Solutions for SM770901 ( ) ( )1 - Time4education.com

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Solutions for SM770901

UNIT – Ι

Exercise – 1 Solutions for questions 1 to 35: 1. Given: 5a = 3b = 225c

Let 5a = 3b = 225c = k (k > 0) 5a = k ⇒ 5 = k1/a 3b = k ⇒ 3 = k1/b And 225c = k ⇒ (152)c = k ⇒ 15 = k1/2c ⇒ 3 Consider 15 = k1/2c 3 × 5 = k1/2c k1/b × k1/a = k1/2c ⇒ k1/b+1/a = k1/2c

c21

b1

a1 =+

or c21

abba =+

2c = ba

ab+

c = )ba(2

ab+

Choice (3)

2. Given: a =

∞+

++

........5

5

55

55

5

i.e., a = a5

5+

5a + a2 – 5 = 0

∴ a = 2

)5)(1(455 2 −−±−

a = 2

535 ±−

Since a cannot be negative,

a = 2

553 − Choice (2)

3. Let 5x = a. ⇒ a2 – 16a – 225 = 0 a2 – 25a + 9a – 225 = 0 (a – 25) (a + 9) = 0

a = 25 ∵ a cannot be negative

⇒ 5x = 52 ⇒ x =2 Choice (1)

4. Let 4x = (0.008)y = 10z = k 4x = k ⇒ 22x = k ⇒ 2 = k1/2x –––––––– (1) (0.08)y = k ⇒ (0.2)3y = k ⇒ 5–3y = k ∴ 5 =k-1/3y ––––––– (2) 10z = k ⇒ (2 × 5)z = k From (1) and (2)

( )y3/1x2/1 kk − = k1/z

y3/1x2/1 kk − = k1/z

⇒ y31

x21

z1 −=

x21

z1

y31 =+

Choice (4)

5. Given: xx4)625(

xy8)125(

[ ] 2y2)25)(3125(

⇒ (54)4x×x. (53)8y × x . (55,52×2)y2

⇒ 2y9xy242x16 5.5.5

⇒ 2)y3x4(2y9xy242x16 55 +++ = Choice (2)

6. A =

347

1

−

By rationalising, we get the denominator

A = 7 + 4 3 ⇒ A2 = 97 + 56 3

B = 35192

1

−

By rationalising, we get the denominator

B = 2 3519 + ⇒ B2 = 151 + 20 75 Similarly, we get

C2 = 49 + 20 6 89 < C2 < 109 –––––– (1) 291 < B2 < 311 ––––– (2) and 153 < A2 < 209 –––– (3) From (1), (2) and (3) B > A > C. Choice (2)

7. 827

57

+

By rationalizing the denominator = ( )22333 −

= 2639 − Choice (3)

8. (a + b) = ( )( )

( )( )711

711

711

711

−

+++

−

= ( ) ( )

711711711

22

−++−

= ( )

47112 +

= 9

Choice (2)

9. a = 608 + ⇒ a = 5 + 3 + 2 15

= ( ) ( ) ( ) ( )3522

32

5 ++ ⇒ 35a +=

Now ( )35

1

a

1

+=

By rationalizing the denominator, we get

( )

235

3535

a

1 −=−−=

∴ 2

3535

a

1a

−−+=+ = ( )

2335 +

Choice (2)

10. d = ( )1.....35

35

+

−

By rationalizing the denominator

⇒ d = 2

1528 − ⇒ d = 154 −

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From (1)

35

35d1

−

+=

By rationalising the denominator

1542

1528d1 +=+=

Now 8d1

d =+

Taking the cube on both sides 33

8d1

d =

+

512d1

d3d

1d

33 =

+++

( ) 48883512d

1d

33 =−=+ Choice (3)

11. Let ∞+++= .........121212a ⇒ a = a12 +

Taking the square on both sides, we get a2 = 12 + a ⇒ a2 − a − 12 = 0 ⇒ (a − 4) (a + 3) = 0 ⇒ a = 4 or a = − 3 But a is positive, ∴ a = 4 Choice (3)

12. 27

x

81

x 3= ⇒

27

81

x

x

31

21

=

⇒ 21

61

3x = ⇒

6

21

3x

= ⇒ x = 27

Choice (4)

13. (i) Consider 31

21

3and2

( ) ( )61

61

321

822 ==

( ) ( )61

61

231

933 ==

∴ 21

31

23 >

(ii) Let us consider 51

21

5and2

( ) 101

32101

5221

2 =

=

( ) 10/12510/1255

1

5 =

=

∴ 51

21

52 >

⇒ 51

21

31

523 >> Choice (4)

14. 241253 + = 625 + = 23 +

Choice (2)

15. 40602410 +++

= 10215262532 +++++

= ( ) ( ) ( ) ( )( ) ( )( ) ( )( )2523523222

52

32

2 +++++

= 532 ++ Choice (1)

16. ( ) ba21139 37

+=+

( ) ba21139 31

+=+ = ( )3136332229 +++

= ( ) ( ) ( ) ( ) 31

223232332332

⋅+⋅++

= ( ) ba232331

3+=+=

+

∴ a > b ∴ a = 3, b = 2 Choice (2)

17. ( ) ( ) 1062562512x12x

=++−−−

It may be observed that 625

1625

+=−

∴ ( ) ( ) 101x

6251x

62522

=−

++−

− can be written as

( )( )

10

625

1625

12x

12x=

+

++−

−

Let ( ) t62512x

=+−

∴ 10t1

t =+

∴ t2 − 10t + 1 = 0

t = 2

410010 −± = 625 ±

∴ ( ) 62562512x

±=+−

( ) ( )112x625625 +=+

−

∴ x2 − 1 = 1 ∴ x = 0 and

( ) ( ) ( ) 112x625625625

−−+=+=+

∴ x2 − 1 = − 1

∴ x = ± 2

∴ x = 0, ± 2 Choice (4)

18. 2yx

yx

yx

yx22

22

22

22=

−++

+−

⇒ ( )

2yx

yx244

44=

−+

⇒ 2

1

yx

yx44

44=

−+

Now 22

1

yx

yx

yx

yx44

44

44

44+=

−++

+−

⇒ ( )

2

3

yx

yx288

88=

−+

⇒ 3

22

yx

yx88

88=

+−

Choice (1)

19. 487487 −−+ = 122712247 −−+

= ( ) ( ) 3232323222

+−+=−−+

= 32 Choice (1)


20. ( )625212225 −=−

= ( ) ( ) ( )222232232322 −⋅=

⋅⋅−+

= ( )2324 −⋅ Choice (4) 21. (i) log81729 = log10x

⇒ x10log81log

729log =

⇒ 2

3

9log

9log= log10x ⇒ log10x =

23

100010x 23

== . Choice (4) (ii) log10x + log102 = 1 ⇒ log102x = 1 ⇒ 2x = 101 ∴ x = 5. Choice (1) (iii) log |x + 1| + log |x − 1| = log 3 ⇒ log (|x + 1| |x − 1|) = log 3 ⇒ log |x2 − 1| = log 3 ⇒ x2 − 1 = 3 ∴ x = ± 2. Choice (4) (iv) log x + log 3 = log 15 ⇒ log 3x = log 15 ⇒ 3x = 15 ∴ x = 5. Choice (2) (v) 1)1x(log

3−=−

x − 1 = 21

3−

∴ 3

11x += Choice (1)

(vi) log (2x + 3) − log 3 = log (x − 3) ⇒ log (2x + 3) = log (3 (x − 3) ⇒ 2x + 3 = 3x − 9 ∴ x = 12. Choice (3)

(vii) log40.0625 = log x ⇒ xlog16

1log4 =

⇒ log4 4−2 = log x ⇒ − 2 = log10x ∴ x = 10−2 = 0⋅01. Choice (3) 22. log (x + 29) − 1 = log (2x + 1) ⇒ log10(x+29) = log10 ((2x + 1) × 10) ⇒ x + 29 = 20x + 10 ⇒ −19x = −19 ∴ x = 1. Choice (1)

23. ( )( ) ( )27loglog2logloglog 331

2723

31 =

= 33

31 3loglog

= ( ) ( ) 13log3log3log 3313−=−=− . Choice (1)

24. log(x – 3) + log(x + 3) = 4

⇒ log2((x − 3) (x + 3)) = 4 ⇒ ( ) 16log2log9xlog 2

42

22 ==−

⇒ x2 − 9 = 16 ⇒ x2 = 25 ⇒ x = ± 25 But x = −5 does not define the argument of the log. ∴ x = 5 Choice (1)

25. 53log2

53log2333 =

+ Choice (2)

26. abclogclog

abclogblog

abclogalog 333

++

= .3abclog

)cba(log3

abclog

clog3blog3alog3 =⋅⋅=++

Choice (3)

27. 3

1

xlogxlog

xlogxlog3

xlogxlog

xlogxlog

ba

b31

a

ba

31

b3

a =+

=+⋅

= ablog

1

alogblog

1

blog

1

alog

1

blog1

.alog

1

xxx

xx

xx =+

=+

= logabx Choice (1)

28. 25log5logx32x

25

2 =+

−

⇒ ( ) 2x3x5log 252 =−

⇒ 5 (x2 − 3x) = ( )252 ⇒ x2 − 3x = 4

⇒ x2 − 3x − 4 = 0 ⇒ (x − 4) (x + 1) = 0 ∴ x = − 1 or x = 4 but, x2 − 3x > 0 ∴ x < 0, x > 3 So x = − 1 or 4 But x = 4 satisfies the original equation. Choice (2) 29. ( )( ) 0xlogloglog 432 = ⇒ ( ) 1

2432logxlogloglog =

⇒ ( ) 3log1xloglog 343 ==

⇒ log4x = 3 ⇒ x = 43 = 64 Choice (3)

30. Let, ( )sayklog7

clog3

blog2

alog ===

⇒ log a = 2 logk = log k2 ∴ a = k2 Similarly, b = k3 and c = k7

Now, 24 akbc ==

∴ a2 = b

c Choice (4)

31. 2 + log3x = 2 log3y

⇒ log3x − 2 log3y = − 2 ⇒

2y

x

3log = − 2 ⇒ 22

3y

x −=

∴ 9y

x2

= . Choice (3)

32. log4 (log3 |x|) < 1 ⇒ log4 (log3 |x|) < log44 ⇒ log3 |x| < 4 ⇒ log3 |x| < log33

4 ⇒ |x| < 34

⇒ |x| < 81. Choice (3)

33. log56 = log5 (2 × 3) = 5log

3log2log5log

)32log( +=×

= 1132513010301

477103010302log1

3log2log ⋅=⋅−

⋅+⋅=−

+

Choice (4)


34. Let cyxzlog

xzylog

zyxlog =

−=

−=

−(say)

⇒ log x = c (y − z) ⇒ x log x = cx (y − z) Similarly log y = c (z − x) ⇒ y log y = cy (z − x) and log z = c (x − y) ⇒ z log z = cz (x − y) ∴ x log x + y log y + z log z = c (xy − zx + yz − xy + zx − yz) = 0 ⇒ log (xx⋅ yy⋅ zz) = 0 ∴ xx⋅yy⋅ zz = 1 Choice (4)

35. nlog

1.........

nlog

1

nlog

1

200632

++

= logn2 + logn3 + logn4 + …..logn2006 = logn(2 × 3 × 4 × 5 ………..× 2006)

= logn (2006)! ⇒ logn(2006!) = 1. (∵ n = 2006!) Choice (4)

Exercise – 2 Solutions for questions 1 to 40: 1. The quadratic equation whose roots are α and β is

x2 − x (α + β) + αβ = 0

(i) Given roots are a + b, a – b. α = a + b β = a − b

∴ The required equation is x2 – 2ax + a2 − b2 = 0. Choice (2)

(ii) The given roots are

α = 23

5,23

5 −=β+

∴ The required equation is

x2 –

−++

23

523

5 x

+

⇒ 04

97x10x2 =+−

⇒ 4x2 − 40x + 97 = 0. Choice (3)

(iii) The given roots are a + b – c, a – b + c x2 − x (a + b − c + a − b + c) + (a + b − c) (a − b + c) = 0

⇒ x2 − 2ax + a2 − (b − c)2 = 0. Choice (3) 2. (i) x2 − 2x + 4 = 0 b2 − 4ac = 4 − 4 (1) (4) = − 12 < 0 ∴ b2 – 4ac < 0 The roots are imaginary. Choice (4) (ii) 4x2 − 3x − 2 = 0 b2 − 4ac = 9 − 4 (4) (−2) 9 + 32 = 41 > 0 The roots are real and unequal. Choice (3) (iii) 4x2 − 20x + 25 = 0 b2 − 4ac = (20)2 − 4 (4) (25) 400 − 400 = 0 The roots are real and equal. Choice (1)

3. (i) x2 − 5x + 3 = 0 Let α and β be the roots of the given equation.

∴α + β 515

ab =

−=−=

α β = 313

ac == Choice (1)

(ii) 4x2 + 8x + 12 = 0

248 −=−=β+α

34

12 ==αβ . Choice (2)

(iii) ax2 + (a + b) x + ab = 0

α + β = ( )

aba +−

ba

ab ==αβ Choice (3)

4. x2 − px + 3 = 0 let α, β be the roots of the equation ∴β = 3α.

∴α + 3α =

−−1p

i.e., 4α = p ………. (1) Also, α .3α = 3 ⇒ 3α2 = 3 ⇒ α2 – 1 α = ± 1

∴ Substituting the value of α in (1). We get p = ± 4. Choice (2) 5. x2 + 2x + a = 0 Let α,2α be the roots of the equation. 2α + α = – 2 ⇒ 3 α = – 2

⇒ α = 32−

-------- (1)

Also, 2α.α = a ⇒ 2α2 = a……… (2) From (1) and (2), we get

α = 98

Choice (4)

6. x2 – 5x + 6 = 0

α + β = ;515 =

−− αβ = 6.

(i) β

+α

22=

αβα+β

2

= .35

65

2 =

Choice (4)

(ii)

α+β+

β+α 11

= ( )

β+

α+β+α 11

( )αβ

β+α+β+α = .6

3565

15 =+

Choice (3)

(iii) 22

11

β+

α =

( )( )

( )22

2

22 2

αβαβ−β+α=

αββ+α

= ( )

3613

361225

6

6252

2

=−=− Choice (1)

(iv) 33

11

β+

α =

33

33

βαα+β

= ( ) ( )

( )33 3

αββ+ααβ−β+α

= ( )( )3

3

6

5635 − =

216

35

216

90125 =−.

Choice (2)


(v) 22 α

β+βα

= ( )2

33

αββ+α

= ( ) ( )

( )23 3


= 3635

3690125 =−

. Choice (2)

7. (i) xx6 =− Taking the square on both sides 6 − x = x2 ⇒ x2 + x − 6 = 0 ⇒ x2 + 3x − 2x − 6 = 0 ⇒ (x + 3) (x − 2) = 0 ⇒ x = 2, − 3. But x = – 3 does not satisfy the equation

∴ x = 2 Choice (1) (ii) 2x4

− 9x2 − 5 = 0 ⇒ 2x4 − 10x2 + x2 − 5 = 0

⇒ 2x2 (x2 − 5) +1 (x2 − 5) = 0 ⇒ (x2 − 5) (2x2 + 1) = 0

⇒ x2 = 5 or 21

x2 −=

⇒ 5x ±= or x = ±2

1 Choice (3)

(iii) 2x23

1x23 =

−+− ⇒ x2321x23 −=+−

⇒ x232x24 −=− ⇒ x23x2 −=−

⇒ (2 − x)2 = 3 − 2x ⇒ x2 + 4 − 4x − 3 + 2x = 0 ⇒ x2 − 2x + 1 = 0 ⇒ (x − 1)2 = 0 ⇒ x = 1. Choice (1) 8. 2x2 − px + 2 = 0 Since the roots are equal, its discriminant will be zero.

p2 − 4 (2) (2) = 0 p2 = 16 p = ± 4. Choice (4) 9. x2 + ax + 5 = 0 let α and β be the roots of the quadratic equation. α + β = – a ; αβ = 5 Given α + β = αβ a = – 5. Choice (3) 10. Given: 32x + 1 − 10.3x + 3 = 0 32x.3 – 10.3x +3 = 0 let 3x = a 3a2 − 10a + 3 = 0 ⇒ 3a2 − 9a − a + 3 = 0 ⇒ 3a (a − 3) − 1(a − 3) = 0

⇒ (a − 3) (3a − 1) = 0 ⇒ a = 3 or 31

a =

⇒ 3x = 31 or 31

3x = = 3–1

⇒ x = 1 or x = −1. Choice (4) 11. 52x + 1 − 5x + 3 − 5x + 25 = 0 52x.5 – 5x.53–5x + 25 = 0 Put 5x = a 5a2 − 125a − a + 25 = 0 ⇒ 5a (a − 25) −1 (a − 25) = 0 ⇒ (a − 25) (5a − 1) = 0

⇒ a = 25 or 51

a = ⇒ 5x = 25 or 51

5x =

5x = 52 or -5x = 5–1 x = 2 or x = −1. ∴ x ∈ {2, –1} Choice (1) 12. (i) x2 − 11x + 30 < 0 ⇒ x2 − 6x − 5x + 30 < 0 ⇒ (x − 6) (x − 5) < 0 ⇒ 5 < x < 6.

Choice (1)

(ii) x2 + 10x + 25 ≥ 0 x2 + 5x + 5x + 25 ≥ 0 x (x + 5) + 5 (x + 5) ≥ 0 (x + 5)2 ≥ 0

For any value of x, (x + 5)2 is always greater than or equal to ‘0’

x ∈ R. Choice (3) (iii) x2 − 14x + 45 > 0 ⇒ x2 − 9x − 5x + 45 > 0 ⇒ (x − 9) (x − 5) > 0 ⇒ x< 5 or x>9.

Choice (1) (iv) x2 + 15x + 44 < 0 x2 + 11x + 4x + 44 < 0 (x + 11) (x + 4) < 0 − 11 < x < − 4. Choice (4) 13. x2 − 8x + 15 = 0 Since α and β are the roots of the given equation, α + β = 8 and αβ = 15

αβ+

βα 22

= αβ

β+α 33

=( ) ( )

αββ+ααβ−β+α 33

= ( )( )

15

815383 − =

15

152

15

360512 =−

Choice (1) 14. 3x2 + 5x − 2 = 0 Since α and β are the roots of the equation,

α + β = – 35

and αβ = 32−

33

11

β+

α =

( )333

αββ+α

=( ) ( )

( )33 3


3

3

32

35

32

335

−

−

−−

−

8

215

27827

90125

278

930

27125

=−

+−=

−

−+−

Choice (3) 15. Given: α and β are the roots of the equation x2 + ax + b = 0 ∴α + β = −a and αβ = b The quadratic equation whose roots are α + β and αβ is x2 − x (α + β + αβ) + (α + β) αβ = 0 x2 − x (− a + b) − ab = 0 x2 + x (a − b) − ab = 0 Choice (4)

16. Let yax

aaxx 22=

−+−

x2 − ax + a2 = yx − ay ⇒ x2 − x (a + y) + a2 + ay = 0 Since x is real, discrrminant is greater than or equal to zero. ⇒ (a + y)2 − 4 (a2 + ay) ≥ 0 ⇒ a2 + y2 + 2ay − 4a2 − 4ay ≥ 0 ⇒ y2 − 2ay − 3a2 ≥ 0 ⇒ y2 − 3ay + ay − 3a2 ≥ 0 ⇒ y (y − 3a) + a (y − 3a) > 0 ⇒ (y + a) (y − 3a) ≥ 0 ⇒ y ≤ −a or y ≥ 3a. Choice (3)


17. If 235 + is one of the roots then the other root will be

its conjugate, i.e. 5 – 3 2

∴ The equation whose roots are 5 + 3 2 and

5 – 3 2 is

( ) ( )( ) 0235235235235xx2 =−++−++−

⇒ x2 − 10x + 7 = 0. Choice (2) 18. The given equation is x2 + px + q = 0. Let the roots be 2α, 2α + 2. The sum of the roots is 2α + 2α + 2 = − p 4α = − p − 2

( )

42p +−=α

The product of the roots is 2α (2α + 2) = q

( ) ( )

q22

2p2

2p =

++−+−

⇒ ( )

q2

2p2

2p =

−+ ⇒ p2 − 4 = 4q

⇒ p2 − 4q = 4 i.e. discriminate = 4. Choice (1)

19. Let α and β be the roots of the equation 2k2 + 2ax + c = 0 Given: α: β = 1: 2 ⇒ 2α = β

α + β = k2a2−⇒ 3α =

ka−

α =k3a−

……. (1)

αβ =k2

c⇒ 2α2 =

k2c

…… (2)

Substituting the value of α in the above equation,

22

k3a

−= .

k2c

k2c

k9

a22

2

= 4a2 = 9ck Choice (3)

20. Given: (a + c − b) x2 − 2cx + (b + c − a) = 0 The sum of the coefficients is ‘0’.

One root of the equation is 1. Let α be the other root.

α.1 = bca

acb

−+−+

α = .bcaacb

−+−+

Hence the roots are .bcaacb

−+−+

and 1

Choice (3) 21. x2 − 3x + 9 = 0 If α and β are the roots of the equation x2 – 3x + 9 = 0 α + β = 3 αβ = 9

2

1

2

12

2

1

2

1

2 βα+β+α=

β+α = ( )2

1923 +

⇒ 63

2

2

1

2

1

+=

β+α ⇒ 9

2

2

1

2

1

=

β+α

⇒ 321

21

±=β+α Choice (3)

22. 35 + is one root of the equation x2 – kx + p = 0, then

35 − is also a root of the equation.

The sum of roots = ( ) ( ) k3535 =−++ ⇒ 10 = k

The product of the roots = ( ) ( ) p3535 =−+ p = 22 k2 − 3p = (10)2 – 3(22) = 100 − 66 = 34. Choice (2) 23. x2 + 4x + 8 = 0 α + β = −4, αβ = 8

β+αβα+β+α=

β+α 3

1

3

1

3

1

3

13

3

1

3

1

3

β+α+−=

β+α 3

131

313

31

31

834

Let a31

31

=β+α

a3 = − 4 + 6a ⇒ a3 − 6a + 4 = 0 ⇒ (a – 2) (a2 + 2a − 2) = 0

⇒ a = 2 or 2

842a

+±−=

⇒ a = 2 or a = 2

322 ±−

⇒ a = 2 or 31a ±−=

31or13or231

31

−−−=β+α . Choice (4)

24. The quadratic equation whose roots are reciprocals of

the roots of the equation x2 + 3x – 5 = 0 is

05x3

x

12

=−+ ⇒ 1 + 3x − 5x2 = 0

⇒ 5x2 − 3x − 1 = 0. ∴ The required equation is 5x2 – 3x – 1 = 0.

Choice (1)

25. ax2 + bx + b = 0. Let α, β be the roots of the equation.

α : β = m : n and α + β = ab−

, αβ = a

b

4

1

4

1

mn

nm

+

=

41

41

αβ+

βα

=

( )41

21

21

αβ

β+α

= ( )

( )4

1

2

1

2

1

2

αβ

βα+β+α

=

41

21

ab

ab

2ab

+

−

=

41

21

ab

ab

2ab

−

=

41

41

ab

ab

2ab

−

= ab

2 −

Choice (4)


26. The expression ax2 +bx + c has minimum value when a > 0 and maximum value when a < 0.

(i) 2x2 − 5x + 1 Here, the coefficient of x2, (i.e. 2) > 0. The expression has minimum value.

The minimum value is .a4

bac4 2−

( )( ) ( )

( ) 817

8258

245124 2 −=−=−−

.

The minimum value is .817−

Choice (2)

(ii) 2x − x2 + 3

Here, the coefficient of x2 is negative, so the expression has maximum value

∴ Its maximum value is a4

bac4 2−

( )

( ) 44

164

41214

2314 2==

−−−=

−−−

∴ The maximum value is 4. Choice (3)

(iii) (2x − 1)

− x23

⇒ (2x − 1) ( )x2321 −

⇒ ( )x23x4x621 2 +−− ⇒ ( )3x8x4

21 2 −+−

⇒ 23

x4x2 2 −+−

Here, the coefficient of x2 is negative, so it has maximum value.

Its maximum value is a4

bac4 2−

=

( )( ) 2

1

8

4

8

1612

24

42

324 2

=−−=

−−=

−

−

−−.

The maximum value is 21

. Choice (4)

27. Let the number be x.

Given: 2xx =−

⇒ x2x =− ⇒ (x − 2)2 = x ⇒ x2 + 4 − 4x = x ⇒ x2 − 5x + 4 = 0 ⇒ (x − 1)(x − 4) = 0 ⇒ x = 1 or x = 4. But x = 1 does not satisfy the equation, so x = 4

Choice (2) 28. Let α and β be the roots of the equation

037x35x2 =++

The sum of the roots = 35− .

The product of the roots = 37 . Choice (1)

29. Given: x2 − x − 56 < 0 ⇒ x2 − 8x + 7x − 56 < 0 ⇒ x (x − 8) + 7 (x − 8) < 0 ⇒ (x − 8) (x + 7) < 0 ⇒ − 7 < x < 8 Choice (1) 30. Given: 2x2 + 3x + 2

The roots of 2x2 + 3x + 2 = 0 are complex numbers so, the expression 2x2 + 3x + 2 is always positive. Choice (1)

31. x2 − 9x + 14 The roots of the equation x2 – 9x + 14 = 0 are 2 and 7. The coefficient of x2 is positive.

∴ When x∈ (2, 7) The given quadratic expression is negative.

Choice (1) 32. Let α and β be the roots of the equation.

x2 − 2px + q2 = 0, then

α + β = 2p ⇒ p2

=β+α = k (Given)

Let ϒ and δ be the roots of the equation. x2 − 2qx + 4p = 0 γδ = 4p

The geometric mean )l( = γδ ⇒ l = p4 ⇒ l2 = 4p

⇒ l2 = 4k. Choice (4) 33. Let ‘α’ be the common root of the two equations. ⇒ pα2 + 2qα + r = 0 …….. (1) and pα2 + 2rα + q = 0…….. (2) 2qα + r = 2rα + q

2α(q – r) = q – r ⇒ α = 2

1

Substituting the value of α in (1) we get,

0rq4

p =++ ⇒ p + 4q + 4r = 0. Choice (3)

34. Let α and β be the roots of the equation x2 – mx + n = 0 α + β = m; αβ = n Given α – β = k ∴ (α + β)2 = (α – β)2 + 4αβ. ⇒ m2 = k2 + 4n.

Choice (1)

35. Let α and β be the roots of the equation, mx2 − ax + 1 = 0

Then ma=β+α ;

m1=αβ

The roots of the equation nx2 − px + 1 = 0 will be

βα1

,1

.

np11 =

β+

α;

n11 =

αβ⇒

n1

m =

np=

αββ+α

⇒ mn = 1

np

m

ma

=1

⇒ a = n

p

p = an Choice (2) 36. x2 − 9x + 20 = 0 ⇒ x2 − 5x − 4x + 20 = 0 ⇒ (x − 5) (x − 4) = 0 ⇒ x = 5 or 4

One third of the above roots is 35

,34

∴ The equation whose roots are is3

5and

3

4

x2 − x (α + β) + αβ = 0 ⇒ 03

5

3

4

3

5

3

4xx2 =×+

+−

9x2 − 27x + 20 = 0.

Alternate method:

The required equation is (3x)2 – 9(3x) + 20 = 0 ⇒ 9x2 – 27x + 20 = 0 Choice (3)


37. The required equation is

072x

52x

2

=+

−

⇒ 072x5

4x2

=+−

⇒ x2 – 10x + 28 = 0 Choice (2) 38. Given: 3 |x|2 − 11 |x| + 4 = 0 When x > 0, 3x2 − 11x + 4 = 0 and when x< 0 3x2 + 11x + 4 = 0 3x2 – 11x + 4 = 0 has two real roots and

3x2 + 11x + 4 = 0 has two real roots. ∴ The total number of real roots is 4. Choice (3) 39. ax2 + bx + 1 = 0 a, b ∈ {1, 3, 5, 6} When b2 – 4a > 0. then the given equation has real

roots. The possible pairs of the form (b,a) that satisfies the above relation is (3,1) ,(5, 1), (6,1), (3,5), (6,3), (5,6), (6,5), (6,6) and (5,5). ∴ The number of quadratic equations whose roots are

real numbers is 9. Choice (2) 40. The roots of the equation x2 − 2kx + k2 + k − 3 = 0

are real , if the discriminant is non-negative. ⇒ (− 2k)2 − 4 (1) (k2 + k − 3) ≥ 0 4k2 − 4k2 − 4k + 12 ≥ 0 −4k ≥ − 12 k ≤ 3. ….. (1) The roots of the equation x2 – 2kx + k2 +k –3 = 0

= ( )

2

3kk4k4k2 22 −+−±

2

k412k2x

−±= ⇒ x =k ± k3 −

Since the roots are less than 3

k ± k3 − < 3 (3 – k) < (3 – k)2

(3 – k) (2 – k) > 0 k < 2 or k > 3 k < 2 ……. (2) From (1) and (2) we have k < 2. Choice (3)

Exercise – 3

Solutions for questions 1 to 45:

1. ,31

11131

12 +=− 31

132

1331

12 +=− and so on.

So the 43rd term = ( ) 6731

114311 =

+×−+ .

Choice (2) 2. Given: t6 = 17 and t13 = 31 17 = a + (6 − 1) d ⇒ a + 5d = 17 ----- (1) 31 = a + (13 − 1) d ⇒ a + 12d = 31 --- (2) Solving (1) and (2), we get a = 7, d = 2. ∴ t17 = 7 + (17 − 1) × 2 = 39. Choice (3) 3. According to problem 7t7 = 13t13 ; d = 6 7 (a + (7 − 1) d) = 13 (a + (13 − 1) d) ⇒ 7 (a + 6d) = 13 (a + 12d) ⇒ 6a + 114d = 0 6a + (114)6 = 0 a = − 114 ∴ t29 = a + 28d t29 = −114 + 28 × 6 = 54 Choice (1)

4. Let a and d be the first term and the common difference of the series.

∴ S1 = a + (n + 1 − 1) d ⇒ S1 = a + nd S2 = a + 2nd and S3 = a + 3nd

Now 231 Snd2a

2

SS=+=

+

∴ S1, S2 and S3 are in AP. Choice (1) 5. Let a and d be the first term and the common difference

of the A.P. tn = a + (n − 1) d = 4n + 5 a + (n − 1) d = 5 + 4 (n − 1) + 4 = 9 + (n − 1) × 4 ∴ a = 9, d = 4. Choice (4) 6. Let a be the first term and d be the common difference of the series. Given: a1 + a4 + a7 + ….. + a16 = 231. ⇒ a + (a + 3d) + (a + 6d) + (a + 9d) + (a + 12d) + (a + 15d) = 231 6a + 45d = 231 ∴ 2a + 15d = 77 → (1) Now a1 + a6 + a11 + a16 = a + a + 5d + a + 10d + a + 15d = 2 (2a + 15d) = 2(77) = 154 ∴ a1 + a6 + a11 + a16 = 154 Choice (2)

7. The given series is ....,510,10,52,2

.5252

tt

r ; 2 a1

2 ====

Let nth term be 50. tn = 50

arn-1 = 50 ⇒ ( ) 5052 1n =× −⇒ 22

1n

5255 ==−

22

1n =− ∴ n = 5 Choice (2)

8. Let a be the first term and r be the common ratio of the series. Given: t7 = 2

t13 = 321

t7 = a r7 − 1 = 2 ------------ (1)

t13 = ar13 − 1 = 321

-------- (2)

(1) ÷ (2) gives

∴ r6 − 12 =

3212

= 64 ⇒ r− 6 = (2)6 = 6

21

−

∴ 21

r =

By substituting the value of r in (1), we get

a 221

6

=

⇒ a = 27 = 128

∴ t17 = ar16

∴ t17 = 512

1

22

1128

21

297

167 =

××=

×

Choice (4) 9. Let the sum of n terms be 7161. ∴ Sn = 7161 The given series is in GP with the common ratio = 2

Sn = 1r

)1r(a n

−−

⇒( )

71611212

7n

=−−

∴ 2n = 10241102317

7161 =+=+ = 2n = 210

∴ n = 10 Choice (1)


10. Let the series be a, ar, ar2, ar3, …… ∞ Given a = 3 (ar + ar2 + ar3 + …..)

r1

r31

−= ⇒ 1 − r = 3r

∴ r = 41

Choice (2)

11. Given: a = − 1, tn = arn − 1 = 19683 ; Sn = 14762 ∴ rn − 1 = − 19683, rn = − 19683r Sn = 14762

∴ ( )

147621r

1ra n

=−

− ⇒ − 1(− 19683r − 1)

= 14762r − 14762 4921r = − 14763

492114763

r−=

r = – 3 Choice (1)

12. Given: Sm = 3m2 + 7m = m (3m + 7) = )m614(2m +

= ( )( ) ( )[ ]61m202m

61m6142m −+=+−+

= ( )[ ] ( )[ ]61m10102

m61m102

2

m −++=−+×

Sm = [ ]m1 tt2m +

∴ tm = 10 + (m − 1) 6 Hence tn + 1 = 10 + 6n Alternate method :

Given: Sm = 3m2.+7m ∴ Sn = 3n2 + 7n ; Sn + 1 = 3(n+1)2 + 7(n+1) We know that tn = Sn – Sn−1

∴ (n+1)th term i.e., tn+1 = Sn+1 −Sn = 3(n+1)2 + 7(n+1) − (3n2+7n) = 3(n2 + 2n + 1) + 7n + 7 − 3n2 − 7n = 3n2 + 6n + 3 + 7n + 7 − 3n2 − 7n = 6n + 10. Choice (4)

13. Let the three terms of the GP be aranda,ra

.

