Pioneer’s...Pioneer’s Sample Paper Solution 11th (Non – medical) Examination Centre: Pioneer Education, Sector – 40-D General Instructions:- The question paper contains 60

Pioneer’s Sample Paper

Solution

11th (Non – medical) Examination Centre: Pioneer Education, Sector – 40-D

General Instructions:-

The question paper contains 60 objective multiple choice questions.

There are three Sections in the question paper consisting of Section – A

MATHEMATICS (1 to 60)

Each right answer carries (4 marks) and wrong (–1marks)

The maximum marks are 180.

Maximum Time is 1Hr. 30 Min.

Give your response in the Answer Sheet provided with the Question Paper.

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Section – A {Mathematics}

1. The set A {x :|2x 3| 7} is equal to the set

(a) D {x :0 x 5 7} (b) B {x : 3 x 7} (c) E {x : 7 x 7} (d) C {x : 13 2x 4}

Ans. (a)

Solution:

Given, set A = {x :|2x 3| 7}

Now, 2x 3 7 7 2x 3 7

7 3 2x 7 3 10 2x 4

5 x 2 0 x 5 7

2. If 2x

f(x) ,x 1,x 1

then the value of for which f(a) a, where a 0, is

(a) 1

1a

(b) 1

a (c)

11

a (d)

11

a

Ans. (c)

Solution:

2αa

f a a aa 1

2 2 1α. a a a α 1

α

3. If 1 2z 2 2(1 i) and z 1 i 3, then 2 31 2z z is equal to

(a) 128i (b) 64i (c) –64i (d) –128i

Ans. (d)

Solution:

Given, 1 2z 2 2 1 i and z 1 i 3

2 2 2 3 2 21 3z z [2 2 1 i ] [1 i 3] 8 1 i 2i

33[1 i 3 3.1.i 3 1 i 3 ]

= 8 1 1 2i [1 3 3 i 3 3i 1 i 3 ]

= 8 2i [1 9] 16i 8 128 i




4. If and are two different complex numbers with | | 1, then 1

is equal to

(a) 1/2 (b) 0 (c) –1 (d) 1

Ans. (d)

Solution:

Given, β 1

β α β α

1 aβ ββ aβ

[ 1 ββ]

=

1 β α 11 1

β 1β α

[ z z ]

5. The value of 2

1 i 3

11

i 1

is

(a) 20 (b) 9 (c) 5/4 (d) 4/5

Ans. (d)

Solution:

Let

2

1 i 3 2i 1 i 3 2i 3 4i1 i 3z

3 4i 25i 2

i 1

Now, |1 i 3||2i||3 4i| 2 2 5 4

z25 25 5

6. Argument of the complex number 1 3i

2 i

is (In degrees is)

(a) 450 (b) 1350 (c) 2250 (d) 2400

Ans. (c)

Solution :

1 3i 1 3i 2 i

2 i 2 i 2 i

= 22 i 6i 3i

1 i4 1




Argument of 1 01 3i 1arg of 1 i tan 225

2 i 1

[since, the given complex number lies in IIIrd quadrant]

7. The value of 4 4(1 3 i) (1 3 i) is

(a) –16 (b) 16 (c) 14 (d) –14

Ans. (a)

Solution:

4 4 4 421 3 i 1 3 i 2ω 2ω

1 3i 1 3iω and ω

2 2

= 2

8 4 3 2 316ω 16ω 16 ω ω 16 ω ω

= 216 ω ω 16 1 16 [ 21 ω ω 0]

8. If and are the roots of the quadratic equation 2x x 1 0, then the equation, whose roots are

19 and 7 , is

(a) 2x x 1 0 (b) 2x x 1 0 (c) 2x x 1 0 (d) 2x x 1 0

Ans. (d)

Solution:

Given equation is x2 + x + 1 = 0, 2α ωand β ω will satisfy the given equation .

Now, 19 19 7 14 2α ω ω, β ω ω

Required equation is

2 2 3x ω ω x ω 0

2x x 1 0

9. Let the two numbers have arithmetic mean 9 and geometric mean 4. Then, these numbers are the

roots of the quadratic equation

(a) 2x 18x 16 0 (b) 2x 18x 16 0 (c) 2x 18x 16 0 (d) 2x 18x 16 0

Ans. (b)

Solution:

Given, arithmetic mean = 9 and geometric = 4.

