xi/fi (i = 1,2,.....n) where fi is the frequency of the variable xi,
then the claculation of A.M. is counted as A.M.=n
1iii N/xf where N = f1+f2.....+fn
If w1,w2.....,wn be the weight assigned to the n values x1,x2,......xn then weighted A.M. is counted
by n
1ii
n
1iii
w
xw
* If G1,G2 are the G.M’s of two series of sizes n1 & n2 respectively, then the geometric mean
(G.M.) of the combined series is counted by log(G.M.)=21
2211
nnGlognGlogn
. On the basis of
above information answer the following questions.
1. If the mean of a set of observations x1,x2......,x2......,xn is x then the mean of observations xi+4i,i = 1,2,3,......, n isa. 1n2x b. 1n4x c. n4x d. nx
2. The A.M. of n numbers of series is x . If the sum of first (n–1) terms is m then nth number isa. m–x b. m–xn c. mn–x d. nm–xn
3. The mean of a set of n numbers is x . If each number is divided by 4 then new mean is
a. x b. 4x c. 4x
d. x4
419
4. The weighted A.M. of first n natural numbers whose weights are squares the correspondingnumbers, is equal to
a. 21nn
b. 1nn1n2
23
c. 1nn23
d. 1n21nn
23
5. Consider the series 1,4,16,64,256,......4n, then which of the following is not true?
a. A.M.= 1n31–4 1n
b. G.M. = 2n
c. H.M. = 1–41n4.3
1n
n
d. A.M=G.M.=H.M.
6. Let G be the G.M. of the product of (r+1) sets of observation with, G.M., G1,G2,......Gr,Gr+1respectively, then the vaue of G is
a.1r
1iiGlog b.
1r
1iiG c.
1r
1iiGlog d. none of these
7. The mean value of ,21C,.........
7C,
5C,
3C,
1C 20
206
204
202
200
20
equals
a.77x3
220
b.77x3
219
c.21220
d.21219
8. Let x be the variate which assumes the values 0,1,2,3,4,......n with frequenciesqn,nC1pqn–1,nC1p
2qn–1..........pn such that p+q = 1,then mean value of the frequency distributionis
a. npq b. np c. npq d. n2p2q2
9. The mean of the divisors of 360 which are odd isa. 15 b. 11 c. 13 d. 9
10. The ratio of the mean of the cubes of first n natural numbers to the means of the cubes of first(n+1) natural nubers is given bya. (n+1) : (n+2) b. (n+1)2 : n+2 c. n(n+1) : (n+2)2 d. none of these
'Note : Questions with * have more than one correct option'
ANSWERS
1. b 2. a 3. c 4. c
5. c 6. d 7. b 8. d
9. b 10. b 11. b 12. b
13. b 14. b
15. (1) a (2) b (3) c (4) d (5) d (6) b (7) a (8) b (9) c (10) c
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