Given: 154r1r1

a =

++ → (1)

and 6776r

aararaa

r

a =

⋅+⋅+⋅ ,

⇒ 67761rr

1a2 =

++ → (2)

2 ÷ 1 gives

441546776

a ==

Substitute the value of ‘a’ in (1)

154r1r1

44 =

++

∴ r2 + r + 1 = r44

154 × r27=

2r2 + 2r + 2 =7r ⇒ 2r2 − 5r + 2 = 0 ⇒ 2r2 − 4r − r + 2 = 0

⇒ (2r − 1) (r − 2) = 0, ⇒ 21

r = or 2

∴ The numbers are, 22, 44, 88 or 88, 44, 22. Choice (3)

14. Let a and b be the two numbers. Then a − b = 10 --- (1)

and 182

ba =+⇒ a + b = 36 --- (ii)

Solving (1) and (2), we get a = 23, b = 13 The product of the numbers = 23 × 13 = 299 Choice (2) 15. Let a be the first term and r be the common ratio of the series.

given t3 = ar2 = 41

∴ The first five terms of the series are a, ar, ar2, ar3, ar4 a ar ar2 ar3 ar4 = a5 r(0 + 1 + 2 + 3 + 4)

= a5r10 = (ar2)5 = 5

4

1

. Choice (1)

16. The maximum value of the sum of the 37 + 33 + 29 +

25 + … is possible up to the positive term because it is in decreasing AP with common difference as − 4.

Let tn be the last positive term Tn = 37 + (n − 1) (− 4) > 0

∴ 14

37n +< ∴ n < 10⋅25

∴ n = 10 ∴ 10th term the is last positive term

S10 = ( )[ ] 190)4(1103722

10 =−×−+× .

Choice (4) 17. Let a and r be the first term and the common ratio of the given GP.

Given a = 8 and t5 = 21

= ar5 − 1 = 21

8 × r4 = 21

4

4

21

161

r

== ∴ 21

r ±=

We know that the sum of the infinite terms of G.P. is

= r1

a−

when r = 21

,

16

21

1

8S =

−=∞ and

when r = 21− ,

3

161228

21

1

8S =

+×=

−−=∞

∴ The sum will be either 16 or 3

16.

Choice (4) 18. Let one of the GP be a1, a1r1, a1r1

2, a1r13……. and

another GP be a2, a2r2, a2r22, a2r2

3, ……..

According to the problem, ( )( ) ,

rr

aa

,rara

,aa

2

2

1

2

1

22

11

2

1

,........r

r

a

a3

2

1

2

1


which is also a GP with the first term 2

1

aa

and common

ratio as

2

1

r

r. Choice (3)

19. Let the four numbers be a− 3d, a − d, a + d, a + 3d accordingly, a − 3d + a − d + a + d + a + 3d = 4a = 12 ∴ a = 3 Again, (a − 3d)2 + (a − d)2

+ (a + d)2 + (a + 3d)2 = 536 ⇒ 4a2 + 20d2 = 536

⇒ 2520

34536d

22 =×−= ⇒ d = ± 5

∴ The series is −12, −2, 8, and 18 or 18, 8, − 2, − 12 so ∴ The least number is − 12. Choice (1) 20. Given: a, b and c are in AP. 2b = a + c Consider a + b − c + b + c − a = 2b = 2 [c + a − b] (Q c + a − b = 2b − b = b) ∴ a + b − c, c + a − b, b + c − a are in AP. Choice (1)

21. The AM of isb1

,a1

ab2

ab2

b1

a1

+=+

,

but given AM is nn

1n1n

ba

ba

++ −−

∴ab2

ba

bab

ba

a

nn

nn

+=+

+

( ) ab2ba

baab

abbann

nn +=+

+ ⇒ 2an b + 2abn = (an + bn) (a + b)

⇒ an (2b − a − b) + bn (2a − a − b) = 0 ⇒ an (b − a) = bn (b − a) ⇒ (b − a) (an − bn) = 0 Q a ≠ b, ∴ The above is possible only when n = 0. ∴ n = 0 Choice (1) 22. Given: a, b and c are in H.P,

c1

,b1

,a1

⇒ are in A.P. c

abc,

babc

,a

abc⇒ are in A.P.

⇒ bc, ca, ab are in A.P. Choice (2) 23. Let a and b be the two numbers, then

HM = 524

baab2 =+

→ (1)

GM = 6ab =

⇒ ab = 36 (Substitute the value of ab in (1))

524

ba362 =

+×

24

5362ba

××=+ = 15

∴ AM = .572

152

ba ⋅==+ Choice (4)

24. Let a1, d1 be the first term and the common difference of

the first series and a2, d2 be the first term and the common difference of the second series .

]d)1n(a2[2n

Sand],d)1n(a2[2n

S 222n111n −+=−+=

Given: 5n38n2

S

S

2n

1n

++=

5n38n2

]d)1n(a2[2n

]d)1n(a2[2n

22

11

++=

−+

−+

5n38n2

d2

)1n(a(2

d2

)1n(a(2

22

11

++

=

−+

−+

Let 1k2n1k2

1n −=⇒−=−

2k66k4

5)1k2(38)1k2(2

d)1k(a

d)1k(a

22

11

++=

+−+−

=−+−+

⇒

1n33n2

2n66n4

d)1n(a

d)1n(a

22

11

++=

++=

−+−+

⇒ Choice (4)

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

According to the problem, ( )( )[ ]d1ma22m

Sm −+=

and ( )( )[ ]d1na22n

Sn −+= .

Given: ( )( ) ( )( )d1na22n

d1ma22m

SS nm −+=−+⇒=

⇒ 2a (m − n) + d (m2 − n2) − d (m − n) = 0 ⇒ (m − n) [2a + (m + n − 1) d] = 0 ⇒ [2a + (m + n − 1) d ] = 0 (Q m ≠ n)

⇒ ( )[ ] 0d1nma22

nm =−++

+

⇒ ( )[ ] .0d1nma22

nmS nm =−+++=+ Choice (4)

26.

−=× 12

161

61

1261

−=× 18

1121

61

18121

−=× 24

1181

61

24181

( ) ( )( )

−+

=+ n

11n

1361

1n6n61

( )1n6n61

....2418

11812

1126

1Sn +⋅

++×

+×

+×

=

−=n1

61

61

−==∞→∞→

∞ n1

61

61

LimSLimSn

nn

= .361

061

61 =

−×

Choice (4)

27. Given: GP with the first term as 1 and the common ratio as 5

∴ ( )

10001r

1raS

n

n <−

−=

⇒ ( )

100015151 n

<−−

⇒ 5n < 4001 55 = 3125 < 4000, 56 = 15625 > 4001 ∴ n = 5. ∴ The greatest value of n is 5. Choice (3)


28. Let a be the 1st term and d be the common difference. Given : t13-t5 =3t3

( )[ ] ( )[ ] ( )( )d13a3d15ad113a −+=−+−−+

⇒ 8d = 3a + 6d ⇒ 2d = 3a

∴ 32

da = Choice (3)

29. Let the work be finished in n days when the workers started dropping, so the total number of workers who worked all these n days is = 120 + 115 + 110 + 105 + 100 + ……n terms (for n days)

∴ ( )[ ] [ ]n52452n

)5(1n21202n

w −=−×−+×= --- (1)

Had the workers not been dropped then the work would have been finished in (n − 5) days with 120 workers. Each worker finishes in 120 (n−5) days i.e. w = 120 (n − 5) --- (ii) from (1) and (2)

∴ ( ) ( )5n120n52452n −=−

⇒ n2 − n − 240 = 0 ⇒ (n + 15) (n − 16) = 0 ∴ n = 16 days. (Q n cannot be negative) Choice (2) 30. The nth term of the series 1 + (1 + a) r + (1 + a + a2) r2 + …….. is

( ) ( )

∑ ∑−−==⋅

−−=

∞

=

−∞

−

1n

1nn

n1n

n

n r1a1a

tS,r1a1a

t

nn

1nr)1a(

r)1a(1 −∑−

=∞

= =

)1a(r1−

( )

∑ ∑−∞

=

∞

=1n 1n

nn rar

= ( )( )

( )( ) ( )1arr1ar11ar

r1r

ar1ar

1ar1

−−−−=

−−

−−

= ( )( )r1ar11

−−. Choice (3)

31. Let d be the common difference of the given series then ⇒ a2 − a1 = a3 − a2 = a4 − a3 = an − an−1 = d

−=

−

−=

2121

12

1221 a1

a1

d1

aaaa

aa

1aa1

(Q a2 − a1 = d)

−=

−−

=3232

23

2332 a1

a1

d1

aa

aa

aa1

aa1

Similarly,

−=

4343 a1

a1

d1

aa1

−=

⋅ −− n1nn1n a1

a1

d1

aa1

Adding the above equations, we get

∴

−=

⋅++++

− n1n1n433221 a1

a1

d1

aa1

...aa1

aa1

aa1

= ( )

n1

1n

aaaa

d1 −

( )( )

n1n1

11

aa1n

aaad1na

d1 −=−−+= . Choice (1)

32. Let the three numbers of AP be a − d, a and a + d Accordingly, a − d + a + a + d = 3a = 9. ∴ a = 3 Again, (a − d)3 + a3 + (a + d)3 = 3a3 + 6ad2 = 729 ⇒ 3a (a2 + 2d2) = 729

⇒ 2d2 = (729 / 3 × 3) − 32 = 72 ⇒ d = ± 6 ∴ The numbers are − 3, 3, 9 or 9, 3, − 3 ∴ The least number is −3. Choice (3) 33. Given: (a + b − c)2,(b + c−a)2 and (c + a − b)2 are in A.P, (b + c − a)2 − (a + b − c)2 = (c + a − b)2 − (b + c − a)2 ⇒ (b + c − a + a + b − c) (b + c − a − a − b + c) = (c + a − b + b + c − a) (c + a − b − b − c + a) ⇒ 2b 2 (c − a) = 2c 2 (a − b) ⇒ b c − ab = ac − bc ⇒ 2bc = ac + ab = a (b + c)

∴ cb

bc2a

+= ⇒ b, a, c are in HP Choice (2)

34. Let S = 1 + 4x + 8x2 + 13x3 + 19x4 + …→ (1) xS = x + 4x2 + 8x3 + 13x4……….→ (2) (1) – (2) gives, (1 − x) S = 1 + 3x + 4x2 + 5x3 + 6x4 + ……. Let M = (1 − x) S − 1 = 3x + 4x2 + 5x3 + 6x4 + …→ (3) xM = 3x2 + 4x3 + 5x4 + …→ (4) − − − − (3) – (4) gives (1 − x)M = 3x + x2 + x3 + x4 + ……….

⇒ (1 − x) M = ( ) ( )x1x2x3

x1x

x322

−−=

−+

( )2

2

x1

x2x3M

−−= ⇒ (1 − x) S − 1

= ( )2

2

x1

x2x3

−−

( )( ) ( )3

22

2

x1

1xxx1

1x1

x2x3

S−

+−=−

+−−

= . Choice (3)

35. Let S = 0⋅6 + 0⋅66 + 0⋅666 + 0⋅6666 + … up to n terms

S = ( )termsnupto.....99909909096 +⋅+⋅+⋅

=

+−+−+− termsnupto........10

11

10

11

101

196

32

++++−+++=

n32 10

1..

10

1

10

1101

)nterm..111(96

=

−

−

−

101

1

10

11

101

n96 n

=9

)101(96

9n6 n−−−

= )101(816

9n6 n−−− Choice (1)

36. The nth term of the series is

( )

( )( )

( ) 61n2

21nn

2

61n21nn

n.....3212n......321

t2222

n+=

+⋅

++

=++++++++=

∴ ( )∑ +∑ ====

n

1n

n

1nnn 1n2

61

tS

( )

++⋅=Σ+Σ= n2

1nn2

61

)1n2(61

= 6

n2n2 + Choice (2)


37. The nth term of the first series = tn = 45 + (n − 1) × (− 4) ------ (i) The nth term of the second series = tn = 5 + (n − 1) × 4 --- (ii) Accordingly, 45 + (n − 1) (−4) = 5 + (n − 1) × 4 ⇒ 45 − 5 = 8 (n − 1) ⇒ n − 1 = 5 ∴ n = 6 Choice (1) 38. Let A = {2, 4, 6, 8, ………98, 100} B = {7, 14, 21, 28, 35, ………98} The sum of the multiples of 2 = 2 + 4 + 6 + ……..+ 100

= ( ) 255010022

50 =+×

The sum of the multiples of 7 = 7 + 14 + 21 + …… + 98 = 7 (1 + 2 + 3 + …14)

7352

15147 =××

The sum of the multiples of 2 and 7 = 14 + 28 + 42 + 56 + 70 +84 + 98 = 392.

∴ The required sum = 2550 + 735 – 392 = 2893. Choice (1)

39. 33

3

1 = .

3

36

36()63(

36

63

1 −=−+

−=+

96

1

+ =

369 −

…………….

And 729726

1

+ 3726729 −=

Adding all the terms, we get

729726

1....

96

1

63

1

3

1

+++

++

++

3

726729.....69363 −++−+−+

.9327

3729 === Choice (3)

40. Let the sides a − d, a and a + d form a right angled

triangle (a − d)2 + a2 = (a + d)2 ⇒ a2 = (a + d)2 − (a − d)2

= 4ad ∴ a = 4d ⇒ the sides are 3d, 4d and 5d. ∴ base: perpendicular : hypotenuse = 3 : 4 : 5. Choice (2) 41. Let x, y and z be a − d, a and a + d respectively.

Then,

+−+

+xyz

yzxz2xylog

xzzx

log

( ) ( )

+−++−

dadadada

log +

( ) ( )( ) ( )

( ) ( )

+−+++−−−

daadadaadada2ada

log

( )

−+++−−+

− 22

2222

22 daa

adad2a2adalog

da

a2log

+

− 22 da

a2log ( )

− 22

2

daa

d2log

( ) ( )

−⋅−=

2222

2

daada

ad4log

ablogblogalog( =+Q )

= ( )

( )

−=

−22222

2

da

d2log2

da

d2log

= ( ) ( )( )( )

−+−−+dadadada

log2 = zx

xzlog2

−. Choice (4)

42. Given abcd = 16 AM (a, b c, d) ≥ GM(a, b, c, d)

∴ 4 abcd4

dcba ≥+++⇒ a + b + c + d ≥ 8164 4 ≥×

∴ The minimum value of (a + b + c + d) = 8. Choice (3) 43. Since the pth term of an H.P is q, then the pth term of

A.P. is q1

, and qth term of H.P. is p . The qth term of

A.P. is q1

.

Let a, d be the first term and the common difference of the A.P. respectively.

∴ tp = a + (p – 1) d; q1

= a + (p –1) d ––––––– (1)

tq = a + (q – 1) d p1

= q + (q – 1) d –––––– (2)

Solving (1) and (2) for a and d, we get a = pq1

and

d = pq1

∴ The (p + q)th term of the A.P.

tp+q = a + (p + q – 1) d = pq1

+ pq

1qp −+ =

pqqp +

∴ The (p + q)th term of H.P. is qp

pq+

Choice (2)

44. Given: a, b and c are in A.P. and a + c = 2b. Also given a + b + c = 27 3b = 27; b = 9 ∴ a + b + c = 27 a + c = 27 – b = 27 – 9 a + c = 18 The possible values of (a, c) are (1, 7) (2, 16)…….. (17, 1) i.e. 17 possibilities. ∴ The number of triplets (a, b, c) that satisfies a + b + c = 27 is 17. Choice (3) 45. Given that a, H1, H2 ………., H20 and b are in H.P

⇒ b1

,H

1,.........

H

1,

a1

201

are in A.P.

Let a1

be x and the common difference be d.

a1

= x, 1H

1 = x + d,

20H

1 = x + 20d and

b1

= x + 21d

20

20

1

1

20

20

1

1

H

1b1

H

1b1

H

1a1

H

1a1

bH

bH

aH

aH

−

+

+−

+

=−

++

−

+

= d

d20xd21xd

dxx +++=−

++

= 40d

d40d

d41x2dx2 ==++−− Choice (4)


Exercise – 4 Solutions for questions 1 to 45: 1. Let f(x) = x4 + 9x2 + kx − 6 When f(x) is divided by x + 2, then the remainder is f(– 2) Given: remainder = 4. f (− 2) = (− 2)4 + 9 (− 2)2 + k (− 2) − 6 = 4 16 + 36 − 2k − 6 = 4 52 − 6 – 4 = 2k 2k = 42 k = 21. Choice (2) 2. Let f(x) = x3 − 2x + k x = 3 Given: x – 3 is a factor of f(x) ⇒ f(3) = 0

f(3) = 33 − 2 (3) + k = 0 27 − 6 + k = 0 21 + k = 0 k = − 21. Choice (1) 3. Let f(x) = x2 − 7x + 12 Given: x + k is a factor of f(x) ⇒ f((– k) = 0 (− k)2 + 7k + 12 = 0 k2 + 7k + 12 = 0 k2 + 4k + 3k + 12 = 0. (k + 4) (k + 3) = 0 k = − 3 or − 4. Choice (4) 4. Let f(x) = 2x4 + x3 − x2 − x − k Given: x – 1 is a factor of f(x) ⇒ f(1) = 0 f (1) = 2 (1)4 + (1)3 − (1)2 − (1) − k = 0 2 + 1 − 1 − 1 − k = 0 k = 1. Choice (4) 5. Let f(x) = x3 + 2x2 − k2x − 2 Given: f(x) is divisible by x – k f (k) = k3 + 2k2 − k3 − 2 = 0 2k2 − 2 = 0 k2 = 1 k = ± 1 Choice (1) 6. Let f(x) = x2 − 2x + 3

f(x) is divided by 2x – k, the remainder is f

2k

Given: f

2k

= 2

32k

22k

2k

f2

+

−

=

⇒ 23k

4k2

=+−

⇒ k2 − 4k + 12 = 8 ⇒ k2 − 4k + 4 = 0 ⇒ (k − 2)2 = 0 ⇒ k = 2. Choice (2) 7. Let f(x) = x3 + 3x2 + 4x + 1 x = − 3 f(– 3) = (− 3)3 + 3 (− 3)2 + 4 (− 3) + 1 = − 27 + 27 − 12 + 1 = − 11. Choice (1) 8. Let f(x) = 2x5 + 3x2 + 4x + 4 x = − 1 f(– 1) = − 2 + 3 + 4(− 1) + 4 = 1 Choice (2) 9. Let f(x) = x3 + 3x2 + 4x + 1 x = − 1

f (− 1) = −1 + 3 − 4 + 1= − 1 ∴ – 1 should be subtracted. Choice (1) 10. Let f(x) = px2 + qx + r When x = 0, f(x) = 8 f(0) = p (0) + q (0) + r = 8

r = 8 ------ (1) When x = 1, f(x) = 0 f(1) = p (1) + q (1) + r = 0 p + q + 8 = 0 p + q = − 8 --- (2) When x = − 2, f(x) = 6 f (− 2) = p (− 2)2 + q (− 2) + r = 6 4p − 2q = − 2 2p − q = −1 --- (3) Adding (2) and (3) 3p = − 9 p = − 3 p + q = − 8 − 3 + q = − 8 q = − 5 2p + q + r = 2 (− 3) − 5 + 8 = − 3 Choice (4) 11. Let f(x) = x3 + ax2 + bx + 1 x2 − 3x + 2 = 0 (x − 1) (x − 2) = 0 x = 1 or 2 f(1) = 1 + a + b + 1 = 0 a + b = − 2 Choice (1) 12. Let f (x) = 3x3 + 4x2 + 5x + 1 ⇒ x + 1 = 0 x = − 1 f (− 1) = 3 (− 1)3 + 4 (− 1)2 + 5 (− 1) + 1 = − 3 + 4 − 5 + 1 = – 3 ∴ − 3 should be subtracted. Choice (4) 13. x3 + ax2 + 26x − 6b = 0 Let the roots be n, n + 1and n + 2 The sum of the roots = n + (n + 1) + (n + 2) = – a 3n + 3 = – a

The sum of the product of the roots taken two at a time = n (n + 1) + (n + 1) (n + 2) + (n + 2) n = 26 n2 + n + n2 + 2n + n + 2 + n2 + 2n = 26 3n2 + 6n + 2= 26 3n2 + 6n − 24 = 0 3n2 + 12n − 6n − 24 = 0 3n (n + 4) − 6 (n + 4) = 0 (n + 4) (3n − 6) = 0 n = − 4 or n = 2 The product of the roots = n (n + 1) (n + 2) = 6b

If n = − 4, − 12 + 3 = – a a = 9 − 4 × − 3 × − 2 = 6b − 24 = 6b b = − 4 If n = 2, 6 + 3 = – a a = – 9 2 × 3 × 4 = 6b b = 4 a + b = 9 − 4 = 5 or a + b = – 9 + 4 = – 5. Choice (4) 14. x3 − 12x2 + 44x − 48 = 0

Let α, β and γ be the roots of the given equation α + β + γ = 12 αβ + βγ + γα = 44 αβγ = 48 The roots are in A.P 2β = α + γ α + β + γ = 12 3β = 12 β = 4 α + γ = 8 ------- (1) αγ = 12 (α – γ)2 = (α + γ)2 − 4αγ = 82 − 4(12) = 16 α – γ = ± 4 ------ (2) If α − γ = 4, adding (1) and (2), 2 α = 12 α = 6. ∴ γ = 2


If α − γ = − 4, it similarly follows that, α = 2 and γ = 6. In either case, β = 4 ∴ In either case, the roots are 2, 4 and 6.

Choice (3) 15. 3x5 − 9x4 + 27x3 + 5x + 1= 0

Σα1α2 = 9327 = Choice (4)

16. x3 + 9x2 + 4x + 5 = 0 α + β + γ = − 9 αβ + βγ + γα = 4 αβγ = − 5

Σ α1

=γ

+β

+α

111=

54

−=

αβγγα+βγ+αβ

.

Choice (2) 17. x3 − 5x2 + 4x + 9 = 0

∑αβ1

= αγ

+βγ

+αβ

111=

αβγγ+β+α

From the equation, we have, α + β + γ = 5 αβ + βγ + γα = 4 αβγ = – 9

9

5−

=αβγ

γ+β+α. Choice (1)

18. x3 − 5x2 + 7x − k = 0, α + β + γ = 5

αβ + βγ + γα = 7 αβγ = k

γ

+β

+α

111= 1=

αβγαβ+γα+βγ

1k7 = k = 7 Choice (2)

19. x3 − ax2 + bx + k = 0

∑ ∑ α=αβ1

γ+β+α=γα

+βγ

+αβ

111

γ+β+α=αβγ

γ+β+α

αβγ = 1 – k = 1. ⇒ k = – 1 Choice (2) 20. x3 − 7x − 6 = 0 α + β + γ = 0 αβ + βγ + γα = − 7 Σα2 = α2 + β2 + γ2 = (α + β + γ)2 − 2 (αβ + βγ + γα) = 0 − 2 (− 7) Σα2 = 14. Choice (3) 21. x3 − 16x2 + ax + k = 0 Let the roots be α, 3α and 4α The sum of the roots = α + 3α + 4α = 16 8α = 16 α = 2 ∴ The roots are 2, 6 and 8 The product of the roots 2 × 6 × 8 = 96 ⇒ – k = 96

k = – 96 a = 2 × 6 + 6 × 8 + 8 × 2 a = 76 a + k = 76 − 96 = – 20. Choice (2)

22. Given: the roots of x3 + 3x2 + 5x + 6 = 0 is α, β γ. Let the roots of the new equation be 3α, 3β, and 3γ

f

3x

. = 0

f

3x

= 063x5

9x3

27x 23

=+++

⇒ x3 + 9x2 + 45x + 162 = 0 Choice (4)

23. Given: The roots of the equation are 4and32 +

Since, 32 + is a root, its conjugate is also a root of

the equation. Thus, 2 – 3 is also a root.

α + β + γ = 843232 =+−++

αβ + βγ + γα = 173481348 =−+++

∴ The equation is αβγ = 4 x3 − 8x2 + 17x − 4 = 0. Choice (2)

24. Let 21and21,35,35 +=δ−=γ+=β−=α

α + β + γ + δ = 12 Σ αβ = αβ + γδ + (α + β) (γ + δ) = 22 − 1+ (10) (2) = 41 Σαβγ = αβ (γ + δ) + γδ (α + β) = 22 (2) + − 1 (10) = 44 − 10 = 34 αβγ δ = − 22 ∴ The equation is x4 − 12x3 + 41x2 − 34x − 22 = 0 Choice (3) 25. x4 + 4x3 + 8x2 + 1 = 0

We substitute 2x

for x

012x

82x

42x

234

=+

+

+

014x

88x

416x 234

=++⋅+

x4 + 8x3 + 32x2 + 16 = 0. Choice (1) 26. x4 + ax3 + bx2 + cx + d = 0

Substitute x1

for x.

0dx1

cx1

bax1

x1

234

=+

+

+

+

1 + ax + bx2 + cx3 + dx4 = 0 dx4 + cx3 + bx2 + ax + 1 = 0. Choice (2)

27. x3 + 5x2 + 3x + 1 = 0 Let α be one root of the equation then α + 2 will be the other root of the required equation. (x + 2)3 + 5 (x + 2)2 + 3 (x + 2) + 1 = 0 x3 + 6x2 + 12x + 8 + 5 (x2 + 4x + 4) + 3x + 6 + 1 = 0 x3 + 11x2 + 35x + 35 = 0. Choice (4) 28. x3 + ax2 + bx + a = 0 x3 + bx2 + ax + b = 0 Let α be the common root ⇒ α3 + aα2 + bα + a = 0 α3 + bα2 + aα + b = 0 − − − − --------------------------------- (a − b) α2 + (b − a) α + a − b = 0 (a − b) [α2 − α + 1] = 0 a − b can be 0. Choice (3) 29. x3 − 5x2 + 3x + 9 = 0 Let α, α and β be the roots of the equation α + α + β = 5 2α + β = 5 --- (1)


(α) (α) (β) = − 9 α2β = − 9 --- (2) (α) (α) + (α) (β) + (α) (β) = 3 α (α + 2β) = 3 --- (3)

From (1) α (α + 2 (5 − 2α)) = 3 α2 + 10α − 4α2 = 3 3α2 − 10α + 3 = 0 3α2 − 9α − α + 3 = 0 (α − 3) (3α − 1) = 0

α = 3 or 31

As the roots are integers, α = 3 ∴ β = − 1 Choice (2) 30. x3 − 7x2 + 14x − 8 = 0 Let α, β and γ be the roots of the equation Let β = 2α α + β + γ = 7 α + 2α + γ = 7 3α + γ = 7 --- (1) γ = 7 − 3α αβ + βγ + γα = 14 2α2 + 3αγ = 14

α (2α+3γ) = 14 --- (2) αβγ = 8 2α2γ = 8 --- (3) From (1) and (2), α (2α + 3 (7 − 3α)) = 14 α (2α + 21 − 9α) = 14 7α2 − 21α + 14 = 0 α2 – 3α + 2 = 0

(α – 2) (α –1) = 0 α = 2 or α = 1

If α = 1, then β = 2 and γ = 4 If α = 2, then β = 4 and γ = 1 Choice (1)

31. x3 + 3ax2 + 3bx + r = 0 Let α, β and γ be the roots of the equation Let α, β and γ be in A.P 2β = α + γ --- (1) α + β + γ = −3a --- (2) From (1) and (2) 3β = − 3a ⇒ β = − a Since β is the root of the equation, – a3 + 3a (– a) + 3b(– a) + r = 0 – a3 + 3a2 – 3ab + r = 0 r = 3ab – 2a3. Choice (1) 32. x3 − 6x2 + 11x − 6 = 0

Let the roots be α, β and γ. Given: the product of two roots is 6. Let α > β

α + β + γ = 6 --- (1) αβ + βγ + γα = 11 --- (2) αβγ = 6 --- (3)

Let 6γ be 6 and αβ be 6 γ = 1 α + β = 6 − 1 = 5 ------ (1) (α − β)2 = (α + β)2 − 4αβ = 52 − 4 (6) = 1

α − β = 1 (∵ α > β) ----- (2)

α + β = 5 Adding (1) and (2) 2α = 6

α = 3 ⇒ β = 2 ∴ The roots are 1, 2 and 3. Choice (2)

33. 2x3 − 9x2 + 12x − 4 = 0 Let α, α and β be the roots of the equation

α + α + β = 29

α2 + 2αβ = 2

12= 6

α2β = 24

= 2

2α + β = 29

⇒ β α−= 229

⇒ α(α + 2β) = 6 ⇒ α(α + 9 − 4α) = 6 ⇒ α (− 3α + 9) = 6 ⇒ − 3α2 + 9α − 6 = 0 ⇒ 3α2 − 9α + 6 = 0 ⇒ α2 − 3α + 2 = 0

⇒ α2 − 2α − α + 2 = 0 ⇒ α (α − 2) − 1(α − 2) = 0 (α − 1) (α − 2) = 0 α = 1 or 2 But α = 1 does not satisfy the equation ∴ α = 2,

21

429 =−=β

∴ The roots of the equation are

2, 2 and 21

.

The sum of the equal roots = 2 + 2 = 4 Choice (1) 34. The equations in choices (1) and (3), satisfies the

condition f (x) =

x1

f Choice (4)

35. 2x3 + 4x2 + 8x + 4 = 0 Let α, β and γ be the roots of the equation.

α + β + γ = 224 −=−

αβ + βγ + γα = 428 =

α2 + β2 + γ2 = (α + β + γ)2 − 2(αβ + βγ + γα) = (− 2)2 − 2(4) = 4 − 8 = − 4. Choice (2) 36. x3 + 3x2 − 4x + 5 = 0 Let α, β and γ be the roots of the equation α + β + γ = − 3 αβ + βγ + γα = − 4 αβγ = − 5 (α + β + γ) (α2 + β2 + γ2 − αβ − βγ − γα) = α3 + β3 + γ3 − 3αβγ (− 3) ((α2 + β2 + γ2) − (− 4)) = α3 + β3 + γ3 − 3 (− 5) ⇒ − 3 (α2 + β2 + γ2) − 12 = α3 + β3 + γ3 + 15 ⇒ − 3 [(α + β + γ)2 − 2(αβ + βγ + γα)] = α3 + β3 + γ3 + 15 + 12 ⇒ − 3 [(− 3)2 − 2 (− 4)] = α3 + β3 + γ3 + 27 ⇒ −27 − 24 − 27 = α3 + β3 + γ3 ⇒ α3 + β3 + γ3 = − 78. Choice (3) 37. x3 − 3x2 + 3x − 1 = 0 Let α, β and γ be the roots of the equation above α + β + γ = 3 αβ + βγ + γα = 3 αβγ = 1

The sum of the roots of the required equation = αβ + βγ + γα = 3

The sum of the product of the two roots = αβ2γ + βγ2α + γα2β = αβγ (α + β + γ) = 1(3) = 3 The product of the roots = αβ βγ γα = α2β2 γ2 = (1)2 = 1 The required equation is x3 − 3x2 + 3x – 1 = 0


Alternate method: Let α, β, γ be the roots of the equation. α + β + γ = 3; αβ + βγ + γα = 3 ; α β γ = 1

Let y be αβ ⇒ y = γ

αβγ = γ = 1/y

Since γ is the root of the equation y3

y

3

y

123

+− – 1 = 0

⇒ y3 – 3y2 + 3y – 1 = 0 ∴ The equation is x3 – 3x2 + 3x – 1 = 0 Choice (3) 38. The roots of any equation having rational coefficients

will always occur as conjugate pairs if any of them is irrational or complex. ∴ The lowest possible degree = 4

i2and3 − are roots of the equation, 3− and − 2 i are also roots of the equation.