Let the two numbers be x1 and x2.




Given, 1 2x x

92

and 1 2x .x 16

1 2x x 18 and 1 2x . x 16

Hence, required equation is

x2 – (sum of roots) x + (product of roots) = 0

x2 – 18x + 16 = 0

10. If and are the roots of the quadratic equation 2x 4x 3 0, then the equation, whose roots are

2 and 2 , is

(a) 2x 12x 35 0 (b) 2x 12x 33 0 (c) 2x 12x 33 0 (d) 2x 12x 35 0

Ans. (d)

Solution:

Given, α and βand the roots of equation x2 + 4x + 3 = 0.

α β 4 and αβ 3

Now, 2α β α 2β 3 α β 12

and 2 22α β a 2β 2α 4αβ αβ 2β

= 2 2 2

α β αβ 2 4 3 35

Hence, required equation is

x2 – (sum of roots)x + (product of roots) = 0

x2 + 12x + 35 = 0

11. In PQR, R .2

If tan

P

2

and tanQ

2

are the roots of ax2 + bx + c = 0, a 0, then

(a) b = a + c (c) b = c (c) c = a + b (d) a = b + c

Ans. (c)

Solution:

Here, P Q b

tan tan2 2 a

and P Q c

tan tan2 2 a

..(i)

Also, P Q R π

2 2 2 2 P Q R π

P Q π

2 4

πR , given

2




P Qtan tan

P Q 2 2tan 1 1P Q2 2 1 tan tan2 2

bb ca 1 1

c a a1a

[from Eq. (i)]

c = a + b

12. A value of b for which the equations x2 + bx –1 = 0, x2 + x + b = 0 have one root in common, is

(a) – 2 (b) –i 3 (c) i 5 (d) 2

Ans. (b)

Solution :

We know that, if a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 have a common real root, then

(a1c2 – a2c1)2 = (b1c2 – b2c1) (a1b2 – a2b1)

Hence, x2 + bx – 1 = 0 and x2 + x + b have a common root.

2 21 b b 1 1 b

b2 + 2b + 1 = b2 – b3 + 1 – b

b3 + 3b = 0 b(b2 + 3) = 0

b 0, i 3

13. Let and be the roots of equation px2 + qx + r = 0, p 0. If p, q and r are in AP and 1 1

=4, then the

value of | | is

(a) 61

9 (b)

2 17

9 (c)

34

9 (d)

2 13

9

Ans. (d)

Solution:

Given, α and β are roots of px2 + qx + r = 0, p 0.

q r

α β , αβp p

..(i)

Since, p, q and r are in AP.

2q = p + r ..(ii)

Also, 1 1 α β

4 4α β αβ




q 4r

α β 4αβp p

[from Eq. (i)]

q = – 4r

On putting the value of q in Eq. (ii), we get

2(–4r) = p +r p = – 9r

Now, q 4r 4r 4r

α βp p 9r 9

and r r 1

αβp 9r 9

2 2 16 4 16 36

α β α β 4αβ81 9 81

2 52 2

α β α β 1381 9

14. In a triangle, the lengths of two larger sides are 10 cm and 9 cm. If the angles of the triangle are in AP,

then the length of the third side is

(a) 5 6 (b) 5 6 (c) 5 6 (d)5 6

Ans. (d)

Let the figure, A > B > C

Since, angles are in AP.

C θ d, B θ and A θ d

As 0A B C 180

0θ d θ θ d 180

0 03θ 180 θ 60

Now, in ABC 2 2 2a c b

cos B2ac




2 2 20 10 x 9

cos 602 10 x

2 2

01 100 x 9 1cos 60

2 20x 2

2x 10x 19 0

10 100 76 10 24

x 5 62 2

15. If S1, S2 and S3 are the sum of n, 2n and 3n terms respectively of an arithmetic progression, then

(a) S3 = 2(S1 + S2) (b) S3 = S1 + S2 (c) S3 = 3(S2 –S1) (d) S3 = 3(S2 + S1)

Ans. (c)

Solution :

Let a and d be the first term and common difference respectively.

