Appendix BMock Test 1

PaPer 1Section a: Only One Option Correct Type

1. For qp

ŒÊËÁ

ˆ¯̃

02

, , the value of definite integral

ln ( tan tan )10

+Ú qq

x dx is equal to

(a) q ln (sec q) (b) q ln(cosec q)

(c) q ln 22

(d) 2q ln sec q

2. The focal length of a mirror is given by 1 1 2v u f

- = .

If errors made in measuring u and v are a, then the relative error in f is

(a) 2a

(b) a1 1u v

+ÊËÁ

ˆ¯̃

(c) a1 1u v

-ÊËÁ

ˆ¯̃

(d) None of these

3. If the function f (x) = 2 tan x + (2a +1) loge | sec x | + (a –2) x is increasing on R, then

(a) a Œ (1/2, •) (b) a Œ (–1/2, 1/2) (c) a = 1/2 (d) a Œ R 4. 3x + 4y – 7 = 0 and 3x – 4y –7 = 0 are equations of

asymptotes of a hyperbola H. From a point P(3, 4), pair of tangents are drawn to hyperbola H in such a way that both tangents touch the same branch of hyperbola H. Then its eccentricity is

(a) 43

(b) 73

(c) 53

(d) 54

5. A parabola touches two given straight lines originat-ing from a given point. The locus of the mid point of the portion of any tangent, which is intercepted between the given straight lines, is a/an

(a) parabola (b) ellipse (c) straight line (d) hyperbola

6. If x x2

42- + -

p sin = | x2 – 2 | + | sin x | + p4

, then

(a) x Œ( , )0 2 (b) x Œ -( , )2 2

(c) x ŒR (d) x Œ -( , )2 0

7. If A and B are different matrices satisfying A3 = B3 and A2B = B2A, then

(a) det (A2 + B2) must be zero. (b) det (A – B) must be zero. (c) det (A2 + B2) as well as det (A – B) must be zero (d) At least one of det (A2 + B2) or det (A – B) must

be zero. 8. A fair dice is thrown 3 times. The probability that the

product of the three outcomes is a prime number is

(a) 124

(b) 136

(c) 132

(d) 18

9. The number of real solution of equation 16 sin–1 x tan–1 x cosec–1 x = p3 is/are (a) 0 (b) 1 (c) 2 (d) infinite 10. Let a, b, c are non-zero constant numbers. Then

Limr Æ •

cos cos cos

sin sin

ar

br

cr

br

cr

- equals

(a) a b cbc

2 2 2

2+ -

(b) c a bbc

2 2 2

2+ -

(c) b c abc

2 2 2

2+ -

(d) Independent of a, b and c

Section B: One or More Options Correct Type

11. Given 2 functions f and g which are integrable on every interval and satisfy

(i) f is odd, g is even (ii) g(x) = f (x + 5), then (a) f (x – 5) = g(x) (b) f (x – 5) = – g(x)

(c) f t dt g t dt( ) ( )0

5

0

5

5Ú Ú= -

(d) f t dt g t dt( ) ( )0

5

0

5

5Ú Ú= - -

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B.2 Mathematics

12. A continuous function f (x) on R Æ R satisfies the relation f (x) + f (2x + y) + 5xy = f (3x – y) + 2x2 + 1 for " x, y Œ R, then which of the following hold(s) good?

(a) f is many-one (b) f has no minima (c) f is neither odd nor even (d) f is bounded 13. The equation x2 – 4x + a sin a = 0 has real roots (a) for all values of a (b) for all values of a provided –p/4 < a < p/4 (c) for all values of a ≥ 4 provided p £ a £ 2p (d) for all a provided | a | £ 4 14. Which of the following statement(s) is/are not correct? (a) If the roots of a quadratic equation are imagi-

nary, then these are conjugates of each other. (b) If a continuous function is strictly monotonic,

then it is differentiable. (c) If f (x) is periodic, then | f (x) | is also periodic. (d) sin x/(2x – 2np – p) and cos x tan x/(2x – 2np

Рp), where n ΠZ, are identical functions. 15. If the curve y = ax1/2 + bx passes through the point

(1, 2) and lies above the x-axis for 0 £ x £ 9 and the

area enclosed by the curve, the x-axis and the line x = 4 is 8 sq. units, then

(a) a2 + b2 = 6 (b) a/b = 1 (c) a – b = 4 (d) ab = –3

Section C: Integer Value Correct Type

16. The value of Limn

k n

n

kÆ •=

+

 1

2

21( )

is

17. If the slope of the curve y = axb x-

at the point (1, 1)

is 2, then the value of a + b is 18. Given

a = 3i + j + 2k,

b = i – 2j – 4k are the posi-

tion vectors of point A and B. Then the distance of point –i + j + k from the plane passing through B and perpendicular to AB is

19. If z = x + iy (x, y Œ R, x π –1/2), then the number of values of z satisfying | z |n = z2 | z |n–2 + z | z |n–2 + 1, (n Œ N, n > 1) is

20. If x, y, z are distinct positive numbers such that

xy

yz

zx

+ = + = +1 1 1 , then the value of xyz is

PaPer 2Section a: One or More Options Correct Type

1. For natural numbers m and n, if (1 – y)m (1 + y)n = 1 + a1 y + a2 y

2 + L, and a1 = a2 = 10, then (a) m < n (b) m > n (c) m + n = 80 (d) m – n = 20 2. Let P(x) = x2 + bx + c, where b and c are integer. If

P(x) is a factor of both x4 + 6x2 + 25 and 3x4 + 4x2 + 28x + 5, then

(a) P(x) = 0 has imaginary roots (b) P(x) = 0 has roots of opposite sign (c) P(1) = 4 (d) P(1) = 6 3. Let tan x – tan2 x > 0 and | 2 sin x | < 1. Then the int-

ersection of which of the following two sets satisfies both inequalities?

(a) x > np + p/6, n ΠZ (b) x > np Рp/6, n ΠZ (c) x < np Рp/4, n ΠZ (d) x < np + p/4, n ΠZ

4. A bag initially contains one red and two blue balls. An experiment consisting of selecting a ball at random, noting its colour and replacing it together with an additional ball of the same colour. If three such trials are made, then

(a) probability that atleast one blue ball is drawn is 0.9.

(b) probability that exactly one blue ball is drawn is 0.2.

(c) probability that all the drawn balls are red given that all the drawn balls are of same colour is 0.2.

(d) probability that atleast one red ball is drawn is 0.6. 5. If a, b, c are non-zero real numbers such that

bc ca abca ab bcab bc ca

= 0, then

(a) 1 1 1 0a b c

+ + = (b) 1 1 1 02a b c+ + =

w w

(c) 1 1 1 02a b cw w+ + = (d) None of these

6. 2007201 + 2019201 – 1982201 – 2044201 is divisible by (a) 74 (b) 50 (c) 1850 (d) 2013

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Appendix B: Mock Tests B.3

(a) 122

(b) 123

(c) 233

(d) 522

Paragraph for Questions 13 and 14

Consider a function f defined by f (x) = sin–1 sin x x+ÊËÁ

ˆ¯̃

sin ,2

" x Π[0, p], which satisfies f (x) + f (2p Рx) = p, " x Π[p, 2p] and f (x) = f (4p Рx) for all x Π[2p, 4p], then. 13. If a is the length of the largest interval on which f (x)

is increasing, then a- (a) p/2 (b) p (c) 2p (d) 4p 14. If f (x) is symmetric about x = b, then b = (a) a/2 (b) a (c) a/4 (d) 2aParagraph for Questions 15 and 16Straight line x + y = 18 and common tangents of the curve

S1 = x2 + y2 – 28x + 160 = 0 and S2 = x y2 2

64 36+ – 1 = 0

form a triangle ABC in the first quadrant and S3 = 0, the circumcircle of DABC. 15. The coordinates of the center of circle inscribed in

triangle ABC are

(a) [ , ]12 2 2 10 2 2- -

(b) [ , ]12 2 10 2- -

(c) [ , ]12 2 2 10 2 2+ +

(d) [ , ]12 2 10 2+ +

16. Equation of S3 is (a) x2 + y2 – 16x – 20x + 156 = 0 (b) x2 + y2 – 10x – 8x + 156 = 0 (c) x2 + y2 – 8x – 10x + 156 = 0 (d) x2 + y2 – 20x – 16x + 156 = 0

Section 3: Matching List Type

17. The graphs of f and g are given in Fig. B.1. Use them to evaluate each limit.

-1

-1-2 1 2

1

2

x

y

O

y f x= ( )

7. If the equation sin2 x – a sin x + b = 0 has only one solution in (0, p), then which of the following state-ments are correct?

(a) a Œ (– •, 1) » (2, •) (b) b Œ (– •, 0) » (1, •) (c) a = 1 + b (d) None of these 8. Let a, b, and g be some angles in the 1st quadrant

satisfying tan (a + b) = 15/8 and cosec g = 17/8, then which of the following holds good?

(a) a + b + g = p (b) cot a cot b cot g = cot a + cot b + cot g (c) tan a + tan b + tan g = tan a tan b tan g (d) tan a tan b + tan b tan g + tan g tan a = 1

Section B: Paragraph TypeThis section contains 4 paragraphs each describing theory, experiment, date etc. Eight questions relate to four para-graphs with two questions on each paragraph. Each question of paragraph has only one correct answer along the four choice (a), (b), (c) and (d).Paragraph for Questions 9 and 10A perpendicular is drawn from a fixed point (3, 4 ) to a variable line having x-intercept unity. P (x, y) = 0 represents the locus of the foot of perpendicular drawn from (3, 4) to the variable line, which is a circle. Then 9. The equation of the circle is: (a) x2 + y2 = 4 (b) (x – 2)2 + (y – 2)2 = 5 (c) (x – 3)2 + (y – 3)2 = 4 (d) (x – 3)2 + (y + 3)2 = 5 10. A tangent is drawn to P(x, y) = 0 from the origin, then

the length of tangent is

(a) 2 (b) 1 (c) 3 (d) 2Paragraph for Questions 11 and 12

Let a point P whose position vector is r xi y j zk= + +

is called lattice point if x, y, z ΠN. If atleast two of x, y, z are equal then this lattice point is called isosceles lattice point. If all x, y, z are equal, then this lattice point is called equilateral lattice point. 11. The number of lattice points on the plane

r i j k.( )+ + = 10 are

(a) 36 (b) 45 (c) 84 (d) 120 12. If a lattice point is selected at random from lattice

points which satisfy r i j k.( ) ,+ + £ 11 then the

probability that the selected lattice point is equilateral given that it is isosceles lattice point is

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B.4 Mathematics

-1

-1-2 1 2

1

2

x

y

O

y g x= ( )

Fig. B.1

Column I Column II(a) lim ( ( ))

xf g x

Æ1(p) 1

(b) lim ( )x

f xÆ

-2

3 2 (q) does not exist

(c) lim ( )( )

( ) ( )x

f xg x

f x g xÆ

+ÊËÁ

ˆ¯̃0

(r) 0

(d) lim ( ) ( )( ) ( )x

f x g xf x g xÆ +

-+1

3 (s) 2

18. Match the locus of z given by equation in Column I with Column II.

Column I

(a) arg zz

2

211

-+

Ê

ËÁˆ

¯̃ = 0; z π ± i, ± 1

(b) || z – cos–1 cos 12 | – | z – sin–1 sin 12 || = 8 (p – 3) (c) z2 + k1 = i | z1 |

2 + k2 ; k1 π k2 Œ R – {0} and z1 is fixed π 0

(d) z z- - + + - =- -1 13

13 2

11 1sin cos p

Column II (p) Portions of a line (q) Point of intersection of hyperbola (r) Pair of open rays (s) line segment 19. The angles of a triangle are in the ratio 2 : 3 : 7.

Column I Column II

(a) If smallest side = 2, then (p) largest side = 3 1+

(b) If smallest side = 2 , then (q) largest side = 6 2+

(c) If s = + +3 3 2, then (r) largest side = 1

(d) If D = -( )/ ,3 1 4 then (s) largest side = 2 2 3+

20. A bag contains 14 balls which are either white or black balls. (all number of white and black balls are equally likely). Five balls are drawn at random from the bag without replacement

Column I Column II(a) Probability that all the five balls are

black is equal to(p) 3/13

(b) If the bag contains 11 black and 3 white balls, then the probability that

all five balls are black is equal to

(q) 1/6

(c) If all the five balls are black then the probability that the bag contains 11 black and 3 white balls is equal to

(r) 6/65

(d) Probability that three balls are black and two are white is equal to

(s) 3/65

Mock Test 2

PaPer 1Section a: Only One Option Correct Type

1. If f (x) = ( ) , ( )( )

( )x n ff

n n

n

-¢

=-

=’ 51

1

50 5151

then

(a) 5050 (b) 11275

(c) 15050

(d) 1275

2. Through the focus of the parabola y2 = 2px (p > 0), a line is drawn which intersects the curve at A(x1, y1)

and B(x2, y2). The ratio y yx x

1 2

1 2 equals

(a) 2 (b) –1 (c) –4 (d) some function of p

3. Limx

k

n k k k kk kÆ •

-

=

+ - + ++

Ê

ËÁ

ˆ

¯˜Âcos

( ) ( ) ( )( )

1

2

1 1 1 21

is

equal to

(a) p6

(b) p4

(c) p3

(d) p2

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Appendix B: Mock Tests B.5

12. Consider the binomial expansion of xx

n

+Ê

ËÁˆ

¯̃1

2 4,

n ΠN, where the terms of the expansion are written in decreasing powers of x. If the coefficients of the first three terms form an arithmetic progression, then which of the following is/are true?