Let ,3,3 −=β=α

γ = 2 i and δ = – 2 i α + β + γ + δ = 0 αβ + βγ + γδ + δα + βδ + αγ = αβ + γδ + (α + β) (γ + δ) = − 3 + 4 + (0) (0) = 1 αβγ + βγδ + γδα + δαβ = αβ (γ + δ) + γδ (α + β) = 0 + 0 = 0 αβγδ = (− 3) (4) = − 12 x4 − x3 (α + β + γ + δ) + x2 (αβ + βγ + γα + αδ + αγ + βδ) − x (αβγ + βγδ + γδα + δαβ) + αβγδ = 0 x4 − x3 (0) + x2 (1) − x (0) − 12 = 0 x4 + x2 − 12 = 0. Choice (3) 39. x3 − 3x2 – 3x − 1 = 0

Given that 32 − is a root of the equation

⇒ 32 + is also a root of the equation

Let the third root be α

The sum of the roots = 33232 =α+−++

4 + α = 3 α = − 1 Choice (4) 40. x4 – 10x3 + 29x2

− 22x + 4 = 0

Since 53 + is a root of the equation, 3 − 5 is also a root of the equation. Let the other roots be α and β with α > β the sum of the

roots = 105353 =β+α+−++

α + β = 4 ….. (1) The product of the roots

= ( )( ) 45353 =αβ−+

44=αβ = 1

(α – β)2 = (α + β)2 − 4αβ (α – β)2 = 42 − 4 = 12

α − β = 32 (α > β) …. (2) Adding (1) and (2),

2α = 324 +

32 +=α ⇒ 32 −=β . Choice (4)

41. 2x2 + 2 |x| + 1 = 0 |x| = ± x 2x2 + 2x + 1 = 0 b2 − 4ac = 4 − 4 (2) (1) = − 4 < 0

∴ The roots are imaginary ∴ The number of real roots are zero. Choice (1)

42. 081

6x

x41

x 23 =−+−

The roots which are n times the roots of the equation above will satisfy

081

nx

61

nx

41

nx

23

=−

+

−

08

nn.x

61

nx41

x3

223 =−+−

If n = 12,

08)12(

)12(6x

)12(4x

x3

22

3 =−+−

x3 − 3x2 + 24x − 216 = 0 Choice (2) 43. x3 − 9x2 + 26x − 24 = 0 Let α, β and γ be the roots of the equation. Let α: β = 1: 2

21=

βα

⇒ β = 2α

α + β + γ = 9 3α + γ = 9 --- (1) αβ + βγ + γα = 26 2α2 + 2αγ + γα = 26 2α2 + 3γα = 26 α(2α + 3γ) = 26 --- (2) αβγ = 24 2α2γ = 24. --- (3) from (1) we have γ = 9 − 3α, substituting this in (2), we have α(2α + 3γ) = 26 α(2α + 27 − 9α) = 26 − 7α2 + 27α = 26 7α2 − 27α + 26 = 0 7α2 − 13α − 14α + 26 = 0 α (7α − 13) − 2 (7α − 13) = 0 (α − 2) (7α − 13) = 0

α = 2 or 7

13

But 7

13 satisfies the equation.

α = 2 ⇒ γ = 3 The roots are 2, 4 and 3. Choice (4)

44. x3 + 3ax2 + 3bx + c = 0

Substitute x1

for x.

0cxb3

x

a3

x

123

=+++ ⇒ cx3 + 3bx2 + 3ax + 1 = 0 ….. (1)

⇒ The roots of (1) are in A.P Let the roots be α − β, α and α + β

The sum of the roots = α − β + α + α + β = cb3−

cb3

3−=α

cb−=α --- (2)

Since α is the root of (1)

∴ c

−+

−+

−cb

ac3cb

b3cb

23

+ 1 = 0

–cab3

c

b3

c

b2

3

2

3−+ + 1 = 0

2b3 – 3abc + c2 = 0. Choice (4)


45. The given roots are i2and23 +− The conjugates of each of the given roots are also roots of the required equation.

∴ i2andi2,23,23 −++− are the roots of the required equation. The required equation is

(x − (3 − 2 )) (x − (3 + 2 )) (x − (2 + i)) (x − (2 − i)) = 0 i.e. (x2 − 6x + 7) (x2 − 4x + 5) = 0 i.e. x4 − 10x3 +36x2 − 58x + 35 = 0 Choice (1)

Exercise – 5

Solutions for questions 1 to 100: 1. A1 × 5 has only one row. ∴ It is a row matrix Choice (3) 2. Clearly, the given matrix is a diagonal matrix. Choice (1) 3. Only choice (4) is the Identity matrix. Choice (4) 4. Trace of a matrix = Sum of the principal diagonal

elements of a square matrix = 1 – 1 + 5 = 5. Choice (1)

5. Trace is defined only for square matrices. Choice (4) 6. Clearly if the order of BA is ℓ × n then the order of (BA)T

is n × ℓ. Choice (2) 7. The given two matrices are not equal because a12 = 0

in 1st matrix and a12 = 3 in second matrix. ∴ We cannot compare. ∴ The given data is inconsistent. Choice (4) 8. We can find determinant only for square matrices.

∴ p = q. Choice (3) 9. Determinant of new matrix = 2 times the determinant of

original matrix. Choice (3) 10. We know that, the determinant of a matrix is the sum of

products of elements of any row or column with the corresponding co-factors. Choice (2)

11. In option (3), each element of 2nd row is multiplied by

the corresponding co-factors, so its sum is ∆ but not zero.

∴ Option 3 is false Choice (3)

12. Adjoint of the matrix

dc

ba is

−−ac

bd

Adjoint of the given matrix is

−−

23

14.

Choice (4)

13. Given,

x2

8x is singular

⇒ x2 – 16 = 0 ⇒ x = ± 4 Choice (4)

14. Consider, A2 = Ι=

=

−

−10

01

10

21

10

21

A2 = Ι ⇒ A is an involuntary matrix. Choice (3)

15. Det A = 2 + 4 ≠ 0 ∴ Rank of A is 2 Choice (2) 16. Since the normal form contains I4, rank is 4.

Choice (2) 17. It is clear that crammer’s rule is applicable when m = n

and co-efficient matrix is non-singular. Choice (4) 18. The system of equations are said to be consistent if

there exists at least one solution. Choice (1) 19. Clearly option 3 is false. Choice (3)

20. Given AT = B ⇒

=

−dc

ba

43

12

⇒ a = 2, c = 3 Choice (1) 21. Two matrices are comparable if they are of the same

order. ∴ m = r, n = s Choice (2) 22. AB and BA both are defined. Choice (1) 23. Since A and B are square matrices of the same order,

all the three options are true. Choice (4) 24. Trace is defined only for square matrices. ∴ Option 3 is false. Choice (3)

25. AB = ( )

−−0

1

2

321 = (– 1 × 2 + 2 × – 1 + 3 × 0 = [– 4]

Choice (2)

26. AB = ( )zyx

cfg

fbh

gha

= ( )czfygxfzbyhxgz hy ax ++++++

ABC = ( )zcfygxfzbyhxgz hy ax ++++++

z

y

x

= ax2 + by2 + cz2 + 2hxy + 2gxz + 2fyz Choice (1)

27. A = 21

(A + B + A – B)

=

=

−−

+

− 4

2

60

010

21

1

1

32

16

5

1

32

14

21

∴ A =

230

105 Choice (4)

28. A = ( )C3B251 −

=

−−−

−

−−

=1020

1010

5

1024

612

104

42

5

1

∴ A =

24

22 Choice (3)

29. Given,

++

=

−−−

2a2

43x

15z

16y47

⇒ x + 3 = 7, 4y – 16 = 4, z – 5 = 2 and a + 2 = – 1 x = 4, y = 5, z = 7, a = – 3 Choice (3)


30. Given,

−−−

−

xx22

02x

x3x2

122

101

212

= I3 × 3

⇒ – 4x + x + 4 = 1 ⇒ – 3x = – 3 ⇒ x = 1 Choice (2) 31. Aij = (– 1) i+j aij

Co-factor of the element 6 (a32) is (– 1)I + 2 21

42

−−−

= (– 1)3 + 2 44 −− = – (– 8) = 8 Choice (2)

32. Given, A =

=

=⇒

00

01

00

01

00

01A

00

01 2

Also B =

=

=⇒

10

00

10

00

10

00B

10

00 2

A + B =

10

01 A – B =

−10

01

(A + B )2 =

=

10

01

10

01

10

01

A2 + B2 =

10

01

∴ (A + B)2 = A2 + B2

Now (A – B)2 =

−10

01

−10

01 =

10

01 = A2 + B2

Q AB =

00

00, we can have (A + B)2

= A2 + B2 + 2AB All the three options are true. Choice (4)

33.

y

x

bh

ha =

++

byhx

hyax

Now (x y)

y

x

bh

ha

(x y)

++

byhx

hyax

= ax2 + hxy + hxy + by2 = ax2 + 2hxy + by2 Choice (3) 34. (A + B)3 = ( )( )[ ]BABA ++ (A + B)

= (A2 + AB + BA + B2) (A + B) = A3 + (AB) A + (BA) A + A2 B + (AB) B + (BA) B + B3 = A3 + (ABA) + BA2 + B2 + B2A + A2 B + AB2 + BAB + B3 Choice (4)

35. Clearly, we know that matrix multiplication is not

commutative hence first three options are false and also we know that matrices satisfy distributive property.

∴ P(Q + R) = PQ + PR Choice (4)

36. A – 2Ι =

−−−

−=Ι−

122

021

132

3B,

122

101

212

(A – 2Ι) (B – 3Ι) =

=

−−−

−

100

010

001

122

021

132

122

101

212

= 1 Ι3 × 3

Choice (2)

37. A – Ι =

−−014

103

101

A(A – Ι) =

−−114

113

102

−−014

103

101

=

−−−

−

515

2110

212

Choice (4)

38. Given: A =

122

101

212

; A2 =

122

101

212

122

101

212

=

748

334

769

f(A) = A2 – 4A – 3Ι =

748

334

769

– 4

122

101

212

–

300

030

003

=

−−−−

040

100

122

Choice (3) 39. (A + B)2 = A2 + B2 is possible when AB = – BA

AB =

−12

ba

−14

11

=

−−+

=

+−−+

32

bab4a

1242

bab4a

BA =

−

− 12

ba

14

11 =

+−−+

1b42a4

1b2a

AB = – BA

−−+−+−−−

=

−−+

1b42a4

1b2a

32

bab4a

– 4a + 2 = – 2 – 4b – 1 = 3 – 4a = – 4 – 4b = 4 a = 1 b = – 1 Choice (2) 40. Given: AI = – A ⇒ A is a skew-symmetric matrix in a

skew-symmetric matrix all the diagonal elements are equal to zero. ∴ x = 0 Choice (4)

41. Given: AT = A ⇒ a + b = 3 and a – b = 1 Solving the above equations, we have a = 2, b = 1 a – 2b = 2 – 2(1) = 0 Choice (1)

42. Given: A =

ωωω22

1 AT =

ωωω

2

21

AAT =

ωω+ωω+ωω+=

ωωω

ωωω

432

322

2

2

22 2

111

But we know, 1 + ω + ω2 = 0 and ω3 = 1

⇒ AAT =

ωω−ω−ω−

2 Choice (3)


43. AAT =

=

αα−αα

ααα−α

10

01

cossin

sincos

cossin

sincos = Ι2 × 2

Choice (4) 44. The addition or subtraction of matrices is possible only

when they are of the same order, here 2A is of 2 × 2 and BT is of 3 × 2.

∴ 2A – BT is not possible. Choice (4) 45. Clearly options (1) and (2) are true. Choice (4) 46. In the given matrix 2R3 = R1. We know that in any square matrix, if two rows are

equal or one is the multiple of the other, then its determinant is zero. Choice (3)

47. Clearly C3 = – C2 ∴ The determinant value is zero. Choice (1)

48. M32 = 41

03

− = 12 Choice (2)

49. Aij = (–1)i + j mij

A23 = (–1)2 + 3 83

52

−−

= – (–1) = 1 Choice (1)

50. We know, (AT)T = A and adj(adjA) = A for individual

matrices. det (KA) =Kn detA ∴ Option 3 is false for any matrix. Choice (3)

51. Given: A =

25

13

Det A = 3(2) – 5(1) =1.

A = 1

We know that, A =

dc

ba then

A–1 =

−−ac

bd

Adet1

∴ A–1 =

−−35

12 Choice (2)

52. A = xq – py

Q Given xq ≠ py, so A is non-singular

∴ A–1 =

−−

−=

xp

yq

pyxq1

adjAA1

Choice (4)

53. adj (A) =

θθθ−θ

cossin

sincos Choice (1)

54. A = 36 – 36 = 0

∴ A is a singular matrix. ∴ A–1 does not exist ∴ A–1 cannot be determined. Choice (4) 55. The given statement is true. Choice (1) 56. Expanding along the 1st row we have 2(6 + 4) + 1(– 9) + 4(– 12) = 20 – 9 – 48 = – 37

Choice (2)

57. R1 → R1 + R2 + R3 ~

1

1

111

2

2

222

ωωωω

ω+ω+ω+ω+ω+ω+

1

1

000

2

2

ωωωω= = 0

(Q 1 + ω + ω2 = 0) Choice (1)

58. 2

2

2

cc1

bb1

aa1

R2 → R2 – R1 ~ 22

22

2

acac0

abab0

aa1

−−−−

= (b – a) (c – a)

ac10

ab10

aa1 2

++

= (b – a) (c – a) (c + a – b – a) = (b – a) (c – a) (c – b) = (a – b) (b – c) (c – a) Choice (3)

59.

bcabca

abcabc

cabcab

−−−−−−−−−

R1 + R2 + R3 ~

acabca

abcabc

000

−−−−−− = 0

Choice (4)

60. C1 + C2 + C3 ~

1ababbcac1

1caacbabc1

1bcacbcab1

+++++++++

= (1 + ab + bc + ac)

1ab1

1ca1

1bc1

= 0

(∵ c1 ≈ c3) Choice (1)

61. ~

( )( )( ) ( )( )

( ) ( )( )!2n3n)4n(!2n)3n()!2n(

!1n2n)3n(!1n2n!)1n(

!n1n)2n(!n)1n(!n

++++++++++++

+++

~ n! (n + 1)!(n + 2)!

( )( )( )( )( )( )3n4n3n1

2n3n2n1

1n2n1n1

+++++++++

R2 → R2 – R1 and R3 → R3 – R2

~ n! (n + 1)! (n + 2)!

( )( )( )( ) 23n10

22n10

1n2n1n1

++

+++

= n! (n + 1)! (n + 2)! {2(n + 3 – n – 2)} = 2 n! (n + 1)! (n + 2)! Choice (2)

R3 → R3 – R1


62. R1 → R1 + R2 + R3 ~

( ) ( ) ( )

accbba

cbbaac

cba2cba2cba2

++++++

++++++

R2 → R2 – R1 and R3 → R3 – R1

∼ 2

bac

acb

cbacbacba

−−−−−−

++++++

R1 → R1 + R2 + R3 = 2

bac

acb

cba

Choice (2)

63. Put x = 0 ⇒

0cb

c0a

ba0

−−−

= abc – abc = 0 Choice (1) 64. R2

→ R2 – R1

R3 → R3 – R1 ~

1086

432

4x2x1x +++

R3 → R3 – 3R2

210

432

4x2x1x

−−

+++

= (x + 1) {– 6 + 4} – (x + 2) {– 4} + (x + 4) {– 2} = – 2x – 2 + 4x + 8 – 2x – 8 = – 2 Choice (3) 65. Given: A is a square matrix of order K. We know, det (KA) = Kr det(A), where r is order of ‘A’. ∴ KK = 27 ⇒ 33 = 27 ∴ K = 3 Choice (4) 66. R2 → R2 – R1

R3 → R3 – R1 ~

666

333

200320022001

Q R3 = 2R2 ⇒ det value = 0 Choice (1)

67. Given:

121

3x1

242

has an inverse

⇒

121

3x1

242

≠ 0

Clearly R1 = 2R3 ⇒ det = 0 for any value of x. ∴ for no value of x the inverse exists ∴ The set of real values of x is null set. Choice (4) 68. For the given matrix, the inverse does not exist. ⇒ If it is singular ⇒ its determinant value = 0 ⇒ 1(2x – 56) – 5(4 – 42) + 7(8 – 3x) = 0 ⇒ x = 10 Choice (3) 69. The given matrix satisfies A = – AT. ∴ it is skew-symmetric. Choice (2) 70. Clearly Idempotent matrices do not satisfy the relation

A2 = Ι . Choice (3)

71. A2 =

−−−

−−

−−−

−−

321

321

321

321

321

321

= O3 ×3

∴ A is nilpotent. Choice (2)

72. Clearly, the given matrix satisfies Aθ = ( )TA = A

⇒ A is skew -Hermition matrix. Choice (4)

73. A–I =

−−

− ac

bd

bcad1

=

−+ 23

41

1221

=

−23

41

141

Choice (1)

74. Clearly the product of the given matrix with the matrix in

2nd option results in a unit matrix. ∴ The inverse of the given matrix is option – 2.

Choice (2) 75. Q The order of matrix A is 3 × 3 and we do not know

the value of its determinant. We can say the rank of the matrix is ≤ 3. Choice (4)

76. In the given matrix, clearly R2 = – 2R1 and R3 = – 5R1 ⇒ det of the given matrix = 0 and det of all 2 × 2 minors

also becomes zero. ∴ Rank is 1. Choice (2) 77. Since the given matrix is of order 2 × 3 and there exits a

minor of order 2 × 2 whose value is not zero. ∴ The rank of the matrix is 2. Choice (2)

78. Given: A = ( )321− and B =

5

4

3

and AB = (20) is of

order 1 × 1. ∴ The rank of AB is 1. Choice (2) 79. If the rank of the matrix is ‘n’, then the linear

homogeneous system of equations in n variables will have only one solution, x = 0

∴ The rank of the matrix = n Choice (4) 80. The given system of non-homogeneous equations is

consistent only when the rank of coefficient matrix A is the same as the rank of the augmented matrix.

∴ ρ(A) = ρ [A B] Choice (1) 81. In the given equations, we notice the rank of augmented

matrix and the rank of coefficient matrix and the number of unknowns equal the number of equations.

∴ There exists a unique non-zero solution. Choice (1) 82. The characteristic equation of the given matrix is

λ−−

λ−12

41 = 0 ⇒ – (1 – λ2) – 8 = 0 or λ2 = 9

⇒ λ = ± 3 for λ = –3, the eigen vectors are

( )

=

−−

− 0

0

x

x

10

013

12

41

2

1

⇒ 4x1 + 4x2 = 0 and 2x1 + 2x2 = 0 ⇒ x1 = – x2 ∴ The eigen vectors corresponding to λ = – 3 are

x =

−=

−=

1

1x

x

x

x

x2

2

2

2

1 Choice (2)


83. Given A =

=

100

010

001

A,

001

010

1002

We notice A3 = A2A = A, A4 = (A2)2 = Ι. ∴ A4 – A3 – A2 = Ι – A – Ι = –A Choice (3)

84. Clearly the rank of the matrix

111

111

111

is one

∴ m = 1

The determinant of the matrix

001

010

100

= (– 1) = – 1 ∴ The rank of the matrix is n = 3 ∴ 3 = 1 + 2 ⇒ n = m + 2 Choice (2)

85. The characteristic equation of the given matrix is

Ιλ−A = 0

⇒

λ−λ−

λ−

987

654

321

= 0

⇒ λ3 – 15λ2 – 18λ = 0 ⇒ λ = 0 is a root of it. Choice (4) 86. Since the eigen values of an upper triangular matrix are

the diagonal elements of the matrix. ∴ The eigen values of the given matrix are 2, 1 and 3. Choice (3) 87. The determinant of the given matrix ≠ 0. ∴ The rank of the matrix = 3 Nullity = order – rank = 3 – 3 = 0 Choice (3) 88. In the given matrix, there exists a. 3 × 3 minor whose

determinant is non zero. Example

321

203

301

−= – 4 + 18 ≠ 0

∴ The rank of A = 3 Choice (2)

89. R1 → R1 + R2

i320

1i34

00i64

−+

= (4 + 6i) {– 3 + 3} = 0 = x + iy

⇒ x = 0, y = 0 ⇒ x2 + y2 = 0 Choice (1) 90. Augmented matrix of the given system of equations is

2151413

1876

2321

R2 – 6R1

R3 – 13R1 ~

−−−−−−2424120

111050

2321

clearly observing

R2 and R3, we can notice rank of A is not the same as rank of [A B]

⇒ The system of equations are inconsistent. Choice (3)

91. For the equations to have a unique solution.

0A ≠ ⇒ 0

101

3k1

120

≠−

⇒ – 2(– 1 – 3) + 1 (– k) ≠ 0

⇒ 8 – k ≠ 0 ⇒ k ≠ 8 Choice (2)

92. A = – 7

We know, A(adjA) = (adjA) A = A Ι

∴ A(adjA) = – 7Ι Choice (3) 93. Standard result Choice (4) 94. If the given equations have a non-zero solutions

⇒ A = 0 ⇒

k23

23k

112

−−

−− = 0 ⇒ k2 – 8K + 7 = 0

⇒ (k – 1) (k – 7) = 0 ⇒ k = 1, 7 ∴ If k = 1 or 7, then the rank of A is less than the

number of variables ⇒ The equations have non-zero solution. Choice (4) 95. The characteristic equation of the given matrix is

0

221

131

122

0A =λ−

λ−λ−

=Ιλ−

⇒ λ3 – 7λ2 + 11λ – 5 = 0 The characteristic roots are 1, 1, 5 ∴ The characteristic roots of A–1 are reciprocals of 1, 1, 5. They are 1, 1, 1/5. Choice (3) 96. We know that by caley Hamilton theorem every square

matrix satisfies its own characteristic equation. The characteristic equation of the given matrix A is

Ιλ−A = 0 ⇒

λ−λ−

λ−

312

132

226

= 0

⇒ λ3 – 12λ2 + 36λ – 32 = 0 ⇒ A3 – 12A2 + 36A – 32Ι = 0 Choice (4)

97. Clearly the characteristic equation of the given matrix

is option (4) Choice (4)

98. Given A =

79

172, we know that every square Matrix

can be expressed as the sum of symmetric and skew symmetric matrices,

i.e., A = 21

(A + AT) + 21

(A – AT)

Here, P = 21

(A + AT) is symmetric,

Q = 21

(A – AT) is skewsymmetric

∴ P = 21

(A + AT)

= 21

+

717

92

79

172=

1426

264

21

=

713

132

Choice (1) 99. Since, all of them are the standard properties of

determinants, all the statements are true. Choice (4)


100. If the system of equations have no solution, then det A = O

Here, A =

−

−

111

k31

324

detA = 4(– 3 – k) – 2(– 1 – k) – (– 3) O = – 12 – 4k + 2 + 2k + 6 – 2k – 4 = 0 2k = – 4 k = – 2. Choice (1)

Exercise – 6 Solutions for questions 1 to 45: 1. We know that in the expansion of (x + y)n,

Tr+1 = nCr xn–r.yr

T5 = T4+1 = 7C4 x7– 4

4

3y2

−

81y16

.x354

3= = 43yx81

560

Choice (2) 2. Here, n = 8 which is even.

∴ There exists only one middle term which is

128

12n tt

++= = t5

t5 = 8C4(2x)8-4.48

44

2 x81

1120

x.81

x16.70

x3

1 ==

−= 4x

811120 −

Choice (3) 3. If Tr+1 has the coefficient of xm in the expansion

n

qp

x

1x

+ , then r = qpmnp

+−

Here, n = 6, p = 2, q = 3, m = −3. ∴ r = 32

)3(26+

−−×= 3

∴ T4 contains x–3

∴ T4 = 6C3(x2)6–3

39

63

3x

2720

x.27

x20

x3

1 −−=−=

−

∴ The coefficient of x–3 is = 2720−

Choice (4)

4. Since ‘n’ is odd, there exists two middle terms; they are

T5 and T6.

T5 = 9C4bax126

axb

bax

45

=

−

and

T6 = 9C5ax

b126ax

bbax

54 −=

−

Choice (1)

5. If the independent term of the expansion n

qp

x

1x

+ is

Tr+1, then r = 2

21

2.10qp

np

+=

+= 8

∴ T8+1 = T9 is independent

and T9 = 10C8(x2)10-8

8

2

1

x

2

− = 45 × 28

Choice (3)

6. Given: The independent term of the expansion

9

2

x1

kx

− is 2

567−.

If Tr+1 is independent, then r = qp

np+

,

here, n = 9, p = 2, q = 1

∴ r = 3

29 ×= 6 ⇒ T7 = 9C6 (+k)3 =

2567−

⇒7x8x9x23x2x567

k2

567)k(

321789 33 −=⇒

−=+××××

∴ k3 = 3

23

−⇒ k =

23−

Choice (2)

7. If Tr+1 is numerically greatest then p = 1|x|

|x|)1n(

++

Here, given (3 − 2x)6 = 36

6

x32

1

−

∴ n = 6, |X| = 32

x.

Given: x = 2 ⇒ |X| = 34

∴ p = 1

34

34

)16(

+

+= 4

Q p is an integer T4 and T5 are numerically greatest.

T3+1 = 36.6C3 3

x32

−

,34

34

34

3x2x14x5x6

.36

−

−

−= ( )2x =Q = − 33 × 44 × 5

∴ The value of the numerically greatest term of the expansion is 33 × 44 × 5 Choice (3)

8. Given: T3 is numerically greatest

⇒ 1

x54

2.C

x54

2.C1

T

Tand1

T

T1

71

8

26

28

3

4

2

3 ≥

⇒≤≥

⇒

∞∪

−∞−∈⇒≥ ,75

75

,x1x57

and

1x54

x54

2C

x54

2C1

TT

26

28

35

38

3

4 ≤=

⇒≤

∴ The range of x is x ∈

∪

−−45

,75

75

,45

Choice (4)

9. The number of terms of the expansion (ax + by + cz)n is

given by 2

)2n)(1n( ++.

Here, n = 15 ∴ The number of terms in the given expansion is

2)215)(115( ++

= 136 Choice (4)


10. Given (1 + x)nn

x1

1

−

= (nC0 + nC1x +….+ nCrxr +….+ nCnx

n)

−++−n

n1

n0

n

x

1)1(...

x1

CC

In the above product x-n is obtained by multiplying 1st term in the first expansion with the last term of the

second expansion = nnn

n0

n x.)1(x

1.)1(C −−=− = − 1

(∵ n is odd) Choice (2) 11. The number of terms in the expansion (x + y)n + (x − y)n

is 2n + 1 when n is even and

21n +

when n is odd.

Here, n = 20

∴ The Number of terms = 2

20 + 1 = 11 Choice (3)

12. In the given expansion T6 and T7 will have x5 and x6 and given coefficients of T6 and T7 are equal numerically.

⇒ nC5(3)n–5 6

6n6

n5

21

.)3(C21

=

−

nC5(3) = nC6 21

61

5n6

61

C

C

6n

5n

=−

⇒=

⇒ n = 41. Choice (4)

13. Given:

−

−=

−

+x31

x2x9

1x4

x31

x2x31

x26

22

76

Clearly, the product does not have constant or independent term of x. ∴ The coefficient of independent term is 0 Choice (4)

14. The general term of the expansion is Tr+1

= 160Cr

r

41)r160(

52

b3a

−

= r4r

5r2

64

r160 3)b(aC

−

Clearly, Tr+1 is independent of radical sign if

4

rand

5

rare integers, where 0 ≤ r ≤ 160

∴ r = 0, 20, 40, 60, 80, 100, 120, 140, 160 ∴ 9 terms will have rational powers. Choice (3)

15. Let ( )813 + = Ι + F where Ι ∈ N and 0 < F < I

Clearly, Ι is the greatest integer less than or equal to

( )813 +

Let H = ( )813 − , then 0 < ( )813 − < 1; ⇒ 0 < H < 1

I + F + H = ( ) ( )881313 −++

= 2 ( ) ( ) ( )

+++−−

...3C83C83C848

428

28

0

= an even integer. ∴ F + H = 1. Since Ι is an interger and 0 < F < 1; 0 < H < 1, ∴ Ι + 1 = 2[81 + 28.27 + 70.9 + 28.3 + 30] ⇒ I = 3103 Choice (2)

16. The ratio of the coefficients of T4 and T6 in the

expansion6

2x

32x

− is 6C3 ( ) 55

633

)3(21

C:321 −

−

⇒ 20 533

)3(21

6:)3(21 −

−

⇒ 5 : 54 Choice (3)

17. (x + 1)(x + 2)(x + 3)….(x + 25) = x25 + (1 + 2 + … + 25)x24

+ (1.2 + 2.3 + … + 24.25)x23+ … + 1.2.3….25

∴ The coefficient of x24 = 1 + 2 + 3 +…+ 25 =2

)125(25 +

= 325 Choice (2) 18. Put x = y = 1 in (3x + 2y)8 = 58 ∴ The sum of coefficients is 58 Choice (2) 19. Put x = 1 in (1 + 2x − x2)10 = 210 ∴ The sum of coefficients is 210 Choice (1) 20. The sum of coefficients in the given binomial expansion is 2n

Given: 2n = 1024 ⇒ n = 10

The greatest coefficient exists in th

12n

+ term, i.e. T6

The coefficient of T6 is 10C5 Choice (4)

21. The sum of coefficients of even powers of x in (1 + x)n is 2n-1

Here, n = 30. ∴ The required sum of coefficients is 229 Choice (3)

22. 4

1

4

1

4

14

811

1)81()181(80

−=−=

= 3

+

++− .........

811

!2

141

41

4x811x1

12

=3

−+− ....!2)324(

5.1

324

11

2= 3 (1 − 0.003 – 0.00001}

= 3 × 0.99705 = 2.9912 Choice (2)

23. (1 + 0.01)8 + (1 − 0.01)8 = 2(8C0 + 8C2(1)6(0.01)2+ 8C4(1)4(0.01)4+ 8C6(1)2 (0.01)6 + (0.01)8} = 2(1 + 28(0.0001) + 70(0.00000001) + …) = 2(1 + 0.0028) = 2.0056 Choice (1)

24. 2

122

12 )9x()16x( +−+ =

21

2212

12

21

9x

1916x

116

+−

+

+−

+2x9

x13

2x16x

1422

(Neglecting higher order

terms in the expansions because given x is small)

24x

12x9

x32x16

x41

222

−=−+ Choice (3)

25. The expansion 5

2

3x5

1

−

− is valid only when

13x5

113x5 <<−⇒<

5

3x

5

3 <<−⇒ ⇒ |x| <

53

Choice (4)


26. Given: 51n − 1 ⇒ (50 + 1)n − 1; clearly ∀ n > 1, n ∈ N 51n − 1 is always divisible by 5, 2 and 6. ∴ All the above are true. Choice (4)

27. (1 − x + x2)–2 = ( ) ( ) 23223

x1x1x1x1 −

−

++=

++

= (1 + 2x + x2)(1 − 2x3 + 3x6 − 4x9 + 5x12 +……) The coefficient of x6 in the above expansion is = 3 Choice (4)

28. (1 − 2x + 3x2 − 4x3 + …) 2

1

= ( )212)x1( −+

(∵ (1 + x)–2 = 1 – 2x + 3x2 + …… 1)

= (1 + x)–1 = 1 − x + x2 − x3 + x4 − …. ∴ The Coefficient of x4 is 1 Choice (2) 29. Let (r + 1)th term of the expansion is negative in the

expansion (1 + x) 2

5

⇒ Tr+1 = 2x.!r

1r25

...225

125

25

+−

−

−

The above is negative, if 25 − r + 1 < 0 ⇒ − r <

27−

r > 27⇒ r = 4 ⇒ T5 is the first negative term

T5 = 4x..!4

325

225

125

25

−

−

−= 4x

1285−

Choice (3)

30. 11n − 10n − 1= (10 + 1)n − 10n − 1 ⇒ nC01

n + nC110 + nC2102 + … + nCn10n − 10n − 1 ⇒ nC2102 + nC3103 + … + nCn10n = 102(nC2 + nC310…+ nCn10n-2 ) Which is of the form 102.l where l = nC2 + nC310 +…

nCn10n-2 which is always divisible by 100. Choice (3)

31. The coefficients of xn in the expansions (1 + x)2n and

(1 + x)2n-1, a = 2nCn and b = 2n−1Cn

∴ a = n.nn2

n2∠−∠

∠and b =

n.n1n21n2

∠−−∠−∠

1n.1n.n

n21n2n.1n

xn.n

1n2.n2ba −∠

−∠=

−∠∠−∠

∠∠−∠=

2ba = ⇒ a = 2b Choice (2)

32. We know, C0 + C2 + C4 + … = C1 + C3 + C5 + …

i.e., The sum of even coefficients = sum of odd coefficients = 2n–1 Choice (2)

33. We know C1 + 2C2 + 3C3 + … + nCn= n2n-1 adding C0, i.e. 1 on both sides we get C0 + C1 + 2C2 + 3C3 + … + nCn = n.2n-1+1

Choice (4) 34. We know that aC0 + (a + d)C1 + (a + 2d)C2 + (a + nd)Cn

= (2a + nd)2n-1 Here, a = 2, d = 3. Substituting, we have

2C0 + 5C1 + 8C2 + …+ (3n + 2) Cn = (4 + 3n)2n-1

Choice (1)

35. 2C0 − 3C1 + 4C2 − 5C3 + … +(−1)n(n + 2) Cn

= ∑ ∑ ∑ −+−=+−=

n

0n

n

0

n

0n

nn

nn

n C)1(2nC.)1(C)2n()1(

= 0 + 0 = 0 Choice (1) 36. (C0 + C1) (C1 + C2) . (Cn–1 + Cn) =

+

+

+

+

−−

1n

n1n

2

32

1

21

0

10

C

C1C.......