1

nS [2a n 1 d]

2

2

nS [2a 2n 1 d]

2

and 3

3nS [2a 3n 1 d]

2

Clearly, 2 1

3n3 S S [4a 4nd 2d 2a nd d]

2

= 3

3n 3n[2a 3nd d] [2a 3n 1 d] S

2 2

16. If a1,a2,….,an are in AP with common difference d 0, then (sin d)[sec a1 sec a2 + sec a2 sec a3 + ……+ sec

an–1 sec an] is equal to

(a) cot an – cot a1 (b) cot a1 – cot an (c) tan an – tan a1 (d) tan an – tan an–1

Ans. (c)

Solution:

Given, 2 1 3 2 4 3 n n 1d a a a a a a .... a a

1 2 2 3 n 1 nsin d [seca seca seca seca ... seca seca ]

= 1 2 2 3 n 1 n

sin d sin d sin d....

cos a cos a cos a cos a cos a cos a

= 2 1 3 2 n n 1

1 2 2 3 n 1 n

sin a a sin a a sin a a....

cos a cos a cos a cos a cos a cos a




= tan a2 – tan a1 + tan a3 – tan a2 + …. + tan an – tan an–1 s

= tan an – tan a1

17. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,…., is

(a) 207179 10

81 (b) 207

99 109

(c) 207179 10

81 (d) 207

99 109

Ans. (c)

Solution:

2 3

7 77 777S .... upto 20 terms

10 10 10

= 7 9 99 999

.... upto 20 terms9 10 100 1000

= 2 3

7 1 1 11 1 1 ... upto 20 terms

9 10 10 10

= 7

9(1[ + 1 + … + upto 20 terms]

– 2 3

1 1 1... upto 20 terms

10 10 10

=

201 1

110 107

2019 1

10

=

207 1 1

209 9 10

= 20

207 179 1 1 7179 10

9 9 9 10 81

18. The value of x which satisfies 21 cosx cos x ........8 64 in , is

(a) ,2 3

(b)

3

(c) ,

2 6

(d) ,

6 3

Ans. (b)

Solution:

Given, 2

1

1 cos x cos x ..... 2 21 cos x8 8 8 8

[ cos x 1, as for cos x = 1 + 1 + 1 + ….. 2, and cos x 1 ]




1

2 2 cos x 11 cos x

1 π

cos x x2 3

19. How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts

with 67 and no digit appears more than once?

(a) 335 (b) 336 (c) 337 (d) 338

Ans. (b)

Solution :

Since, telephone number start with 67, so two digits is already fixed. Now, we have to do arrangement

of three digits from remaining eight digits.

Possible number of ways = 8P3

=

8! 8!8 7 6 366 days

8 3 ! 5!

20. If Pm stands for mmP , then 1 + 1P1 + 2P2 + 3P3 + ……+ n . Pn is equal to

(a) n! (b) (n + 3)! (c) (n + 2)! (d) (n + 1)!

Ans. (d)

Solution:

1 + 1. P1 + 2. P2 + 3. P3 + … + nPn

= 1 + 1. (1!) + 2. (2!) + …. + n. (n!)

= 1 + (2 – 1)1! + (3 – 1)2! + … + [(n + 1) – 1]n!

= 1 + 2! – 1! + 3! – 2! + … + (n + 1)! = n! = (n + 1)!

21. In how many different ways can the letters of the word MATHEMATICS be arranged?

(a) 11! (b) 11!/2! (c) 11!/(2!)2 (d) 11!/(2!)3

Ans. (d)

Solution:

In the word ‘MATHEMATICS’ the letters are 2A, C, E, H, I, 2M, S, 2T.

Total number of different words =

3

11! 11!

2!2!2! 2!

22. The number of four-letter words that can be formed (the words need not be meaningful) using the

letters of the word ‘MEDITERRANEAN’ such that the first letter is E and the last letter is R, is

(a) 11!

2!2!2! (b) 59 (c) 56 (d)

11!

3!2!2!

Ans. (b)




Solution :

Word ‘MEDITERRANEAN’ has 2A, 2E, 1D, 1I, 1M, 2N, 2R, 1T.

Out of four letters E and R is fixed and rest of the two letters can be chosen in following ways :

Case I Both letters are of same kind i.e. 3C2 ways, therefore number of words = 32

2!C 3

2!