(a) Total number of terms in the expansion of the binomial is 8.

(b) Number of terms in the expansion with integral power of x is 3.

(c) There is no term in the expansion which is independent of x.

(d) Fourth and fifth are middle terms of the expan-sion.

13. If xa

yb

xc

yd

+ = + =1 1and intersect the axes at four

concyclic points and a2 + c2 = b2 + d2, then these lines can intersect at (a, b, c, d > 0)

(a) (1, 1) (b) (1, –1) (c) (2, –2) (d) (3, 3) 14. A forecast is to be made of the results of five cricket

matches, each of which can be a win or a draw or a loss for the Indian team.

Let p = number of forecasts with exactly 1 error q = number of forecasts with exactly 3 errors r = number of forecasts with all 5 errors then the incorrect statement is (a) 2q = 5r (b) 8p = q (c) 8p = 5r (d) 2(p + r) > q 15. A tangent drawn to the curve y = f (x) at P(x, y) cuts

the x-axis at A. Now a perpendicular drawn from P(x, y) to the x-axis meets the x-axis at B. If B is the midpoint of OA (O being the origin), and f (1) = 1, then

(a) General point on the curve is (t, 1/t), where t is a real parameter.

(b) y = f (x) is a circle. (c) Tangent at P(1, 1) is x + y = 2. (d) f ¢(2) = –1/4.

Section C: Integer Type

16. The value of Limt

x t xt

dxÆ

+ -Ú00

2sin ( ) sin

p

is

4. If f ≤(x) > 0 and f ¢(1) = 0 such that g(x) = f (cot2 x + 2 cot x + 2), where 0 < x < p, then the interval in which g(x) is decreasing is

(a) (0, p) (b) pp

2,Ê

ËÁˆ¯̃

(c) 34p

p,ÊËÁ

ˆ¯̃

(d) 0 34

, pÊËÁ

ˆ¯̃

5. A continuous and differentiable function y = f (x) is such that its graph cuts line y = mx + c at n distinct points. Then the minimum number of points at which f ≤(x) = 0 is/are

(a) n – 1 (b) n – 3 (c) n – 2 (d) Cannot say

6. The number of terms in xx

33

100

1 1+ +ÊËÁ

ˆ¯̃

is

(a) 300 (b) 200 (c) 100 (d) 201 7. The set of all values of a for which ax2 + (a – 2)x – 2

is negative for exactly two integral x is (a) (0, 2) (b) (1, 2) (c) (1, 2) (d) (0, 2) 8. If the fundamental period of the function f (x) =

sin cos2 2a ax x+ is p/8, then the value of a is (a) ± 4 (b) ± 2 (c) ± 8 (d) ± 1

9. x f x f x

x f x

◊ ¢ -

◊Ú ( ) ( )

( )

24

dx equals

(a) x2 f (x) + c (b) | x | f (x) + c

(c) 2 f xx

c( )+ (d)

2 f xx

c( )

+

10. In which one of the following intervals, the inequality sin x < cos x < tan x < cot x can hold good?

(a) (0, p/8) (b) (3p/4, p) (c) (5p/4, 3p/2) (d) (7p/4, 2p)

Section B: One or More Options Correct Type

11. Let f (x) = e dt xt tx

- >Ú [ ] ( ),00

where [x] denotes great-

est integer less than or equal to x is (a) continuous and differentiable " x Π(0, 3) (b) continuous but not differentiable " x Π(0, 3) (c) f (1) = e (d) f (2) = 2(e Р1)

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B.6 Mathematics

19. If 2 11 03 4

1 8 101 2 59 22 15

-

-

È

Î

ÍÍÍ

˘

˚

˙˙˙

=- - -

- -È

Î

ÍÍÍ

˘

˚

˙˙˙

A , then the sum of

all the elements of matrix A is

20. Let f (x) = sin23 x – cos22 x and g(x) = 1 12

+ tan–1 | x |,

then the number of values of x in interval [–10p, 8p] satisfying the equation f (x) = sgn(g(x)) is

17. If a, b and c are distinct positive real numbers such that a + b + c = 1, then the least integral value of ( )( )( )( )( )( )1 1 11 1 1

+ + +- - -

a b ca b c

is

18. The number of elements in the domain of the func-

tion f (x) = sin [ ] [ ]- -Ê

ËÁˆ

¯̃+ + -1

2 23

x x x x , where [◊]

represents greatest integer function, is

PaPer 2Section a: One or More Options Correct Type

1. Let z be a complex number satisfying equation zp = z q, where n, m ΠN, then

(a) If p = q, then the number of solutions of equa-tion will be infinite.

(b) If p = q, then the number of solution of equation will be finite.

(c) If p π q, then the number of solution of equation will be p + q + 1.

(d) If p π q, then the number of solution of equation will be p + q.

2. If xx

dx f x g x c4

61 11

123

++

= = +- -Ú tan ( ) tan ( ) , then

(a) both f (x) and g(x) are odd functions. (b) g(x) is monotonic function. (c) f (x) = g(x) has no real roots.

(d) f xg x

dxx x

c( )( )

= + +Ú 1 33

3. A tangent is drawn at any point (x1, y1) other than vertex on the parabola y2 = 4ax. If tangents are drawn from any point on this tangent to the circle x2 + y2 = a2 such that all the chords of contact pass through a fixed point (x2, y2), then

(a) x1, a, x2 are in GP. (b) y1/2, a, y2 are in GP. (c) –4, y1/y2, x1/x2 are in GP. (d) x1x2 + y1 y2 = a2

4. Let PM be the perpendicular from the point P(1, 2, 3) to x–y plane. If OP makes an angle q with the positive direction of the z-axis and OM makes an angle f with the positive direction of x-axis, where O is the positive direction of x-axis, where O is the origin, then (q and f are acute angles)

(a) tan /q = 5 3 (b) sin sin /q f = 2 14

(c) tan f = 2 (d) cos cos /q f = 1 14 5. If the side AB

of an equilateral triangle ABC lying

in the x–y plane is 3i, then the side CB

can be

(a) - -32

3( )i j (b) 32

3( )i j -

(c) - +32

3( )i j (d) 32

3( )i j +

6. In a precision bombing attack, there is a 50% chance that any one bomb will strike the target. Two direct hits are required to destroy the target completely. The number of bombs which should be dropped to give a 99% chance of better of completely destroying the target can be

(a) 12 (b) 11 (c) 10 (d) 13 7. The value(s) of p for which the equations

ax2 – px + ab = 0 and x2 – ax – bx + ab = 0 may have a common root, given a, b are non-zero real num-bers, is(are)

(a) a + b2 (b) a2 + b (c) a(1 + b) (d) b(1 + a)

8. Let f (x, t) = x t x tt x t x

( ),( ),

,- £- <

ÏÌÓ

11

where t is a continu-

ous function of x in [0, 1].

Let g x f t x t dt( ) ( ) ( , ) .=Ú f0

1Then

(a) g(0) = 1 (b) g(0) = 0 (c) g(1) = 1 (d) g(x) = f (x)

Section B: Paragraph Type

Paragraph for Questions 9 and 10The curve y = (x – a)(x2 + bx + c) meets the x-axis at a point P(p, 0). A straight line through P meets the curve at two more points A and B. Tangents drawn to the curve at P meet the curves at C. (b2 – 4c < 0)

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Appendix B: Mock Tests B.7

9. The abscissa of C is (a) – a – b (b) – a + b (c) a – b (d) a + b 10. Number of points on the curve where the slope of the

tangent is same as slope of AB is (a) 0 (b) 1 (c) 2 (d) 3Paragraph for Questions 11 and 12Let f (x) be one-one and onto function, such that | f (x) – f –1(x) | > 0 for x Œ (0, 2) » (2, 4) given f (2) = 2, f (4) = 0

and f (0) = 4. Let g(x) = 16 2- x . Given f (x) < g(x) " x Œ (0, 4) and graph of y = f (x) and y = f –1(x) are symmetrical about the line x + y = 4.

11. If f (x) – f –1(x) < 0 for x Œ (0, 2) and f x dx( ) ,=Ú 50

2

then f x dx-Ú 1

2

4

( ) is

(a) 2 (b) 1 (c) 5 (d) 7 12. If f (x) – f –1(x) > 0 for x Œ (0, 2) and

( ( ) ( ))f x f x dx- -Ú 1

0

2

= 2, f x dx-Ú 1

2

4

( ) = 4, then

g x f x f x dx( ) max ( ( ), ( ))- -Ú 1

0

4

is

(a) 4p – 12 (b) 8p – 1 (c) 4p – 10 (d) 8Paragraph for Questions 13 and 14A function f (x) having the following properties: (i) f (x) is continuous except at x = 3. (ii) f (x) is differentiable except at x = –2 and x = 3 (iii) f (x) = 0, lim

x Æ 3f (x) Æ – •, lim

x Æ -•f (x) = 3, lim

x Æ •f (x) = 0

(iv) f ¢(x) > 0 " x Œ (– •, – 2) » (3, •) and f ¢(x) £ 0 " x Œ (– 2, 3)

(v) f ≤(x) > 0 " x Œ (– •, – 2) » (– 2, 0) and f ≤(x) < 0 " x Œ (0, 3) » (3, •)

13. Maximum possible number of solutions of f (x) = | x | is

(a) 2 (b) 1 (c) 3 (d) 4 14. f (x) + 3x = 0 has five solution if (a) f (– 2) > 6 (b) f ¢(0) < – 3 and f (– 2) > 6 (c) f ¢(0) > – 3 (d) f ¢(0) > – 3 and f (– 2) > 6Paragraph for Questions 15 and 16There are three curve defined here in Argand plane:

C z z195

15

2: - + - =

C z i2 24

: arg ( )- = -p

C z i3 1 2: - - = 15. The number of points common to curves C1, C2 and

C3 is/are (a) 0 (b) 1 (c) 2 (d) None of these 16. Curves C4, C5, C6 are the reflections of the curves C1,

C2 and C3, respectively, about the real axis. Then

(a) C z z495

15

2: - + - =

(b) C z i5 3 34

: arg ( )- =p

(c) C z i6 1 2: - - = (d) None of these

Section C: Matching List Type

17. Column I Column II

(a) Sum of coefficients in expansion of (px + qy – 5rz + 6t)3

(p) 0

(b) Coefficient of x103 in (1 + x + x2 + x3 + x4)199(x – 1)201 is

(q) 100

(c) Remainder when 22003 is divided by 17 is

(r) 27

(d) Remainder when (11111 … 1001 times) is divided by 1001 is

(s) 8

18. A tangent having slope -43

touches the ellipse x2

18

+ y2

32 = 1 at point P and intersects the major and

minor axes at A & B respectively, O is the centre of the ellipse

Column I Column II(a) Distance between the parallel tangents

having slopes -43

, is

(p) 24

(b) Area of DAOB is (q) 7/24

(c) If coordinates of p are (l, m), then the value of l ¥ m is

(r) 48/5

(d) If equation of the tangent intersecting posi-tive axes is lx + my = 1, then l + m is equal to

(s) 12

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B.8 Mathematics

19. Column I Column II

(a) The number of integer lying in the domain

of the function f (x) = log .0 55 2-Ê

ËÁˆ¯̃

xx

(p) 1

(b) Number of positive roots of the equation (x – 1) (x – 2) (x – 3) + (x – 1) (x – 2) (x – 4) + (x – 2) (x – 3) (x – 4) + (x – 1) (x – 3) (x – 4) = 0 is/are

(q) 2

(c) If f (x) = ex " x Œ [0, 1] and f (1) – f (0) = f ¢(c). where c Œ (0, 1) then ln (ec + 1) is equal to

(r) 3

(d) Number of values of x satisfying the equa-

tion tan tan- --ÊËÁ

ˆ¯̃

+ =1 12 110

12 4

xx

p(s) 4

20. Match the following:

Column I Column II

(a) x x

xdxln

( )1 2 20 +

•

Ú (p) 0

(b) ln (tan ( ))/

x dx0

2p

Ú (q) –2/5

(c) x dx

x x3 61

1

1+ +-Ú (r) –1

(d) [ ]x dx-Ú 1

1, where [◊] represents

greatest integer function

(s) –4/5

Mock Test 3

PaPer 1Section a: Only One Option Correct Type

1. A curve passes through the point (2a, a) and is such that the sum of subtangent and abscissa is a. Its equa-tion is

(a) (x – a)y2 = a3 (b) (x – a)2 y = a3

(c) (x – a) y = a2 (d) none of these 2. A function f(x) satisfies

f (x) = sin x + 0

x

Ú f ¢(t) (2 sin t – sin2t) dt. Then f (x) is/are

(a) xx1- sin

(b) sinsin

xx1-

(c) 1- coscos

xx

(d) tansin

xx1-

3. The number 916238457 is an example of nine-digit number which contains each of the digit 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. Number of such numbers are

(a) 2268 (b) 2520 (c) 2975 (d) 1560 4. Let z1, z2, z3 be complex numbers such that z1 + z2

+ z3 = 0 and | z1 | = | z2 | = | z3 | = 1. Then z z z12

22

32+ + is

(a) greater than zero (b) equal to 3 (c) equal to zero (d) equal to 1

5. P and Q are two points on the upper half of the ell-

ipse xa

yb

2

2

2

2+ = 1. The center of the ellipse is at the

origin O and PQ is parallel to the x-axis such that the triangle OPQ has the maximum possible area. A point is randomly selected from inside of the upper half of the ellipse. The probability that it lies outside the triangle is