C

C1C

C

C1C

C

C1C

⇒

+

−−−+

−++n1

1...)1n(n!3

!2).2n)(1n(n1

2.n)1n(n

1)n1(

C0. C1. C2 …….Cn–1

⇒ ( ) ( ) ( ) ( )

n1n

...3

1n2

1n1

n1 ++++ C0C1C2…Cn-1

⇒ C0C1C2…Cn-1.Cn!n

)1n( n+ ( )1Cn =Q Choice (3)

37. The sum of coefficients in the expansion (a0 + a1x + a2x

2 + anx

n)m is obtained by putting x = 1 in (1 + 2x − 3x2 + 4x3 − x4)10 = (1 + 2 − 3 + 4 − 1)10 = 310

∴ The sum of coefficients is 310 Choice (3) 38. The sum of coefficients of even powers of x is given by

2

)1(f)1(f −+ ⇒ f(1) = 310 and f(− 1) = (− 9)10 = 910 = 320

The sum of coefficients of even powers of x is

2

33 2010 +

=2

)31(3 1010 + Choice (2)

39. The sum of coefficients of odd powers of x in the

expansion is given by 2

)1(f)1(f −−

Here, f(1) = 0 and f(− 1) = 48. ∴ The sum of coefficients of odd powers of

x is2

40 8−= − 215

Choice (4)

40. Given: 1 + ...81

!3)41)(21(1

81

!2)21(1

81

32

+

+++

++

which is of the form

1 + ...qx

!3)q2p)(qp(p

qx

!2)qp(p

qx

p32

+

+++

++

Comparing, we have p = 1, q = 2 and 81

qx = ⇒ x =

4

1

∴ It can be written as 2

1

2

1

43

41

1

−−

=

− = 3

2

Choice (3) 41. Given: The series can be written as

1 + ...91

!3)62)(32(2

91

!2)32(2

92

32

+

+++

++

Here, p = 2, q = 3, 91

qx = ⇒ x =

3

1

∴ The series can be written as 3

2

31

1

−

− = 3

2

23

Choice (2)


B A

C

B A

42. 1 − ...41

!35.3.1

41

!23.1

41

32

+

−

+

Here, p = 1, q = 2 and 41

qx = ⇒ x =

2

1

∴ = 32

or23

21

12

1

2

1 −−

=

+ Choice (1)

43. Given: 36.27.18.9

10.7.4.127.18.97.4.1

18.94.1 +−

1 −9

11...

9

1

3

7.4.1

9

1

2

4.1

9

132

+−+

−

+

Here, p = 1, q = 3, 91

qx = ⇒ x =

31

⇒ 9

8

4

3

9

8

3

4

9

11

3

11

31

31

31

−

=−

=+−

+

−−

Choice (4)

44. From the 4th option we can notice

...327

!3)41)(21(1

327

!2)21.(1

327

.1132

+

+++

+++

Here, p = 1, q = 2, 327

qx = ⇒ x =

167

∴ Its sum is 21

21

21

916

169

167

1

=

=

+−−

= 34

Choice (4) 45. Given: (1 + 4x + 12x2 + 32x3 + … )–3 = (1 + 2(2x) + 3(2x)2 + 4(2x)3 + …)–3

We know that, 1 + 2x + 3x2 + 4x3 + …. = (1 − 2x)–2

The given expansion can be written as

( ){ } 632 )x21(x21 −=−−−

4th term of the expansion will have x3 ∴ T3+1 = 6C3(− 2x)3 = 20(− 8x3)

∴ The coefficient of x3 is −160 Choice (4)

Exercise – 7 Solutions for questions 1 to 17: 1. In the given set, there are only 4 elements. They are 1, {1, 2}, {3, 4} and 5. Choice (3) 2. Standard result: (B − A) ∪ B = B Choice (2) 3. A ∪ B = {2, 3, 4, 5, 6, 7, 8} and C = {2, 3, 4, 5, 6} ∴ (A ∪ B) ∩ C = {2, 3, 4, 5, 6} Choice (4) 4. Standard result: (A – B) ∩ A = A – B Choice (3) 5. By demogan's law, A – (B ∪ C) = (A – B) ∩ (A – C) Choice (3) 6.

In the figure = X and = Y

⇒ Y – X = Y Choice (1)

7. Standard result: A ∆ B = (A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B)

Choice (4)

8.

From the figure A – B = and B – C = ∴ (A – B) ∩ (B – C) = φ Choice (4) 9. From demorgan's laws, we have (A ∩ B)c = Ac ∪ Bc

∴ Option (3) is false. Choice (3)

10. Standard result:

A ∆ B = (A – B) ∪ (B – A) Choice (4) 11. We know that, A ∩ Ac = φ and A ∪ Ac = µ ∴ µ – (A ∪ Ac) = φ and (A – B) ∩ (A ∩ B) = φ but

A – Ac = A Choice (1) 12. n(A ∩ B) is maximum, when A ⊂ B. ∴ A ∩ B can have at most 5 elements Choice (3) 13. (0, 0), (1, 1) only satisfies both y = x2 and y = x ∴ n (A ∩ B) = 2 Choice (3)

14. When x = 1, 2x = 2, and x + x1

= 2 and

when x = – 1, 2x = – 2 and x + x1

= – 2

∴ n (A ∩ B) = 2 Choice (2) 15. When B ⊇ A ⇒ A ⊆ B and C ⊆ A ⇒ C ⊆ B Choice (4) 16. P = Q = R Choice (3) 17. A – B, A ∪ B, B – A are mutually disjoint sets. Choice (1) Solutions for questions 18 to 20: 18. A = {a, b, c, d}

The number of subsets of A containing a but not d is 22 = 4 Choice (2)

19. The number of ways of selecting r elements from

n elements is nCr The number of subsets that contain exactly 3 elements

is 4C3 i.e., 4. Choice (3) 20. Since A contains 4 elements, the number of subsets

that contain at least one element is 12n− i.e., 24 – 1 = 15

Choice (1)


n(µ) = 20 n(A) = 12 n(B) = 8

8 4 4

n(C) = 10 n(µ) = 20 n(V) = 12

5 5 7

n(µ) = 250

n(T) = 110

n(H) = 110

20 + x

40 + x

n(C) = 140

60 – x x

x 40 – x 50 – x

Solutions for questions 21 to 23: 21. Given: A – B = C – B and A ∩ B = C ∩ B ⇒ A = C Choice (3) 22. The number of proper subsets of any given set having n

elements is given by 2n – 1

∴The required number of proper subsets = 24 – 1 = 15 Choice (2) 23.

The number of persons in group A = 20 × 53

= 12

The number of persons in group B = 20 × 5

2= 8

The number of persons in both the groups = 20 × 51

= 4

n(A ∪ B) = n (A) + n (B) – n (A ∩ B) ⇒ n (A ∪ B) = 12 + 8 – 4 = 16

The number of persons belong to neither of the groups ⇒ {n (A ∪ B)}1 = n(µ) – n(A ∪ B) = 20 – 16 = 4 Choice (2)

Solutions for questions 24 to 26: 24.

⇒ n ( C ∪ V) = 20 – 3 = 17

n (C ∩ V) = n (C ) + n (V) – n (C ∪ V) = 10 + 12 – 17 = 5 ∴ The number of persons who play both the games = 5 Choice (4) 25. The number of persons who play exactly one game

= n (C ∪ V) – n(C ∩ V) = 17 – 5 = 12 Choice (1)

26. The number of persons who play at least one game is n (C ∪ V) = 17 Choice (2)


Let x be the number of families which subscribe for all the three news papers, i.e. n (C ∩ T ∩ H) = x

n(C ∪ T ∪ H) = 250 – 20 = 230 n(C ∪ T ∪ H) = n(C) + n (T) + n (H) – n(C ∩ T) –

n(C ∩ A) – n(T ∩ A) + n(C ∩ T ∩ H) 230 = 140 + 110 + 110 – 60 – 40 – 50 + x 20 = x Choice (4) 28. The number of families which subscribe for exactly one

newspaper = 60 + 20 + 40 = 120 (from figure) Choice (3)

29 The number of families which subscribe for exactly two

newspaper = 40 + 20 + 30 = 90 Choice (1) 30. The number of families which subscribe for at least two

news papers = 40 + 20 + 30 + 20 = 110 Choice (3) Solutions for questions 31 to 55: 31. The number of elements in A × B = 4 × 2 = 8 ∴ The number of relations from A to B is the same as

the number of subsets of A × B i.e.,= 28 Choice (2) 32. When n (A × B) = n (B × A) first three options need not be true. Choice (4) 33. If a set (A) has n elements, then the number of relations

from A to A is 2n2 , i.e.

242 = 216. Choice (4)

34. Clearly R is symmetric and anti-symmetric.

Choice (3) 35. Since n (A) = 6, then the number of elements in identity

relation on set A is 6 Choice (2) 36. If A = {1, 3, 5, 7}, then the maximum number of

elements in any relation is n (A × A) i.e., 4 × 4 = 16 Choice (3) 37. The relation > satisfies only transitive property. Choice (3) 38. The relation ⊇ does not satisfy symmetric property. Choice (2) 39. If R = R–1, then R is symmetric relation.

Choice (3) 40. Clearly, the given relation satisfies reflexive and

transitive properties. Choice (2)


A B

C

A B

C

A B

C

41. The relation x divides y satifies reflexive property since any number divides itself and also it is transitive since if x divides y and y divides z, then x divides z. But it will not satisfy symmetric since if x divides y then y need not be divisible by x. Choice (4)

42. Since the sum of a number added to itself, need not be

equal to 10, so it is not reflexive. Clearly R is symmetric.

If (4, 6) ∈ R; (6, 4) ∈ R but (4, 4) ∉ R Since 4 + 4 ≠ 10. ∴ R is not transitive. Choice (2)

43. Clearly > satisfies only transitive property Choice (4) 44. Clearly x2 + 2x, y2 + 2y are equal for all pairs of (x, y)

and we know that it is an equivalence relation. Choice (4) 45. x – x ≤ 5 is true for all x = y. ∴ R is reflexive.

x – y ≤ 5, then y – x ≤ 5 ∴ R is symmetric. ∴ R is reflexive as well as symmetric. Choice (3) 46. A matrix need not be conformable for multiplication with

itself. ∴ R is not reflexive. If matrix A is conformable for multiplication with B it is

not necessary that B is conformable for multiplication with A is possible ∴ R is not symmetric.

Clearly R is not transitive. Choice (4) 47. Clearly, ≥ is a partial ordered Relation. Hence, it does not satisfy symmetry property. Choice (2) 48. Since R is reflexive, the minimum number of ordered

pairs in R is x. ∴ y ≥ x is always true. Choice (1) 49. Clearly, R1 ∩ R2 is equivalence on A since both R1 and

R2 are equivalence on A Choice (1) 50. We know that, if n(A) = m, then the cardinality of p(A) = 2m. ∴ Given p(A) = 512, 2m = 512 ⇒ m = 9. ∴ The cardinality of A = 9. Choice (2) 51. The Venn diagram of (A ∪ C) – B is as follows. `

The above shaded region is also equal to (A – B) ∪ (C – B) Choice (4) 52. The venn diagram of A – (B – C) is as follows

The venn diagram of (A – B) – (A – C) is

∴ A – (B – C) ≠ (A – B) – (A – C) ∴ Option 3 is not true. Choice (3) 53. Case1: If all the images are different from set Q we

have to take any 3 elements. This can be done in 7C3 ways.

Case2: If exactly two elements are same, the third

element can be taken from 6 elements in 6 ways. The first two elements can be taken in 7 ways. ∴ Total number of ways = 7 × 6 = 42 Case3: If all the three elements are equal, we take

one elements from 7 elements, this can be done in 7 ways.

The total number of possible ways = 7C3 + 42 + 7 = 35 +

42 + 7 = 84 = 9C3 Choice (3) 54. Given: n (T) = 72, n(M) = 60; n(k) = 45, n(T ∩ M) = 48, n(K ∩ T) = 37, n(m ∩ k) = 36 and n(M ∩ T ∩ K) = 30. n(M ∪ K ∪ T) = n(M) + n(T) + N(K) – n(M ∩ K) – n (K ∩ T) – n(T ∩ M) + n(m ∩ T ∩ K) = 72 + 60 + 45 – 48 – 37 – 36 + 30 = 86. ∴ The number of students who can speak neither of the languages is = 125 – 86 = 39. Choice (2) 55. Let there be ‘m’ elements in S. It is given that each element of S belongs to exactly 8 of

Ai s and exactly 6 of the Bj s. The total number of elements (Not necessarily distinct

as used as Ai s) in Ai is = 20 × 4 = 80 = 8 × k ⇒ k = 10. The total number of elements (not distinct as used Bj s) in Bj S = p × 2 = 6 × 10 ⇒ p = 30. Choice (1)

Exercise – 8

Solutions for questions 1 to 55: 1. Given: A = {1, 3, 5, 7} and B = {2, 4, 5, 6, 7} Clearly, option 3 satisfies the conditions of a function from A → B Choice (3) 2. Choice (1) when x = 2, y = 1 As 1 ∉ A, f is not a function from A to A Choice (2) When x= 2, y = 4 When x= 3, y = 3 When x= 4, y = 2 All posible values of f are there in A ∴ f2 is a function in A to A. Choice (2) 3. The number of mappings from A → B when n(A) = p and n(B) = q is {n(B)}n(A) = qp Here p = 3 and q = 5 ∴ qp = 53 Choice (1)


4. We know that, the number of one-one functions possible from set A to set B when n(A) = m and n(B) = N where n > m is nPm. Here, n = 5 and m = 4

∴ The number of one-one functions possible is 5P4 = 5! = 120 Choice (3) 5. We know that, the number of onto functions from A → B

when n(B) < n(A), where n(B) = q and n(A) = p are qp − {qC1(q − 1)p − qC2(q − 2)p + qC3(q − 3)p ….}

Here, p = 4, and q = 3, ∴ The required number = 34 − (3C1(3 − 1)4− 3C2(3 − 2)4) = 81 − (3 × 24 − 3(1)4) = 81 − 45 = 36 Choice (3) 6. The number of onto mappings from set A to set B when n(A) < n(B) is zero. Choice (4) 7. The number of bijections from set A to set A is (n(A))! Given: n(A) = 4 ∴ The number of bijections = 4! = 24 Choice (2) 8. Clearly, f(x) is real, ∴ – 8 x ≥ 0 i.e. x ≤ 0 x ∈ R– U {0} ∴ The domain is R– ∪ {0} Choice (3)

9. |3x − 2| = 0 ⇒ x = 32

∴ f(x) is not defined when x = 32

∴ The domain is

−

32

R Choice (4)

10. f(x) is defined only, when 2x2 + 7x + 5 > 0 ⇒ 2x2 + 2x + 5x + 5 > 0 ⇒ 2x(x + 1) + 5(x + 1) > 0

⇒ (x + 1)(2x + 5) > 0 ⇒ x ∉

−−1,

25

⇒ The domain is ( )∞−∪

−∞− ,125

, Choice (3)

11. The domain of 4x3

1−

is x ≠ 34

and

The domain of 1x2 − is x ≥ 2

1

∴ The domain is x 21≥ , x

34≠ Choice (2)

12. Given: The function is not defined when 2x2 − 11x − 30 = 0

⇒ (2x − 15)(x + 2) = 0 ⇒ x = 2

15, x = − 2

The domain is

−− 2,

215

R . Choice (1)

13. The given function is of the form f(x) = bxax

−−

, where

a < b, The domain of any function in this form is ]( )),b(a, ∞∪−∞ here, a = 3, b = 7

∴ The domain is ]( )),7(3, ∞∪−∞ Choice (3)

14. If f(x) is a function, then f(x) + f(− x) is always an even function Choice (2) 15. Going by the choices, it can be verified that only choice (4) satisfies the given condition. Choice (4)

16. f(p).f

p1

= f(p) +

p1

is possible if f(p) = pn ± 1

Here, given f(5) = 124 = 53 – 1 ⇒ f(x) = xn – 1 ∴ f(9) = 93 – 1 = 728. Choice (3) 17. Given: f(x) = 2x + 2–x and g(x) = 2x − 2–x

f(x).g(y) + f(y) g(x) = (2x + 2–x)( 2y − 2–y)+ (2y + 2–y)( 2x − 2–x) ⇒ 2x+y − 2x−y + 2−x+y − 2−x−y + 2x+y − 2-x+y − 2 −x−y + 2–y+x ⇒ 2(2x+y − 2−(x+y)) = 2g(x + y) Choice (3)

18. Given: f(x+ y) = f(x) + f(y)

∑=

10

1i)i(f = f(1) + f(2) + f(3) + f(4) + … + f(10)

Given, f(1) = 2 f(2) = f(1 + 1) = f(1) + f(1) = 2 + 2 = 2.2= 4 f(3) = f(1 + 2) = f(1) + f(2) = 2 + 4 = 2.3 = 6 f(10) = f(1 + 9) = f(1) + f(9) = 2 + 18 = 2.10

∴ ∑=

10

1i)i(f = 2(1 + 2 + 3 + … + 10)

= 2

1) 2(10)(10 + = 110 Choice (2)

19. The domain of Sin-1x is [−1, 1] ⇒ −1 ≤ 3

x25 − ≤ 1

⇒ − 3 ≤ 5 − 2x ≤ 3 ⇒ − 8 ≤ −2x ≤ − 2 ⇒ 2 ≤ 2x ≤ 8 ⇒1 ≤ x ≤ 4 ∴ The domain of the given function is [1, 4] Choice (3)

20. |x|7

|x|5

−−

≥ 0, ∴ 5 – |x |and 7 – |x| ≥ 0 or 5 – |x| ≤ 0 and

7 – |x| ≤ 0 Clearly, 5 − |x| and 7 − |x| are positive in the interval [−5, 5] and both are negative in the intervals (− ∞, −7) and in (7, ∝ ). ∴ The domain is [− 5, 5]∪ (− ∝ , – 7) ∪ (7, ∝ ) Choice (2)

21. Given: f(x) = x85x −+−

f(x) is defined only, when x − 5 ≥ 0 and 8 − x ≥ 0 ⇒ x ≥ 5 and x ≤ 8 ⇒ x ∈ [5, 8] ∴ The domain is [5, 8] Choice (4)

22. The domain of x21− is 1 − 2x ≥ 0 ⇒ −2x ≥ −1

⇒ 2x ≤ 1 ⇒ x ≤ 2

1 → (1)

The domain of

−−

34x5

cos 1 is −1 ≤ 3

4x5 − ≤ 1

⇒ −3 ≤ 5x − 4 ≤ 3 ⇒ 1 ≤ 5x ≤ 7 ⇒ 51 ≤ x ≤

57 → (2)

From (1) and (2),

≤≤∩

≤

57

x51

21

x =

21

,51

Choice (3)

23. log )x2log()x3)(x3(logx2x9 2

+−+−=+−

log )x3()x3( +− is defined only when

(3 − x)(3 + x) > 0 ⇒ x ∈ (− 3, 3). log (2 + x) is defined only when 2 + x > 0 ⇒ x > − 2

∴ Both 2x9 − and log (2 + x) are defined when

x ∈ (− 2, 3) ∴ Domain is (− 2, 3) Choice (2)


24.

+−

gfg2f2

(3) = { }

13

1816

94

9x24

)3(g)3(f

)3(g2)3(f 22 −=+−=

+−

= 132− Choice (4)

25. Clearly |2x − 1| is always 0 or positive for x ∈ R ∴ 5 − |2x − 1| is always less than or equal to 5. ∴ The range is (− ∝ , 5] Choice (4) 26. We know − 1 ≤ cos 2x ≤ 1 − 1 ≤ −cos 2x ≤ 1 or 2 ≤ 3 − cos 2x ≤ 4

⇒ 41 ≤

x2cos31

−≤

21

∴ The range is

21

,41

Choice (3)

27. f(− 2.5) + f(− 1.5) + f(− 3.5) = (− 2.5 − 1) − 1.5 + (– 3.5 – 1) = − 9.5. Choice (3)

28. The range is c ± 22 ba + , Here c = 18, a = 15, and b = 8

The range is

+++− 2222 81518,81518 i.e. [1, 35]

Choice (4)

29. Sin2x + cos4x = 2

2x2cos1

2x2cos1

++−

⇒ 8

x4cos81

x2cos21

41

x2cos21

21 ++++−

81

87 += cos4x

We know that, The range is –1 ≤ cos4x ≤ 1

The range is

=

+−+ 1,43

)1(81

87

),1(81

87

=

1,

43

Choice (1)

30. Let1xx

1xx2

2

+++−

= y

⇒ x2 − x + 1 = x2y + xy + y ⇒ x2(y − 1) + x(y + 1) + (y − 1) = 0 If this equation has real roots ⇒ (y + 1)2 − 4(y − 1)2 ≥ 0 ⇒ y2 + 2y + 1 − 4y2 + 8y − 4 ≥ 0 ⇒ 3y2 − 10y + 3 ≤ 0 or (3y − 1)(y − 3) ≤ 0

⇒ y ∈

3,

31

∴ The range is

3,

31

Choice (1)

31. Let1x

x2 +

= y → y2(x2 + 1) = x2 or x2(y2 − 1) + y2 = 0

for its roots to be real b2 − 4ac ≥ 0 must hold true i.e. − 4(y2 − 1)y2 ≥ 0 y2(y2 − 1) ≤ 0 ∴ y2 – 1 < 0 …. –1 < y < 1 ∴ The range is (− 1, 1) Choice (1) 32. Given: f(x) = 2x2 − 1 where x ∈ {− 2, − 1, 0, 1, 2} f(− 2) = 2(− 2)2 − 1 = 7 = f(2) and f(0) = −1 f(− 1) = 2(− 1)2 − 1 = 1 = f(1) ∴ The range is {− 1, 1, 7} Choice (4)

33. We know that, ∀ x ∈ R,x

|x|2= − 2 and

∀ x ∈ R+ x

|x|2= 2 ∴ The range is {− 2, 2}

Choice (2)

34. Clearly, 2x1

2

+is positive ∀ x ∈ R. Hence, it is not onto

and x = − 1 and 1 have the same image. ∴ f(x) not one-one and also it is not onto.

∴ 2x1

2

+is not bijective. Choice (4)

35. Let y = 2x31x2

+−

⇒ 3xy + 2y = 2x − 1 or x(3y − 2) = − 2 y − 1

⇒ x = y321y2

−+ ∴ f-1(y) =

y32y21

−+

∴ Inverse of f is x32x21

−+

Choice (1)

36. Let x = 100

1, f 4

1001

2f100

1 =

−+

⇒

+

100199

f100

1f = 4

Let x = 100

2, f 4

1002

2f100

2 =

−+

⇒ 4100198

f100

2 =

+

Let x = 10099

⇒

−+

10099

2f10099

f = 4

⇒

+

100101

f10099

f = 4

Let x = 100

100⇒

−+

100100

2f100100

f = 4

⇒

+

100100

f100100

f = 4 ⇒

100100

f = 2

∴ The value of

++

+

100199

f.........100

2f

1001

f

= 4 × 99 + 2 = 396 + 2 = 398. Choice (4) 37. f(− x) = 2(− x)3 + 3 sin 2(− x) + 4 tan 3(− x) = − (2x3 + 3 sin 2x + 4 tan 3x) ⇒ − f(x) ∴ f(− x) = −f(x) ∴ f(x) is an odd function. Choice (2) 38. f(x) = x2 − 2x + 3 = (x − 1)2 + 2 Given: domain = [1, ∝ ) ⇒ The range is [2, ∝ ) Choice (2)

39. Given: f(x) = xx

xx

22

22−

−

−+

⇒ f(−x)

= ( )( )xx

xx

xx

xx

22

22

22

22−

−

−

−

−−+=

−+

= − f(x)

∴ f(x) = − f(x) ⇒ f is odd function. Choice (1) 40. Given: f(x) is even ⇒f(− x) = f(x) Let g(x) = sin(f(x) − 2x) g(– x) = sin {f(− x) − 2(− x)} = sin{f(x) + 2x} which is neither g(x) nor − g(x) ∴ g(x) is neither even nor odd. Choice (4)


41. By finding f-1, we notice f-1(x) = f(x).

∴ f-1(2) = f(2) = 21

21

+−

=3

1−.

Alternate method: Let f–1 (s) = y 2 = f(y)

2 = y1y1

+−

2 + 2y = 1 – y 3y = – 1 y = – 1/3 ∴ f–1 (2) = –1/3. Choice (3) 42. Since A has 6 elements and ‘f’ is an onto function, the

number of elements B can have is at most ‘6’. Choice (2) 43.

∴ fog(x) = {(1, − 1), (2, 1), (3, 2), (4, 3)} Choice (3) 44. gof(− 2) = g[f(− 2)] = g{3 − 2(− 2)} = g(7) = 3(7) + 1 = 22 Choice (1) 45. hofog(− 1) = hof[g(− 1)] = hof[3 − 2(− 1)] = h{f(5)} ⇒ h{2(5) + 1} = h(11) = 112 = 121 Choice (4)

46. Given: f(x) = 3x22x3

−+

⇒ f-1 = 3x22x3

−+

∴ f–1 (x) = f(x) ∴ f–1 0 f(x) = (x) f–1 0 f(0) = 0. Choice (1) 47. We notice that fof(x) = x, fofof(x) = f(x), ∴ fofofof(x) = x

∴ fofofofof(x) = f(x) =x1x1

+−

∴ fofofofof(−3) = 2

4−

= −2 Choice (4)

48. gof(x) = g[f(x)] = g(x+1) = 2(x + 1)2 − 3

Let 2(x + 1)2 − 3 = y ⇒ x = 2

3y + − 1

∴ (gof)-1(x) = 12

3x−

+

and (gof)-1(5) = 12

35−

== 2 − 1 = 1 Choice (2)

49. Given: f(x) = ( )p1pxa − where p ∈ N

fof(x) = p

1

p

p

1p

p

1

p )xaa()xa(a +−=

−− = x

Choice (1)

50. Given: ef(x) = x1x1

−+

⇒ f(x) = log x1x1

−+

F (a)1 + (b) = log

−++

−+

b1b1

loga1a1

= log

−+

−+

b1b1

a1a1

=

−−++++

baab1baab1

log

=

+−−+

++++

ab1baab1

ab1baab1

log

=

++−

+++

ab1)ba(

1

ab1ba

1log =

++ab1ba

f .

Choice (3)

51. We know that, )x(f is defined when f(x) ≥ 0; and log

f(x) is defined f(x) > 0.

6x2x

log 5.0 +−

is defined

log0.5 6x2x

+−

> 0 and 6x2x

+−

> 0

log0.5 6x2x

+−

> log0.51 and

2)6x(

0)6x()2x(

+>+−

6x2x

+−

< 1 and (x – 2) (x + 6) > 0

)6x(

06x2x+

<−−− and x < – 6 or x > 2.

6x

8+−

< 0

∴ (– ∞, – 6) ∪ (2, ∞) –––– (1)

6x

8+

> 0

(x + 6) > 0 x > – 6 x ( – 6, ∞) –––––––– (2) From (1) and (2), the domain is (2, ∞)

But when x = 7, then function 4x

12 −

is not defined.

∴ The domain of the given function is (2, 7) ∪ (7, ∞) Choice (3)

52. f(x) = max (4x + 3, 5 – 6x) Let 4x + 3 > 5 – 6x 10x > 2. x > 1/5

∴ When x > 51

, f(x) = 4x + 3

∴ When x > 51

, f(x) > 5

19

Let 4x + 3 < 5 – 6x 10x < 2 x < 1/5

for x < 51

, f(x) > 5

19

∴ The minimum value f(x) is 5

19 Choice (1)

2 4 5 6

1 2 3 4

–1 1 2 3

g f


X � X

–1 27

∞ –∞

X

–13

� X

–8 ∞ –∞

53. We must have,

– 1 ≤ 6

x3 − ≤ 1

– 6 ≤ 3 – x ≤ 6 – 6 – 3 ≤ – x ≤ 6 – 3 – 9 ≤ – x ≤ 3 – 3 ≤ x≤ 9 x ≤ 9.

The domain of Sin–1

−6

x3 is – 9 ≤ x ≤ 9

)x6(log

1−

is defined

When 6 – x > 0 and 6 – x ≠ 1 x < 6 x ≠ 5

∴ The domain of )x6(log

1−

is (– ∞, 6) – {5}.

The domain of Sin–1

−6

x3 + [log (6 – x)]–1 is

[– 9, 5) ∪ (5, 6) Choice (4) 54. logax is defined if x > 0

x2x9 2

−−

> 0,

9 – x2 > 0 and 2 – x > 0 0)3x()3x( <+− x∈ (– 3, 3) and x < 2

The domain of

−−

x2x9

logcos2

is (– 3, 2).

Choice (4)

55.

−2

xx3log

2

is defined only when

2

xx3 2− > 0 and log

−2

xx3 2 ≥ 0

⇒ log

−2

xx3 2 ≥ log 1

2

xx3 2− ≥ 1

3x2 – x2 > 0 and 3x – x2 ≥ 2 x2 – 3x < 0 and – x2 + 3x – 2 ≥ 0 x ∈ (0, 3) x2 – 3x + 2 ≤ 0 x ∈ [1, 2]

∴ The domain is [1, 2] Choice (2)

Exercise – 9 Solutions for questions 1 to 25: 1. (i) Given: 4x + 3 > 6x + 7 ⇒ 4x − 6x > 7 − 3 ⇒ − 2x > 4 ⇒ x < − 2 Choice (3)

(ii) Given: 7x − 5 > 4x + 13 ⇒ 7x − 4x > 13 + 5 ⇒ 3x > 18 ⇒ x > 6 ∴ x ∈ (6, ∞) Choice (2)

(iii) Given: 6x − 9 ≥ 3x + 5 6x − 3x ≥ 5 + 9 3x ≥ 14

x ≥ 3

14 Choice (3)

(iv) Given: 7x + 5 ≤ 3x − 11 ⇒ 7x − 3x ≤ − 11 − 5 ⇒ 4x ≤ − 16 ⇒ x ≤ − 4 ∴ x ∈ (− ∞, − 4] Choice (4)

(v) Given: 10x − 13 > 7x + 9

⇒ 10x − 7x > 9 + 13 ⇒3x > 22 ⇒ x > 3

22

∴ x∈

∞,3

22 Choice (2)

2. Given: 6x + 9 < 3x + 5 ; 4x + 7 > 2x − 5 6x − 3x < 5 − 9 ; 4x − 2x > −5 − 7 3x < − 4 ; 2x > − 12

⇒x < − 34

…..(1) ; x > − 6 ……(2)

∴ From (1) and (2)

we have x ∈

−−3

4,6

Choice (3) 3. Given: 5x + 7 − 2x2 > 0 ⇒ 2x2 − 5x − 7 < 0

⇒ 2x2 − 7x + 2x − 7 < 0 ⇒ x(2x − 7) + 1(2x − 7) < 0 ⇒ (x + 1) (2x − 7) < 0

So, the critical points are − 1, 27

Now, when x = 0, the inequality is true.

∴ When x ∈

−2

7,1 the inequality is true.

Hence, the required solution set is

−2

7,1

Choice (1)

4. Given: 48x

7x >+−

⇒ 048x

7x >−+−

⇒ 08x

)8x(47x >+

+−− ⇒ 0

8x

32x47x >+

−−−

⇒ 08x39x3 >

+−−

⇒ 08x

)13x(3 >++−

⇒ 08x13x <

++

⇒ ( )( )

( )0

8x

8x13x2

<+

++

⇒ (x + 13) (x + 8) < 0 (Q (x + 8)2 > 0 for x ≠ 8) The critical points are −13 and − 8 When x = 0, the inequality is not true. ∴ The solution set is (− 13, − 8) Choice (3)

5. Given: 012x7x

4x5x2

2

≤+−++

⇒ ( )( )( )( ) 0

3x4x

1x4x≤

−−++

⇒ ( )( )( )( )

( ) ( )0

3x4x

3x4x1x4x22

≤−−

−−++


� � X X X

−4 −1 3 4 ∞ –∞

−2 1 ∞ –∞

�

1

X � X X

∞ –∞ –6 –1 7

⇒ (x + 4)(x + 1) (x − 4) (x − 3) ≤ 0 The critical points are − 4, − 1, 3 and 4

When x = 0, the inequality is not true. Hence, when x ∈ [– 4, − 1] or (3, 4) the inequality is true. ∴ The required solution set [– 4, − 1] ∪ (3, 4).

Choice (1)

6. Given: 2x2 + 10x + 17 < 0 ⇒ x2 + 5x + 2

17 < 0

⇒ x2 + 5x + 4

25 –

217

425 + < 0 ⇒

49

25

x2

+

+ < 0

Since 2

2

5x

+ is always greater than 0 for any value of x,

There is no real value of x such that 2x2 + 10x + 17 < 0. Choice (4)

7. We know that ∴ |2x – 3| > 0 for every real value of x. ∴ |2x – 3| + 17 ≥ 17 ∴ The minimum value of |2x + 3| + 17 is 17. Choice (2) 8. Given: |8 − x| > 4x − 5

Case (1): When x < 8, |8 − x| = 8 − x ∴ |8 − x| > 4x − 5 ⇒ 8 − x > 4x − 5 ⇒ 8 + 5 > 4x + x

⇒ 13 > 5x ⇒ x < 5

13 which agrees with x < 8.

Case (2): When x > 8, |8 − x| = x − 8

∴ |8 − x| > 4x − 5 ⇒ x − 8 > 4x − 5 ⇒ −8 + 5 > 4x – x ⇒ −3 > 3x ⇒ 3x < −3 ⇒ x < − 1 which does not agree with x > 8.