Case II Both letters are of different kinds i.e. 8C2 ways,

therefore number of words = 8C2 2! = 56

Hence, total number of words = 56 + 3 = 59

23. If a polygon of n sides has 275 diagonals, then n is equal to

(a) 25 (b) 35 (c) 20 (d) 15

Ans. (a)

Solution:

A polygon of n sides has number of diagonals

= n n 3

2752

[given]

n2 – 3n – 550 = 0

(n – 25) (n + 22) = 0

n = 25 [ n 22]

24. In the expansion ofn

3

2

1x ,n N,

x

if the sum of the coefficients of x5 and x10 is 0, then n is equal to

(a) 25 (b) 20 (c) 15 (d) none of these

Ans. (c)

Solution:

r

n rn 3

r 1 r 2

1T C x

x

= r rn 3n 3r 2r n 3n 5r

r rC x 1 x C x 1

For coefficient of x5 and x10, substitute 3n – 5r = 5 and 10 respectively, we get coefficient of

3n 5

5 n5

3n 5

5

x C 1

and coefficient of x10 + 3n 10

.n5

3n 10

5

C 1

.

coefficient of x5 + Coefficient of x10 = 0

n 3n 5 3n 10

n3n 5 5 53n 10

5

C1 C 1

5




3n 5 3n 5

n n5 5

3n 5 3n 10

5 5

C 1 C 1

n n3n 5 3n 10

5 5

C C

3n 5 3n 10

n5 5

6n 15

n5

6n – 15 = 5n

n = 15

25. The equation of the base BC of an equilateral ABC is x + y = 2 and A is (2, –1). The length of the side of

the triangle is

(a) 2 (b) 1/2

3

2

(c) /2

1

2

(d) /2

2

3

Ans. (d)

Solution:

Length of perpendicular from A(2, – 1) to the line x + y – 2 = 0 is

2 1 2 1AD

1 1 2

In ABD, 0ADcos 30

AB

1/2

1 3 2 2AB

2 32 AB 3

26. If C is the reflection of A (2, 4) is X-axis and B is the reflection of C in Y-axis, then |AB| is equal to

(a) 20 (b) 2 5 (c) 4 5 (d) 4




Ans. (c)

Solution:

Here, coordinates are A(2, 4), B(–2, – 4) and C(2, – 4).

Now, 2 2

AB 2 2 4 4 16 64 80 4 5

27. The circumcentre of the triangle with vertices (8, 6), (8, –2) and (2, –2) is at the point

(a) (2, –1) (b) (1. –2) (c) (5, 2) (d) (2, 5)

Ans. (c)

Solution :

Let the vertices of a triangle are A(8, 6), B(8, - 2) and C(2 – 2).

Now, AB = 2 2 2(8 8) (16 2) 0 8 8

BC = 2 2

2 8 2 2 36 0 6

and 2 2 2 2CA 8 2 6 2 6 8

= 36 64 100 10

Now, AB2 + BC2 = (8)2 + (6)2 = 64 + 36 = 100 = AC2

So, ABC is a right angled triangle and right angled at B.




We know that, in a right angled, the circumcentre of the mid-point of hypotenuse.

Mid-point of AC = 8 2 6 2

, 5,22 2

Hence, the required circumcentre is (5, 2).

28. If orthocentre and circumcentre of a triangle are respectively (1, 1) and (3, 2), then the coordinates of

its centroid are

(a) 7 5

,3 3

(b) 5 7

,3 3

(c) (7, 5) (d) none of these

Ans. (a)

Solution:

We know that, centroid divides the line segment joining orthocentre and circumcentre in the ratio 2 :

1. Since, the coordinates of orthocentre are (1, 1) and (3, 2), respectively.

The coordinates of centroid are

2.3 1.1 2.2 1.1 7 5, ,

2 1 2 1 3 3

29. A straight line perpendicular to the line 2x + y = 3 is passing through (1, 1). Its y-intercept is

(a) 1 (b) 3 (c) 2 (d) 1/2

Ans. (d)

Solution:

A straight line perpendicular to 2x + y = 3 is

2y – x + c = 0 …(i)

Since, it passes through (1, 1).

2(1) – 1 + c = 0 c = – 1

On putting c = – 1 in Eq. (i), we get

2y x 1 0 2y x 1 y x

11/ 2 1

So, the y-intercept is 1

2.

30. If a ray of light along x + 3y 3 gets reflected upon reaching X-axis, then the equation of the

reflected ray is

(a) y x 3 (b) 3y x 3 (c) y 3x 3 (d) 3y x 1

Ans. (b)

Solution:

Take any point B(0, 1) on given line.