(a) pp-1 (b) 2 1

2p

p-

(c) pp-1

2 (d) p

p-1

4 6. If x, y ΠR and satisfy the equation xy(x2 Рy2) = x2 + y2

where x π 0, then the minimum possible value of x2 + y2 is

(a) 1 (b) 2 (c) 4 (d) 8

7. If f (x) = [ ] sin[ ]

[ ][ ]

[ ]

x xx

x

x

2

0

0 0

+π

=

ÏÌÔ

ÓÔ

for

for

where [x] denotes the greatest integer less than or equal to x, then lim

xÆ0 f (x) equals:

(a) 1 (b) 0 (c) –1 (d) None of these 8. Matrices of order 3 ¥ 3 are formed by using the ele-

ments of the set A = {–3, –2, –1, 0, 1, 2, 3}. Then the

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Appendix B: Mock Tests B.9

probability that matrix is either symmetric or skew symmetric is

(a) 17

176 3+ (b) 1

717

179 3 6+ -

(c) 17

173 9+ (d) 1

717

173 6 9+ -

9. Limn

k

n

n k nƕ

-

=

ÊËÁ

ˆ¯̃ + ( )

Ê

ËÁˆ

¯̃Âtan

tan1

1

1 11

has the value equal

to

(a) 1 12

+ ln(cos )

(b) 1 12

+ ln(sin )

(c) 1 1 12

- +ln(sin cos )

(d) 1 1 12

+ +ln(sin cos )

10. If f (x, y) = x2 + y2 Р2xy (x, y ΠR) and the matrix A is given by

A = f x y f x y f x yf x y f x y f x yf x y f

( , ) ( , ) ( , )( , ) ( , ) ( , )( , ) (

1 1 1 2 1 3

2 1 2 2 2 3

3 1 xx y f x y3 2 3 3, ) ( , )

È

Î

ÍÍÍ

˘

˚

˙˙˙

such that

trace (A) = 0, then (a) det(A) ≥ 0 (b) det(A) £ 0 (c) det(A) = 0 (d) None of these

Section B: One or More Options Correct Type

11. If f : R Æ R be a continuous function such that f (x)

= 21

t f t dtx

( )Ú , then which of the following does not

hold good? (a) f (p) = ep 2

(b) f (1) = e (c) f (0) = 1 (d) f (2) = 2 12. Let (a – 1)(x2 + 3x + 1)2 – (a + 1)(x4 – x2 + 1) £ 0

" x ΠR. Then which of the following is/are correct?

(a) a Œ-È

ÎÍ

˘

˚˙

13

43

,

(b) Largest possible value of a is 3 . (c) Number of possible integral values of a is 3. (d) Sum of all possible values of a is 0. 13. If planes

r i j k q r i a j k q( ) , ( )+ + = + + =1 22 and r ai a j k q( )+ + =2

3 intersect in a line, then the value of a is

(a) 1/4 (b) 1/2 (c) 1 (d) 2

14. If A = 13

1 2 22 1 2

2

1-È

Î

ÍÍÍ

˘

˚

˙˙˙

= -

a bA ATand then,

(a) a = –2 (b) a = 2 (c) b = –1 (d) b = 1

15. The value of a for which x x xx x x

a3 2

3 26 11 6

10 8 30- + -+ - +

+

= 0 does not have real solution is (a) –10 (b) 12 (c) 5 (d) –30

Section C: Integer Value Correct Type 16. If f : (0, •) Æ (0, •) satisfies f (x f (y)) = x2ya(a Œ R),

then find a. 17. If a, b, and c are real numbers such that a2 + 2b = 7,

b2 + 4c = –7, and c2 + 6a = –14, then find the value of (a2 + b2 + c2)/2.

18. Let an = 16, 4, 1, ... be a geometric sequence. Define Pn as the product of the first n terms. Then find the

value of pn

nn =

•Â 1

4.

19. Find the distance of the point P(3, 8, 2) from the line 12

1 14

3 13

2( ) ( ) ( )x y z- = - = - measured parallel

to the plane 3x + 2y Р2z + 15 = 0. 20. Consider the equation x2 + 2x Рn = 0, where n ΠN and

n Π[5, 100]. Find the total number of different values of n so that the given equation has integral roots.

PaPer 2

Section a: One or More Options Correct Type

1. If g(x) = f xx a x b x c

( )( ) ( ) ( )

,- - -

where f (x) is a

polynomial of degree < 3, then

(a) g x dxa f a x ab f b x bc f c x c

( )( ) log( ) log( ) log

Ú =---

111

∏ +

1

1

1

2

2

2

a a

b b

c c

k

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B.10 Mathematics

(b) dg xdx

a f a x a

b f b x b

c f c x c

a a

b b

c

( )( ) ( )

( ) ( )

( ) ( )

=

-

-

-

∏

-

-

-

1

1

1

1

1

2

2

2

2

2

2 cc 1

(c) dg xdx

a f a x a

b f b x b

c f c x c

a a

b b

c

( )( ) ( )

( ) ( )

( ) ( )

=

-

-

-

∏

-

-

-

1

1

1

1

1

1

2

2

2

2

2

cc2

(d) g x dxa f a x ab f b x bc f c x c

( )( ) log( ) log( ) log

Ú =---

111

∏ +

a a

b b

c c

k

2

2

2

1

1

1

2. If A and B are two events such that P(A) = 3/4 and P(B) = 5/8, then

(a) P(A » B) ≥ 3/4 (b) P(A¢ « B) £ 1/4 (c) 3/8 £ P(A « B) £ 5/8 (d) 1/8 £ P(A « B¢) £ 3/8 3. If (5, 12) and (24, 77) are the focii of a hyperbola

passing through the origin, then

(a) e = 386 12/ (b) e = 386 13/ (c) LR = 121/6 (d) LR = 121/3

4. y = ae–1/x + b is a solution of dydx

yx

= 2 . Then

(a) a ΠR (b) b = 0 (c) b = 1 (d) A takes finite number of

values 5. If AB = A and BA = B, then (a) A2B = A2 (b) B2A = B2

(c) ABA = A (d) BAB = B 6. If the parabola y = (x2 + 4(b – c)x + 4a2)/4 touches

the x-axis, then the line ax + by + c = 0 always passes through the fixed point/points

(a) (1, 1) (b) (–1, –1) (c) (–1, 1) (d) (1, –1)

7. Let f (x) = 31 2

1

tx

tdt

+Ú , x > 0, then

(a) for 0 < a < b, f (a) < f (b) (b) for 0 < a < b, f (a) > f (b)

(c) f (x) + p/4 < tan–1 x, "x ≥ 1 (d) f (x) + p/4 > tan–1 x, "x ≥ 1 8. If 10! = 2p3q5r7s, then (a) 2q = p (b) pqrs = 64 (c) number of divisors of 10! is 270. (d) number of ways of putting 10! as a product of

two natural numbers is 135.

Section B: Paragraph TypeParagraph for Questions 9 and 10Let angles a, b, g of a triangle satisfy the relation,

sin sin sina b a g a-ÊËÁ

ˆ¯̃

+-Ê

ËÁˆ¯̃

+ ÊËÁ

ˆ¯̃

=2 2

32

32

.

9. The largest angle of triangle is (a) 70° (b) 100° (c) 110° (d) 130° 10. The triangle is (a) acute-angled (b) right-angled (c) isosceles (d) scaleneParagraph for Questions 11 and 12

Let b xx

b xx x x

b Rcoscos

sin(cos sin ) tan

,2 2 1 32 2-

=+

-Œ

11. Equation has solutions if

(a) b Œ - •ÊËÁ

ˆ¯̃

- -ÏÌÓ

¸˝˛

, , ,12

1 0 13

(b) b Œ - •( ) - -ÏÌÓ

¸˝˛

, , ,1 1 0 13

(c) b Œ R – -ÏÌÓ

¸˝˛

1 0 13

, ,

(d) None of these 12. For any value of b for which the equation has solution,

then the number of solutions for x Œ(0, 2p) are always (a) infinite (b) depends upon the value of b (c) 4 (d) none of theseParagraph for Questions 13 and 14The base of pyramid is rectangular, three of its vertices of the base are A(2, 2, –1), B(3, 1, 2) and C(1,1,1) (points

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Appendix B: Mock Tests B.11

may or may not be in order). Its vertex at the top is

P 4 263

103

, ,- -ÊËÁ

ˆ¯̃

and fourth vertex of the base is D.

13. Coordinates of D are (a) (4, 0, 2) (b) (4, 2, 0) (c) (2, 0, 4) (d) (0, 2, 4) 14. Volume of the pyramid is (in cubic units) (a) 20 (b) 10 (c) 40 (d) 30Paragraph for Questions 15 and 16Consider two circles C1 : x

2 + y2 = r12 and C2 : x

2 + y2 = r22

(r1 > r2). Let A be a fixed point on circle C1 say (r1, 0) and B be a variable point on circle C2. The line BA meets circle C2 again at C. Then 15. The maximum value of BC2 is (a) 4r1

2 (b) 4r22

(c) 4r22 + 4r1

2 (d) None 16. The value of OA2 + OB2 + BC2 is (a) [5r2

2 – 3r12, 5r2

2 + r12]

(b) [4r22 – 4r1

2, 4r22]

(c) [4r12, 4r2

2] (d) None of these

Section C: Matching List Type 17. Match the following:

Column I Column II(a) Common normals to the parabola y2

= 4ax and x2 = 4ay is/are (p) x = a

(b) The locus of P point is, if tangents from P to the parabola y2 = 4ax intersects the co-ordinate axes in concyclic points

(q) x + y = 3a

(c) The locus of P, if tangents from it to the parabolas y2 = 4a (x + a) and y2 = 8a (x + 2a) are perpendicular is

(r) x + 3a = 0

(d) The chord of contact of a point P w.r.t. the parabola y2 + 4ax = 0 subtends right angle at the vertex. The locus of P is

(s) x = 4a

18. Match the following:

Column I Column II

(a) If the vectors a , b , c

form sides BA

, CA

, AB

of DABC, then

(p) a c b c c a. . .= =

Column I Column II

(b) If a , b , c are forming

three adjacent sides of regular tetrahedron, then

(q) a b b c c a. . .= = = 0

(c) If a b¥ =

c , and b c¥ =

a , then

(r) a b b c c a¥ = ¥ = ¥

(d) If a , b , c are unit vectors

and a b c+ + = 0, then

(s) a b b c c a. . .+ + = - 3

2

19. Match the following:

Column I Column II(a) If Langrange’s mean value theorem is

applicable in [–2, 2] for the function

f (x) = mx c x

e xx

+ <≥

ÏÌÔ

ÓÔ

,,

00

then the value of

m + 3c is

(p) 2

(b) If the ends of latus rectum of parabola are (2, 6) and (6, 2) and the equation of the possible directrix is x + y = li where i = 1, 2. Then the value of l1 + l2 is

(q) 8

(c) The maximum value of f(x) = 2x3 – 3x2 – 12x in [–2, 5/2] is

(r) 4

(d) If lim ( ) ( )n

n n

n

Æ•

-+

-+

+

Ê

Ë

ÁÁÁÁ

ˆ

¯

˜˜˜˜

12 1

14 2

1

2 2

L

is

equal to k.p4

, then k is

(s) 16

20. Match the following:

Column I Column II

(a) If 1, a, a2, L a19 are 20th roots of unity, then 1 + a10 + a20 + L + a190 is equal to

(p) 4

(b) Number of integral values of x satisfying the inequality (log2x)2 + log2 0.03125x + 3 < 0

(q) 2

(c) If l is the maximum value of the function

f(x) = x2e–2x, x > 0, then 12

ÊËÁ

ˆ¯̃

ln ( )l

is

(r) 0

(d) Number of points on x2

9 – y2 = 1 from

which pair of perpendicular tangents can be drawn to parabola y2 = – 12x is

(s) 1

(Contd.)

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B.12 Mathematics

Solutions of Mock Test 1

fi 2a –1 = 0 fi a = ½

4. Ans (c): m m1 234

34

= - =,

Since tangents are drawn on same branch and point P(3, 4) lies in obtuse angle between the asymptotes, therefore

tan q2

34

=

\ e ba

2 12

43

- = = =cot q

\ e =53

5. Ans (c): Let the parabola is y2 = 4ax. a = at1t2, b = a(t1 + t2), Q ∫ (at1t3, a(t1 + t3)), R ∫ (at2t3, a(t2 + t3))

Let T ∫ (h, k). Then h = at3

2(t1 + t2) = bt3

2 and

2k = a(t1 + t2) + 2at3 = bb

+4ah

Therefore, the locus is 4ax – 2by + b2 = 0, which is a straight line.