∴ The solution set is x < 5

13

i.e., x∈

∞−5

13, Choice (1)

9. Given: x = |a| b and b ≥ 2 a + xb = a + |a| b2 and a – xb = a - |a|b2

As |− 6| ≥ 1, the magnitude of |a| b2 is greater than that of a. The sign of either expression is determined by the sign of this term. i.e. a + |a| b2 ≥ 0 and a − |a| b2 ≤ 0 The equality holds when a = 0. Choice (4)

10. Given: 2x2 + |4x −9| = 7 (1)

Case (1); Let x <49

Then, |4x - 9| = − (4x − 9). So, (1) is reduced to 2x2 − 4x +9 = 7 ⇒ 2x2 – 4x + 2 = 0 ⇒ x2 − 2x + 1 = 0 ⇒ (x − 1)2 = 0

⇒ x = 1 which agrees with x < 49

Case (2); Let x > 49

Then, |4x − 9| = 4x – 9

So, (1) reduces to 2x2 + 4x − 9 = 7 ⇒ 2x2 + 4x −16 = 0 ⇒ x2 + 2x − 8 = 0 ⇒ (x + 4) (x − 2) = 0

⇒ x = − 4, 2 These values do not agree with x > 49

Hence, the solution of the given equation is x = 1 Choice (1) 11.

When x < − 2, x – 1 > 3 So, x + 2 + x − 1 > 3 When x > 1, x + 2 > 3 and so x + 2 + x − 1 > 3. When – 2 ≤ x ≤ 1, x + 2 + x − 1 = 3

Hence, there is no real number x such that x + 2 + x − 1 < 2. Choice (4)

12. Given: (x2 + 5x – 6) (x2 – 6x – 7) < 0 (x2 + 6x − x − 6) (x2 − 7x + x − 7) < 0 (x + 6)(x − 1) (x + 1) (x − 7) < 0 The critical points are − 6, − 1, 1 and 7.

When x = 0, the inequality is not true. ∴ The solution set is, x ∈ (− 6, − 1) ∪ (1, 7).

The integral values of x in the above solution set are − 5, – 4, − 8, − 3, − 2, 2, 3, 4, 5 and 6.

∴ The required number of integral values is 9. Choice (1) 13. Given: |3x − 5| ≥ 10 We know that, if |x| ≥ a then x ≤ − a or x ≥ a ∴ |3x − 5| ≥ 10 ⇒ 3x − 5 ≤ –10, or 3x − 5 ≥ 10 ⇒ 3x ≤ − 5 − or 3x ≥ 15

⇒ x ≤ − 35

or x ≥ 5

∴ The solution [ )∞∪

−∞− ,535

,

∴ i.e., R −

− 5,35

Choice (2)

14. Given: (x2 + x + 1) x+2 < 1 ⇒ (x + 2) log (x2 + x + 1) < log 1 ⇒ (x + 2) log (x2 + x + 1) < 0 ⇒ x + 2 < 0 and log (x2 + x + 1) > 0 (or) x + 2 > 0 and

log (x2 + x + 1) < 0 Let x +2 < 0 and log (x2 + x + 1) > 0 ⇒ x < − 2 and x2 + x + 1 > 1 ⇒ x < − 2 and x2 + x > 0 ⇒ x < − 2 and x (x + 1) > 0 ⇒ x < − 2 and x < − 1 or x > 0 ⇒ x < − 2 ---------------------- (1)

Now, let x + 2 > 0 and log (x2 + x + 1) < 0 ⇒ x > − 2 and x2 + x + 1 < 1 ⇒ x > − 2 and x2 + x < 0 ⇒ x > − 2 and x(x + 1) < 0 ⇒ x > − 2 and x ∈ ( − 1, 0) ⇒ x ∈ (− 1, 0) ---------------- (2)

∴ From (1) and (2), the required solution set is

(− ∞, − 2) ∪ (− 1, 0). Choice (2)

4

9


15. We know that AM(x, y) ≥ GM(x, y).

So,

≥

2222 y

1,

x

1GM

y

1,

x

1AM

2222 y

1.

x

12

y

1

x

1 ≥+

xy2

y

1

x

122

≥+ ………………(1)

Similarly, we get

yz2

z

1

y

122

≥+ ………………(2)

zx

2

x

1

z

122

≥+ ………………(3)

Adding the three inequalities, we get

zx1

yz1

xy1

z

1

y

1

x

1222

++≥++

∴ Option (1) is true. We know that A.M.(x, y, z) ≥ H.M.(x, y, z)

z1

y1

x1

33

zyx

++≥++

(x + y + z)

++

z1

y1

x1

≥ 9

∴ (x + y + z) (yz + zx + xy) ≥ 9xyz Option (2) is also true.

We know that, A.M.(x3, y3, z3) ≥ G.M.(x3, y3, z3)

⇒ 3 333333

zyx3

zyx ≥++

⇒ x3 + y3 + z3 ≥ 3xyz

∴Option (3) is also true. Choice (4) 16. Given: x + y + z = 2 We know that, A.M.(x, y, z) ≥ G.M.(x, y, z)

3 xyz3

zyx ≥++

3 xyz3

2 ≥

xyz278 ≥

∴ xyz ≤ 278

Choice (3)

17. Given: a2 + b2 = 6 ; c2 + d2 = 6

∴ a2 + b2 + c2 + d2 = 12 We know that,

bd2

db;ac

2ca 2222

≥+≥+

Adding the two inequalities, we get

bdac2

dcba 2222

+≥+++

6 ≥ ac + bd (or) ac + bd ≤ 6

Similarly, we can show that cd2

dc;ab

2ba 2222

≥+≥+

ab + cd ≤ 6 and

bc2

cb;ad

2da 2222

≥+≥+ ⇒ ad + bc ≤ 6.

Choice (4)

18. Given: a, b and c are positive real numbers. ∴ a2 + b2 ≥ 2ab ; b2 + c2 ≥2 bc ; c2 + a2 ≥ 2ac ⇒ a2 + b2 + b2 + c2 + c2 + a2 ≥ 2ab + 2bc + 2ca

⇒ 1cabcabcba 222

≥++++

Since a, b and c are the lengths of the sides of triangle, a − b < c ; b − c < a ; c − a < b ⇒ a2 + b2 − 2ab < c2 b2 + c2 − 2bc < a2 and c2 + a2 − 2ca < b2 a2 + b2 − c2 < 2ab, b2 + c2 − a2 < 2bc, c2 + a2 − b2 < 2ca Adding these inequalities, we get

a2 + b2 + c2 < 2ab + 2bc + 2ca

2 cabcab

cba 222

<++++

Hence, 1 ≤ 2 cabcabcba 222

<++++

Choice (2)

19. We know that |a| − |b| ≤ |a − b| and |a − b| ≤ |a| + |b| ∴ |a| − |b| ≤ |a − b| ≤ |a| + |b| i.e., q ≤ r ≤ p Choice (3) 20. Let x3 + 1 > x2 + x ⇒ x3 − x2 − x + 1 > 0 ⇒ x2 (x − 1) −1(x − 1) > 0 ⇒ (x − 1) (x2 − 1) > 0 ⇒ (x − 1)2 (x + 1) > 0 ⇒ x +1 > 0 and x ≠ 1 [Q (x – 1)2 ≥ 0 x ∈ R] ⇒ x > 1 and x ≠ 1⇒ x ∈ (−1. 1) ∪ (1, ∞) ∴ The solution set is (−1, 1) ∪ (1, ∞) Choice (2) 21. We know that, (n!)2 ≥ nn So, all the given options are false. Choice (4)

22. Given: 2 − n1

< x ≤ 4 + n1

n = 1 ⇒ 2 − 1 < x ≤ 4 + 1 1 < x ≤ 5

x = 2 ⇒ 2

9x

2

3 << and so on

The range of x is (1, 5] Choice (1) 23. We know that,

2ba

ab ≥+ ,

2ca

ac ≥+ and

2bc

cb ≥+

Adding these inequalities, we get

⇒ 6cb

ca

bc

ba

ac

ab ≥+++++

i.e., 6c

bab

caa

cb ≥+++++

∴ The minimum value is 6. Choice (4) 24. As x2y3 = 108,

32

3y

2x

=

108yx 32

= 1

x + y = 3y

3y

3y

2x

2x ++++

Consider the product of all the factors which is constant.

When all these are equal, the sum of the factors, i.e. x + y will be the minimum.

3y

3y

3y

2x

2x ====


= 108108

532

yx5

32

22= = 1

∴ 12x = and

3y

= 1, i.e., x = 2, y = 3

∴ The minimum value of x + y is 5. Choice (3) 25. As x2y = 24 (given),

)y4(2x3

2

= 9x2y = 216

3x + 4y = 2x3

2x3 + + 4y

The product is constant; So, the sum will be minimum, when all the factors are equal.

∴ 3 y42x3

2x3

y42x3

2x3 ===

yx9y42x3

2x3 2=== = 3 249 ×

= 3 216 = 6.

∴2x3

= 6;

4y = 6

⇒ y = 46

x = 4, y = 23

Hence the minimum value of 3x + 4y

= 3 × 4 + 4 × 23

= 12 + 6 = 18. Choice (1)

Exercise – 10

Solutions for questions 1 to 25: 1. The sentence in choice (1) cannot be judged as either

true or false as we have no information about ‘He’. The sentence in choice (2), cannot be judged either true

or false, as the information about the good books is not known.

The sentence in choice (3), will get both the true values as both true and false, so it is not a statement.

The statement in choice (4) is false, hence it is a statement. Choice (4)

2. The symbolic form of not “p” and not “q” is ~p ∧ ~q

Choice (3) 3. Given: p : A is beautiful, q : A is intelligent.

So, ~p : A is ugly. Now, the symbolic form of ‘A is both beautiful and intelligent’ is (p ∧ q). Hence, the symbolic form of “A is ugly or A is both beautiful and intelligent” is “~p ∨ (p ∧ q)” Choice (4)

4. Given: ~p ∨ q ∨ r Now, from choice (4), if p is false, then ~p is T. ⇒ ~p ∨ (q ∨ r) = T ∨ (q ∨ r) = T Hence, ~p ∨ (q ∨ r) = T. Choice (4) 5. The given statement is ~p ∧ q ∧ ~r if p = T, then ~p is F. Also ~p ∧ q ∧ ~r is always false (Q F ∧ p ≡ F) Choice (1)

6. The given statement is p ∨ (p ∧ q)

Truth table:

p q p ∧ q p ∨ (p ∧ q)

T T T T

T F F T

F T F F

F F F F

From the above truth table, we can observe that the first and fourth columns are identical.

∴ p ∨ (p ∧ q) ≡ p Choice (2) 7. The given statement is ~p ⇒ (p ∧ (~q)) The statement in choice (2), i.e. F if p = F Here, If p = F ⇒ ~p = T Clearly p ∧ (~q) be F. ∴ The truth value ~p ⇒ (p ∧ ~q) is F. Choice (2) 8. Given, (p ∨ (~q)) ∧ r

Now, ~(p ∨ (-q)) ∧ r. = ~(p ∨ (~q)) ∨ ~r (Demorgan’s law) = (~p ∧ q) ∨ ~r (Demorgan’s law) Choice (3)

9. Given: p: Charminar is in Hyderabad. (T) q: Golconda is in Delhi. (F) We know that, p.q is true if the truth values of both the

statements are not same. Hence, the truth value of p.q is T. Choice (3) 10. The symbolic form of the given statement “If she is beautiful, then she is not intelligent” be p ⇒ q We know that ~(p ⇒ q) ≡ p ∧ (~q) Hence the negation of given statement is “She is beautiful and she is intelligent”. Choice (3) 11. The given statement is “All friends of Pasha are good”.

Its negation is ‘Some friends of Pasha are not good’. (or) There is at least one friend of Pasha who is not good. Choice (4)

12. The negation of the statement “some apples in the

market are not delicious” is “All apples in the market are delicious”. Choice (1) 13. We know that, inverse of p ⇒ q is ~p ⇒ ~q. Hence, the

inverse of p ⇒ ~q is ~p ⇒ ~(~q), i.e. ~p ⇒ q. Choice (1) 14. The given statement is ~p ⇒ ~q We know that the converse of p ⇒ q is q ⇒ p ∴ The converse of ~p ⇒ ~q is ~q ⇒ ~p. Choice (3) 15. We know that, contrapositive of p ⇒ q is ~q ⇒ ~p ∴ The contrapositive of p ⇒ ~q is ~(~q) ⇒ ~p i.e., q ⇒ ~p. Choice (1) 16. The inverse of statement p ⇒ q is ~p ⇒ ~q. ∴ Clearly choice ‘3’ the conditional statement. Choice (3)

17. Conditional : p ⇒ q Converse : q ⇒ p

Now, q ⇒ p = (~q) ∨ p (∵ p ⇒ q ≡ (~p) ∨ q)

= p ∨ (~q) (commutative property) = ~p ⇒ ~q the inverse of p ⇒ q is ~p ⇒ ~q

Hence, the converse of a conditional statement is always equivalent to its inverse. Choice (2)


X1

X2

A B

C2 C1

• • •

P

18. Given, ~p ⇒ ~p ∨ q We know that, contrapositive of p ⇒ q is ~q ⇒ ~p ∴ The contrapositive of ~p ⇒ ~p ∨ q is ~(~p ∨ q)

⇒ ~(~p) i.e., p ∧ (~q) ⇒ p. Choice (4) 19. Given:

p ∧ (~p ∨ q) = (p ∧ ~p) ∨ (p ∨ q)

= f ∨ (p ∧ q) (∴ p ∧ ~p is a contradiction) ∴ p ∧ (~p ∨ q) ≡ p ∧ q Choice (1)

20. The given statement is (~p) ∨ (p ∧ q) = (~p) ∨ p) ∧ (~p ∨ q) (distributive property) = t ∧ (~p ∨ q) (~p ∨ p is a tautology, i.e. t)

As ‘t’ takes only true (T), value so we can say that t ∧ p ≡ p ∴ t ∧ (~p ∨ q) ≡ ~p ∨ q Choice (3) 21. Consider, choice, (1) i.e. p ∨ (~p ∨ ~q)

= (p ∨ ~p) ∨ ~q (Associative property) = t ∨ ~q (p ∨ ~p is a tautology ‘t’) We know that, ‘t’ takes only true values. ∴ t ∨ (~q) = t. Hence, p ∨ (~p ∨ ~q) is a tautology. Choice (1)

22. We know that, p ∨ (~p) is a tautology.

p ∧ (~p) is a contradiction. P ⇒ p ∨ q is a tautology. Now, truth table of p ∧ (~p ∨ q) is

p q ~p ~p ∨ q p ∧ (~p ∨ q)

T T F T T

T F F F F

F T T T F

F F T T F

By observing the last column of the truth table, we can say that p ∧ (~p ∨ q) is a contingency. Choice (4)

23. Consider the statement in choice (2), i.e. (~p) ∧ (p ∧ (~q) = (~p ∧ p) ∧ (~q) (Associative property) = f ∧ (~q) (∵ ~p ∧ p is a contradiction, i.e. f)

= f (f ∧ p ≡ f) ∴ (~p) ∧ (p ∧ (~q)) is a contradiction.

Choice (2)

24. Let the given statement “Mobile is good if and only if

battery backup is good” be p ⇔ q. Now, ~( p ⇔ q) = (~ p) ⇔ q (or) p ⇔ (~q) ∴ The negation of the given statement is

Mobile is not good if and only if battery backup is good. (or) Mobile is good if and only if battery backup is not good. Choice (3)

25. The given statement is “4 + 8 = 18 ⇔ 2 is prime” 4 + 4 = 18 is F and ‘2 is prime’ is T. ∴ p ⇔ q is false if p and q have different truth values. Hence, the truth value of the given statement is false.

Choice (2)

Exercise – 11

Solutions for questions 1 to 25: 1. The given LPP (linear programming problem) is in

canonical form as the objective function is of maximization type and all constraints are of the form ‘≤’ with non-negative restriction to variables.

In a LPP, canonical form is a general form as well. Choice (4) 2. All the characteristics given in choices 1, 2, 3 are

required for a LPP to be called a standard LPP. Choice (4)

3. We get an unbounded solution while solving a given

LPP using graphical method if no corner point in the convex region gives optimal solution (standard result).

Choice (2) 4. Simplex method uses standard form of the given LPP to

solve the problem. Choice (3) 5. If two feasible regions are available then the effective

feasible region will be the intersection of the two regions. (Standard Result). Choice (3)

6. Convex set has a property that the line segment joining

any two points in the set lies entirely within or on the set. Consider option (3),

Here, part of the line lies outside the figure. Choice (3)

7. In all the diagrams given, it is possible to find a line

segment joining two points not being completely contained within or on it. So none of these are convex.

Choice (4) 8. A sector of a circle can be represented as (I) (II)

(I) Minor Sector (II) Major Sector In case of a major sector, we do not get a convex set. Choice (2) 9. The feasible region has came out to be disjoint sets.

So no region is common. Hence no solution. Choice (3) 10. The union of 2 convex sets may not be a convex set.

Consider two circles C1 and C2

The line segment joining A and B is not completely contained in C1 ∪ C2 Choice (2)


X1

(0, 4) D

E (0, 2) (2,4/3)

C

B A (3, 0)

(6, 0)

O

X2

(0, 0)

(2, 0) 0

x 2

x1

(0, 2)

(5, 0)

x2

0 x1

(0, 6)

(0, 2)

(2, 0) (5, 0) x

0

X1 O (0, 0)

X2

A

B

X1 − 2x ≤ 1 X1 +2X ≥3

(0, 4) C

(0, 3) D

(0, 2) E

(0, 0) O (3, 0) A

(0, 4) B x1

x2

11. The feasible region is bounded by the lines x1 = 10 and x2 = 5. Also (0, 0) is in feasible region. Accordingly x1 ≤ 10 and x2 ≤ 5. Choice (2)

12. Graph in choice (4) satisfies the given constraints.

Choice (4)

13. The lines that demarcate the constraints are

19

x

3

xand1

4

x

8

x 2121 =+=+ or x1 + 2x2 = 8 and 3x1 + x2 = 9

The corresponding inequalities are x1 + 2x2 ≤ 8 and 2x1 + x2 ≤ 9. Choice (3)

14. The region represented by 3x1 + x2 ≤ 6 is

The region represented by 2x1 + 5x2 ≤ 0 is The intersection of these two regions is

Choice (2) 15. At (8, 12) z = 176 − 216 = − 40 At (16, 0) z = 352 At (0, 0) z = 0 At (0, 20) z = − 360 Min z = − 360 at (x1, x2) (0, 20) ∴ Optimal solution is z = – 360 and solution set is (0, 20). Choice (4) 16. Feasible region is ABCD At (2, 0), z = 8 At (0, 2) z = 6 At (3, 0) z = 12 At (0, 4) z = 12 Min z = 6 ∴ Optional solution is z = 6 at x1 = 2, x2 = 0. Choice (3)

17. On plotting the constraints in the graph the feasible region is unbounded.

Choice (3) 18. We first plot the region that is satisfied by the

constraints of LPP as shown below:

The corner points in the feasible region are (0, 0), (3, 0) (0, 2) and (2, 4/3).

At (3, 0) z = 3 At (0, 2) z = 12 At (2, 3/4) z = 6½ Choice (2)

Note: The same can be solved by using simplex method. 19. Using graph method: -

OAEO is convex set. Min is at (0, 0) z = 0 Maximum is at (3, 0), and z = 15 Choice (3)


C (6,

(0, 6) (0, 15) E

O (9, 0) A (10, 0)

X2

(0, 30)

(0, 25)

O

C (10, 10)

(15,0) A

B (50/3, 0)

D (0, 8)

C

E (0, 3)

0

A (4, 0) B (6, 0)

D

x2

x2

(10/3, 4/3)

(0,

0 A(6, 0) B (9, 0)

C

x2

x1

20. Using graphical method: -

Z = 3x1 + 2x2 ECD is the feasible region.

At (0, 3) z = 6,

At3

38z,

34

,3

10 =

At (0, 8), z = 16. ∴ Minimum is 6 at (0, 3). Choice (3) 21. Let x1 be the number of units of the first product and

x2 be the number of the units of second product. The given information can be represented as max.profit = 100x1 + 120x2

subject to 2x1 + 4x2 ≤ 12 (gold constraints) x1 + 3x2 ≤ 9 (silver constraints)

x1, x2 ≥ 0 (non-negative constraints) Using graphical method:

The corner points in the feasible region are (0, 3), (6, 0) and (0, 0). The value of maximum profit is attained at (6, 0) as Rs.600. Choice (2)

22. Let x1 be the weight the first metal and x2 be the weight

of the second. As LPP pertains to cost, it is a minima problem.

The LPP may be stated as min. cost = 100(x + 2y) + 150(2x + 3y) Subject to: X1 + 2x2 ≥ 10 (Metal 1 constraint) 2x1 + 3x2 ≤ 18 (Metal 2 constraint) and x1, x2 > 0

ECD is the feasible region. But as x1, x2 > 0 We cannot take (0, 5) and (0, 6) as optional solution.

∴ C (6, 2) is the optimum solution and the optimum cost is z = 100(10) + 150(18) = 3700 Choice (4)

23. The LPP can be defined as min. cost = 20x1 + 8x2

subject to 3x1 + 2x2 ≥ 50 2x1 + x2 ≥ 30 x1 ≥ 0, x2 ≥ 0

At (0, 30) cost = Rs.240 (50/3/0) cost Rs.334 (10, 10) lost = 280

∴ 50/3 and 0 packets can be purchased from x1 and x2 respectively. Choice (4)

24. Since both the constraints are of type ‘≥’ we require surplus variable to make it of type ‘=’ and artificial variable Choice (2)

25. for ‘≤’ we require only slack variable for ‘≥’ surplus and

artificial variable and for ’=’ only artificial variable is required. Thus a total 5 additional variables is required.

Choice (3)

UNIT – ΙΙ

Exercise – 1 Solutions for questions 1 to 50: 1. The radius of the circle ‘r' is the distance between the

centre of the circle to any point on the circle.

∴ r = 22 9)(63)(5 +++ = 22564 + = 289

∴ r = 17 The diameter of the circle = 2r = 2 x 17= 34 units. Choice (2)

2. Given: The distance between (2, 3) and (– 4, k) is 2 ;10

22 k)(34)(2 −++ = 2 10

36 + (3 – k)2 = 4 x 10 (3 – k)2 = 4 ⇒ 3 – k = ± 2 3 – k = 2 or 3 – k = – 2 ⇒ k = 1 or k = 5 ∴ k = 1 or 5 Choice (3)

3. Distance from (0, 0) to (– 2, 4) is 22 4)2( +− = 20

Distance from (0, 0) to (– 1, –1) is 22 11 + = 2

Distance from (0, 0) to (3, 0) is 22 03 + = 3

Distance from (0, 0) to (0, 4) is 240 + = 4

∴ (– 1, – 1) is nearest to origin. Choice (2)


A

E

C D B

(−1, 2) (4, −2)

(3, −4)

F

4. Given: A (1, 2), B (3, 4) and C (5, 2)

AB = 22 2)(4)1(3 −+− = 8

BC = 22 )4(23)(5 −+− = 8

AC = 0 )1(5 2 +− = 16

AB = BC and AC2 = AB2 + BC2 ⇒ A, B, C form a right-angled isosceles triangle. Choice (2) 5. Given: A (– 1, 4), B (– 3, 0), C (1, – 2), D (3, 2)

AB = 20 , BC = 20 , CD = 20 , AD = 20 and

AC = 40 , BD = 40 Clearly AB = BC = CD = AD The length of diagonal AC = length of diagonal BD ∴ ABCD is a square. Choice (4) 6. Let A (3, 2), B (– 5, 4), C(5, – 3) AB2 = 68 BC2 = 149 AC2 = 29 Here BC2 > AB2 +AC2

∴ The given points are the vertices of an obtuse-angled triangle. Choice (2)

Note: if AB2 < BC2 +AC2 AC2 < AB2 + BC2

BC2 < AB2 +AC2, then ABC forms an acute-angled triangle.

7. The ratio in which the point (x, y) divides the line segment joining the points (x1, y1) and (x2, y2) is –(x – x1) : (x – x2) or – (y – y1) : (y – y2) – (5 – 10) : (5 – 3) = 5 : 2 ∴ (5, 10) divides the given line segment in the ratio

5 : 2 internally. Choice (1) 8. Given: points are (– 2, 3), (5, 1) and ratio is 2 : 1.

The required point is given by

−−

−−

nmnymy

,nmnxmx 1212 .

−−

−−−

123.11.2

,12

)2(15.2 = (12, –1) Choice (3)

9. The ratio in which X – axis divides the line segment

joining the points (x1, y1) and (x2, y2) is – y1 : y2. = – 4 : 7 ⇒ 4 : 7 externally Choice (2) 10. If P, Q are the points of trisection of the given line

segment, then P and Q divides the line segment in the ratio 1 : 2 or 2 : 1 respectively.

Using

++

++

nmnymy

,nmnxmx 1212 , we get

P = (1, 2) and Q (– 1, – 3) Choice (3) 11. If P, Q are points of trisection of the line segment

AB, then AB = 3PQ, here P(– 1, 2) and Q (4, – 2).

= 3

+ 22 43 = 3 x 5 = 15 units Choice (2)

12. The mid-point of AB is the same as mid-point of PQ.

The mid-point of PQ is

−−2

42,

217

= (3, – 1).

Choice (1) 13. We know that, the ratio in which (x, y) divides the line

segment joining the points (x1, y1) and (x2, y2) is

– (x – x1) : (x – x2) or – (y – y1) : (y – y2). The ratio in which (3, 5) divides the line segment joining the points (1, 7) and (6, 2) is – (3 – 1) : ( 3 – 6) = 2 : 3. This implies harmonic conjugate of (3, 5), divides (1, 7) and (6, 2)in the ratio – 2 : 3.

The point which divides (1, 7) and (6, 2) in the ratio – 2 :3 is (– 9, 17) Choice (1) 14.

If D, E, F are the mid-points of the sides of the triangle ABC, then the area of ∆ ABC = 4 (area of ∆ DEF)

= 4.2241

2413

21

+−−

−−+ = 2 16–30 = 28 sq. units

Choice (1) 15. If G is the centroid of ∆ ABC, then the area of ∆ ABC = 3.area

of ∆ GAB. Given: A (3, –4), B (7, 2) and G (4, –1).

∴ The area of ∆ ABC = 3.1247

2473

21

+−−−−

= 181223 −

= 3. 621

= 9 sq. units Choice (2)

16. If the points (x1, y1), (x2, y2) and (x3, y3) are collinear, then

the area of the triangle formed by these points is zero.

⇒ 0yyxx

yyxx

3232

2121=

−−

−−

Given: (2, 3) (–4, 5) and (k, 6) are collinear.

065k4

5342=

−−−

−+⇒ ⇒ – 6 – 8 – 2k = 0.

∴ k = – 7 Choice (2) 17. If (x1, y1), (x2, y2), (x3, y3) and (x4, y4) are the vertices of

a quadrilateral, then the area of the quadrilateral

= 2

1

4242

3131

yyxx

yyxx

−−−−

Given: (2, 3), (– 1, 4), (5, – 6) and (7, 2) are the vertices of a quadrilateral.

Area = 21

2471

6352

−−−

+− = 726

21 +− = 33 sq. units

Choice (1) 18. If D, E, F are the mid-points of the sides of the triangle

ABC, then the centroid of ∆ ABC is the same as the centroid of ∆ DEF.

∴ The centroid of ∆ ABC =

++++3

yyy,

3

xxx 321321

Given: D (– 2, 5), E (4, – 1) and F (7, 2)

The centroid of ∆ ABC =

+−++−3

215,

3742

= (3, 2) Choice (4)


A

C B

c b

a (1, – 1) (4, −1)

(1, 3)

D C

B A (−1, 4)

(1, – 2)

A

B C

B

A

C

E F

D

(−1, 2) (x, y1)

(x2, y2) (3, 4)

(x3, y3) (4, 2)

(x4, y4)

19.

The points form a right–angled triangle with its right angle at B (1, 2) and A (– 4, – 1), C (4, – 3) are the ends of hypotenuse.

The orthocentre of ∆ ABC is B (1, 2). (i.e., the vertex of right angle).

The circumcentre of ∆ ABC is the mid-point of the hypotenuse,

i.e., mid-point of AC. Circumcentre S is (0, – 2) ∴ The distance between orthocentre and circumcentre is

BS = 161+ = 17 units. Choice (2) 20.

Given: A (1, 3), B (4, – 1) and C (1, – 1)

a = BC = 9 = 3

b = AC = 16 = 4

c = AB = 25 = 5

The incentre of ∆ ABC where A (x1, y1), B (x2, y2) and C(x3, y3) is

++++

++++

cba

cybyay,

cba

cxbxax 321321

Given: A (1, 3), B (4, – 1), C(1, – 1) and a = 3, b = 4 and c = 5

∴ The incentre is (2, 0). Choice (3) 21. Given: A (1, 3), B (4, –1) and C (1, – 1) a = BC = 3 b = AC = 4 c = AB = 5 The excentre opposite to vertex A is

++−++−

++−++−

cba

cybyay,

cba

cxbxax 321321

= (3, – 3) Choice (1) 22. Given: A (– 1, 4) and C(1, – 2). The length of the diagonal

AC = 22 2)(41)(1 +++ = 40

The area of square =( ) ( )

240

2diagonal

22

= = 20 sq. units

Choice (4)

23. Given: A (– 2, 5), B (4, – 1) and C (7, 2).

AB = 6 ,2 BC = 3 2 and AC = 3 10 and AC2= AB2 + BC2. ∴∆ ABC is a right-angled triangle with the right angle at B. ∴ The orthocentre of ∆ ABC is B (4, – 1) and the circumcentre is the mid-point of the hypotenuse = mid-point of AC

∴ S =

27

,25

∴ Nine point circle centre is the mid-point of Orthocentre and Circumcentre.

i.e.

+−+

227

1,

225

4

∴ N is

45

,4

13. Choice (3)

24. Given: vertices of the triangle are (3, – 4), (–1, 2) and (4, 2)

Centroid (G) is

++−+−3

224,

3413

= (2, 0)

Given: The orthocentre O (3, 4/3) ___________|______|______|_ O N G S ON : NG : GS = 3 : 1 : 2 Ratio in which S divides OG is OS : GS externally 6 : 2, i.e. 3 : 1 externally O(3, 4/3), G (2, 0) ratio 3 : 1 externally

∴ S =

−=

−−32

,23

23/40

,2

36

Alternate method: Circumcentre S(x, y) is equidistant from all the vertices.

⇒ SA = SB = SC. SA = 22 )4y()3x( ++−

SB = 22 )2y()1x( ++− and

SC = 22 )2y()4x( ++−

On solving the above, we get

S =

−32

,23

Choice (2)

25.

Given: D (3, 4), E (4, 2) and F (– 1, 2) are the mid-points of the sides of ∆ ABC. We know FDEA is a parallelogram and A is the fourth vertex of the parallelogram.

x4 = x1 + x3 – x2 y4 = y1 + y3 – y2 ∴ Vertex A is (0, 0) Choice (2)


B

Y

A O

P (x, y)

(0, b)

(a, 0)

26. Let P (x, y) be a point on the locus and given PA = 5 and A (3, 4) PA2 = 25 (x – 3)2 + (y – 4)2 = 25 x2 + y2 – 6x – 8y = 0 Choice (1) 27. Let P (x, y) be a point on the locus. Given: PA = PB, where A (2, – 2) and B(– 2, 2). PA2 = PB2, (x – 2)2 + (y + 2)2 = (x + 2)2 + (y – 2)2 ⇒ x – y = 0 Choice (2) 28. Let P (x, y) be a point on the locus. Given: PB = 2PA and A(1, 1), B(4, 4) PB2 = 4PA2; (x – 4)2 + (y – 4)2 = {4 [(x – 1)2 + (y – 1)2]} ⇒ x2 + y2 = 8 Choice (2) 29. Let P (x, y) be a point on the locus. Given: PA – PB = 2, where A(0, 3), B(0, – 3). PA2 = (2 + PB)2 x2 + (y – 3)2 = 4 + x2 + (y + 3)2 + 4PB ⇒ – 3y – 1 = PB

(3y + 1)2= PB2 ⇒ (3y + 1)2 = x2 + (y + 3)2 = x2 – 8y2 + 8 = 0 Choice (1) 30. Let P (x, y) be a point on the locus. Given: ∠ APB = 90° AB2 = AP2 + PB2 (QAPB is a right-angled triangle) 20 = 2x2 + 2y2 + 10

x2 + y2 = 5 Choice (3) 31. Let P (x, y) be any point on the locus. Given: PA2 + PB2 = 2c2 x2 + (y – 2)2 + x2 + (y + 2)2 = 2c2

x2 + y2 + 4 – C2 = 0 Choice (2) 32. Let P(x, y) be any point on the locus.

Given: area of ∆ POB = 21 area of ∆POA

21

13

yx

21

.21

32

yx

−−=

−;

4(– 3x – 2y)2 = (– x + 3y)2 35x2 + 54xy + 7y2 = 0 Choice (4) 33. Let P(x, y) be a point on the locus. Given: A(– 2, 1),

B(4, – 1), C(1, 0). Given: PA2 + PC2 = 3PB2 (x + 2)2 + (y – 1)2 + (x – 1)2 + y2 = 3 {(x – 4)2 + (y + 1)2} x2 + y2 – 26x + 8y + 45 = 0 Choice (2) 34.

AB is a rod of length ℓ. It meets the axes at A(a, 0) and B(0, b) Let P(x, y) be the mid-point of AB.