Equation of AB’, y – 0 = 1 0x 3

0 3

3 y x 3

3 y x 3

31. If X = n{4 3n 1 : n N} and Y = {9 n 1 : n N} , where N is the set of natural numbers, then X Y

is equal to

(a) N (b) Y – X (c) X (d) Y

Ans.(d)

Solution:

Set X contains elements of the form

4n – 3n – 1 = (1 + 3)n – 3n – 1

= n n n 1 n 2

n 1 23 C 3 ..... C 3

= n 2 n n 1 nn 1 29 3 C 3 ... C

Set X has natural numbers which are multiples of 9 (not all)

Set Y has all multiples of 9

X Y Y

32. If is cube root of unity then 200

200

1tan

4

equals

(a) 1 (b) 1

2 (c) 0 (d) None of these

Ans. (a)

Solution:

200

200

1tan

4




= 2tan tan4 4

( 21 0 )

= 3

tan 14

.

33. Number of irrational terms in the expansion of 60

5 102 3 are

(a) 54 (b) 61 (c) 30 (d) 31

Ans. (a)

Solution :

Given 601 160

5 10 5 102 3 2 3

Now L. C. M. of 5 and 10 is 10

Number of rational terms let us write

60 r r1 160 5 10

r 1 rT C 2 3

= r r

1260 5 10

rC 2 3

As 0 r 60

r = 0, 10, 20, 30, 40, 50, 60

Number of rational terms are 7

Number of irrational terms equals to

Total Number of terms – Number of rational terms

= 61 – 7 = 54

34. If is a root of 225cos 5 cos 12 0, ,2

then sin2 is equal to

(a) 24

25 (b) –

24

25 (c)

13

18 (d) –

13

18

Ans. (b)

Solution:

Since, is a root of 252cos 5cos 12 0.

225 cos 5 cos 12 0

5 cos 3 5 cos 4 0




4 3cos ,

5 5

But 2

ie, in second quadrant.

4

cos5

3sin

5

Now, sin 2 2 sin cos

= 3 4 24

25 5 25

35. Let , be such that 3 . If sin 21 27sin and cos cos ,

65 65 then the value of cos

is2

(a) 3

130 (b)

3

130 (c)

6

65 (d) –

6

65

Ans. (a)

Solution:

Given that,

21

sin sin65

..(i)

and 27

cos cos65

..(ii)

On squaring and adding Eqs. (i) and (ii), we get

2 2 2 2sin sin 2 sin sin cos cos 2cos cos

= 2 2

21 27

65 65

2 2 cos cos sin sin

= 441 729

4225 4225

1170

2[1 cos ]4225

2 1170cos

2 4 4225




2 9cos

2 130

3

cos2 130

33

2 2 2

36. Let a1, a2, a3, … be terms of an AP. If 2

1 2 p 62

1 2 q 21

a a ... a ap, p q, then equals

a a .... a q a

(a) 7

2 (b) 2

7 (c) 11

41 (d) 41

11

Ans. (c)

Solution:

Given that, 2

1 2 p

21 2 q

a a ... a p

a a ... a q

21

2

1

p[2a p 1 d]

p2 ,q q[2a q 1 d]2

where d be a common difference of an AP.

1

1

2a d pd p

2a d qd q

12a d p q 0

1

da

2

Now, 6 1

21 1

a a 5d

a a 20d

=

d5d

112d 4120d2

37. Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is

(a) a function (b) transitive (c) not symmetric (d) reflexive

Ans. (c)

Solution:

Given, R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}.

(a) Since, (2, 4) R and (2, 3) R R. So, R is not a function.




(b) Since, (2, 3) R but (3, 2) R . So, R is not symmetric.

(d) Since, (1, 1), (2, 2), (3, 3), (4, 4) R . So, R is not reflexive.

Hence, the option (c) is correct.

38. Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon.

If n 1 nT T 10, then the value of n is

(a) 7 (b) 5 (c) 10 (d) 8

Ans. (b)

Solution:

Given, n

n 3T C

n 1n 1 3T C

n 1 n

n 1 n 3 3T T C C 10

(given)

n n n2 3 3C C C 10

n2C 10

n = 5

39. If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0, a, b, c R, have a common root, then a : b : c is

(a) 1 :2 : 3 (b) 3 : 2 : 1 (c) 1 : 3 : 2 (d) 3 : 1 : 2

Ans. (a)

Solution:

Given equations ARE

x2 + 2x + 3 = 0 ..(i)

and ax2 + bx + c = 0 ..(ii)

Since, Eq. (i) has imaginary roots.