6. Ans (d): | x + y + z | = | x | + | y | + | z | fi x, y, z are of same sign.

Thus, x2 Р2 < 0 and sin x < 0 fi x Π-( , )2 0 7. Ans (d): A3 = B3 ...(1) and A2B = B2A ...(2) (1) Р(2) gives A3 РA2B = B3 РB2A A2(A РB) = РB2(A РB) fi (A2 + B2) (A РB) = 0 det (A2 + B2) (A РB) = 0 det (A2 + B2) . det (A РB) = 0 fi (d) 8. Ans (a): Possible outcomes are 11 2 (C);113 (C);115 (C)

2 3 5p p p= = =

Hence, P(A) = 9216

124

=

9. Ans (b): Domain is x = ± 1. However, only x = 1 satisfies.

PaPer 1Section a

1. Ans (a): I = ln ( tan tan ( ))10

+ -Ú q qq

x dx

fi I = ln tan (tan tan )tan tan

11

0

+-

+ÊËÁ

ˆ¯̃Ú q q

q

qx

xdx

= ln tantan tan

11

2

0

++

Ê

ËÁˆ

¯̃Ú qq

q

xdx

fi I = ln ( tan ) ln ( tan tan )1 12

0 0

+ - +Ú Úq qq q

dx x dx

fi I = 2q ln sec q – I fi 2I = 2q ln sec q fi I = q ln sec q 2. Ans (b): We have

1 1 2v u f

- =

fi dv u

df

1 1 2-Ê

ËÁˆ¯̃

=ÊËÁ

ˆ¯̃

fi - + = -1 1 22 2 2v

dvu

duf

df

fi 1 1 22 2 2v u f

df-ÊËÁ

ˆ¯̃

=a [Q du = dv = a]

fi a1 1 1 1 2

2v u v u fdf+Ê

ËÁˆ¯̃

-ÊËÁ

ˆ¯̃

=

fi a1 1 2 2

2v u f fdf+Ê

ËÁˆ¯̃

¥ = Q1 1 2v u f

- =È

ÎÍ

˘

˚˙

fi dff u v

= +ÊËÁ

ˆ¯̃

a1 1

Thus, relative error in f = a1 1u v

+ÊËÁ

ˆ¯̃

.

Hence, (b) is the correct answer. 3. Ans (c): f (x) is strictly increasing ⇒ f ¢(x) ≥ 0 fi 2 sec2 x + (2a +1) tan x + (a – 2) ≥ 0, " x Œ R fi 2 tan2 x + (2a +1) tan x + a ≥ 0, " x Œ R fi (2a +1)2 – 8a £ 0 fi (2a –1)2 £ 0

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Appendix B: Mock Tests B.13

10. Ans (c): Let 1/r = x so that as r Æ •, x Æ 0.

Limx

ax bx cxbx

bxcx

cxbc xÆ

-0 2

cos cos .cossin . sin . .

= 10 2bc

ax bx cxxx

LimÆ

-cos cos .cos

= 120bc

a ax b bx cxc cx bx

xxlim

sin sin cossin cos

Æ

- ++ ◊

= 12

2 2 2

bxb c a( )+ -

Section B 11. Ans (b, c): g(x) = f (x + 5) from (ii) fi g(–x) = f (–x + 5) fi g(–x) = – f (x – 5) from (i) Thus, choice (b) is true.

Further I = f t dt( )0

5

Ú

\ I = g t dt( )-Ú 50

5

(Q f (t) = g(t – 5))

= g t dt( )50

5

-Ú (Q g is even)

Therefore, choice (c) is correct. 12. Ans (a, b): Let 2x + y = 3x – y fi 2y = x fi y = x/2 Put y = x/2 f (x) + f (5x/2) + 5x2/2 = f (5x/2) + 2x2 + 1 \ f (x) = 1 – (x2/2) 13. Ans (c, d): x2 – 4x + a sin a ∫ (x – 2)2 + 4 – a sin a The roots are real if a sin a £ 4. As | sin a | £ 1, the roots are real for all values of a if

| a | £ 4. For p £ a £ 2p, sin a £ 0. Even if a ≥ 4, a sin p £ 4

and the roots are real. 14. Ans (a, b): If the coefficients are imaginary, then the

imaginary roots may not be conjugate of each other. Hence, Statement (a) is false.

The continuous monotonic function may not be dif-ferentiable. Hence, Statement (b) is false.

Statement (c) is true

cos tan sinx xx n

xx n2 2 2 2- -

=- -p p p p

for all value of x π np

+ p2

(which are not in the domains of both the sides).

Hence, Statement (d) is true. 15. Ans (c, d): Since the curve y = ax1/2 + bx passes

through the point (1, 2), 2 = a + b (1) By observation, the curve also passes through (0, 0).

Therefore, the area enclosed by the curve, x-axis and x = 4 is given by

A ax bx dx= + =Ú ( )/1 2

0

4

8

fi 23

82

16 8a b◊ + ◊ =

fi 23

1a b+ = (2)

Solving (1) and (2), we get a = 3, b = –1.

Section C

16. Ans (2):

f(n) = 1 1

2

1

22 2 2n n n+

++

++

( ) ( )...

++ +

1

2 12n n

The terms of the sequence are decreasing and the number of terms are (2n + 2).

2 2

2 1

2 22 2

n

n nf n n

n

+

+ +£ £

+( )

Now Lim Limn n

n

n n

nn

nn n

Æ • Æ •

+

+ +=

+ÊËÁ

ˆ¯̃

+ +

2 1

2 1

2 1 1

1 2 12

2

( ) = 2

Similarly Lim Limn n

n

n

nnÆ • Æ •

+=

+=

2 1 2 1 22

( ) ( )

\ Limn

f xÆ •

=( ) 2

17. Ans (1): We have y = ax/(b – x)

fi dydx

b x a axb x

abb x

= - - --

=-

( ) ( )( ) ( )

12 2

dydx

abb

ÈÎÍ

˘˚̇

=-

=( , ) ( )1 1

212 (given) (1)

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B.14 Mathematics

Since the curve passes through the point (1, 1), there-fore,

1 = a/(b – 1) fi a = b – 1 (2) On putting a = b – 1 in Eq. (1), we get

( )( )b bb

--

11 2 = 2 fi b = 2 fi a = 2 – 1 = 1

Hence, a = 1, b = 2

18. Ans (5): AB i j k

= - = - - -b a 2 3 6 Equation of the plane passing through B and perpen-

dicular to AB

is

( ) r OB AB- ◊ = 0

fi r ◊ (2i + 3 j + 6k) + 28 = 0

Hence, the required distance from c = – i + j + k is

c i j k

i j k◊ + + +

+ += - + + +( )2 3 6 28

2 3 62 3 6 28

7

= 5 units

19. Ans (1): The given equation is | z |n = (z2 + z) | z |n – 2 + 1 fi z2 + z is real

fi z z z z2 2+ = +

fi ( ) ( )z z z z- + + =1 0

fi z z x z z x= = + + π π -as 1 0 1 2( / )

fi The given equation reduces to xn = xn + x | x |n – 2 + 1 fi x | x |n – 2 = –1 fi x = –1. So, the number of solution is 1. 20. Ans (1):

x y y zyz

y z z xzx

z x x yxy

- = - - = - - = -, ,

\ ( ) ( )( ) ( )( )( )( )

x y y z z x x y y z z xxyz

- - - = - - -2

fi xyz = 1

PaPer 2Section a 1. Ans (b, c): (1 – y)m (1 + y)n

= (1 – mC1 y + mC2 y2 – L) (1 + nC1 y + nC2 y

2 + L)

= + - +-

+-

-ÏÌÓ

¸˝˛

+1 12

12

2( ) ( ) ( )n m m m n n mn y L

Given: a1 = 10 fi a1 = n – m = 10 (1)

a2 = m n m n mn2 2 22

10+ - - - =

fi (m – n)2 – (m + n) = 20 fi m + n = 80 (2) 2. Ans (c): Since P(x) divides into both of them, hence P(x) also divides (3x4 + 4x2 + 28x + 5) – 3(x4 + 6x2 + 25) = – 14x2 + 28x – 70 = – 14(x2 – 2x + 5 which is a quadratic. Hence, P(x) = x2 – 2x + 5 fi P(1) = 4 3. Ans (a, d): tan x – tan2 ¥ > 0 fi tan x(tan x – 1) < 0 fi 0 < tan x < 1 or 0 < x < p/4 or np < x < np + p/4, n Œ Z (generalizing)

sin x < 12

fi - < <12

12

sin x

or –p/6 < x < p/6 or –p/6 + np < x < p, n Œ Z (generalizing) Then the common values are np + p/6 < x < np + p/4. 4. Ans (a, b, c, d): (a) P(E1) = 1 – P(RRR)

= 1 13

24

35

0 9- ¥ ¥ = .

(b) P(E2) = 3P(BRR) = 3 23

14

25

0 2¥ ¥ ¥ = .

(c) P(E3) = P(RRR/(RRR » BBB))

= P RRRP RRR P BBB

( )( ) ( )+

= 0 1

0 1 23

34

45

.

. + ¥ ¥

= 0 10 1 0 4

0 2.. .

.+

=

(d) P(E4) = 1 – P(BBB) = 1 25

- = 0.6

5. Ans (a, b, c): We have bc ca abca ab bcab bc ca

= 0

fi (ab)3 + (bc)3 + (ca)3 – 3(ab)(bc)(ca) = 0 or ab + bc + ca(abw + bcw2 + ca) (abw2 + bcw + ca) = 0

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Appendix B: Mock Tests B.15

or ab + bc + ca = 0, abw + bcw2 + ca = 0, abw2 + bcw + ca = 0

or 1 1 1 0 1 1 1 02a b c c a b+ + = + + =, ,

w w

1 1 1 02c a b+ + =

w w 6. Ans (a, b, c, d): Odd + Odd – Even – Even = Even So the given number is divisible by 2. Also 2007201 – 1982201 is divisible by 2007 – 1982

= 25 2044201 – 2019201 is divisible by 2044 – 2019 = 25 Hence, the given number is divisible by 25. Also 2007201 – 2044201 is divisible by 2007 – 2044 = –37

and 2019201 – 1982201 is divisible by 2019 – 1982 = 37 Hence, the Given number is divisible by 37. So the given number is divisible by 2 ¥ 37, 2 ¥ 25,

2 ¥ 25 ¥ 37 = 74, 50, 1850. 2007201 + 2019201 is divisible by 2007 + 2019 = 4026 1982201 + 2044201 is divisible by 1982 + 2044 = 4026 So, the given number is divisible by 4026 = 2 ¥ 2013 [\ an + bn has a factor a + b if n is odd] 7. Ans (a, b, c): sin2 x – a sin x + b = 0 has only one

solution in (0, p). So, sin x = 1 gives one solution and sin x = a gives

other solution such that a > 1 or a £ 0. Hence, (sin x – 1)(sin x – a) is the same equation as

sin2 x – a sin x + b = 0 fi 1 + a = a and a = b or 1 + b = a and b > 1 or b £ 0 or b Œ (–•, 0) » (1, •) and a Œ (–•, 1) » (2, •) 8. Ans (b, d): tan (a + b) = 15/8 and tan g = 8/15 \ a + b + g = p/2 fi (b) and (d)

Section B Ans 9. (b), 10. (c): Let the variable line AC be x + ay – 1 = 0 which passes

through the point A(1, 0). Let the foot of perpendicular on variable line from

B(3, 4) is C(h, k). Now BC ^ AC \ (slope of BC) ¥ (slope of AC) = –1

or kh

kh

--

ÊËÁ

ˆ¯̃

--

ÊËÁ

ˆ¯̃

= -43

01

1

or x2 + y2 – 4x – 4y + 3 = 0 or (x – 2)2 + (y – 2)2 = 5

Thus, the length of tangent from origin is S1 3= .

Ans 11. (a), 12. (b): Let r xi y j zk= + +

r i j k.( )+ + = 10

\ x + y + z = 10 fi x ≥ 1, y ≥ 1, z ≥ 1 Therefore, the number of latice points = 9C2 x + y + z £ 11 For equilateral latice points, x = y = z = 1, 2, 3 So, three cases are possible. For isosceles lattice point, x = y π z

1 1 9

5 5 1

¸˝Ô

˛Ô

15 cases

1 1 8

4 4 2

¸˝Ô

˛Ô

12 cases

1 1 72 2 54 4 1

¸˝Ô

˛Ô

9 cases

1 1 62 2 43 3 2

¸˝Ô

˛Ô

9 cases

1 1 52 2 33 3 1

¸˝Ô

˛Ô

9 cases

1 1 4} 3 cases

1 1 32 2 1

¸˝˛

6 cases

1 1 2} 3 cases Total number of isosceles lattice points which are not

equilateral points = 15 + 12 + 9 + 9 + 9 + 3 + 6 + 3 = 66 Therefore, total number of isosceles latice point = 66 + 3 = 69

Therefore, required probability, P = 369

123

=

Ans 13. (c), 14. (b):

g(x) = x x+ sin2

increasing function of x

range 02

, pÈÎÍ

˘˚̇

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B.16 Mathematics

\ f (x) = x x+ sin ,2

x Π[0, p]

p £ t £ 2p, then f (t) + f (2p – t) = p

f (t) + 2 22

p p- + -t tsin ( ) = p

f (t) + p – t t2 2

-sin = p

f (t) = t t+ sin2

f (x) = x x+ sin2

p £ x £ 2p

f (x) = x x+ sin2

for 0 £ x < 2p

f (x) = f (4p Рx) for x Π[2p, 4p] f (x) is symmetric about x = 2p.

O 2p 4p

Fig. B.2

a = 2p – 0 = 2p from graph b = a Ans 15. (a), 16. (d): From the graph (Fig. B.3), y = 6 and x = 8 are the

common tangents.

C(8,10)A(8,6) B(12,6)

y

x

Fig. B.3

15. Area of DABC = 8 sq. unit 2s = 4 + 4 + 4 2

Radius rs

= =+

=+

= -D 8

4 4 22

1 22 2 1

Co-ordinate of centre are ( , )7 2 2 5 2 2+ +

16. Centre of circumcircle is (10, 8) and radius 2 2 . Therefore, equation of circle is x2 + y2 – 20x – 16y + 156 = 0

Section C 17. Ans: (a Æ q); (b Æ s); (c Æ r); (d Æ p) (a) lim

x Æ -1 f (g(x)) = f (2–) = 2

limx Æ +1

f (g(x)) = f (1–) = 1

Therefore, limit does not exist.