Then, =

2b

,2a

(x, y)

a = 2x, b = 2y OAB is a right-angled triangle, ∴ AB2 = OA2 + OB2 ⇒ l2 =a2 + b2

l2 = (2x)2 + (2y)2 4x2 + 4y2 = l2 Choice (3)

35. Let x = a secθ, y = b tan θ

sec θ = a

x, tan θ =

b

y

But, sec2θ – tan2 θ = 1 2

2

2

2

b

y

a

x −⇒ =1 Choice (4)

36. Let x = 2+3 sinθ and y = 2 + 3 cos θ

sinθ = 3

2x − and cos θ =

3

2y −

We have sin2θ + cos2θ = 1

13

2y3

2x22

=

−+

−

(x – 2)2 + (y – 2)2 = 9, which is a circle Choice (1) 37. Translation equations are: X = x – α, Y = y – β Given: (x, y) = (2, 3), (α, β) = (– 4, 5) X = 6, Y = – 2 ∴ The required point is (6, – 2) Choice (2) 38. Translation equations are: x = X + α, y = Y + β Given: (X, Y) = (– 2, – 1), (∝ , β) = (– 3, 1) ∴ The required point (x, y) = (– 5, 0) Choice (2) 39. Given: (x, y) = (– 1, 5), (X, Y) = (3, 1) We know α = x – X, β = y – Y ∴ (α, β) = (– 4, 4) Choice (3) 40. Given: (α, β) = (– 1, 1) and f(x, y) = 2x2 – xy + y2 – 4x +

7y – 5 = 0 The transformed equation is f (X + α, Y + β) = f(X – 1), (Y+1) =0 2(X – 1)2 – (X – 1) (Y + 1) + (Y + 1)2 – 4(X –1) + 7(Y + 1) – 5 = 0

2X2 – XY + Y2 – 9X + 10Y + 10 = 0 Choice (1)

41. Given: (α, β) = (1, 1) and f(X, Y) = 2X2 – 3XY – Y2 – 5 = 0 The original equation is f(x – ∝ , y – β) = f(x – 1, y – 1) = 0 2(x – 1)2–3(x – 1)(y – 1) – (y – 1)2 – 5 = 0 2x2 – 3xy – y2 – x + 5y – 7 = 0 Choice (4) 42. In the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,

after the transformation to remove the first degree terms, the point to which the origin is to be shifted is

−−

−−

22 hab

afgh,

hab

bghf

Comparing the given equation, we have a = 2, b = – 1, c = – 7, 2h = – 3, 2g = –1, 2f = 5. On substitution, we get the point as (1, 1). Choice (4) 43. Rotation equations are X = x cos θ + y sin θ, Y = – x sinθ

+ y cos θ Given: (x, y) = (2√3, 2) and θ = 30° ∴ The point in the new system is (4, 0) Choice (3) 44. The rotation equations are x = X cos θ – Y sin θ,

y = X sin θ + Y cos θ

Given: (X, Y) = (2 2 , – 2 ) and θ = 45°. ∴ The original coordinates are (3, 1) Choice (2) 45. Rotation equations are X = x cos θ + y sinθ and

Y = – x sin θ + y cos θ

Given: (x, y) = (– 2 2 , 5 2 ) and θ = – 45° (Q rotation is clockwise so ‘θ’ is negative)

∴ The coordinates of the point in the new system are (– 7, 3). Choice (1)


A B

C D

E

(1, –1)

(3, 4)

46. Given: f (x, y) = x2 – xy + y2 – 5 2 x = 0 The new equation is f(X cosθ – Y sinθ, X sinθ + Y cos θ) = 0 Given: θ = 45°

∴ The new equation is f

+−2

YX,

2

YX= 0

⇒ 22

2

YX

2

YX

2

YX

2

YX

++

+

−−

−

−−2

YX25 = 0

⇒ x2 + 3y2 – 10x + 10y = 0 Choice (1) 47. Given: f(X, Y) = X2 + Y2 – 2 = 0

The original equation is f(x cos θ + y sin θ, – x sin θ + y cos θ) = 0

where θ = 60° ⇒ f

+−+2

yx3,

2y3x

=0

i.e., 22

yx32

y3x22

=

+−+

+

⇒ x2+y2 = 2 Choice (3) 48. If ‘θ’ is the angle of rotation so that the transformed

equation does not contain ‘xy’ term, then

θ = 21 tan–1

− bah2

.

Here, a = 2, 2h = 3 , b = – 1

∴ θ = 21 tan–1

3

1= ⇒ θ = 15° or Πc/12

Choice (2) 49. The given equation is x2 + y2 – 8x – 6y – 10 = 0 Here a = 1, b = 1, h = 0, 2g=– 8, 2f =– 6, c = –10.

The point to which origin is to be shifted to remove x, y terms in the transformed equation is

−−

−−

22 hab

afgh,

hab

bghf.

Substituting the values, we get the point as (4, 3) The transformed equation is given by f(x + 4, y + 3) = 0 ⇒ (x + 4)2 + (y + 3)2 – 8(x + 4) – 6(y + 3) – 10 = 0 ⇒ x2 + y2 = 35 Choice (2) 50. The given four points form a square. ∴ The required equation is the equation of the diagonal

passing through the points (6, 6) and (– 1, 3).

y – 6 = 73

−−

(x – 6)

7y – 42 = 3x – 18 3x – 7y + 24 = 0 Choice (4)

Exercise – 2 Solutions for questions 1 to 55: 1. The given points are (–1, 2) and (3, –4)

Slope = 12

12

xxyy

−−

= 1324

+−−

∴ The slope of the line = 23−

Choice (2)

2. The given equation of the line is 2x + 2 3 y + 3 = 0

We know that the slope of the line ax + by + c = 0 isb

a−

∴ The slope of the line 2x + y32 + 3 = 0 is = 32

2− =

3

1−

tan θ = 3

1−⇒ θ = 150° or

65 cπ

Choice (2)

3. Given: θ = 135° and the point as (– 1, 2) The equation of the line with inclination θ and passing

through the point (x1, y1) is y – y1 = tan θ (x – x1) ⇒ y – 2 = tan 135° (x + 1) y – 2 = – (x + 1) y – 2 = – x – 1 x + y – 1 = 0 Choice (1) 4. We know that the equation of the line passing through

(x1, y1) and parallel to the x - axis is y = y1 The equation of the required line is y = 4 or y – 4 = 0 Choice (2) 5. The equation of the line joining the points (– 4, – 1) and

(– 3, 1) is y – y1 = 12

12

xx

yy

−−

(x – x1)

y + 1 = 43

11

+−+

(x + 4)

y + 1 = 2(x + 4) ⇒ 2x – y + 7 = 0 But it is given that the ordinate of a point on the line is 3 Substituting y = 3 in the line equation, we have 2x – 3 + 7 = 0 x = – 2 ∴ The abscissa of the required point is – 2

Alternate method: Let the required point be (x, 3) A(– 4, – 1), B(– 3, – 1) and C(x, 3) are collinear. The slope of AB = the slope of AC

)3(x

13)4(3

)1(1−−

−=−−−

−− ⇒

3x2

12

+= ⇒ x + 3 = 1

⇒ x = – 2 ∴ The abscissa is – 2. Choice (3) 6. The equation of the line with slope m and x - intercept a is y = m (x – a) Given: m = – 2 and a = 3 The equation of the line is y = – 2(x – 3) ⇒ 2x + y = 6

Choice (4) 7. The equation of the line parallel to ax + by + c = 0 and

passing through the point (x1, y1) is a (x – x1) + b (y – y1) = 0 The given line is 2x – 3y + 5 = 0 and it passes through

the point (– 3, 4) The required equation of the line is 2 (x + 3) – 3(y – 4) = 0 ⇒ 2x – 3y + 18 = 0 Choice (1) 8. Given: – p2x + 2y – 3 = 0 and – 4px + 2y – 1 = 0 are

parallel If a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are

parallel, then 2

1

2

1

bb

aa =

22

p4p2

=−−

⇒ ⇒ p (p – 4) = 0

⇒ p = 0 or p = 4 Choice (2) 9.


0 A B

C

y

x

θ θ

A

B

O

b

a

y

x

A B

P

Q

C midpoint

l

In a square, diagonals bisect each other at right angles ⇒ E is the mid point of AC and BD and BD⊥ AC

E =

23

,2 , the slope of AC = 2

5

31

41 =−−−

The slope of BD = – 2/5 The equation of BD is

−23

y = 52−

(x – 2)

⇒ 4x + 10 y = 23 Choice (3) 10. If two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are

perpendicular to each other then a1a2 + b1b2 = 0 Given: – (2 λ + 1) x + y + 3 = 0 and – (λ – 1) x + y – 4 = 0 are perpendicular to each other,

⇒ (2 λ + 1) (λ – 1) + 1 = 0 ⇒ 2 λ2 – λ = 0 ⇒ λ = 0 or λ = 1/2 Choice (1) 11.

Let PQ be the perpendicular bisector of AB and Let PQ meet AB at d. ∴ C is the mid point of AB and PQ ⊥ AB

Let A = (– 1, 3), and B = (3, 5) ⇒ C = (1, 4) and the slope of AB = 1/2

Slope of PQ = – 2 ∴ The equation of perpendicular bisector of AB is y – 4 = – 2 (x – 1) 2x + y – 6 = 0 Choice (1)

12. The given lines are 2x + 3 y – 5 = 0 and 3 x – 2y + 1 = 0. These two lines satisfy the condition a1a2 + b1b2 = 0 ∴ The given lines are perpendicular to each other.

Choice (2) 13. From the options, x intercept of 3x + 5y = 2 is 2/3 which

is of the least value. Choice (2) 14.

Let A B C be the triangle with base BC. AB would lie along x - axis ⇒ its slop is zero. The given triangle ABC is isosceles.

Let ∠ BAC = θ the ∠ ABC = θ ∴ AC makes an angle θ with the x–axis BC would make an angle of 180° – θ with the x–axis. The sum of the slopes is tan (180 – θ) + tan θ + tan (180 – ((180 – θ + θ)) = – tan θ + tan θ = 0.

∴ The sum of the slopes of the sides is zero. Choice (3)

15. The given lines are 2x – 3y + 6 = 0 and 3x + y + 7 = 0 Their slopes are m1 = 2/3,.and m2 = – 3 if θ is the acute angle between the lines having slopes

m1 and m2, then tan θ = 21

21

mm1mm

+−

= 3x

32

1

332

−

+

tan θ = 3

11 Choice (4)

16. The point of intersection of the given lines a1x + b1y + c1 = 0

and a2x + b2y + c2 = 0 is

−−

−−

1221

2121

1221

1221

babacaac

,baba

cbcb

The given lines are 2x – y + 5 = 0 and x + y + 1 = 0 Substituting the values in the above formula, we get point of intersection as (– 2, 1) Choice (4)

17.

The area of the triangle formed by the line by

ax + = 1

with coordinate axes as 21

ab where a and b are the

intercepts of the line. Given equation of the line is 3x – 2y + 12 = 0

x – intercept = 3

12a.c =−

= – 4

y – intercept = 2

12bc

−−=−

= 6

∴ The area of ∆ OAB = 1/2 4× – 6 = 12 sq. units Choice (3) 18. The distance from (x1, y1) to a line ax + by + c = 0 is

22

11

ba

cbyax

+

++

The given point is (2, 3) and the line is 3x – 4y – 7 = 0

We get the required distance as 5

13 units

Choice (1) 19. If the image of the point (x1, y1) w.r.t. the line ax + by + c = 0

is (h, k), then 221111

ba

)cbyax(2b

yka

xh

+++−=−=−

Given: (x1, y1) = (– 1, 3) and line equation is 2x – y – 5 = 0

i.e., 2

1h + =

13k

−−

= – 2( )

5532 −−−

⇒ h = 7, k = – 1

∴ The image is (7, – 1) Choice (2) 20. Given: the line has the equation 3x – 2y + 7 = 0 and the

points are (– 1, 4) and (2, 5) Substituting (– 1, 4) in the equation, we have

3 (– 1) – 2 (4) + 7 = – 4 ≠ 0 But – 3 – 8 +7 < 0 ______________(1) Substituting (2, 5) in the equation, we have 3 (2) – 2 (5) + 7 3 > 0____________(2) (1) and (2) have opposite signs ∴ The given points lie

on either side of the given line Choice (2)


4x –

7y +

1=

0

A

B C

y = 5 – x

7x +4y – 15= 0

Required line

x – y3 + 7 = 0

30°

30°

y

x

21. Given: 2x – my + 1 = 0 ; 3x – ny + 1 = 0 and 4x – py + 1 = 0 are concurrent

1p4

1n3

1m2

−−−

⇒ = 0 ⇒ 2(– n + p) + m (3, – 4) +

1 (– 3p + 4n) = 0 ⇒ – 2n + 2p – m – 3p + 4n = 0. ⇒ 2n = m + p ∴ m, n, p are in A.P Choice (1) 22. The given equations of the sides of the triangle are 4x – 7y + 10 = 0___________(1) 4y + 7x – 15 = 0____________(2) y = 5 – x The gradients of (1) and (2) have a product of – 1. ∴ (1) and (2) are perpendicular. from (1) and (2) a1a2+b1b2 = 0 The given equations represent a right angle triangle. The orthocentre of a right angle triangle is the vertex of

the right angle. ∴ Orthocentre is the point of intersection of (1) and (2)

i.e. (1, 2) Choice (2) 23.

We know that the circumcentre of a right angle triangle is the mid point of the hypotenuse i.e. the mid point of AC The point of intersection of 4x – 7y + 10 = 0 and y = 5 – x is

A =

1130

,1125

The point of intersection of 7x + 4y – 15 = 0 and y = 5 – x is

C =

−320

,35

Circumcentre is the mid point of AC =

33155

,3310

Choice (1) 24.

The slope of the line x – 3 y + 7 = 0 is 3

1

tan θ =3

1 ⇒ θ = 30°

The required line makes an angle of 30° with the given line.

There are two possibilities. Possibility 1 :

The required line makes an angle of 60° (30 + 30°) with the

x – axis ⇒ Slope = tan 60° = 3 Given: It passes through point (1, 1)

∴ The equation of the line is y – 1 = 3 (x – 1)

3 x – y – 3 + 1 = 0 Possibility 2 : The required line make an angle of 0° with x – axis. ∴ The slope of the required line is 0 ∴ The equation of the line having a slope of 0 and

passing through the point (1, 1) is y – 1 = 0(x – 1) ⇒ y – 1 = 0 ∴ The equation of the line may be y – 1 = 0

or 3 x – y – 3 + 1 = 0. Choice (3) 25. The given line makes intercepts in the ratio 2 : 3 on the

axes ⇒ If x intercept is 2 k, then y - intercept is 3k.

The equation of the line is 1k3y

k2x =+ . (Intercept form

of the line) It passes through (–1, 2)

⇒ k32

k21 +−

= 1 ⇒ k = 61

The equation of the line is 3x + 2y = 1 Choice (4) 26. Given: The sum of intercepts is zero Let the intercepts

be a and – a

The equation of the line is a

yax

−+ =1 or x – y = a

It passes through (2, –1) ⇒ 2 – (–1) = a a = 3 ∴ The equation of the line is x – y = 3 Choice (1) 27. The point of intersection of the two given lines 2x – y – 4 = 0

and x – 2y – 5 = 0 is (1, – 2). ∴ The equation of the line parallel to ax + by – c = 0 and

passing through (x1 y1) is a (x – x1) + b (y – y1) = 0 The equation of the required line is ⇒ 3 (x – 1) – 1(y + 2) = 0 ⇒ 3x – y – 5 = 0

Choice (2) 28. The point of intersection of the two lines is (1, – 2)

∴ The equation of the line perpendicular to ax + by+ c = 0 and passing through (x1, y1) is b (x – x1) – a (y – y1) = 0

The equation of the line perpendicular to 3x + 5y – 8 = 0 and passing through (1, – 2) is 5 (x – 1) – 3 (y + 2) = 0

i.e. 5x – 3y – 11 = 0 Choice (2)

29. Given: The inclination of the line is 4

3π

θ = 4

3π ⇒ cos θ =

2

1− and sin θ =

2

1. Any point on

the line with inclination θ and at r units from the point p x1, y1 on the line is given by

x = (x1 ± r cos θ); y = (y1 ± r sin θ)

Here, (x1, y1) = (2, 1), r = 2

∴ x = 2 ± 2 .

−2

1 = 1 or 3

and y = 1 ± 2

2

1= 2 or 0

∴ The required points are (1, 2) or (3, 0) Choice (3) 30. 2a + b – c = 0 ⇒ – 2a – b + c = 0 ∴ a (-2) + b (-1) + c = 0 ∴ ax + by + c = 0 passes through (– 2, – 1) Choice (4)


A (1, 0)

D (0, –1)

C (–1,0)

B (0, 1)

y

y1

O x1

x

D

A

C

B

X

Y

y =4

y =3

x =2 x =3

31.

The given equation represents four lines which are

x + y = 1, – x + y = 1, x – y = 1 and – x – y = 1 The figure formed by these lines is a square with side

AB = 2 Area = (side)2 = (AB)2 = 2 sq units Choice (1)

32. Given: x + y – 3 = λ and 5x – y – 7 = 3λ Eliminating λ from the above equations, we have

x + y – 3 = 3

7yx5 −−⇒ 2x – 4y + 2 = 0

or x – 2y + 1 = 0 Choice (2) 33. The given curve is 5x2 + 12xy – 8y2 + 8x – 4y + 12 = 0 –→ (1) The line is x – y = k

1k

yx =− –––→ (2)

Homogenising the curve (1) with the help of the line (2)

5x2 + 12xy – 8y2 + (8x – 4y)

−k

yx + 12

2

kyx

−= 0

Since this represents a pair of lines equally inclined to the co-ordinate axes, coefficient of xy = 0

{Lines are equally inclined to the axes if the axes are its angular bisectors}.

∴ 12 – 0k

24k4

k8

2=−− ⇒ k2 – k – 2 = 0

⇒ (k – 2) ( k + 1) = 0 ⇒ k = 2 or k = – 1 Choice (2) 34. If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair

of lines, then ∆ ≡ abc + 2fgh – af2 – bg2 – ch2 = 0 Here, a = 12, b = 2, c = 12, 2h = – 10, 2g = 2k, and 2f = – 5 ∴ ∆ = 2k2 + 25 k + 77 = 0 None of the first three choices satisfies the equation

given above. Choice (4) 35. If θ is the angle between the pair of lines ax2 + 2hxy +

by2 = 0, then tan θ = baabh2 2

+−

Here, 2h = – 2 , a = 2, b = –1

∴ tan θ = 10 θ = tan-1 10 Choice (4)

36. If m1 and m2 are the slopes of the lines represented by

the equation ax2 + 2hxy + by2 = 0, then m1 + m2 = b

h2−;

m1m2 = a / b

Given: m1 = 1/2 m2 or 2m1 = m2

⇒ m1 + 2m1 = b

h2−

⇒ m1 = b3h2−

_______________(1)

and m1 . 2m1 = ba

⇒ m 21 =

b2a

______________(2)

substituting m in (1) in equation (2),

b2a

b3h2

2

=

−

b2a

b9

h42

2

= 8h2 = 9 ab Choice (3)

37. Given: x2 – 5x + 6 = 0 and y2 – 7y + 12 = 0 (x – 2) (x – 3) = 0 and (y – 3) (y – 4) = 0 x = 2 or x = 3 and y = 3 or y = 4

x = 2θ, x = 3y = 3θ and y = 4 form a square of side 1 unit ∴ The area of the square = 1 sq. unit Choice (2) 38. The given equation 4x2 + 12 xy + 9y2 – 6x – 9y +1 = 0 represents a pair of parallel lines, so they never intersect. Choice (4) 39. The point of intersection of the pair of lines 12x2 + 7xy –

12y2 = 0 is the origin. The lines represented by the pair of lines are perpendicular.

Also the point of intersection of 12x2 + 7xy – 12y2 – x +

7y – 1 = 0 is obtained by

−−

−−

22 hab

afgh,

hab

bghf

∴ Here, a = 12, b = – 12, c = –1, 2h = 7, 2g = – 1, and 2f = 7

The point of intersection is

−257

,25

1

Clearly, the figure formed by the given equation of lines is a square and the distance from the origin to their

point of intersection

−257

,25

1 gives the length of the

diagonal (d)

∴ d = 252

The area of the square with diagonal d is 252

22d2

×=

= 251

= 0.04 sq. units

Alternate method: Since the quadrilateral is a square, the diagonal not

passing through the origin is x – 7y + 1 = 0


The distance of the diagonal from (0, 0) to this diagonal

is 25

1 which is half of the diagonal

Diaonal = 25

2

Area = 251

2)Diagonal( 2

= = 0.04 sq units

Choice (3) 40. If ax2 + 2hxy + by2 = 0 represents a pair of perpendicular lines, then the coefficient of x2 + coefficient of y2 = 0 ⇒ 3 + k = 0 ⇒ k = – 3 Choice (2) 41. The co efficient of x2 + coefficient of y2 = 0. The angle

between the pair of bisectors of the angle between the given pair of lines is always 90°

θ = π / 2 Choice (3) 42. If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair

of parallel lines, h2 = ab _______ (1) and af2 = bg2 ____________ (2) Here, a = 9, b = b, c = –12, 2h = –24, 2g = c, and

2f = 16 substituting these in (1) and (2), we get b = 16 and c = – 12 Choice (4) 43. The distance between the parallel lines represented by

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is given by

2)ba(a

acg2

+−

Here, a = 4, c = 1, 2g = – 6, and b = 9.

On substitution, we get 135

units. Choice (4)

44. The area of the triangle formed by the pair of lines

ax2 +2hxy + by2 = 0 with line ℓx + my + n = 0 is

22

22

bmh2am

abhn

ll +−−

________(1)

Here, ℓ = 2, m = – 1, n = 1, a = 1, 2h = – 7, and b = 12

Area = 701

sq. units. Choice (3)

45. The product of the perpendiculars from ( α, β) to the

pair of lines ax2 + 2hxy + by2 = 0 is given by

22

22

h4)ba(

bh2a

+−

β+αβ+α

Here, (α, β) = (1, 1), a = 1, b = 3, and 2h = – 6

364

)1(3)1)(1(61 2

++−

= 10

1

102

2 = units

Choice (2) 46. 6x2 – xy – 5y2 = 0 ______________ (1) 3x2 – 5xy + py2 = 0 ______________ (2) ⇒ (x – y) (6x + 5y) = 0 Given: One line is common in (1) and (2) If x – y = 0 is a common line, then x = y satisfies (2), ⇒ 3y2 – 5y2 + py2 = 0 y ≠ 0 ⇒ p = 2 If 6x + 5y = 0 is common in both equations,

6x = – 5y ⇒ y = 5

x6−

3x2 – 5xy + py2 = 0

3x2 – 5x

−5x6

+ p

2

5x6

− = 0

3x2 + 6x2 + 25

x36 2

p = 0

25

p36 = – 9

p = 36

259 ×− = 4

25− .

∴ p = 4

25− or 2. Choice (1)

47. The area of an equilateral triangle formed by ax + by +

c = 0 with the pair of lines (ax + by)2 – 3 (bx – ay)2 = 0 is

given by )ba(3

c22

2

+ a = 1, b = 1, and c = – 1

∴ The area = 32

1 sq. units Choice (3)

48. The angle between the lines in each pair is 90°____ (1)

Q coefficient of x2 + coefficient of y2 = 0 ∴ The figure is either of a rectangle or a square. The point of intersection of 12x2 + 7xy – 12y2 = 0 is (0, 0) and 12x2 + 7xy – 12y2 – x + 7y – 1 = 0 is

−−

−−

22 hab

afgh,

hab

bghf Here a = 12, b = – 12, c = – 1, 2h = 7,

2g = – 1, and 2f = 7 we get

the point of intersection as

−257

,25

1

The slope of the diagonal joining (0, 0) and

−257

,25

1is – 7

The equation of one of the diagonal is obtained by substituting one of the given equations from the other, i.e. x – 7y + 1 = 0

Its slope is 71

(– 7) )71

( = 1

∴ x – 7y + 1 = 0 must be perpendicular to the other diagonal.

The given figure is a square. Choice (4) 49. Homogenising the equation 3x2 + 5xy – 3y2 + 2x + 3y = 0 using the equation 3x – 2y = 1, we get 3x2 + 5xy – 3y2 + 2x (3x – 2y) + 3y (3x – 2y) = 0 x2 ( 3 + 6) + xy ( 5 – 4 + 9) + y2 ( –3 – 6) = 0 9x2 + 10 xy – 9y2 = 0 In the above equation, coefficient of x2 + coefficient of y2 = 0

⇒ ∠ AOB = c

2π or 90° Choice (4)

50. Homogenising x2 + y2 = c2 using the line ℓx + my + 1 = 0

we have x2 + y2 = c2 ( – (lx + my))2 i.e. x2 ( 1 – l2c2) – 2c2lmxy + y2 (1 – c2m2) = 0 The given angle between the above pair of lines is 90° ⇒ coefficient of x2 + coefficient of y2 = 0 ⇒ 1 – ℓ 2c2 + 1 – c2m2 = 0 2 – c2 (ℓ2 + m2) = 0 or c2 (ℓ2 + m2) = 2 Choice (1) 51. Homogenising 25x2 + 25y2 = 4 using 3x + 4y = 1 we

have 25x2 + 25y2 = 4 (3x+ 4y)2 ⇒ 11x2 + 96xy + 39y2 = 0


if θ is the angle between the pair of lines

ax2+2hxy+by2=0 then cos θ = 22 h4)ba(

ba

+−

+

Here, a = 11, b = 39, and 2h = 96

cos θ = 21

10050

)96()3911(

391122

==+−

+

⇒ θ = 60° or c

3π Choice (1)

52. Given: x2 – 4xy – y2 = 0 bisects the angle between the

pair of lines. x2 – 2pxy – y2 = 0 ___________(1) The equation of the pairs of bisectors of the angle

between the pair of lines ax2+2hxy+by2 = 0 is given by h (x2 – y2) – xy ( a – b) = 0

Here, a = 1, b = – 1, and 2h = – 4 – 2 (x2 – y2) – xy (1+ 1) = 0 ⇒ x2 – y2 + xy = 0 ___________ (2)

but (1) and (2) represent the same equation so

comparing like coefficients, we have 1p2

11 −=

⇒ p = – 1/2 Choice (1) 53. Given: The equation of pair of lines is 3x2 – 8xy – 3y2 = 0 If m1, and m2 are the slopes of the lines represented by

ax2 + 2hxy + by2 = 0, then m1 + m2 = b

h2− and m1m2

= ba

Here m1 + m2 = 38

−−

and m1 m2 = –1

m1 – m2 = 212

21 mm4)mm( −+ = )1(438

2

−−

−

The difference of the slopes = m1 – m2 = 10 / 3 Choice (1) 54. The point of intersection of the lines 2x + 3y = 5 and y = 3 is A (– 2, 3) Let P(h, k) is the foot of the perpendicular of the line ℓ from the origin ‘O’

The slope of OP = wk

The slope of PA = h2

k3−−

−

Since PA is perpendicular to OP, slope of PA × slope of OP = – 1

hk

h2k3 ×

−−−

= – 1

3k – k2 = h2 + 2h h2 + k2 + 2h – 3k = 0 ∴ The locus of p(h, k) is x2 + y2 + 2x – 3y = 0. Choice (2) 55. When θ = 45°, the given pair of straight lines becomes

x2

+21

1 + 2xy (1) + 2

y2

= 0 ⇒ 2

3x2 + 2xy +

2y2

= 0

⇒ 3x2 + 4xy + y2 = 0 ………. (1) The equation of the pair of angular bisectors of ax2 + 2h

xy + by2 = 0 is h(x2 – y2) = (a – b) xy ⇒ 2(x2 – y2) = (3 – 1) xy ⇒ x2 – y2 – xy = 0 ……… (2)

Given: y = m1

x bisector of (1)

Therefore (1) satisfies (2)

substituting y = m1

x in (2), we get,

x2 – mx

xm

x2

2×− = 0

m2 – m – 1 = 0 m2 – m – 1 = 0 m2 – m = 1 ∴ The value of 4m2 – 4m is 4. Choice (4)

Exercise – 3 Solutions for questions 1 to 50: 1. Given: 2x2 + 2hxy + by2 – 2x + y – 5 = 0 represents a circle ⇒ coefficient of x2 = coefficient of y2 ⇒ b = 2 and coefficient of xy = 0 ⇒ h = 0 b = 2, h = 0. Choice (2) 2. Given: The equation of the circle 2x2 + by2 – 4x + 8y – 40 = 0 ∴ Coefficient of x2 = coefficient of y2 ⇒ b = 2 Converting it into the standard form by dividing it by 2, we have x2 + y2 – 2x + 4y – 20 = 0 The centre (– g, – f) = (1, – 2)

The radius = )20()2(1cfg 2222 −−−+=−+

r = 5 units. Choice (1) 3. Given: The centre (3, – 1) and a point on the circle (2, – 1) radius = distance between the centre and any point on the circle.

⇒ r = 0)23( 2 +− = 1

The equation of the required circle is (x −3)2 + (y + 1)2 = 12 x2 + y2 – 6x + 2y + 9 = 0 Choice (2) 4. We know that concentric circles differ by constant term. ∴ The required equation of the circle is x2 + y2 + 3x – 6y + k = 0 → (1) Given: It passes through (– 2, 3) Substituting this point in (1), we get, k = 11 ∴ The equation of the required circle is x2 + y2 + 3x – 6y + 11 = 0. Choice (4) 5. Given: lines 2x – 3y = 5 and x – y = 2 are along the

diameters of the circle. ⇒ The point of intersection of them, i.e. (1, – 1) is the centre of the circle. Also given:

area of the circle = 22 sq. units πr2 = 22

⇒ r = 7 units ∴ The equation of the circle with centre (1, – 1) and

radius 7 is (x – 1)2 + (y + 1)2 = 7. Choice (2) 6. Since it is given that the line segment joining the points

(– 3, 3) and (– 1, 1) subtends right angle at every point on the circle,

⇒ the given points are the end points of the diameter of the circle.

∴ The equation of the circle with ends of diameter as (x1, y1) and (x2, y2) is (x – x1) (x – x2) + (y – y1) (y – y2) = 0 Here, (x1 y1) = (– 3, 3), (x2, y2) = (– 1, 1)

On substitution, we get the equation of the circle as (x + 2)2 + (y – 2)2 = 2. Choice (2)

(0,0)

p (h, k) (– 2, 3) ℓ


C (1,1)

X

Y

A(1, 0)

B (0, 1)

O

(a, 0)

Y

(0, b)

X

• C

2b

,2a

O

7. Given: the circle touches the axes at (1, 0) and (0, 1) ⇒ The centre of the circle is (1, 1) and radius = 1 ∴ The equation of the circle is (x – 1)2 + (y – 1)2 = 1 (or) x2 + y2 – 2x – 2y + 1 = 0. Choice (2) 8.

Given: The intercepts on axes are OA = a and OB = b

⇒ The centre of the circle becomes

2b

,2a

The radius of the circle is 4

b4

aOC

22

+=

∴ The equation of the circle is

+=

−+

−4

b4

a2b

y2a

x2222

(or) x2 + y2 − ax − by = 0 Note: It is a standard form. Choice (1) 9. The centre of the circle is (3, − 2) Given: one end of the diameter is (− 1, 1)

If (x1, y1) is the other end of the diameter, then the mid-point of it is the centre.

⇒

+−2

1y,

21x 11 = (3, − 2)

⇒ x1 = 7, y1 = – 5 ∴ The other end of diameter is (7, − 5) Choice (3) 10. To know the position of the point (x1, y1) we have to find S11 value. Given: point (– 1, 5) and the equation of the circle S = x2 + y2 – 3x + 2y – 7 = 0 S11 = 1 + 25 + 3 + 10 – 7 > 0 ⇒ The point lies outside the circle. Choice (2)

Note: 1. If S11 = 0, then the given point lies on the circle. 2. If S11 < 0, then the given point lies in the interior of

the circle. 11. The given parametric equations are x = a + r Cosθ, and y = b + r Sinθ ⇒ Centre is (a, b) and radius is r.

Comparing the given equations, we get the centre of the circle as (4, 3) and radius r = 5. Choice (2)

12. Given centre (2, 1) and the equation of the tangent to the circle is 3x + 2y + 5 = 0 Its radius = perpendicular distance from centre to line

r = 22 23

51x22x3

+

++

r = 13

∴ The equation of the circle with centre (2, 1) and

radius 13 is (x – 2)2 + (y – 1)2 = ( 13 )2 x2 + y2 – 4x – 2y – 8 = 0. Choice (3) 13. Standard result i.e. c2 = b2(1 + m2) Choice (3) 14. Given: The equation of the circle is x2 + y2 – 4x + 6y – 3 = 0 If the line x – 3y + 1 = 0 intersects the circle at the point of intersection circle and line satisfies x = 3y − 1 Substituting in the circle equation, we get 5y2 – 6y + 1 = 0 ⇒ (y – 1) (5y – 1) = 0

y = 1 or y = 51

Substituting y = 1 in x = 3y – 1 we have x = 2 ∴ The point of intersection of the line and circle in the

first quadrant is (2, 1) Choice (4) 15. The equation of tangent to circle S = 0 at point (x1, y1) is given by S1 = 0 Here, (x1, y1) = (– 2, 1) and S = x2 + y2 – 6x – 4y – 13 = 0 The equation of tangent is – 2x + y – 3 (x – 2) – 2(y + 1) – 13 = 0 i.e. 5x + y + 9 = 0. Choice (3) 16. Given: line 5x + y + k = 0 and circle x2 + y2 – 6x – 4y –

13 = 0 has only one point in common. ⇒ The given line is tangent to the circle.