So, Eq. (ii) will also have both roots same as Eq. (i).

Thus, a b c

1 2 3

Hence, a : b : c is 1 : 2 : 3

40. If one root of the equation x2 + px + 12 = 0 is 4, while the equation x2 + px + q = 0 has equal roots, then

the value of q is

(a) 49

4 (b) 12 (c) 3 (d) 4

Ans. (a)

Solution:




Since, one of the roots of equation x2 + px + 12 = 0 is 4.

16 + 4p + 12 = 0

4p = – 28

p = – 7

So, the other equation is x2 – 7x + q = 0 whose roots are equal. Let the roots are and .

Sum of roots = 7

1

7

2

and product of roots = . q

2

7 49q q

2 4

41. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4

from the first five questions. The number of choices available to him is

(a) 140 (b) 196 (c) 280 (d) 346

Ans. (b)

Solution:

The number of choices available to him

= 5C4 × 8C6 + 5C5 × 8C5

= 5! 8! 5! 8!

4!1! 6!2! 5!0! 5!3!

= 8 7 8 7 6

5 12 3 2

= 5 × 4 × 7 + 8 × 7

= 140 + 56 = 196

42. If z2 + z + 1 = 0, where z is complex number, then the value of

2 2 2 2

2 3 6

2 3 6

1 1 1 1z z z ... z is

z z z z

(a) 54 (b) 6 (c) 12 (d) 18

Ans. (c)

Solution :

Given equation is

z2 + z + 1 = 0




1 1 4 1 1

z2 1

21 3z z ,

2

Now, 2 2 2 2 2 2

2 3 4 5 6

2 3 4 5 6

1 1 1 1 1 1z z z z z z

z z z z z z

= (–1)2 + (–1)2 + (1 + 1)2 + (–1)2 + (-1)2 + (1 + 1)2

when, we put either 2z or z , we get the same result

= 1 + 1 + 4 +1 + 1 + 4 = 12

43. Three straight lines 2x + 11y – 5 = 0, 24x + 7y – 20 = 0 and 4x – 3y – 2 = 0

(a) form a triangle

(b) are only concurrent

(c) are concurrent with one line bisecting the angle between the other two

(d) None of the above

Ans. (c)

Solution :

The angle bisector for the given two lines 24x + 7y –20 = 0 and 4x – 3y – 2 = 0, are

24x 7y 20 4x 3y 2

25 5

Taking positive sign, we get

2x + 11y – 5 = 0

This equation of line is already given.

Therefore the given three lines are concurrent with one line bisecting the angle between the other two.

44. Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the

area of the triangle is 1, then the set of values which ‘k’ can take is given by

(a) {1, 3} (b) {0, 2} (c) {–1, 3} (d) {–3, –2}

Ans. (c)

Solution :

Since, A(h, k), B(1, 1) and C(2, 1) are the vertices of a right angled triangle ABC.




Now, AB = 2 2

AB 1 h 1 k

2 2

BC 2 1 1 1 1

and 2 2

CA h 2 k 1

From Pythagoras theorem

AC2 = AB2 + BC2

4 + h2 – 4h + k2 + 1 – 2k

= h2 + 1 – 2h + k2 + 1 – 2k + 1

5 – 4h = 3 – 2h

h = 1 ..(i)

Now, given that area of triangle is 1.

Then, area 1

ABC AB BC2

2 21

1 1 h 1 k 12

2 2

2 1 h 1 k ..(ii)

2

2 k 1 [From Eq. (i)]

4 = k2 + 1 – 2k

k2 – 2k – 3 = 0

(k – 3) (k + 1) = 0

k = –1, 3

Thus, the set of values of k is {–1, 3}.




45. The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept –4. Then, a

possible value of k is

(a) – 4 (b) 1 (c) 2 (d) –2

Ans.(a)

Solution :

Since, slope of

4 3 1PQ

1 k 1 k

Slope of AM = (k – 1)

Equation of AM is

7 k 1

y k 1 x2 2

For y-intercept, x = 0 , y = – 4

7 k 1

4 k 12 2

215 k 1

2 2

2k 1 15

k2 = 16

k = 4

–Good Luck

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