(b) limx Æ 2

f (x) = 2 fi lim ( )x

f xÆ

- =2

3 2 2

(c) limx Æ 0

f (x) = 0 and limx Æ 0

g(x) = finite quantity

\ lim ( )( )

( ) ( )x

f xg x

f x g xÆ

+ÊËÁ

ˆ¯̃

=0

0

(d) lim ( ) ( )( ) ( )

( )x

f x g xf x g xÆ +

-+

=-

+= =

1

3 3 1 11 1

22

1

18. Ans: (a Æ r), (b Æ r), (c Æ q), (d Æ s)

(a) arg zz

2

211

-+

Ê

ËÁˆ

¯̃ = 0 ; z π ± i

zz

zz

z z z z2

2

2

211

11

0 0-+

=-+

fi - = + =,

y = 0, x = 0 Locus of z is portion of pair of lines xy = 0.

Q zz

2

211

0-+

Ê

ËÁˆ

¯̃>

È

ÎÍÍ

˘

˚˙˙

(b) Given || z – cos–1 cos 12 | – | z – sin–1 sin 12 || = 8 (p – 3) Since, | cos–1 cos 12 – sin–1 sin 12 | = 8 (p – 3) Therefore, locus of z is the portion of a line

joining z1 and z2 except the segment between z1 and z2.

(c) z2 – i | z1 |2 = k2 – k1

x2 – y2 + 2 ixy – il1 = l2

x2 – y2 = l2 and xy = l1

2 The locus of z is the point of intersection of

hyperbola. (d) Given,

z z- - + + - =- -1 13

13 2

11 1sin cos p

Since 1 13

13 2

11 1+ + - =- -sin cos p

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Appendix B: Mock Tests B.17

| z – z1 | + | z – z2 | = | z1 + z2 | Thus, the locus of z is line segment joining z1

and z2. 19. Ans: (a Æ q), (b Æ p), (c Æ s), (d Æ r) Angles of the triangle are 30º, 45°, 105°.

Then a b csin sin sin30 45 105∞

=∞

=∞

fi a b c1 2 1 2 3 1 2 2/ / ( )/

= =+

fi a b c k2 2 3 1

= =+

= (say)

Smallest side, a = 2 k and largest side, c = ( )3 1+ k

If a k c= = = +2 2 6 2, ,

If a k c= = = +2 1 3 1, ,

If s k c= + + = = +3 2 3 2 2 2 3, ,

If D =-3 1

4, then from D = 1

2 bc sin A, we have

3 14

12

2 3 1 30-= ¥ ¥ + ∞k k( ) sin

fi k k223 1

2 3 13 1

43 12

=-+

=-

fi =-

( )( )

and hence c = 1 20. Ans: (a Æ q); (b Æ p); (c Æ r); (d Æ q) Let Ei denotes the event that the bag contains i black

and (14 – i) white balls (i = 0, 1, 2, ...14) and A denotes the event that five balls drawn are all black. Then

P(Ei) = 115

(i = 0, 1, ... 14)

P(A/Ei) = 0 for i = 0, 1, 2, 3, 4

P(A/Ei) = i C

C5

145

for i ≥ 5

(a) P(A) = P P( ) ( / )E A Ei ii =Â

0

14

= 115

114

5C (5C5 + 6C5 + L + 14C5)

= 115

16

156

145

CC

=

(b) Clearly P(A/E11) = 11

514

5

313

CC

=

(c) From Baye’s theorem,

P(A/E11) = P PP

( ) ( / )( )

.E A EA

11 11

115

313

16

665

= =

(d) Let B denotes the probability of 3 black and 2 white balls, then

P(B/Ei) = 0 if i = 0, 1, 2 or 13, 14

P(B/Ei) = i iC C

C3

142

145

-

for i = 3, 4, ..., 12

\ P(B) = P P( ) ( / )E B Ei ii =Â

0

14

= 115

114

5.

C[3C3.

11C2 + 4C3.10C2 + L+ 12C3.

2C2]

= 500515

1614

5. C=

Solutions of Mock Test 2

PaPer 1Section a

1. Ans (b): f (x) = ( ) ( )x n n n

n

- -

=’ 51

1

50

ln f (x) = n n x nn

( ) ln ( )511

50

- -=

Â

Differentiating both sides with respect to x, we get

¢ = --

=Âf x

f xn n

x nn

( )( )

( )51

1

50

¢ = --

=Âf

fn n

nn

( )( )

( )5151

5151

1

50

=1 + 2 + 3 + L + 50 = 1275

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B.18 Mathematics

or ff

( )( )5151

11275¢

=

2. Ans (c): y2 = 4ax, 4a = 2p > 0 and t1t2 = –1.

Ratio = 4 21 2

212

22

a t ta t t

= – 4

3. Ans (a):

Tk = cos .( )

( ) ( )( )( )

-

++

- + ++

Ê

ËÁ

ˆ

¯˜

1 1 11

1 1 21k k

k k k kk k

Let x = 1k

and y = 1

1k +

1 1 11

22- = -

+y

k( )

=+ -+

=++

( ) ( )kk

k kk

1 11

21

2

Tk is in the form of

cos .- + - -( )1 2 21 1xy x y = cos–1(y) – cos–1(x) (Q y < x)

Tk = cos cos- -

+ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

1 111

1k k

Substituting k = 2, 3, 4, ...

Sum = Limn nÆ •

- -

+ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

cos cos1 111

12

= cos ( ) cos- -- ÊËÁ

ˆ¯̃

1 10 12

Sum = p p p2 3 6

- =

4. Ans (d): g(x) = f (cot2 x + 2 cot x + 2) fi g¢(x) = f ¢(cot2 x + 2 cot x + 2) ◊ {–2 cot x cosec2 x – 2 cosec2 x} for g(x) to be decreasing, g¢(x) < 0 fi f ¢{(cot x + 1)2 + 1} ◊ (–2 cosec2 x) (cot x + 1) < 0 fi f ¢{(cot x + 1)2 + 1} ◊ (cot x + 1) > 0 …(i) {as f ≤(x) > 0 fi f ¢(x) is increasing, then f ≤(cot x + 1)2 + 1} > f ¢(1) = 0

" ŒÊËÁ

ˆ¯̃

» ÊËÁ

ˆ¯̃

x 0 34

34

, ,p pp

Thus, equation (i) holds, if cot x + 1 > 0

fi cot ,x x> - " ŒÊËÁ

ˆ¯̃

1 0 34p

5. Ans (c): From LMVT, there exists atleast (n – 1) point where f ¢(x) = m.

fi $ atleast (n – 2) points where f ≤(x) = 0 (using Rolle’s theorem)

6. Ans (d):

1 1 1 133

100100

13

3+ +ÊËÁ

ˆ¯̃

È

ÎÍ

˘

˚˙ = + +Ê

ËÁˆ¯̃

xx

C xx

+ +ÊËÁ

ˆ¯̃

+ + +ÊËÁ

ˆ¯̃

1002

33

2100

1003

3

1001 1C xx

C xx

L

= (1 + r) + A1x3 + A2x

6 + L + A100(x3)100 + B1

13x

+ L + B100 13

100

xÊËÁ

ˆ¯̃

All other terms obtained by the combination of x3 and 1/x3 well get converted into a term involving x3 or 1/x3 and hence it will be present among above terms.

So number of terms = 1 + 100 + 100 = 201 7. Ans (b): f(x) = ax2 + (a – 2) x – 2 = (ax – 2) (x + 1) f (0) = – 2 and f (–1) = 0 If a is negative, then the expression becomes negative

for infinite values of x. Therefore, it must be positive. Expression to be negative for exactly two integral values of x

O 1-1 22a

Fig. B.4

So 2 2 1a

a£ fi ≥

and 2 1 2a

a> fi <

\ a Π[1, 2) 8. Ans (a): f (x) = | sin a x | + | cos a x |

Period of f (x) = pa

p2 8

= (given)

fi a = ± 4 9. Ans (d): Given integral

= xf x f xx f x

dx x f x xf x x

f x xdx¢ -

= ¢ -ÚÚ ( ) ( )( )

( ( ) ( ))/

( )/

2 22

2 4

2

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Appendix B: Mock Tests B.19

Let t = f xx

dtdx

x f x x f xx

( ) ( ) ( )2

2

42

fi = ¢ -

Therefore, the required integral is

= 1 221 2

tdt t c

f xx

c= + = +Ú / ( )

10. Ans (a): In 2nd quadrant, sin x < cos x is false as sin x is positive and cos x in negative.

In 4th quadrant, cos x < tan x is false as cos x is posi-tive and tan x in negative.

In 3rd quadrant, i.e., 54

32

p p,ÊËÁ

ˆ¯̃

, if tan x < cot x,

fi tan2x < 1, which is false. Hence, (a) can be correct.

Now sin x < cos x is true in 04

, pÊËÁ

ˆ¯̃

and tan x < cot x

is also true. Further, cos x < tan x as tan x = (sin x)/(cos x) and

cos x < 1.

Section B

11. Ans (b, d):

We have f (x) = e dt e dtt tx

tx

-Ú Ú=[ ] { } ,0 0

So f (x) =

e dt x

e dt e dt x

e dt e dt

tx

t tx

t t

0

0

11

1

0

11

1

2

0 1

1 2

Ú

Ú Ú

Ú

Œ

+ Œ

+

-

-

if

if

( , )

( , )

ÚÚ Ú+ Œ

È

Î

ÍÍÍÍÍÍÍÍÍÍÍ

-e dt xtx

2

2

2 3if ( , )

⇒ f (x) =

e x

e e x

e e x

x

x

x

- Œ

- + - Œ

- + - Œ

-

-

1 0 1

1 1 1 2

2 1 1 2

1

2

if

if

if

( , )

( ) ( ) ( , )

( ) ( ) ( , 33)

Ï

ÌÔÔ

ÓÔÔ

Clearly f (x) is continuous " x > 0 but not differen-tiable " x > N.

Also f (2) = 2(e – 1) = 0 = 2(e – 1)

12. Ans (b, c): x xn

1 2 1 412

/ /+ÊËÁ

ˆ¯̃

-

T C x xrn

r r

n r r

+ = ¥ ¥- -

12 4

12

Coefficient of the first three terms are nC0, nC1

12

¥ ,

nC2 212

¥

\ n n nC C C0 2 114

2 12

+ ¥ = ¥ ¥

fi 1 18

+-

=n n n( )

fi n n n( ) ( )-= -

18

1

fi n = 8 (as n π 1)

\ T C x xr r r

r r

+-= ¥ ¥

-

18

82 4

12

= ¥ ¥-( )8 41

2

34C xr r

r

Terms of x with integer power occur when r = 0, 4, 8. Thus, three terms.

Hence, (b) and (c) are correct. 13. Ans (a, b, c, d):

O(0, 0)

y

x

B b(0, )

D d(0, )

A a( , 0) C c( , 0)

Fig. B.5

Points A, B, C and D are concyclic, then ac = bd. The co-ordinates of the points of intersection of lines

are

ac b dbc ad

bd c abc ad

( ) , ( )--

--

ÊËÁ

ˆ¯̃

Let the co-ordinates of the point of intersection be (h, k). Then

h ac b dbc ad

k bd c abc ad

=-

-=

--

( ) , ( )

Given c2 + a2 = b2 + d2

(Q ac = bd) fi (c – a)2 = (b – d)2 or (c – a) = ± (b – d) Then the locus of the points of intersection is y = ± x.

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B.20 Mathematics

14. Ans (a, b, d): p = 5C4 ◊ 2C1 = 10

q = 5C2(2C1)

3 = 80 r = 5C0(

2C1)5 = 32

fi 2q = 5r, 8p = q, and 2(p + r) > q 15. Ans (a, c, d): From Fig. B.6,

Ay x dy

dxdydx

=- +Ê

Ë

ÁÁÁ

ˆ

¯

˜˜˜

, 0

O B x( , 0) A

P x y( , )

Fig. B.6

As per the given condition,

2x dydx

y x dydx

= - +

fi x dydx

y= -

fi dxx

dyy

= -

fi ln x = –ln y + ln c fi xy = c Also, f (1) = 1 fi c = 1 fi xy = 1

Section C

16. Ans (4): I = Limt

x t xt

dxÆ

+ -Ú00

2sin ( ) sin

p

= Limt

x t t

tdt

Æ

+ÊËÁ

ˆ¯̃

¥Ê

Ë

ÁÁÁÁ

ˆ

¯

˜˜˜˜

Ú 00

2 22 2

cos sinp

= cos x dx0

2p

Ú = 4

17. Ans (9): a = 1 – b – c fi 1 + a = (1 – b) + (1 – c) > 2 ( )( )1 1- -b c

Similarly, 1 + b > 2 ( )( )1 1- -c a

and 1 + c > 2 ( )( )1 1- -a b Required expression > 8.

18. Ans (5): f (x) = sin [ ] [ ]- -Ê

ËÁˆ

¯̃+ + -1

2 23

x x x x

sin- -Ê

ËÁˆ

¯̃1

2 23

x x is defined for

- £-

£ + -1 23

12x x x xand [ ] [ ] defined only

for integral values of x. fi x = – 1, 0, 1, 2, 3 Therefore, total number of element in domain is 5. 19. Ans: Since the produce matrix is 3 ¥ 3 matrix and

the premultiplier of A is a 3 ¥ 2 matrix, A is a 2 ¥ 3 matrix.