⇒ Radius = perpendicular distance from the centre to the line.

i.e., =++ 1323 22

22 15

K23x5

+

++

on simplifying we get k = 9 or – 43. Choice (2) 17. Let the equation of the required circle be x2 + y2 + 2gx + 2fy + c = 0 Given: it is passing through the points (– 2, 3), (1, 2) and (– 1, 1) (– 2, 3) which lie on the circle. ⇒ 4 + 9 – 4g + 6f + c = 0 ⇒ – 4g + 6f + c = – 13 ………… (1) Similarly, (1, 2), (– 1, 1) lie on the circle ⇒ 2g + 4f + c = – 5 ………… (2) and – 2g + 2f + c = – 2 ……..(3)

Solving (1), (2) and (3), we get g =21

, f = 25−

and c = 4

∴ The required equation of the circle is

x2 + y2 + 2 21

x + 2.2

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

Choice (1) 18. Centres C1 and C2 of the given circles are C1 = (2, 3) and C2 = (– 3, – 9) and radii r1 and r2 are respectively 5 and 8.

d = c1 c2 = 13215 22 =+

Clearly, d = r1 + r2 ⇒ The given circles touch each other externally.

Hence 3 common tangents can be drawn. Choice (3)


D

Y

X A B

C

O p1x + q1y + c1 = 0

p2x + q2y + c2 = 0

A B

C

d r

19. The radius of the given circle r = 10 The length of the tangent from a point (x1, y1) to circle

x2 + y2 = r2 is given by 11S = 10

If θ is the angle between tangents drawn from the point (4, – 2) to the circle x2 + y2 = 10, then

θ = 2Tan-1

11S

r = 2Tan-1

10

10 = 2

π4

θ = 2π

Choice (4) 20. Let the equation of the tangent parallel to 4x – 3y – 1

= 0 be 4x – 3y + k = 0 ⇒ radius = perpendicular distance from the centre to line.

25 = 25

k3x32x4 +−

On simplifying, we get (k – 1)2 = 252 k – 1 = ± 25 k = 26 or – 24. ∴ The required equation of the tangent is 4x − 3y − 24 = 0 Choice (2) 21. The point of contact is the foot of the perpendicular from

the centre (3, 2) on the tangent 5x + y – 43 = 0. Let it be (h, k)

∴ 26

)43215(1

2k5

3h −+−=−=− = 1

∴ The point of contact is (8, 3). Alternate method: Let P(x1, y1) be the point of contact, then the equation of

tangent at P to circle S = 0 is S1 = 0 ⇒ xx1 + yy1 − 3(x + x1) − 2(y + y1) − 12 = 0 x (x1 − 3) + y (y1 − 2) − 3x1 − 2y1 − 12 = 0 --- (1) But the tangent is given as 5x + y − 43 = 0 ---- (2) (1) and (2) represent the same, so on comparing the

coefficients of like terms.

we have ( )

4312y3x2

12y

53x 1111

−−+−=−=−

Solving for x1 and y1, we get (8, 3). Choice (3) 22. If a circle touches x-axis at (α, 0), then (x – α)2

= x2 + 2gx + c The given point (3, 0) ⇒ x2 + 2gx + c = (x – 3)2 ⇒ g = – 3, c = 9.

Given: y-interscept = 8 ⇒ 2 8cf2 =− ⇒ 2 89f2 =−

On simplification, we get f = 5 ∴ The equation of the required circle is

x2 + y2 – 6x + 10y + 9 = 0

Alternate method:

∴ The radius = 5 The centre = (3, ± 5) ∴ The required equation is (x − 3)2 + (y m 5)2 = 25 i.e., x2 + y2 − 6x m 10y + 9 = 0 Choice (4) 23.

If two lines intersect coordinate axes at 4 concyclic

points A, B, C and D, then we have OA.OB = OC.OD. ⇒ p1q1 = p2q2 (standard result). Choice (1) 24.

If d is the perpendicular distance from centre C of the circle to chord AB and r is the radius of the circle, then the length of the chord,

AB is given by 2 22 dr −

Here, r = 52014 =++ and

d = 59

5164 =−+

The length of chord AB = 2 859

52

2 =

−

= 54452

= 3458

units Choice (2)

25. The equation of the circle with (1, t) and (t, 1) as ends

points of the diameter is (x – 1) (x – t) + (y – t) (y – 1) + λ (x + y – (1 x t)) = 0 (t, t) lies on it.

⇒ λ = 0 ∴ The equation of the required circle is (x – 1) (x – t) + (y – t) (y – 1) = 0

Clearly (1, 1) lies on it. Alternate method: Let the equation of the circle through the points (1, t)

(t, 1) and (t, t) be x2 + y2 + 2gx + 2fy + c = 0 --- (1) As (1) passes through the given points, we have 1 + t2 + 2g + 2ft + c = 0 --- (2) t2 + 1 + 2gt + 2f + c = 0 --- (3) and t2 + t2 + 2gt + 2ft + c = 0 --- (4) solving (2), (3) and (4), we get

( ) ( )2

t11f,

2t1

g+−=+−= and c = 2t

∴ The circle equation is x2 + y2 − (1 + t) x − (1 + t) y + 2t = 0 or x2 + y2 − x − y − t (x + y − 2) = 0 Clearly for different values of t, x2 + y2 − x − y = 0 and x + y − z = 0 passes through the point (1, 1).

Choice (1)

(3, 0)

(3, 5) 3

4 5

X

Y

O


B

d C1

r1

A

P(x1, y1)

(0, 0)

B A

O c c

26.

In ∆OAB, AB2 = 2c2 ⇒ (2PB)2 = 2c2 ⇒ PB2 = c2/2 In ∆OPB, OP2 = OB2 − (PB)2 = c2 − [c2/2] x1

2 + y12 = c2/2

∴ The required locus is x2 + y2 = c2/2.

Alternate method: Let P(x1, y1) be the mid point of the chord of the circle

x2 + y2 = c2 --- (1) Then the chord equation is S1 = S11 ⇒ xx1 + yy1 = x1

2 + y12 --- (2)

Homogenizing (1) using (2), we have

x2 + y2 = c2

+

+2

1

2

1

11

yx

yyxx

But the given angle between the above pair of lines is 90° ⇒ coefficient of x2 + coefficient of y2 = 0

( ) ( ) 0yx

yc1

yx

xc1

21

21

21

2

221

21

21

2

=+

−++

−

The locus of (x1, y1) is ( )

( )222

222

yx

yxc

+

+ = 2

The or c2 = 2 (x2 + y2) Choice (3) 27.

The equation of the common chord is radical axis

⇒ S − S1= 0 4x – 3y = 0 → ( 1)

C1(– 2, 0), C2

−23

,0 and r1 = 2, r2 = 23

d = perpendicular distance from C1 to AB.

⇒ d = 58

5

8=

−

∴ The length of the common chord is 22 dr2 −

= 22564

4 − = 5

125

6x2 = units. Choice (2)

28. Let the pole be (x1, y1) The equation of polar w.r.t. circle S = 0 is S1 = 0 ⇒ xx1 + yy1 – 3(x +x1) – 4(y + y1) + 5 = 0 x(x1 – 3) + y (y1 – 4) – 3x1 – 4y1 + = 5 = 0 . . . (1) But the polar is given as 3x + 4y – 45 = 0 . . . . (2) Comparing (1) and (2) we have

45

)5y4x3(

4

4y

3

3x 1111

−−+−

=−

=−

Solving the above, we get (x1, y1) = (6, 8) Choice (2)

29. If (x1, y1) and (x2, y2) are conjugate with respect to circle S = 0 then S12 = 0 Given: x2 + y2 = 5 and points (1, 2) and (k, 1) ⇒ k + 2 – 5 = 0 k = 3. Choice (4) 30. If the lines l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0 are conjugate with respect to the circle x2 + y2 = r2 then r2(l1l2 + m1m2) = n1n2 → (1) The given equations of circle and lines are x2 + y2 = 9, 3x – ky – 1 = 0 and 2x + y – 9 = 0 Substituting the values in (1), we have 9(3 x 2 + (– k).1) = – 9 x –1 ⇒ k = 5. Choice (2) 31. Let P(x1,y1) be the pole Given: it lies on circle x2 + y2 = a2 ⇒ x1

2 + y12 = a2 → (1)

The polar of P with respect to circle x2 + y2 = b2 is S1 = 0 ⇒ xx1 + yy1 – b2 = 0 → (2) Given: (2) touches the circle x2 + y2 = c2 If a line y = mx + c touches the circle x2 + y2 = r2, then c2 = r2(1+m2)

(2) can be written as y =1

2

1

1

yb

xy

x+−

⇒

+=

21

212

2

1

2

y

x1c

yb

⇒ b4 = c2(x12+ y1

2)

⇒ b4 = c2a2 or b2 = ac. ∴ from (1) Choice (2) 32. The inverse of a point P is the foot of the perpendicular

of P on to its polar with respect to circle. Polar is S1 = 0 i.e., −x + 3y – 4 = 0 x − 3y + 4 = 0 The foot of perpendicular of (– 1, 3) is

53

10)491(

33k

11h =+−−−=

−−=+

The inverse point is

−56

,52

Alternative method:

Given: point (− 1, 3) and circle x2 + y2 = 4 The point of intersection of polar of P w.r.t to circle and

the line joining P and the centre of the circle is the inverse point of P

The polar of P w.r t. circle x2 + y2 = 4 is S1 = 0 ⇒ − x + 3y − 4 = 0 --- (1) The equation of a line joining P and the centre of the

circle (0, 0) is y = − 3x ⇒ 3x + y = 0 --- (2)

Solving (1) and (2), we have P =

−56

,52

Choice (3) 33. Centre C1 and radius r1 of the circle x2 + y2 – 8x – 6y + 21 = 0 are C1 = (4, 3), r1 = 2 Similarly centre C2 = (0, 1) and r2 = 4 If d is the distance between the centres C1 and C2

d = C1C2 = 20

If θ is the angle between two circles, then

Cosθ = 21

22

21

2

rr2rrd −−

Cosθ = 04.2.2

16420 =−−

⇒ θ = 2π

Choice (4)


34. If two circles x2 + y2 + 2gx + 2fy + c = 0 and x2 + y2 + 2g1x + 2f1y + c1 = 0 cut each other orthogonally, then 2gg1 + 2ff1 = c + c1

Here, g = – 4, f = 3, c = -5 and g1 = 2

k, f1 = – 1, c1 = – 9

on substitution, we get k = 2. Choice (4) 35. Given circles are (x – 4)2 + (y – 3)2 = 4 and

x2 + (y – 1)2 = 16 C1 = (4, 3), r1 = 2 and C2 = (0, 1), r2 = 4

d = C1C2 = 20 Clearly d < r1 + r2 The given circles intersect each other. ∴ Only the external centre of similitude exists and we

know external centre of similitude divides the line of centres in r1 : r2 ratio externally.

C1(4, 3) C2(0, 1) and r1 : r2 = 2 : 4 = 1 : 2 externally ∴ The external centre of similitude =

−−

−−

nmnymy

,nmnxmx 1212 , we get as (8, 5)

Choice (1) 36. Given: circles are x2 + y2 + 2x – 6y + 9 = 0……… (1) x2 + y2 – 2x + 2y – 7 = 0 …………. (2) and x2 + y2 – 14x + 8y + 61 = 0 ………. (3) The radical axis of (1) and (2) is (1) – (2) i.e., 4x – 8y + 16 = 0 or 2x – 4y + 8 = 0…………(4) The radical axis of (2) and (3) is (2) – (3) i.e., 12x – 6y – 68 = 0 or 6x – 3y – 34 = 0 ………….(5)

Solving (4) and (5), we get x = 9

80 and y =

958

∴ The radical centre of the given circle is

958

,9

80

The length of the tangent from radical centre

958

,9

80

to circle (2) is 11S

= 81

88017

9116

9160

813364

816400 =−+−+

The circle with radius 11S - 81

8801 and centre as

radical centre

958

,9

80 interests all the given circles

orthogonally. ∴ The required circle equation is

222

818801

958

y9

80x

=

−+

−

i.e., 9(x2 + y2) – 160x – 116y +107 = 0. Choice (2) 37. If a circle cuts two circles orthogonally, then its centre

lies on their radical axis. The radical axis is –6x – 2x + 6y + 6 = 0 4x – 3y – 3 = 0 The centre lies on 2x + y + 1 = 0 ∴ Centre = (0, – 1) The required equation is x2 + y2 + 2y = 0. Choice (3) 38. Given: The circles’ equations are x2 + y2 – 6x – 4y – 3 = 0 → (1) and x2 + y2 + 10x + 12y + 61 = 0 → (2) The radical axis is given by (2) – (1), i.e. x + y + 4 = 0 → (3)

Clearly the mid point of the line segment joining the points in option (2) satisfies equation (3) ∴ The points in option (2) are the limiting points of the given system of circles. Choice (2) 39. Given: (1, 2) is a limit point. The limiting point circle is (x – 1)2 + (y – 2)2 = 0 i.e. x2 + y2 – 2x – 4y + 5 = 0 → (1) Given: The member of coaxial system is

x2 + y2 + 2x – 6y = 0 → (2) The radical axis of the system is (2) – (1) ⇒ 4x – 2y – 5 = 0 We know one limit point is the image of the other with respect to radial axis. If (h, k) is the other limit point, then by image formulae,

we have, 416

)544(222k

41h

+−−−=

−−=−

⇒ h = 3 and K = 1 ∴ The other limit point is (3, 1) Choice (2) 40. Given: The limiting points of the coaxial system are

(2, 1) and (– 5, – 6) The limiting point circles are (x – 2)2 + (y – 1)2 = 0 x2 + y2 – 4x – 2y + 5 = 0 → (1) and (x + 5)2 + (y + 6)2 = 0 x2 + y2 + 10x + 12y + 61 = 0 . . . . (2) The radical axis is S – S1 = 0 ⇒ (2) – (1) The radical axis is x + y + 4 = 0. Choice (1) 41. If two circles touch each other, then their radical axes

are their common tangents. The equation of radical axis is 2ax – 2by = 0 ⇒ ax – by = 0

The centre C1 ≡ (–a, 0), r1 = ca2 −

∴ caba

a 2

22

2

−=+

−

∴ a4 = (a2 – c) (a2 + b2) = a4 – c(a2 + b2) + a2b2 ∴ c(a2 + b2) = a2b2

∴ c1

b

1

a

122

=+

Alternate method:

C1= (− a, 0), car 21 −= and C2 = (0, − b), cbr 2

2 −=

d = C1C2 = 22 ba + Given: The two circles touch each other. ⇒ d = r1 ± r2 Taking the square on both sides d2 = r1

2 + r22 ± 2r1r2

a2 + b2 = a2 − c + b2 − c ± ( )( )cbca2 22 −−

⇒ ( )( )cbcac 22 −−=

Taking the square again on both sides, we have c2

= a2b2 − c (a2 + b2) + c2 ⇒ c (a2 + b2) = a2b2. Choice (2) 42. Given: The equations of the circles are (x – 1)2 + (y – 1)2 = 3 or x2 + y2 – 2x – 2y – 1 = 0 . . . (1) and (x + 7)2 + (y + 3)2 = 83 or

x2 + y2 + 14x + 6y – 25 = 0 . . . . (2) We know that for intersecting system of circles, common

chord is the radical axis which is given by (2) – (1) ∴ The radical axis (or) common chord is 2x + y – 3 = 0 Choice (3)


43. Given the circles equations are x2 + y2 – 12y + 20 = 0…..(1) 2x2 + 2y2 + 20x + 32 = 0 ……….. (2) and x2 + y2 – 8x + 12 = 0 ………. (3) The radical axis of (1) and (2) is (1) – (2) 10x + 12y – 4 = 0 i.e. 5x + 6y – 2 = 0 ……………. (4) The radical axis of (2) and (3) is (2) – (3), i.e. 18x + 4 = 0

⇒ x = 92−

Substituting x = 92−

in (4), we get y = 27

14

The point of concurrence of radical axes is radical centre

∴ The radical centre of the given circles is

−2714

,92

Choice (4) 44. The radical axis of the two given circles is S – S1 = 0 ⇒ x2 + y2 – 6 – x2 – y2 – 4y + 1 = 0 ⇒ – 4y – 5 = 0 ⇒ 4y + 5 = 0

The equation of the circle through the points of intersections of the two given circles S = 0 and S1 = 0 whose radical axis is L = 0 is S + λL = 0

i.e. x2 + y2 – 6 + λ(4y + 5) = 0 Given: The above equation passes through origin ⇒ – 6 + 5λ = 0

λ = 56

∴ The equation of the required circle is

x2 + y2 – 6 + 56

(4y + 5) = 0

5(x2 + y2) + 24y = 0 Choice (1) 45. We know that the equation of the circle through the

points of intersection of the line L = 6x – 8y – 11 = 0 and the circles = x2 + y2 – 4x + 6y – 12 = 0 is given by S + λL = 0

⇒ x2 + y2 – 4x + 6y – 12 + λ(6x – 8y – 11) = 0 . . . . (1) x2 + y2 + x(– 4 + 6λ) + y(6 – 8λ) – 12 − 11λ = 0

Its centre is

λ−−λ+−−2

)43(2,

2)32(2

If the line is the diameter, then the centre of the circle lies on it ⇒ − 6(– 2 + 3λ) + 8(3 − 4λ) – 11 = 0 12 − 18λ + 24 – 32 λ − 11 = 0 − 50λ = −25

λ = 21

The equation of the required circle is obtained by

substituting the value of λ = 21

in equation (1)

The circle equation is 2x2 + 2y2 – 2x + 4y – 35 = 0 Choice (1) 46. Given: The limiting points of the coaxial system are

(1, 2) and (4, 3) Limiting circles are (x – 1)2 + (y – 2)2 = 0 i.e. x2 + y2 – 2x – 4y + 5 = 0 . . . . (1) and (x – 4)2 + (y – 3)2 = 0 or x2 + y2 – 8x – 6y + 25 = 0 . . . . (2) The radical axis of the system is (2) – (1), ⇒ – 6x – 2y + 20 = 0 3x + y – 10 = 0 The equation of the required circle is of the form S + λL = 0 ⇒ x2 + y2 – 2x – 4y + 5 + λ(3x + y – 10) = 0 If it passes through the point (2, 3), then (2)2 + (3)2 – 2.2 – 4.3 + 5 + λ (6 + 3 – 10) = 0 ⇒ λ = 2 ∴ The equation of the required circle is

x2 + y2 + 4x – 2y – 15 = 0 Choice (2)

47. Given the two circles are x2 + y2 – 2x + 6y + 6 = 0. . . (1) and x2 + y2 – 5x + 6y + 15 = 0 . . . . (2)

We know if two circles touch each other then their common tangent is their radical axis which is given by S – S1 = 0 i.e., (2) – (1)

⇒ − 3x + 9 = 0 ⇒ x – 3 = 0 Choice (3) 48. The system of circles that are orthogonal to the given

system of cycles x2 + y2 – 2λx + c = 0 is given by x2 + y2 + 2ƒy – c = 0

∴ The required orthogonal system is x2 + y2 + 2fy – 10 = 0 Choice (4) 49. Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 It passes through the points (2, 1) and (– 2, – 3) 4g + 2f + c = – 5 ––––––– (1) – 4g – 6f + c = – 13 ––––– (2) The centre (– g, – f) lies on the line 2x + 5y + 11 = 0 – 2g – 5f + 11 = 0 ⇒ 2g + 5f = 11 ––––– (3) Solving (1), (2) and (3), we get g = – 2; f = 3 c = – 3 ∴ The equation of the circle is x2 + y2 – 4x + 6y – 3 = 0 Choice (3) 50. Let the other end of the diameter of the circle be p(h, k). ∴ The length of the diameter

AP = 22 )ah()bk( −+− ––– (1)

The centre of the circle is

==2

bk,

2ah

Since the circle touches the y – axis, the radius of the

circle is 2

ah +.

Its diameter is = (h + a) –––– (2) Since = (1) = (2)

22 )ah()bk( −=− = (h + a)

S.B.S. (k – b)2 + (h – a)2 = (h + a)2 (k – b)2 = 4ah ∴ The locus of p(h, k) is (y – b)2 = 4ax. Choice (4)

Exercise – 4 Solutions for questions 1 to 55: 1. Given: focus(S) = (2, – 1) and directrix x + y = 0

If P(x, y) is any point on the parabola, then by definition we have SP = PM (PM is the perpendicular distance from P to directrix)

⇒ SP2 = PM2

(x − 2)2 + (y + 1)2 = 2

2

yx

+

⇒ x2 − 2xy + y2 − 8x + 4y + 10 = 0 Choice (2) 2. Given: vertex is (− 1, − 2) The length of latusrectum 4a = 4 a = 1 Axis is parallel to x − axis ∴ The equation of the parabola is (y − β)2 = 4a(x − α) Here, (α, β) = (− 1, − 2), 4a = 4 ⇒ (y + 2)2 = 4(x + 1) y2 − 4x + 4y = 0 Choice (1) 3. Given: focus(S) = (2, 1) and the vertex (0, – 1)

If the point of intersection of directrix and axis is (α, β), then the vertex is the mid point of (α, β) and focus (2, 1)

⇒ (0, −1) =

+β+α2

1,

22

∝ = −2, β = − 3

A(a, b)

• r

p(h, k)


The slope of axis is 0211

xxyy

12

12

−+=

−−

= 1

The slope directrix is = − 1 (Q it is perpendicular to axis) ∴ The equation of directrix is y + 3 = − 1(x + 2) i.e., x + y + 5 = 0 Choice (2) 4. Given: The equation of parabola is x2 − 2x − 2y − 5 = 0 i.e., (x − 1)2 = 2(y + 3) ∴ Vertex (α, β) =(1, − 3) and 4a = 2 Focus is (α, β + a)

⇒(1, −3 + )21

= (1, )25−

Choice (3)

5. Given: The equation of the parabola is x2 − 8x + 16y = 0 i.e., (x − 4)2 = −16(y − 1) The length of latusrectum is 4a = 16 Choice (4) 6. Given: The equation of the parabola is y2 − 2y − 8x − 23 = 0 i.e., (y − 1)2 = 8(x + 3) . . . . (1) Vertex (α, β) = (– 3, 1)

Equation (1) is of the form (y − β)2 = 4a(x − α) which represents the equation of the parabola whose axis is parallel to x − axis and passing through vertex (α, β)

∴ The equation of axis is y = β, i.e. y = 1 or y − 1 = 0 Choice (1) 7. The equation of the parabola whose axis parallel to

x − axis is x = l y2 + my + n Given: (2, 1), (4, − 1) and (1, 0) lie on parabola ⇒ 2 = l + m + n. . . . . .(1) 4 = l − m + n. . . . . (2) 1 = n. . . . .(3)

Solving (1), (2) and (3), we get l = 2, m = −1 and n = 1 ∴ The equation of the parabola is x = 2y2 − y + 1 or 2y2 − y − x + 1 = 0 Choice (3) 8. Given: The equation of the parabola is y2 + 6y − 2x + 5 = 0 i.e., (y + 3)2 = 2(x + 2) Which is of the form (y − β)2 = 4a(x − α) Vertex is (α, β) = (− 2, − 3) Choice (2) 9. If P(x1, y1) is a point on the parabola with focus S(a, 0), then the focal distance of point is SP = x1 + a Here a = 1, SP = 3

3 = x1 + 1 ⇒ x1 = 2, then y 21 = 4x1

y 21 = 8

⇒ y1 = ± 8

∴ The point on the parabola is (2, − 8 ) Choice (2) 10. The equation of tangent to the parabola y2 = 4ax in

slope form is y = mx + a/m

Here, m = tan 30° = 3

1and a = 2

The equation of tangent is y =

312

x3

1 +

i.e., x − y3 + 6 = 0 Choice (4) 11. Given: y2 = 4x and a= 1 The ends of latusrectum are (a, ± 2a) ⇒ (1, ± 2) The end of latusrectum in Q4is (1, –2) The equation of tangent at (1, − 2) to parabola is S1 = 0 is − 2y = 2(x + 1) or x + y + 1 = 0 Choice (3)

12. Given: (1, − 1) lies on parabola (y − 1)2 = 4a(x − 2) ⇒ (−1 − 1)2 = 4a(1 − 2) 4a = − 4 ⇒ a = – 1 The equation of parabola is (y − 1)2 = – 4(x − 2) The directrix equation is x – α + a = 0 x – 2 – 1 = 0 ⇒ x = 3 Choice (2) 13. The equation of parabola is x2 + 8x + 12y + 4 = 0 i.e., (x + 4)2 = − 12(y − 1) Vertex is (− 4, 1) The equation of tangent at vertex is y − 1 = 0

Choice (1) 14. Given: y = mx + c is a tangent to the parabola y2

= 4a(x + a) shifting origin to (− a, 0), then the equation of the parabola changes as y2 = 4ax .. . (1) and the line equation changes to y = m(x − a) + c y = mx − am + c . . . . .(2) If (2) is a tangent to (1), then c = a/m i.e., – am + c = a/m

c = am +ma

= a(m + )m1

Choice (3)

15. Given: The mid-point of the chord of the parabola y2 = 8x is (− 1, 1) The equation of chord is S1 = S11

y − 4(x − 1) = 12 − 8(− 1) i.e., 4x − y + 5 = 0 Choice (4) 16. Given: The abscissa of a point on the parabola y2 = 4x is 9, i.e. x = 9

⇒ y = ± 6 The point on the parabola is (9, − 6)

The slope of the tangent at (9, − 6) is

dxdy

= 31

62 −=

−

The slope of normal is 3 The equation of normal at (9, − 6) to parabola is

y + 6 = 3 (x − 9) 3x − y − 33 = 0 Similarly, by taking the point on the parabola as (9, 6)

we get the equation of the normal as 3x + y – 33 = 0. Choice (4) 17. We know that, the locus of the point of the intersection

of perpendicular tangents is directrix. ∴ The equation of directrix is x + 1 = 0 Choice (2) 18. Given: (2, k) and (− 4, 4) are conjugate points with

reference to parabola y2 = 4x, then we know S12 = 0

(k) 4 = 2(2 − 4) ⇒ k ⇒ −1 Choice (3) 19. Given: The two lines 2x + y − 3 = 0 and kx + 2y + 5 = 0

are conjugate with reference to parabola y2 = 4x, then l1 n2 + 12 n1 = 2 am1 m2

Here l1 = 2, l2 = k, m1 = 1, m2 = 2, n1 = − 3, n2 = 5 and a = 1 ⇒ 10 − 3k = 2 (1) (2) ⇒ k = 2 Choice (2) 20. Let P(x1, y1) be pole. The Polar of p with reference to y2 = 4 ax is S1 = 0 ⇒ y y1 − 2a(x + x1) = 0

or y = 1

1

1 yax2

xya2 + . . . . . (1)

If (1) is tangent to y2 = 4 bx, then c = a/m

11

1

y/a2b

yax2 =

⇒ 4a2 x1 = by 21

The locus equation is 4a2x = by2 Choice (1)


21. Given: centre (α, β) = (2, 3) Axis is parallel to y − axis ⇒ Major axis = 2b=12 ⇒ b = 6 Minor axis = 2a = 8 ⇒ a = 4

The equation of ellipse is ( ) ( )

2

2

2

2

b

y

a

x β−+α− = 1

Substituting the values we get, the equation of the

ellipse as ( ) ( )

363y

162x 22 −+−

= 1 Choice (2)

22. Given: focus S = (2, 1), e = 2/3 and directrix is 2x − y +7 = 0

By definition of ellipse, we have SP2 = e2 PM2, where P = (x, y)

∴ The equation of ellipse is

(x − 2)2 + (y − 1)2 =

2

5

7yx294

+−

⇒ 29x2 + 16xy + 41y2 − 292x − 34y + 29 = 0 Choice (1) 23. Given: 2a = 3 (2b) a = 3b

Eccentricity, e = 2

2

2

2

b9

b1

a

b1 −=−

e = 3

22 Choice (2)

24. Given: Minor axis = distance between the foci, i.e. 2b = 2ae ⇒ b = ae

The length of latusrectum = ab2 2

=12

b2 = 6a . . . . (1) We have, b2 = a2(1 − e2) b2 = a2 − a2 e2 b2 = a2 − b2

2b2 = a2

b2 = 2

a2

From (1), 2

a2

= 6a ⇒ a = 12

and b2 = 6a = 72

∴ The equation of ellipse is 1b

y

a

x2

2

2

2

=+

i.e., 72

y

144

x 22

+ = 1 Choice (3)

25. Given: The ellipse’s equation is

( ) ( )

92yx

163yx2 22 −++−+

= 1

The centre is the point of intersection of 2x + y – 3 = 0 and x + y −2 = 0 The centre of the ellipse is (1, 1) Choice (4) 26. Given: The equation is 4x2 − 16x + y2 − 6y – 39 = 0 It can be written as 4(x − 2)2 + (y − 3)2 = 64

or ( ) ( )

643y

162x 22 −+−

= 1

∴ The Centre is (2, 3) Choice (1) 27. The equation of the ellipse whose axes are coordinates

axes is 2

2

2

2

b

y

a

x + = 1

Given: ( )3,3 and (1, 2) lie on the ellipse.

⇒ 22 b

3

a

3 + = 1 . . . . (1)

and 22 b

4

a

1 + = 1 . . . . (2)

Solving (1) and (2) for a2 and b2, we get a2 = 9 and b2 = 9/2

∴ The Equation of the ellipse is 2/9

y9x 22

+ = 1

or x2 + 2y2 = 9 Choice (2)

28. Given: The equation of the ellipse ( ) ( )

1694y

251x 22 −+−

= 1

The centre of the ellipse (α, β) = (1, 4), a2 = 25, b2 = 169 (a < b)

Eccentricity e = 1312

16925

1b

a1

2

2

=−=−

The focii are (α, β ± be)

= (1, 4 ± 13. )1312

⇒ (1, 16) or (1, − 8) Choice (3)

29. Given: a2 = 16 and b2 = 9

⇒ a = 4 and b = 3 If the eccentric angle is θ, then the coordinates of the point on the ellipse are (a cosθ, bsinθ) = (4 cosθ, 3sinθ)

Coordinates of the point required are (4cos(π/2+θ), 3sin(π/2+θ)), (Q The difference between eccentric angle is π/2) i.e., (− 4sinθ, 3cosθ) Choice (4)

30. Given: The quation of the ellipse is 5x2 + 7y2 = 15 and

the point (3, − 1) S11 ⇒ 5(3)2+7(− 1)2 − 15 = 45 + 7 − 15 > 0 i.e. S11 > 0

∴ The given point lies outside the ellipse (or) exterior to the ellipse. Choice (2) 31. Given: The tangent makes an angle 45° with major axis,

i.e. with x − axis ⇒ The slope of the tangent is tan45° = 1

From the equation of ellipse a = 20 , b = 16 The equation of tangent in slope form is

y = mx ± 222 bma +

i.e., y = x ± 20 + 16 ⇒ y = x ± 6 Choice (1) 32. The equation of the line parallel to x + y + 5 = 0 is x + y + k = 0

⇒ y = −x − k . . . . (1)

If (1) is a tangent to the ellipse 9

y16x 22

+ = 1, then it

satisfies the condition c2 = a2 m2 + b2 ⇒ k2 = 16(−1)2 + 9 k = ± 5 ∴ The equation of the tangent is x + y ± 5 = 0 Choice (2) 33. The equation of the normal at any point (x1, y1) to the

ellipse 2

2

2

2

b

y

a

x + = 1 is given by 1

2

1

2

yyb

xxa − = a2 − b2

Here, a2 = 18; b2 = 8 and (x1, y1) = (3, 2)

2y8

3x18 − = 18 − 8

⇒ 3x − 2y = 5 Choice (3)


34. The equation of normal at any point θ on the ellipse

2

2

2

2

b

y

a

x + = 1 is θ

−θ sin

bycos

ax= a2 − b2

Here, θ = Π/4, a2 = 8 and b2 = 4

a = 2 2 , and b = 2

The equation of normal is °

−° 45sin

y2

45cos

x22 = 8 − 4

21

y2

21

x22 − = 4

2x – 2 y = 2.

2 x – y = 2 . Choice (4)

35. If P(x1, y1) be any point on the ellipse 2

2

2

2

b

y

a

x + = 1

and S, S1 are its focii then PS + PS1 = 2a Given: SP = 4 and a2 = 9 ⇒ a = 3 4 + PS1 = 6 PS1 = 2 Choice (2) 36. Let P(x1, y1) be the mid point of the chord of the ellipse

2

2

2

2

b

y

a

x + = 1, then the equation of chord is S1 = S11

2

21

2

21

21

21

b

y

a

x

b

yy

a

xx +=+

Given: a2 = 36 and b2 = 9

9y

36x

9yy

36xx 2

12111 +=+

xx1 + 4yy1 − x 21 − 2

1y4 = 0 . . . . (1)

But the chord’s equation is given as x + 2y − 4 = 0 . . (2) As (1) and (2) represent the same line.