Let A =1 m nx y z

È

ÎÍ

˘

˚˙ , then the given equation becomes

2 11 03 4

11 8 101 2 59 22 15

-

-

È

Î

ÍÍÍ

˘

˚

˙˙˙

È

ÎÍ

˘

˚˙ =

- - -- -

È

Î

ÍÍÍ

˘

˚

˙˙˙

m nx y z

fi 2 3 2

3 4 3 4 3 4

l x m y n zl m x

l x m y n z

- - -

- + - + - +

È

Î

ÍÍÍ

˘

˚

˙˙˙

=- - -

- -È

Î

ÍÍÍ

˘

˚

˙˙˙

1 8 101 2 59 22 15

fi 2l – x = –1, 2m – y = –8, 2n – z = –10, l = 1, m = –2, n = –5 fi x = 3, y = 4, z = 0, 1 = 1, m = –2, n = –5

fi Am n

x y z=

È

ÎÍ

˘

˚˙ =

- -È

ÎÍ

˘

˚˙

1 1 2 53 4 0

20. Ans (9): g(x) = 12

tan–1 | x | + 1 fi sgn (g(x)) = 1

\ sin23 x – cos22 x = 1 fi sin23 x = 1 + cos22 x, which is possible if sin x = 1

and cos x = 0.

fi x n= +22

pp

Hence, - £ + £10 22

8p pp

pn

fi - £ £214

154

n fi –5 £ n £ 3

Hence, the number of values of x = 9.

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Appendix B: Mock Tests B.21

Also, g(x) is monotonic.

Also f xg x

dx x xx

dx( )( )

/Ú Ú=- 1

3

= -ÊËÁ

ˆ¯̃

= - + +Ú 1 1 1 32 4 3x x

dxx x

c

3. Ans (b, c, d): Let (x1, y1) ∫ (at2, 2at) Tangent at this point is ty = x + at2.

Any point on this tangent is h h att

, +Ê

ËÁˆ

¯̃

2

.

Chord of contact of this point with respect to the circle

x2 + y2 = a2 is hx h att

y a++Ê

ËÁˆ

¯̃=

22

or (aty – a2) + +ÊËÁ

ˆ¯̃

=h x yt

0

which is a family of straight lines passing through

the point of intersection of ty – a = 0 and x yt

+ = 0

So, the fixed point is -ÊËÁ

ˆ¯̃

at

at2 , . Therefore,

x at

y at2 2 2= - =,

Clearly, x1x2 = – a2, y1 y2 = 2a2

Also, xx

t yy

t1

2

4 1

2

22= - =,

fi 4 01

2

1

2

2xx

yy

+ÊËÁ

ˆ¯̃

=

4. Ans (a, b, c): Here, let P be (x, y, z). Then, x = r sin q cos f, y = r sin q sin f, z = r cos q fi 1 = r sin q cos f, 2 = r sin q sin f, 3 = r cos q (i) fi 12 + 22 + 32

= r2 sin2q cos2f + r2 sin2f sin2q + r2 cos2q = r2 sin2q (cos2f + sin2f) + r2 cos2q = r2 sin2q + r2 cos2q = r2

fi r = ± 14 fi From (i), we have

sin q cos f = +114

, sin q sin f = 214

,

cos q = 314

(neglecting –ve sign assuming acute angles)

sin sinsin cos

sincos

q fq f

qqq

= = =21

53

and tan

fi tan f = 2 and tan q = 5 3/

PaPer 2Section a

1. Ans (a, c): If p = q, then the equation becomes zp = z p and it has infinite solution because any z Πreal will satisfy it.

If p π q, let p > q, then zp = z q

fi | z |p = | z |q

fi | z |p (| z |p – q – 1) = 0 fi | z | = 0 or | z | = 1 | z | = 0 fi z = 0 + i0 | z | = 1 fi z = eiq

fi e(p + q)qi = 1 fi z = 11/(p + q)

Hence, the number of solution is p + q + 1. 2. Ans (a, c, d):

Let I = ( )( )xx

dx4

611

++Ú

= ( )( )( )

x xx x x

dx2 2 2

2 4 21 2

1 1+ -

+ - +Ú = ( )

( ) ( )x dx

x xx dxx

2

4 2

2

61

12

1+

- +-

+ÚÚ

= 1 1

1 12

1

2

22

2

3 2

+ÊËÁ

ˆ¯̃

- +ÊËÁ

ˆ¯̃

-+ÚÚ x

dx

xx

x dxx( )

In first integral, put xx

t- =1

Therefore, 1 12+Ê

ËÁˆ¯̃

=x

dx dt and in second integral,

put x3 = u.

\ x dx du2

3= , then

I = dtt

duu1

23 12 2+

-+ÚÚ

= tan tan- -- +1 123

t u c

= tan tan ( )- --ÊËÁ

ˆ¯̃

- +1 1 31 23

xx

x c

Here f x xx

g x x( ) ( )= - =1 3and

Both the functions are odd.

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B.22 Mathematics

x

y

M

O

Z

rq

f

P(1, 2, 3)

Fig. B.7

5. Ans (b, d):

Let a i j b i j= - - = -

32

3 32

3( ), ( )

c i j d i j= - + = +

32

3 32

3( ), ( )

AB i

= =3 a (say)

Clearly, a b c d= = = = 3

If a makes angle q and f, y and b with

a b c, , and d , respectively, then

cos /.

qa

a=

◊=

-= -

aa

9 23 3

12

cos /.

fa

a=

◊= =

bb

9 23 3

12

\ f = 60°

cos /.

ya

a=

◊=

-= -

cc

9 23 3

12

cos /.

ba

a=

◊= =

dd

9 23 3

12

\ b = 60° 6. Ans (a, b, d): We have P = probability that the bomb

strikes the target = 1/2. Let n be the number of bombs which should be

dropped to ensure 99% chance or better of completely destroying the target. Then the probability that out of n bombs at least two strike the target is greater than 0.99.

Let X denote the number of bombs striking the target.

Then P(X = r) = nr

r n rn

r

n

C C r12

12

12

ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

-

;

= nr

r n rn

r

n

C C r12

12

12

ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

-

; = 0, 1, 2, ..., n

We should have P(X ≥ 2) ≥ 0.99

fi {1 – P(X < 2)} ≥ 0.99 fi 1 – {P(X = 0) + P(X = 1)} ≥ 0.99

fi 1 1 12

0 99- +ÏÌÓ

¸˝˛

≥( ) .n n

fi 0 001 12

. ≥+ n

n

fi 2n > 100 + 100 n fi n ≥ 11 Thus, the minimum number of bombs is 11. 7. Ans (b, c): x2 – (a + b)x + ab = 0 or (x – a)(x – b) = 0 fi x = a or b If x = a is the root of other equation, then a3 – ap + ab = 0 fi p = a2 + b If x = b is the root of the other equation, then ab2 – pb + ab = 0 or p = a(1 + b) 8. Ans (b, d): f (0, t) = 0 for t ≥ 0, so

g(0) = f t dt( )◊ ◊ =Ú 0 00

1

f (1, t) = t (1 – 1) = 0 for t < 1; so g(1) = 0

Also g(x) = f t t x dt f t x t dtx

x( ) ( ) ( ) ( )- + -Ú Ú1 1

0

1

= ( ) ( ) ( ) ( )x t f t dt x f t t dtx

x- + -Ú Ú1 1

0

1

\ g¢(x) = t f t dt f t dtx

( ) ( )0

1 1

Ú Ú-

g≤(x) = f (x)

Section B

9. Ans (a) 10. Ans (c) y = (x – a) (x2 + bx + c) for y = 0, x = a \ p = a fi The point P is (a, 0). Let the slope of the line through P be m. Therefore, equation of the line is y = m (x – a). It meets the curve in A and B. Therefore, m (x – a) = (x – a) (x2 + bx + c)

dydx

= x2 + bx + c + (x – a) (2x + b)

= x2 + bx + c + 2x2 – 2ax + bx – ab = 3x2 + (2b – 2a) x + c – ab Therefore, slope of the tangent at P(a, 0) is a2 + ab + c. Therefore, equation of the tangent y – 0 = (a2 + ab + c) (x – a)

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Appendix B: Mock Tests B.23

It meets the curve at C. \ (a2 + ab + c) (x – a) = (x – a) (x2 + bx + c) i.e., a2 + ab + c = x2 + bx + c i.e., x2 + bx – a2 – ab = 0 i.e., (x – a) (x + a) + b (x – a) = 0 i.e., (x – a) (x + a + b) = 0 Therefore, absicca of C is – a – b. Since the line through P intersects at two different

points A and B, therefore, b2 – 4 (c – m) > 0 ...(i) Now if the slope of any tangent is m, then 3x2 + (2b – 2a) x + c – ab = m Its D = 4(b – a)2 – 12(c – ab – m) = 4[a2 + b2 – 2ab – 3c + 3ab + 3m] = 4[a2 + b2 + ab + 3m – 3c] ...(ii) From (i) and (ii), we get

D a b ab b> + + -ÈÎÍ

˘˚̇

4 34

2 2 2

= 44

22

a b ab+ +È

ÎÍÍ

˘

˚˙˙

= 4a2 + b2 + 4ab = (2a + b)2 ≥ 0 \ D > 0 Hence, there are 2 points.

11. Ans (b): f x dx( ) =Ú 50

2

= shaded region PSTR

y f x= ( )–1

T M

A3

A2

A4

y f x= ( )

RA1

Q

S

P

Fig. B.8

Area of PQR = 5 – Area QRST = 5 – 4 = 1

= Area RMT = f x dx-Ú 1

2

4

( ) = 1

12. Ans (a): ( ( ) ( ))f x f x dx A A- = fi = =-Ú 11 2

0

2

1 1

f x dx A- = fi = - =Ú 13

2

4

4 4 1 3( )

fi g x f x f x dx( ) max ( ( ), ( ))- -Ú 1

0

4

= 4p – (A1 + A2 + 2A3 + 4) = 4p – (2 + 6 + 4) = 4p – 12. 13. Ans (c):

3

–2 3

(–2, (–2))f

O

Fig. B.9

So three solutions 14. Ans (b):

x = 3

x = – 2

y = 3

(–2, (–2))f

Fig. B.10

Five solution but f (–2) > 6 f ¢(0) < –3

15. Ans (b): C z z195

15

2: - + - =

Here 95

15

85

2- = <

Therefore, locus C1 is ellipse with foci 15

0,ÊËÁ

ˆ¯̃

and

95

0,ÊËÁ

ˆ¯̃

.

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B.24 Mathematics

\ 2 85

2ae a= =and 2

fi e a= =45

1and

Center is the mid point of foci which is (1, 0). Therefore, vertices are (0, 0) and (2, 0).

C2 : Arg (z – 2i) = -p4

represents a ray originating

from (0, 2) and making an angle, if 135° with the positive x-axis.

C3 : | z – 1 – i | = 2 which is the circle having center at (1, 1) and radius

2 .

-1 1 2 3

–1

1

2

3

C1 C3

C2

Fig. B.11

From Fig. B.11, there is only one point common to all curves which is (2, 0).

16. Ans (a): From the above diagram, C1 is symmetrical about x-axis.

So, C4 will be the same as C1.

Section C

17. Ans: (a Æ r); (b Æ p); (c Æ s); (d Æ q) (a) Put px = qy = rz = t =1

(b) Coefficient of x103 in 11

5 199--

Ê

ËÁˆ

¯̃xx

(x – 1)201

= Coefficient of x103 in – (1 – x)2 (1 – x5)199

= Coefficient of x103 in – (1 + x2 – 2x) (1 – 199C1x

5 + 199C2x10 + L) = 0

(c) 22003 = 23 (17 – 1)500 = 8 (17k + 1); k ŒI \ Remainder = 8 (d) Therefore, 111111 is divisible by 1001 com-

pletely (6 one’s) So 6 ¥ 166 = 996 ones are divisible. Now remaining ones are 11111 when divisible

by 1001 gives 100 as remainder 18. Ans: (a Æ r), (b Æ p), (c Æ s), (d Æ q)

(a) y mx a m b= ± +2 2 2

y x= - ± ¥ +43

18 169

32 y x= - ±43

8

0

A

Bx

yP

Fig. B.12

Distance between tangent = 16

1 169

+

= 16 35

485

¥=

(b) y x= - +43

8 A(6, 0) and B(0, 8)

Area of DAOB = 12

6 8 24¥ ¥ =

(c) Point of contact

-+ +

Ê

ËÁÁ

ˆ

¯˜˜

a m

a m b

b

a m b

2

2 2 2

2

2 2 2,

Product of coordinates

= -+

a b ma m b

2 2

2 2 2 = -¥ ¥ -Ê

ËÁˆ¯̃

18 32 43

64 = 12

(d) 4x + 3y = 24 \ l = 424

m = 324

\ l m+ =724

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Appendix B: Mock Tests B.25

19. Ans: (a Æ p); (b Æ r); (c Æ p); (d Æ q)

(a) log .0 55 2 0-Ê

ËÁˆ¯̃

≥x

x

0 5 2 1 53

52

<-

£ fi ŒÊËÁ

ˆ¯̃

xx

x ,

2 is the only integer in the domain. (b) L.H.S. of given equation is f ¢(x) where f (x) = (x – 1) (x – 2) (x – 3)(x – 4) Because f (x) is continuous and derivable and f (1) = f (2) = 0 Therefore, f ¢(x) = 0 has atleast one root in (1, 2). Similarly f ¢(x) = 0 has atleast one root in (2, 3)

and (3, 4). Because f ¢(x) is a cubic function, hence it has

exactly one root in (1, 2), (2, 3), (3, 4). (c) Applying LMVT in [0, 1]

e --

11 0

= ec for some c Π(0, 1)

ec + 1 = e fi ln (ec + 1) = 1

(d)

2 110

12

1 2 110

12

xx

xx

- +

- - . = 1

fi 4x2 – 20x + 9 = 0 or (2x – 9) (2x – 1) = 0

or x = 12

92

,

At x = 12

, tan–1(0) + tan–11 = p4

(true)

At x = 92

,

tan tan tan- - -ÊËÁ

ˆ¯̃

+ = =1 1 145

19

14p (true)

So two values of x satisfy the above equation. 20. Ans: (a Æ p); (b Æ p); (c Æ q); (d Æ r)

(a) I = x xx

dxln | |( )1 2 2

0 +

•

Ú

xz

I zz

zz

dz= =+Ê

ËÁˆ¯̃

¥-

-

•Ú1

1

1 11

1

2

2 2

0

,log ( )

I = log ( )( )

zz

z dz1 2 2

0

+¥

•Ú = x x

xdxln

( )1 2 2

0

+•Ú = -

+

•

Ú x xx

dxln( )1 2 2

0

I + I = 0 or 2I = 0 or I = 0

(b) I = ln tan/

x dx0

2p

Ú = ln cot

/x dx

0

2p

Ú 2I = ln (tan cot )

/x x dx¥Ú0

2p

= 0 or I = 0

(c) I = xdx

x x3 61

1

1+ +-Ú = x x x dx1 6 3

1

1+ -( )

-Ú = x x dx x dx1 6 4

1

1

1

1+ -

Ø-- ÚÚ

Zero

I = - = -Ú2 2 54

0

1x dx /

(d) [ ] ( )x dx dx dx= - + ÚÚÚ --1 0

0

1

1

0

1

1

= – [x]0–1 = – [0 – (–1)] = –1

Solutions of Mock Test 3

PaPer 1Section a

1. Ans (c): Y – y = dydx

(X – x)

Therefore, the coordinates of T are x -ÊËÁ

ˆ¯̃

y dxdy

, 0 .

Therefore, the sum of subtangent and abscissa

= y dxdy

+ x = a

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B.26 Mathematics

\ y dxdy

= a – x

i.e., dxa x-

= dyy

\ –log (a – x) = log y + log C

or cy = 1a x-

Curve passes through (2a, a)

or c = – 12a

Therefore, equation of the curve is (x – a) y = a2. 2. Ans (b): Differnciate both sides w.r.t. x, f ¢(x) = cos x + f ¢(x) (2 sin x – sin2 x)

f ¢(x) = cossin sin

xx x1 22+ -

= cos( sin )

xx1 2-

Integrating,

f (x) = cos( sin )

xdxx1 2-Ú (Put 1 – sin x = t)

= - = + =-

+Údtt t

Cx

C2

1 11 sin

Also f (0) = 0, hence C = –1.

f (x) = 11- sin x

–1 = 1 11- +

-sin

sinx

x = sin

sinx

x1- 3. Ans (b): Six places can be selected in 9C2 ways and 6

can be placed only at 5 places except the right most of other 6 selected. Remaining numbers, i.e., 7, 8, 9 in 3! ways.

Hence, number of number 9C6 . 5 . 3! = 84 . 30 = 2520 4. Ans (c): z z z z z z1 1 2 2 3 31 1 1= = =; ; Given z1 + z2 + z3 = 0 fi z z z1 2 3+ + = 0

or (z1 + z2 + z3)2 = 0

z12Â + 2z1z2z3

1 1 1

1 2 3z z z+ +

È

ÎÍ

˘

˚˙ = 0

or z12Â + 2z1z2z3 [ z z z1 2 3+ + ] = 0

\ z12Â = 0

5. Ans (a):

A = 2a bcos sinq q2

= ab sinq cos q = absin 22

q

Hence, A is maximum if q = p/4 and Amax = ab/2. Also sample space = pab/2.

Therefore, favourable and come = pab2

– ab2

= ab2

(p – 1)

Probability = ab2

. ( )pp-1 2ab

= pp-1

6. Ans (c): If x = r cosq and y = r sin q, then E = x2 + y2 = r2. Hence, we have to minimize r2. Now in the given equation substituting x = r cosq and y = r sinq,

we get r2 = 4 cosec 4q fi r 2 4min = 7. Ans (d): As x Æ 0– (i.e., approaches 0 from the left),

[x] = –1

\ limxÆ -0

f (x) = limxÆ -0

1 11

+ --

sin ( ) = –1 + sin 1

whereas if x Æ 0+, we get [x] = 0. \ f (x) = 0 fi lim

xÆ +0 f (x) = 0

Thus, limxÆ0

f (x) does not exist.

8. Ans (d): For symmetric matrix, each place in upper triangle and leading diagonal can be filled in seven ways. Then the number of symmetric matrices are 76.

For skew symmetric matrix, leading diagonal ele-ments are zero.

Upper triangle elements can be filled in 73 ways.

The required probability is 77

77

17

6

9

3

9 9+ - (as one

matrix is common in both the cases)

9. Ans (d): Lim Limn n

k

n

nn n k nƕ

-

Æ•=

ÊËÁ

ˆ¯̃ + ( )Âtan ·

tan1

1

1 1 11

= dxx1

1 1 120

1

+=

+ +Ú tan

ln(sin cos )

10. Ans (a): tr(A) = (x1 – y1)2 + (x2 – y2)

2 + (x3 – y3)2 = 0

fi x1 = y1, x2 = y2, x3 = y3 (1)

|A| = x x y y x x y y x x y yx x y y x x

12

1 1 12

12

1 2 22

12

1 3 32

22

2 1 12

22

2 2 22 2

- + - + - +- + - 22 2 2

222

2 3 32

32

3 1 12

32

3 2 23

32

3 3

22 2 2

y y x x y yx x y y x x y y x x y y

+ - +- + - + - + 33

2

= x xx xx x

y y yy y y

12

1

22

2

32

3

1 2 3

12

2 32

2 12 12 1

1 1 1---

= 2(x1 – x2)(x2 – x3)(x3 – x1)(y1 – y2)(y2 – y3) (y3 – y1) = 2[(x1 – x2)(x2 – x3)(x3 – x1)]

2 ≥ 0

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Appendix B: Mock Tests B.27

Section B

11. Ans (a, b, c, d): Q f ¢(x) = 2x f (x) fi ln f (x) = x2 + c fi f (x) = ex2 f (x) = lex2

Q f (1) = 0 fi 0 = le fi l = 0 Hence, f (x) = 0, x ΠR 12. Ans (c, d): (a Р1)(x2 + 3x + 1)2

– (a + 1) [(x2 + 1)2 – ( ) ]x 3 2 ≤ 0 or (a – 1) (x2 + 3x + 1)2

– (a + 1) (x2 + x 3 + 1) (x2 – x 3 + 1) ≤ 0 (x2 + 3x + 1) [(a – 1) (x2 + 3x + 1) – (a + 1) (x2 – 3x + 1)] £ 0, " x Œ R or –2(x2 + 1) + 2 3a x £ 0 or x2 – a x3 + 1 ≥ 0, " x Œ R or 3a2 – 4 £ 0 (D £ 0)

or a Œ -È

ÎÍ

˘

˚˙

23

23

,

Therefore, number of possible integral value of a is {– 1, 0, 1} fi 3 fi (c) and the sum of all integral values of a is –1 + 0 + 1

= 0 fi (d) 13. Ans (b, c):

r n q r n q r n q◊ = ◊ = ◊ =1 1 2 2 3 3, , intersect in a line if [ ]

n n n1 2 3 0= . So,

1 1 11 2 1

1

02

a

a a

=

or 0 0 1

1 2 2 1 1

1 1

02 2

- -

- -

=a a

a a a or (1 – 2a) (a2 – 1) – (a – a2) (2a – 1) = 0

or a =12

1,

14. Ans (a, c): AAT = I

fi 13

1 2 22 1 2

2

13

1 22 1 22 2

1 0 00 1 00 0 1

-È

Î

ÍÍÍ

˘

˚

˙˙˙◊

-

È

Î

ÍÍÍ

˘

˚

˙˙˙

=È

Î

Í

a b

a

bÍÍÍ

˘

˚

˙˙˙

or 19

1 2 22 1 2

2

1 22 1 22 2

1 0 00 1 00 0 1

-È

Î

ÍÍÍ

˘

˚

˙˙˙ -

È

Î

ÍÍÍ

˘

˚

˙˙˙

=È

Î

ÍÍÍ

˘

a b

a

b ˚̊

˙˙˙

or 9 0 4 20 9 2 2 2

4 2 2 2 2 42 2

a ba b

a b a b a b

+ ++ -

+ + + - + +

È

Î

ÍÍÍ

˘

˚

˙˙˙

= 9 0 00 9 00 0 9

È

Î

ÍÍÍ

˘

˚

˙˙˙

or a + 4 + 2b = 0, 2a + 2 – 2b = 0 and a2 + 4 + b2 = 9 or a + 2b + 4 = 0, a – b + 1 = 0 and a2 + b2 = 5 or a = –2, b = –1

15. Ans (b, c, d): Let f x a( ) + =30

0

where

f (x) = x x xx x x

x x xx x x

3 2

3 26 11 6

10 81 2 31 2 4

- + -+ - +

=- - -- - +

( ) ( ) ( )( ) ( ) ( )

= ( ) , , ,xx

x-+

π -34

1 2 4

Range of f (x) = R - - -ÏÌÓ

¸˝˛

1 25

16

, ,

So (1) does not have solution if

a30

1 25

16

= - , ,

a = –30, 12, 5

Section C

16. Ans (4): f (0, •) Æ (0, •) f (x f (y)) = x2ya(a Œ R) Put x = 1, we get f ( f (y)) = ya

Put f (y) = 1x

, we get f (1) = 12( ( ))f y

ya

and put y = 1, we get ( f (1))3 = 1 f (1) = 1 For y = 1, we have f (x( f (1))) = x2

f (x) = x2

Thus, a = 4 17. Ans (7): a2 + 2b = 7 b2 + 4c = –7 c2 + 6a = –14 a2 + b2 + c2 + 2b + 4c – 6a = –14 (a + 3)2 + (b + 1)2 + (c + 2)2 – 14 = –14 fi a = –3, b = –1, and c = –2. or a2 + b2 + c2 = 14

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B.28 Mathematics

18. Ans (8): For the GP a, ar, ar2, ... Pn = a, (ar), (ar2), ..., (arn – 1) = anrn (n – 1)/2

S = P arnn

n

n

n=

•-

=

•

 Â=1

1 2

1

( ) /

Now,

ar a r r r rn

n

( ) /-

=

•= + + + + + •È

΢˚ 1 2

11 L

= ar1 -

Given a = 16 and r = 1/4

\ S = 161 1 2- ( / )

= 32

19. Ans (7): Let general point of line be A(2l + 1, 4l + 3, 3l + 2). Let this point lies at the same distance as point P(3,

8, 2) from the plane 3x + 2y – 2z + 15 = 0

Therefore, 3 3 2 8 2 2 1517

. . .+ - +

= 3 2 1 2 3 3 2 3 2 1517

( ) ( ) ( )l l l+ + + - + +

fi 36 = 8l + 20 fi l = 2

P(3, 8, 2)A

Fig. B.13

Therefore, A is (5, 11, 8)

PA = ( ) ( ) ( )5 3 11 8 8 22 2 2- + - + -

= 4 9 36 7+ + = .

20. Ans (8): x2 + 2x – n = 0 fi (x + 1)2 = n + 1 fi x = –1 ± n + 1 Thus, n + 1 should be a perfect square. Since m Œ [5, 100] fi n + 1 fi [6, 101] Perfect square values of n + 1 are 9, 16, 25, 36, 49,

64, 81, 100. Hence, the number of values is 8.

PaPer 2Section a

1. Ans (a, b): By partial fractions, we have

g x f ax a a b a c

( ) ( )( ) ( ) ( )

=- - -

+- - -

+- - -

f bb a x b b c

f cc a c b x c

( )( ) ( ) ( )

( )( ) ( ) ( )

fi g xa b b c c a

( )( ) ( ) ( )

=- - -

1

¥-

-+

--

+-

-È

ÎÍ

˘

˚˙

f a c bx a

f b a cx b

f c b ax c

( ) ( )( )

( ) ( )( )

( ) ( )( )

fi g xa f a x ab f b x bc f c x c

a a

b b

c c

( )( )/( )( )/( )( )/( )

=---

∏111

1

1

1

2

2

2

fi g x dxa f a x ab f b x bc f c x c

( )( ) log( ) log( ) log

Ú =---

111

∏ +

1

1

1

2

2

2

a a

b b

c c

k

and df xdx

a f a x a

b f b x a

c f c x a

a a

b b

c

( )( ) ( )

( ) ( )

( ) ( )

=

-

-

-

∏

-

-

-

1

1

1

1

1

1

2

2

2

2

2

cc2

=

1

1

1

1

1

1

2

2

2

2

2

2

a f a x a

b f b x a

c f c x a

a a

b b

c c

( ) ( )

( ) ( )

( ) ( )

-

-

-

∏

-

-

-

2. Ans (a, b, c, d): A Õ A » B,

fi P(A) £ P(A » B) fi P(A » B) ≥ 34

Also, P(A « B) = P(A) + P(B) – P(A » B) ≥ P(A) + P(B) – 1

= 34

58

1 38

+ - =

Now, A « B Õ B

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Appendix B: Mock Tests B.29

fi P(A « B) £ P(B) = 58

\ 38

58

£ « £P A B( )

Next, P(A « B¢) = P(A) – P(A « B)

fi 34

58

34

38

- £ « ¢ £ -P A B( )

fi P(A « B) = P(B) – P(A¢ « B) (Using (2))

\ 38

58

£ - ¢ « £P B P A B( ) ( )

fi 0 14

£ ¢ « £P A B( )

Hence, all the options are correct. 3. Ans (a, c): | PS1 – PS2 | = 2a

fi ( ) ( )24 0 7 0 12 5 122 2 2 2- + - - + = \ a = 6

Also 2ae = ( ) ( )24 5 12 7 3862 2- + - =

\ e = 38612

LR = 2 2 1 2 6 386144

1 1216

2 2 2ba

a ea

=-

= ¥ -ÊËÁ

ˆ¯̃

=( )

4. Ans (a, b): dydx

yx

dyy

dxx

= fi =2 2

fi ln lnyx

c yc

e y cex x= - + fi = fi =- -1 1 1

Comparing with y = aeР1/x + b, a ΠR, b = 0. 5. Ans (a, b, c, d): We have A2B = A(AB) = AA = A2, B2A = B(BA) = BB = B2, ABA = A(BA) = AB = A, and BAB = B(AB) = BA = B 6. Ans (b, d): Since x-axis touches the given parabola, the equation

14

02 2x b c x a+ - + =( ) must have equal roots.

fi (b – c)2 – a2 = 0 fi (c – a – b)(c + a – b) = 0 Hence, the given line passes through the points

(–1, –1) and (1, –1). 7. Ans (a, d):

¢ =+

> " >f xx

xx

( ) 31

0 02

=+

>+

" >3

11

112 2

x

x xx,

fi ¢ >+Ú Úf x dx

xdx

x x

( )1

21

11

fi f (x) > tan–1 x – tan–1 1 fi f (x) + p/4 > tan–1 x 8. Ans (a, b, c, d):

Exponent of 2 = 102

102

1022 3

ÈÎÍ

˘˚̇

+ ÈÎÍ

˘˚̇

+ ÈÎÍ

˘˚̇

= 5 + 2 + 1 = 8

Exponent of 3 = 103

1032

ÈÎÍ

˘˚̇

+ ÈÎÍ

˘˚̇

= 3 + 1 = 4

Exponent of 5 = 105

ÈÎÍ

˘˚̇

= 2

Exponent of 7 = 107

ÈÎÍ

˘˚̇

= 1

Number of divisors of 10 is (8 + 1) (4 + 1) (2 + 1) (1 + 1) = 270.

Number of ways of putting N as a product of two natural numbers is 270/2 = 135.

Section B

Ans 9. (b), 10. (c): a + b + g = p

sin sin sina b a g a-ÊËÁ

ˆ¯̃

+-Ê

ËÁˆ¯̃

+ ÊËÁ

ˆ¯̃

=2 2

32

32

fi sin( )

sin( )

sin ( )

p b g b p g b g

p b g

- +( ) -ÊËÁ

ˆ¯̃

+- +( ) -Ê

ËÁˆ¯̃

+- +Ê

2 2

3 32ËËÁ

ˆ¯̃

=32

fi cos cos cos ( )22

22

32

32

b g g b b g+ÊËÁ

ˆ¯̃

++Ê

ËÁˆ¯̃

++Ê

ËÁˆ¯̃

=

fi 2 34 4

1 2 34

32

2

cos ( ) cos

cos ( )

b gb g

b g

+ÊËÁ

ˆ¯̃

-ÊËÁ

ˆ¯̃

+ - +ÊËÁ

ˆ¯̃

=

fi 2 34 4

2 34

12

2

cos ( ) cos

cos ( )

b gb g

b g

+ÊËÁ

ˆ¯̃

-ÊËÁ

ˆ¯̃

- +ÊËÁ

ˆ¯̃

=

fi 4 34

4 34 4

1 0

2cos ( )

cos ( ) cos

b g

b gb g

+ÊËÁ

ˆ¯̃

- +ÊËÁ

ˆ¯̃

-ÊËÁ

ˆ¯̃

+ = (1)

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B.30 Mathematics

Since cos ( )34

b g+ÊËÁ

ˆ¯̃

is a real number, D ≥ 0. Hence,

164

16 02cos b g-ÊËÁ

ˆ¯̃

- ≥

cos2

41b g-Ê

ËÁˆ¯̃

≥

cos2

41b g-Ê

ËÁˆ¯̃

=

or b = g (2) Hence, Expression (2) for b = g is

2 34

12

cos ( )b g+ÊËÁ

ˆ¯̃

-ÈÎÍ

˘˚˙ = 0

Hence 2 34

cos ( )b g+ = 1

cos ( )34

12

b g+ =

34

60( )b g+ = ∞

or b + g = 80° or a = 100° or b = 40°, g = 40°, and a = 100°.Ans 11. (a), 12. (c):

b xx

b xx x x

coscos

sin(cos sin ) tan2 2 1 32 2-

=+

-

(1) 2 cos 2x – 1 π 0 fi x π np ±p6

(2) tan x π 0 fi x π ± p2

(3) cos2 x – 3 sin2 x π 0 fi x π n p ± p6

Also 2 cos 2x – 1 = 2 (cos2 x – sin2x) – (cos2 x + sin2 x) = cos2x – 3 sin2 x Now, the given equation reduces to b sin x = b + sin x

fi sin x = bb - 1

\ – 1 £ sin x £ 1

or – 1 £ bb - 1

£ 1

or bb - 1

+ 1 ≥ 0 and bb - 1

– 1 £ 0

or 2 11

bb

--

≥ 0 and 11b-

£ 0

or b Œ - •ÊËÁ

˘˚̇

, 12

» [1, •) and b Œ (– •, 1)

when b = 12

fi sin x = 1 which is not possible.

fi b Œ - •ÊËÁ

ˆ¯̃

- -ÏÌÓ

¸˝˛

, , ,12

1 0 13

For any other value b, sin x takes two values for x Œ(0, p).

Hence, four solutions for x Œ(0, 2p). 13. Ans (b):

A

E

D( , , )x y z

C (1, 1, 1)

B (3, 1, 2)

(2, 2, – 1)

P(4, –26/3, –10/3)

Fig. B.14

A (2, 2, – 1), B(3, 1, 2), C(1, 1, 1)

AB i j k AC i j k BC i k = - + = - - + = - -3 2 2, ,

\ AC ^ BC Mid point of AB = Mid point of CD

x + =12

52

fi x = 4 fi y + =12

32

fi y = 2

z + =12

12

fi z = 0

Therefore, coordinates of D are (4, 2, 0). 14. Ans (a): Equation of the base a(x – 1) + b(y – 1) + C(z – 1) = 0 which also passes through A(2,2,–1) and B(3,1,2). \ a + b – 2c = 0 (i) and 2a + c = 0 (ii) Therefore, c = –2a, b = –5a a (x – 1) – 5a (y – 1) – 2a (z – 1) = 0 \ x – 5y – 2z + 6 = 0 Foot of the normal from P

x y z- =+

-=

+

-4

1

263

5

103

2

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Appendix B: Mock Tests B.31

= -- -Ê

ËÁˆ¯̃

- -ÊËÁ

ˆ¯̃

+

+ +

Ê

Ë

ÁÁÁÁ

ˆ

¯

˜˜˜̃

4 5 263

2 103

6

1 25 4

x y z- =+

-=

+

-= -4

1

263

5

103

22

fi x = 2, y = 43

, z = 23

\ 2 43

23

, ,ÊËÁ

ˆ¯̃

Volume of the pyramid

= 13

(Base area) ¥ Height

= 13

AB AC

¥ ¥ EP

= 13

5 2 120i j k - - ¥

= 13

30 120 = 20 cubic units

15. Ans (b): Coordinate of any point on line AB can be taken as

h = r1 + r cos q

CB

O

C1

C2

A r( , 0)1

X

Y

Fig. B.15

k = 0 + r sin q Therefore, it lies on C2. \ (r1 + r cos q)2 + r2 sin2q = r2

2

r2 + 2 r r1 cos q + r12 – r2

2 = 0 Let AB = rAB and AC = rAC. Therefore, rAB + rAC = –2r1cos q, rAB.rAC = r1

2 – r22

So (BC) = | AC – AB | = | rAC – rAB |

= ( )r r r rAC AB AC AB+ -2 4

= 4 4 412 2

12

22r r rcos q - +

BC = - +4 412 2

22r rsin q

Therefore, for max sin q = 0 BC2

max = 4r22

16. Ans (a): Now OA2 + OB2 + BC2 = r1

2 + r22 + 4r2

2 – 4r12sin2q

= 5r22 + r1

2 – 4r12sin2q

\ 0 £ sin2q £ 1 So OA2 + OB2 + OC2 Œ [5r2

2 – 3r12, 5r2

2 + r12]

Section C

17. Ans: (a Æ q); (b Æ p); (c Æ r); (d Æ s) (a) we get common normal perpendicular to y = x.

So, slope = – 1 fi x + y = 3a (b) Tangent to the parabola y = mx + a/m passes

through the point P(h, k). fi m2h – mk + a = 0. If its roots are m1 and m2,

then m1m2 = +1. Locus is x = a. (c) The tangents are y = m(x + a) + a/m (1)

and y = -1m

(x + 2a) – 2am (2)

Subtracting from (1) to (2), we get x + 3 a = 0. (d) If the chord joining t1 and t2 subtends 90º at

vertex, then t1t2 = –4. point of intersection of tangents is (–at1t2, –a(t1 + t2)). So the locus is x = 4a.

18. Ans: (a Æ r); (b Æ p); (c Æ q); (d Æ s) (a) | | a b¥ = | |

b c¥ = | | c a¥ = 2DABC Also directions of

a b¥ ,

b c¥ and c a¥ are the same.

Hence,

a b¥ = b c¥ = c a¥

(b) For regular tetrahedron all sides are of equal length, hence, | | | | | | a b c= = . Also, all the faces are equilateral triangle.

Therefore, angle between a and b is 60º,

b

and c is 60º, and between a and c is 60º.

Hence, a b b c c a. . .= =

(c) Since

a b¥ = c fi a c^ and b c^ and

b c a¥ = fi

b a^ and c a^ .

Therefore, a ,b , c are mutually perpendicular.

(d) Since a b c+ + = 0

fi a b c a b b c c a2 2 2 2 0+ + + + + =( . . . )

fi a b b c c a. . .+ + = - 3

2

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B.32 Mathematics

19. Ans: (a Æ r); (b Æ s); (c Æ q); (d Æ p)

(a) f (x) = mx c x

e xx

+ <≥

ÏÌÔ

ÓÔ

00

For application of LMV function must be con-tinous and derivable in [–2, 2].

f (0+) = f (0–) fi c = 1 f ¢(0+) = f ¢(0–) fi m = 1 fi m + 3c = 4 (b)

(6, 2)

(4, 4)

(2, 6)

Fig. B.16

Length of latus rectum is 4 2 = 4a fi a = 2 Therefore, equation of directrix is x + y + l = 0 whose distance from (4, 4) is 2 2 .

\ 4 42

+ + l = 2 2 fi l = ± 4 – 8

fi l = –4 or –12 Therefore, directrix are x + y = 4 and x + y = 12. \ l1 + l2 = 4 + 12 = 16 (c) f ¢(x) = 6x2 – 6x – 12 = 0 fi (x – 2)(x + 1) = 0 fi x = –1, 2 f ¢¢(x) = 12x – 6 f ¢¢(–1) = –18 < 0 fi x = –1 is the point of maxima f ¢¢(2) = 18 > 0 fi x = 2 is the point of minima.

Now f (–1) = 8, f (2) = –19, f(5/2) = - 332

,

f (–2) = –5 \ maximum value of f(x) in [–2, 5/2] is 8.

(d) lim( ) ( )

....

n

n n

n n

Æ•

-+

-+

+-

Ê

Ë

ÁÁÁÁ

ˆ

¯

˜˜˜˜

12 1

14 2

12

2 2

2 2

= limn

r

n

rn rƕ= -

 12 2

1

= limn

r

n

n rn

rn

Æ•= -

 1

22

21

= dxx x2 2

0

1

-Ú

= dxx1 1 2

0

1

- -Ú ( )

= sin ( )- -ÈÎ ˘̊10

11x

= 02 2 4

2- -ÊËÁ

ˆ¯̃

= = fi =p p pk k

20. Ans: (a Æ r); (b Æ s); (c Æ p); (d Æ s) (a) 1 + a10 + a20 + L + a190 = 0 as 10 is not an integral multiple of 20 (b) (log2x)2 + log2(0.03125) + log2x + 3 < 0 fi (log2x)2 + log2(1/32) + log2x + 3 < 0 fi (log2x)2 + (log2x) – 2 < 0 fi (log2x + 2) (log2x – 1) < 0 fi –2 < log2x < 1

fi 14

< x < 2

Number of integral solution = 1 (c) f (x) = x2e–2x

f ¢(x) = x2(–2) e–2x + e–2x.2x = 2x e–2x(1 – x) f ¢(x) > 0 "x Œ (0, 1) f ¢(x) < 0 "x Œ (1, •) \ f (x)max = f (1) = e–2 = l

12

12

42

ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

=-ln ( ) ln ( )l e

(d)

y x2= – 12 x = 3

y

(3, 0)x

Fig. B.17

Perpendicular tangents can be drawn to the parabola from points which lies on directrix. There is only one point (3, 0) which lies on its directrix as well as on the hyperbola.

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