∴ ( )4

y4x2y4

1x 2

12111 +==

x1 = 2y1 and 4x1 = − ( x 21 + 4y 2

1 ) Solving the above equations, we get x1 = 2 and y1 = 1 ∴ The mid-point of the chord is (2, 1) Choice (1) 37. Given: (−1/5, 3) and (k, 1/3) are conjugate with

reference to 3x2 + 5y2 = 7 ∴ S12 = 0

3

−51

k + 5(3) (3

1) = 7

53−

k = 2

⇒ k = − 10/3 Choice (1) 38. If the lines l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0 are

conjugate with reference to ellipse 2

2

2

2

b

y

a

x + = 1, then

a2 l1 l2 + b2 m1 m2 = n1 n2

Here, a2 = 7/3, b2 = 7/5, l1 = 2, l2 = k, m1 = 1, m2 = 10, n1 = 4 and n2 = 7

Substituting the values, we have

37

(2) (k) + 57

(1) (10) = 4 x 7

32

k = 2

⇒ k = 3 Choice (3)

39. The pole of the line lx + my + n = 0 with reference to

ellipse 2

2

2

2

b

y

a

x + = 1 is

−−n

mb,

n

la 22

Here, a2 = 16, b2 = 9, l = 3, m = 8 and n = − 24

∴ Pole = =

−−

−−

24)8(9

,24

)3(16(2, 3)

Choice (2) 40. Let P(x1, y1) be the mid-point of the chord to the ellipse

2

2

2

2

b

y

a

x + = 1.

The equation of chord is S1 = S11

⇒ 2

21

2

21

21

21

b

y

a

x

b

yy

a

xx +=+

Given, it passes through the foot of directrix, i.e. (a/e, 0)

⇒ =+21

2

1

b

y.0

a

xea

2

21

2

21

b

y

a

x +

The locus of (x1, y1) is 2

2

2

2

b

y

a

x + =aex

Choice (4)

41. The polar of P (x1, y1) with reference to ellipse 2

2

2

2

b

y

a

x +

= 1 is S1 = 0

y = 1

2

1

12

2

yb

xy

x

a

b +− . . . . (1)

The equation of the auxillary circle is x2 + y2 = a2 If (1) is tangent to it, then c2 = r2 (1 + m2)

Hence, c = 1

2

yb

, r2 = a2, m = 1

12

2

yx

a

b−

+=

21

4

21

42

21

4

ya

xb1a

y

b

The locus of (x1, y1) is a2 b4 = a4 y 21 + b4 x 2

1

∴24

2

4

2

a

1

b

y

a

x =+ Choice (4)

42. The product of perpendiculars from focii of the ellipse

2

2

2

2

b

y

a

x + = 1(a > b) on to a tangent is given by b2

Here, b2 = 9 ∴ The product of the perpendicular is 9 Choice (3) 43. The radius of the circumscribed circle of the ellipse is

half the length of major axis Here, the length of the major axis is 10 ∴ radius = 5 The area of the circle is πr2 = 25π sq. units Choice (4) 44. Given, the length of conjugate axis 2b = 4 ⇒ b = 2

and the distance between focii 2ae = 132

⇒ ae = 13

we know that b2 = a2 (e2 − 1) 4 = a2 e2 − a2 4 = 13 − a2 a2 = 9 ⇒ a = 3

∴ The equation of hyperbola is 4y

9x 22

− = 1

(or) 4x2 − 9y2 = 36 Choice (4)


45. Given: The equation of the hyperbola is

( ) ( )

162y

363x 22 +−−

= 1

Its centre is (α, β) = (3, −2), a2 = 36 and b2 = 16

Eccentricity e = 6132

3616

1 =+

The directrix of the hyperbola is x = α ± a/e

x = 3 ±

6132

6

x = 3 ± 13

18 Choice (1)

46. Given: The eccentricity and focii of the hyperbola are

e = 27

and S (4 – 6 7 , 5) and S1(4 + 6 7 , 5)

SS1 = 2ae = (2)( 76 )

a. 2

7= 76

a = 12 b2 = a2 (e2 − 1)

b2 = 144

−147

b2 = 108 The length of latusrectum of the hyperbola is

12

)108(2ab2 2

= = 18 Choice (2)

47. Given: θ = π/3 ∴ The point on the hyperbola is (a secθ, b tan θ) Here, θ = π /3, a = 3 and b = 2

The point is (3 sec π/3, 2 tan π/3) = (6, 2 3 ) The equation of tangent at any point is S1 = 0

⇒ 4(6x) − 9(2 y3 ) = 36 ⇒ 4x − 3 y3 = 6 Choice (2) 48. The equation of normal at any point (x1, y1) on the

hyperbola 2

2

2

2

b

y

a

x − = 1 is 1

2

1

2

yyb

xxa + = a2 + b2

The equation of normal at (1, 1) on the hyperbola

3/1

y4/1

x 22

− = 1

is 31

41

1

y31

1

x41

+=+

i.e., 3x + 4y = 7 Choice (1) 49. The equation of the pair of asymptotes is (2x + 3y) (2x − 3y) = 0

The equation of asymptotes and the equation of the hyperbola differ by constant.

The equation of the hyperbola is (2x + 3y) (2x − 3y) = k But given vertices of the hyperbola are (± 3, 0) ⇒ (6 + 0) (6 − 0) = k ⇒ k = 36 ∴ The equation of the hyperbola is 4x2 − 9y2 = 36 Choice (1) 50. Given: The equation of the hyperbola is 3xy + 3x + 3y + 1 = 0 or 3xy + 3x +3y + 3 = 2 (3x + 3) (y + 1) = 2 The equation of asymptotes is (3x + 3) (y + 1) = 0 (or) 3xy + 3x +3y + 3 = 0 Choice (3)

51. Let P(x1, y1) be the pole. The polar of P with reference to y2 = 4ax is S1 = 0 yy1 = 2a(x + x1)

⇒ y = 1

1

1 yax2

xya2 + . . . . . (1)

If (1) is a tangent to x2 – y2 = a2, then c2 = a2 m2 − b2

Here, c = 1

1

yax2

, a2 = b2 = a2 and m = 1y

a2

4a2

21

21

y

x=a2

−1

y

a421

2

4x 21 = 4a2 – y 2

1

∴ The locus of P is 4x2+y2 = 4a2 Choice (2) 52. Let P (x1, y1) be the pole.

The polar of P with reference to 2

2

2

2

b

y

a

x − = 1 is S1 = 0

21

21

b

yy

a

xx − = 1 . . . . (1)

If lx + my + n = 0 is normal to the hyperbola 2

2

2

2

b

y

a

x − = 1,

then ( )

2

222

2

2

2

2

n

ba

m

b

l

a +=−

i.e., ( )

( )2222

2

21

2

2

21

2

1

ba

by

b

ax

a

−+=

−

21

6

21

6

y

b

x

a − = (a2 + b2)2

∴ The locus of P(x1, y1) is 2

6

2

6

y

b

x

a − = (a2 + b2)2

Choice (3) 53. If θ is the angle between the pair of asymptotes of the

hyperbola 2

2

2

2

b

y

a

x − = 1, then θ = 2 tan-1 (b/a)

⇒ tan θ/2 = ab

,

⇒ tan2 (θ/2) = 2

2

a

b

⇒ 1+tan2 (θ/2) = 1+2

2

a

b

sec 2 (θ/ 2) = e2 ∴ sec (θ/2) = e Choice (3) 54. The angle between the asymptotes of the hyperbola

2

2

2

2

b

y

a

x − = 1 is 2 tan–1 (b/a).

Given: Hyperbola 9

y25x 22

− = 1

∴ θ = 2 tan–1 53

tan–1

−

×

259

1

53

2 = tan–1

×1625

56

= tan–1

815

.

Choice (2)


θ

a

b

22 ba +

24 25

7 θ

55. Given: The hyperbola is 4x2 – 9y2 = 36

i.e., 4

y9

x 22− = 1

The length of latusrectum = a

b2

2

= 3

42 × = 8/3 Choice (3)

UNIT – ΙΙΙ

Exercise – 1 Solutions for questions 1 to 50:

1. 125 π = 180

125 × = 75° Choice (1)

2. 180

144π× =

54π

Choice (2)

3. Let ∝ , β and γ be the angles of triangle Given: The angles of triangle are in AP ⇒ β = 60° ⇒ α + β + γ = 180° ⇒ α + γ = 120° ……….(1) Given that

⇒ α × 360400

: γ × 180

π= 40 : π ⇒ x

3640α π = γ ×

18040π

⇒ 5α = γ From (1) α + 5α = 120 ⇒ α = 20° ⇒ γ = 5α = 100° The angles in circular measure are

20 ×180

cπ, 60 ×

180

cπ, 100 ×

180π

i.e., ,9

cπ ,

3

cπ,

9

5 cπ Choice (1)

4. 722

180

r

l o

××=θ o13522

718070

165 =××= .

Choice (2)

5. A, B, C and D are angles of a cyclic quadrilateral A + C = 180° and B + D = 180° ⇒ C = 180° - A and D = 180° – B cos A + cos B + cos (180 – A) + cos (180 – B) = cos A + cos B – cos A – cos B = 0 Choice (1) 6.

Given: cot θ = ab

and θ∉ Q1

⇒ θ∈ Q3

∴ sin θ = 22 ba

a

+

−, cos θ =

22 ba

b

+

−

22

22

ba

bacosbsinacosbsina

+

+−=θ+θθ−θ

22

22

ba

ba

+

−−

= 22

22

ba

ba

+−

Alternate method:

cot θ = ab

, ⇒

ab

a

ab

a

cosbsinacosbsina

+

−=

θ+θθ−θ

= 22

22

ba

ba

+−

Choice (4)

7. Given: cosθ = 257− and tanθ =

724

Here, cosθ is negative and tanθ is positive. ⇒ θ lies in the third quadrant.

cscθ = 2425−

cotθ = 247

∴ cscθ + cotθ =2418

247

2425 −=+−

= 43−

Choice (1)

8. tan(A − B) = 3

1

⇒ A − B = 30° sin B = 1 ∴ B = 90° A − B = 30° ⇒ A − 90° = 30°

A = 120°=32π

. Choice (2)

9. 33

Acos23

Asin21Asin21

Acos23

−++

−+

is in the form of

a3 + b3 = (a + b) (a2 – ab + b2) Now a + b =

)Acos23)(Asin21(

Asin41Acos43

Acos23

Asin21Asin21

Acos23 22

−−−+−=

−++

−+

= =−−

+−)Acos23)(Asin21(

)AsinA(cos44 22

0

∴ a + b = 0 ⇒ (a + b) (a2 – ab + b2) = a3 + b3 = 0 Choice (1)

10. Asec

11Acosec

Acot1cosecA

Acot

−+

+

= Asec

1

1Asin

1AsinAcos

1Asin

1AsinAcos

−+

+

= Asec

1Asin1

AcosAsin1

Acos

−+

+

= AcosAcos

2Acos

2

= 2 Choice (3)


11. a2 + b2 + c2 + λ2 = r2 cos2 α cos2 β cos2γ + r2 cos2 α cos2 β sin2 γ + r2sin2α

cos2 β + r2 sin β = r2 cos2 α cos2 β (cos2 γ + sin2 γ) + r2 sin2 α cos2 β +

r2sin2 β = r2 cos2 β (cos2 α + sin2 α) + r2 sin2 β = r2 (cos2β + sin2β) ∴ a2 + b2 + c2 + λ2 = r2 Choice (2)

12. x = θ−θ

2tan12sec

⇒ x = θ

−θθ

2tan1

2tan2sec

x = cosec2θ − cot2θ Similarly y = cosec2θ + cot2θ ∴ xy = cosec22θ − cot22θ ⇒ xy = 1 Choice (2) 13. x = acotθ + b cosecθ → (1) y = acotθ − b cosecθ → (2) Solving (1) and (2) for cotθ and cosecθ

cotθ = b2

)yx(eccosand

a2)yx( −=θ+

Since cosec2θ - cot2θ = 1

⇒( ) ( )

2

2

2

2

a4

yx

b4

yx +−− = 1

Choice (4) 14. sinθ + cosθ = 2 It is possible only when θ = 45° tannθ + cotnθ = tann45 + cotn45 = 1 + 1 = 2

Choice (4) 15. Given A + B = 45° ⇒ B = 45° – A (1 – cot A) (a + cot B) = (1 – cot A) (1 + cot (45 – A)

= (1 – cot A)

−++

1AcotAcot1

1

= )Acot1(

Acot1−−

−[cot A – 1 +1 + cot A] = – 2 cot A

Choice (4)

16. =βα

cottan

k ⇒ tan α tan β = k …………. (1)

βα−βαβα+βα=

β+αβ−α

sinsincoscossinsincoscos

)cos()cos(

⇒ k1k1

)tantan1(coscos)tantan1(coscos

−+=

βα−βαβα+βα

(from (1))

Choice (1) 17. Given: sin 12° sin 48° sin 54°

= °

°°+°°−°°72sin

54sin)1260sin()1260sin(12sin

=°

°°

72sin54sin)12(3sin

41

°

°°

=72sin

36cos36sin281

= 81

72sin.872sin =°

Choice (2) 18. tan 9° + tan (90° – 9) – (tan 27° + tan (90° – 27°) ⇒ tan 9° + cot 9° – (tan 27°+ cot 27°)

⇒

°°+

°°−

°°+

°°

27sin27cos

27cos27sin

9sin9cos

9cos9sin

⇒ °°

°+°−°°

°+°27cos27sin2

)27cos27(sin29cos9sin2

)9cos9(sin2 2222

⇒

−=°

−° 36cos

118sin

12

54sin2

18sin2

⇒ 2

+−+=

+−

− 41515

815

4

15

4= 4

Choice (4)

19. Cos 7π π/7 + cos

72π

+cos7

3π+ cos (π-

73π

) + cos

(π – 72π

) + cos (π – 7π

)

cos 7π

+ cos72π

+ cos7

3π– cos

73π

– cos72π

– cos

7π

= 0 Choice (1)

20. sec2θ ≥ 1

∴ 2)yx(

xy4

+ ≥ 1 ⇒ (x + y)2 − 4xy ≤ 0

⇒ (x – y)2 ≤ 0 ⇒ x = y Choice (1) 21. cos2x – cos4x = sin2 x − (1− sin2 x)2

= sin2 x − (sinx)2 = 0 (∵ sinx + sin2x = 1) Choice (1) 22. sin 81° = sin (90 – 9°) = cos 9°

We know that cos θ = 2

2cos1 θ+

⇒ cos 9° = 2

18cos1 °+ =

24

52101

++

=16

)55)(53(25553 −++−++

=4

5553 −++ Choice (3)

23. Given: sin 10° sin 50° cos 20° = sin 10° sin 50° cos (90° – 70°) = sin 10° sin 50° sin 70° = sin 10° sin (60° – 10°) sin (60° + 10°)

= 41

sin 3.(10) =81

21

41 =

Choice (2)

24. By componendo dividendo rule we have

yxyxyxyx

)BAcos()BAcos()BAcos()BAcos(

+−+−++=

−−+−++

⇒ yx

BsinAsin2BcosAcos2 =

−

cotAcotB = –yx

⇒ tan A tan B = xy−

Choice (1)

25. cos( x + 8x + 27x + ------+n3x) = cos x4

)1n(n 22 +

∴ The required period is 2222 )1n(n

8

4)1n(n

2

+π=

+π

Choice (4)


26. f(x) = cos3x + sin3x cos3x = 4cos3x − 3cosx

cos3x = 4

xcos3x3cos +

sin3x = 3sinx − 4sin3x

sin3x = 4

x3sinxsin3 −

The period of sinx and cosx are 2π, 2π

The period of sin3x =32π

and that of cos 3x = 32π

∴ The period of f(x) = L.C.M.

ππππ32

,32

,2,2

∴ The period of the function is 2π. Choice (2) 27. If A + B = 45° then cot (A + B) = cot 45

AcotBcot

1BCotAcot

+−

= 1

⇒ cot A cot B – 1 = cot B + cot A ⇒ − 1 = cot A + cot B – cot A cot B −2 = cot A (1 – cot B) – 1 (1 – cot B) −2 = (cot A – 1) (1 – cot B) Or (1 – cot A) (1 – cot B) = 2 ∴ (1 – cot 4°) (1 – cot 41°) (1 – cot 5°) (1 – cot 40°) - - - - - - - - - - (1 – cot 22) (1 – cot 23) = 2P i.e. 219 = 2P ⇒ P = 19 Choice (2)

28. cos2 θ + sin4 θ = sin4θ + 1 − sin2θ

+×θ−θ2

222

21

21

sin.2)(sin +1 -2

21

4

3

2

1sin

22 +

−θ

When sin2θ = 0 the expression is maximum

Maximum value 43

21

2

+

− = 1

43

41 =+

Sin2θ = 21

; the expression is minimum.

Minimum values 43

43

0 =+

∴ The range =

1,

43

Choice (1)

29. 4sin(30° + θ) + 3cosθ + 5 = 4[sin30° cosθ + cos30° sinθ] + 3cosθ + 5

= 4

θ+θ sin

23

cos21 + 3cosθ + 5

= 5cosθ + 2 3 sinθ + 5

Maximum value is C + 22 ba + .

= 5 + ( )22 325 + = 5 + 1225 +

= 5 + 37 Choice (1)

30. 1 + 8 sin2x2cos2x2 = 1 + 2 (2sinx2cosx2)2 = 1 + 2 sin2 2x2 = 1 + (1 – cos4x2) = 2 – cos4x2

Minimum value is c – 22 ba + here, c = 2, a = 1, b = 0

∴ The required minimum value is 2 – 1 = 1. Choice (1)

31. sin2 (120° + θ) + sin2 (120° – θ) = 1 – [cos2 (120° + θ) – sin2 ( 120° – θ)] = 1 – { cos 240 cos 2θ }

= 1 + 21

cos θ (∵ cos2 A – sin2 B = cos (A + B) cos (A – B))

The required range = 1 ± 21

ie

23

,21

Choice (3) 32. We know that AM ≥ GM

⇒2

xcot9xtan4 22 + ≥ xcot9.xtan4 22

∴ 4 tan2 x + 9 cot 2 x ≥ 12 ∴ The minimum value of the function is 12. Choice (4)

33. The given expression is 29 sin x + 7 cos x + 4. The maximum value and minimum value are

4 ± 729 + = 4 ± 6 ⇒ 10 and – 2

∴ The range of the given function is

∞∪

−∞ ,101

21

,

Choice (4)

34. sin2x − cos2x = 2

x2cos1−

− cos2x

= 2

x2cos31− The above will be maximum, if cos2x is − 1

2

x2cos31−=

231+

= 2 Choice (2)

35. x2sinx2cos 55 +

It is maximum when 2

nx

π= , or nπ.

⇒ 5 + 1 = 6 or 1 + 5 = 6 ∴ The maximum value is 6. Choice (3) 36. Given: sinθ1 + sinθ2 + sinθ3 = cosθ1 + cosθ2 + cosθ3, Qθ1, θ2, θ3 are in AP, Let θ1 = θ – d; θ2 = θ; θ3 = θ + d Given: θ1 = 2θ3 θ – d = 2(θ + d) d = – θ/3.

∴ sin

θ−θ+θ+

θ−−θ3

sin)sin(3

= cos

θ−θ+θ+

θ−−θ3

cos)cos(3

⇒ sin

θ34

+ sinθ + sin

θ32

= cos

θ34

+cosθ + cos

θ32

⇒ 2 sin θ . cos 3θ

+ sinθ = 2 cosθ cos 3θ

+ cosθ

⇒ sinθ

+θ1

3cos2 = cosθ

+θ1

3cos2

⇒

+θ1

3cos2 (sinθ – cosθ) = 0

Since θ is acute, cos θ/3 ≠ – 1/2


⇒ sinθ = cosθ ⇒ θ = π/4

d = 123π−=θ−

∴ θ1 = θ – d = 3124π=π+π

θ2 = π/4 and θ3 = θ + d = π/4 – π/12 = π/6.

∴θ 1 + θ2 + θ3 =4

321̀

9643

π=π=π+π+π

Choice (3)

37. θ+++ 2cos2222

)cos2(222 2 θ++=

= 2/cos2.22 2 θ+ 2/cos22 θ+=

4/cos2.2 2 θ= = 2 cos θ/4 Choice (3) 38. cos20° cos40° cos80° We know that cosA cos(60° – A) cos(60 + A)

= 41

cos3A. cos20° cos(60 – 20°) cos(60° + 20°)

= 41

cos 3.20 = 41

cos 60° = 81

.

Choice (4)

39. 10cos30sin210cos60sin2

40sin20sin50sin70sin

)4090cos(20sin50sin)7090cos( =

++=

−++−

= 32/12/3 = Choice (2)

40. We know that Cos A + cos B

= cos

+2

BA.cos

−2

BA

⇒ 2 cos α cos ( β + γ) + 2 cos α cos (β – γ) = 2 cos α [cos (β + γ) + cos (β – γ)] = 2 cos α 2 cos β cos γ = 4 cos α cos β cos γ Choice (1) 41. Given: sinα + sin β = a and cos α + cos β = b.

⇒ 2 sin 2

β+αcos

2β−α

= a ………..(1)

and 2 cos 2

β+αcos

2β−α

= b …… .(2)

(1) ÷ (2) ⇒ tan ba

2=

β+α

cos (α + β) = 22

22

2

2

2

2

2

2

ab

ab

b

a1

b

a1

2tan1

2tan1

+−=

+

−=

β+α+

β+α−

Choice (4) 42. Given: tan A/2, tan B/2, and tan C/2 are in AP ∴ tan A/2 – tan B/2 = tan B/2 – tan C/2

⇒2/Bcos2/Bsin

2/Acos2/Asin −

2/Ccos2/Csin

2/Bcos2/Bsin −=

⇒ sin

−2

BA cos

2C

= sin 2

CB − cos

2A

⇒ 2 sin 2

BA +sin

−2

BA= 2 sin

2CB +

sin 2

CB −

⇒ cos B – cos A = cos C – cos B ⇒ cos A, cos B, and cos C are in AP Choice (3)

43. cos2A + cos2B − cos2C − 1 = cos2A − cos2C − (1 − cos2B)

= cos2A − sin2B − cos2C = cos(A + B) cos(A − B) − cos2C (QA + B + C = 180°) = cos(180 − C)cos(A − B) − cos2C = − cosC cos(A − B) + cos(A + B) (cosC) = − cosC 2 sinA sinB = − 2sinA sinB cosC Choice (2)

44. sin2(θ − 45°) + sin2(θ + 15°) − sin2(θ − 15°) = sin2(θ − 45°) + (sin2θ. sin30°)

= θ+θ−°−2sin

21

2)290cos(1

= θ+θ− 2sin21

2sin21

21

= 21

Choice (2)

45. sin25°+ sin35° + cos175°

= 2sin )5180cos(2

10cos

260 °−°−°°

= 2 × 21

cos5 − cos 5° = 0 Choice (1)

46. 21

(2sin25° sin5° − cos20°)

= 21

[cos20° − cos30°] −21

cos20°

= 21 00 20cos

21

23

20cos −

−

= 21

cos20° − 020cos21

43 −

= 4

3− Choice (2)

47. acosα + bsinα = c ⇒ acosα = c − b sinα ⇒ a2cos2α = c2 + b2 sin2 α − 2bcsinα ⇒ a2 − a2sin2α = c2 + b2 sin2α − 2bcsinα ⇒ (b2 + a2)sin2α − 2bc sinα + c2 − a2 = 0 If sinθ1 and sinθ2 are two roots of this equation, then

The sum of the roots = sinθ1 + sinθ2 = 22 ba

bc2

+

Choice (1)

48. ( )

AcosAsin

A2cos1A2cos1

A2sin −+

= Asin

Acos).Asin2(.

)Acos2(

)AcosAsin2( 2

2

= 2sin2A Choice (4) 49. cos 28° + cos 65° + cos 115° + cos 240° + cos 208° +

cos 300° ⇒ cos 28° + cos 65° – cos 65° – cos 60° – cos 28° +

cos 60° ⇒ 0. Choice (4) 50. We know that b2 = c2 + a2 – 2ac cos β Since β is obtuse cos β < 0 ∴ b2 = c2 + a2 + 2ac cos β ⇒ b2 > c2 + a2. Choice (3)


Exercise – 2


1. Given: 4sin2θ = 3 ⇒ sinθ = 2

3±

If sinθ = 23

, then the principal value of θ is 3π

If sinθ = 2

3−, then the principle value of θ is

3π−

Choice (2)

2. Given: cosθ = cos4

5π⇒ cosθ = cos

π+π4

= – sin4

π

⇒ cosθ = cos

π−π4

= cos4

3π

The principal value of θ = 4

3π

The general solution: θ = 2nπ ± 4

3π ∀ n∈ Z.

Choice (2) 3. Given: xsinlogxcoslog xcosxsin + = 2

logsinxcosx + 2xcoslog

1

xsin=

(logsinxcosx − 1)2 = 0 ⇒ xcoslog xsin = 1

⇒ sinx = cosx ⇒ tan x = 1 ⇒ x =4

π Choice (3)

4. Given: 2sin2θ + 3sinθ + 1 = 0

⇒ (2sinθ + 1) (sinθ + 1) = 0 ⇒ sinθ = 2

1− or −1

sinθ = − 1 ⇒ sinθ = sin

π2

3 ⇒ θ =

23π

2sinθ + 1 = 0 ⇒ sinθ = 21−⇒ θ =

611

,6

7 ππ

∴ The number of values of θ in (0, 2π) are 3. Choice (1)

5. Given:23

4xcosxsin3 ⇒=+ sinx + 21

cosx = 2

⇒ sinx cos6π

+ sin 6π

cosx = 2 ⇒ sin (6π

+ x) = 2

As sin (any angle) always lies between − 1 and +1, the above equation can never hold true.

⇒ The given equation has no solution. Choice (4)

6. Given: 2sin2x + 3 cosx + 1 = 0

2(1 – cos2x) + 3 cosx + 1 = 0

⇒ 2cos2x– 3 cosx–3 = 0.

(cosx– 3 ) (2cosx+ 3 ) = 0 ⇒ cosx = 3 or 2

3−

As cosx ≠ 3

⇒ cosx = 2

3− = − cos

6π

= cos(π −6π

) = cos6

5π

∴ The general solution: x = 2nπ ± 6

5π ∀ n∈ Z.

Choice (2)

7. Given: tan(2π

sinθ) = cot

θπ

cos2

⇒ tan (2π

sinθ) = tan (2π −

2π

cosθ)

⇒ 2π

sinθ = nπ + 2π

−

2π

cosθ

∴ sinθ + cosθ = 2 n+1

∴ cos (θ −4π

) = 2

1n2 + = cosx (say)

⇒ θ −4

π = 2kπ ± α ⇒ θ −

4

π = 2kπ ± cos-1

+2

1n2

Only when n = 0 or −1, cos x ≤ 1. ∴ General solution:

θ − 4π

= 2kπ ± 4π

, 2kπ ± 4

3π

∴θ = 2kπ, 2kπ + 2π

, π(2k + 1) or π (2k − 1)

Choice (4) 8. tan3θ − tan2θ = tan2θ − tanθ

θθ−

θθ=

θθ−

θθ

cossin

2cos2sin

2cos2sin

3cos3sin

⇒ θθ

θ−θ=

θθθ−θ

cos2cos

)2sin(

2cos3cos

)23sin(

⇒ sinθ(cosθ − cos3θ) = 0 ⇒ sinθ. 2sinθ sin2θ = 0 ⇒ sinθ = 0 or sin2θ = 0 sinθ = 0 ⇒ θ = nπ

sin2θ = 0

⇒ 2θ = nπ or θ = 2

nπ

But θ = 2

nπ does not satisfy the equation.

Choice (3) 9. Given: (sin3θ + cos3θ) – (1− sinθcosθ) = 0

⇒ (sinθ +cosθ) (sin2θ + cos2θ − sinθcosθ) − (1− sinθ cosθ)=0

⇒ (1 − sinθ cosθ) (sinθ + cosθ – 1) = 0 ⇒ 1− sinθcosθ = 0 or sinθ + cosθ = 1 ⇒ sin2θ = 2 which is impossible

or sinθ + cosθ = 1 i.e.,⇒ 2

1sinθ +

2

1cosθ =

2

1

⇒ cos(θ −4π

) = 2

1

⇒ cos(θ –4π

) = cos4π⇒ θ –

4π

= 2nπ ± 4π

∴ θ = 2nπ + 2π

or 2nπ. Choice (2)

10. Given: −−−++++ |x3cos||x2cos||xcos|18 = 43 = 82 ⇒ 1 + |cosx| + |cos2x| + …. = 2

⇒ |xcos|1

1−

= 2 or 1 − |cosx| = 21

(∵ s∞ of a G.P = r1

a−

) |cosx| = 21

⇒ cosx = ± 21

∴ The values of x that lie between − π and π satisfying

cosx = ±2

1 are ±

3

π and ±

3

2π. Choice (4)


11. Given: cotθ = 3

1 and sinθ =

23−

⇒ The value of θ lying between 0 and 2π satisfying both

equations is θ = 34π

General solution: θ = 2nπ + 34π

= (2n + 1)π + 3π

, ∀

n∈ Z Choice (2) 12. Given: sinx = siny and cosx = cosy ⇒ sinx – siny = 0 and cosx – cosy = 0

2cos 2

yx +sin

2yx −

= 0 → (1) and

2sin2

yx +sin

2yx −

= 0 → (2)

Equating (1) and (2), we have sin2

yx − = 0

The principal value of 2

yx − = 0

The general solution:2

yx − = nπ ⇒ x = 2nπ + y, ∀ n∈ Z

Choice (2) 13. Given: xtanθ + ysecθ = z ysecθ = z – x tanθ or y2sec2θ = z2 + x2 tan2θ – 2xztanθ ⇒ y2(1+ tan2θ) = z2 + x2 tan2θ − 2xztanθ ⇒ (x2 – y2)tan2θ − 2xztanθ + z2 – y2 = 0

Given: tanα and tanβ are the roots of the above equation,

⇒ the sum of the roots = tanα + tanβ = 22 yx

xz2

−

and the product of the roots = tanα tanβ = 22

22

yx

yz

−−

tan(α + β) = βα−β+α

tantan1tantan

=22

22

22

22

zx

xz2

yx

yz1

yx

xz2

−=

−−−

−

Choice (4) 14. Given: tanθ + tan4θ + tan7θ = tanθ tan4θ.tan7θ tanθ + tan4θ = − (1 − tanθtan4θ)tan7θ

⇒ θθ−θ+θ

4tantan14tantan

= − tan7θ ⇒ tan5θ = − tan7θ

The general solution: 5θ = nπ − 7θ i.e., θ = 12

nπ, ∀ n∈ Z

Choice (3) 15. The given: sinθ + cosθ = 2

θ+

θ

2

cos

2

sin2 = 2

2 sin

π+θ4

which cannot exceed 2 .

∴ No solution exists for the given equation. Choice (4) 16. Given: sinθ + sin2θ + sin3θ = cosθ + cos2θ + cos3θ sinθ + sin3θ + sin2θ = cosθ + cos3θ + cos2θ 2sin2θcosθ + sin2θ = 2cos2θcosθ + cos2θ sin2θ(2cosθ+1) = cos2θ (2cosθ + 1) ⇒ 2cosθ + 1 = 0 or tan2θ = 1

⇒ θ = 2nπ ± 3

2π or 2θ = nπ +

4π

Put ‘n’ = 0, 1, 2 and 3. The solutions of θ lying between

0 and 2π are 8π

, 8

5π,

32π

, 8

9π,

34π

and8

13π

Choice (2) 17. Given: sin6x + cos6x = a ⇒ (sin2x + cos2x)3 – 3sin2xcos2x (sin2x + cos2x) = a

⇒ 1 − 43

sin22x = a ⇒ sin2x = ± 23

a1−

⇒ 1 – a ≥ 0 and –1 ≤ 23

a1− ≤ 1

⇒ a ≤ 1 and 0 ≤ 3

a1− ≤ 41

i.e. a ≥ 41

.

⇒ 41

≤ a ≤ 1 Choice (2)

18. Given: tan ax – tan bx = 0

⇒bxcosbxsin

axcosaxsin − = 0 ⇒ sin[(a – b)x] = 0

General solution: x is x(a – b) = nπ

⇒ x = ba

n

−π

for n ∈ Z.

Putting n = 0, 1, 2 and 3, we have

x = ba −

π,

ba2−π

, ba

3−π

……. which forms an A.P with

the common difference ba −

π. Choice (3)

19. Given: cos2x + asinx = 3a – 17 1 – 2sin2x + a sinx = 3a – 17 2sin2x – a sinx + 3a – 18

sinx =)2(2

)18a3(8aa 2 −−± =

4

144a24aa 2 +−±

sinx = ( )

4

12aa 2−±

sinx = 4

12a2 −

sinx = 4

12a2 −

The equation has a solution. ⇒ –1 ≤ 4

12a2 − ≤ 1

– 4 ≤ 2a – 12 ≤ 4 8 ≤ 2a ≤ 16 4 ≤ a ≤ 8. Choice (1)

20. Given: tan2θ – 2 3 tanθ – 1 = 0

tanθ = 2

41232 +±

= 2

432 ±

32 + or 23 −

tanθ = 32 +

The principle value α is 75°

∴ The general solution of θ is = nπ + 125π

Choice (2)


15 17

8

α

∝

2x1+ x

1

Exercise – 3


Let Tan-1x = α ⇒ tanα = x

∴ cos (tan–1 α ) cosα = 2x1

1

+

Choice (2)

2. Cos-1

−2

3= π − Cos-1

23

= π −6π

=6

5π

Choice (4)

3. Tan-1

31

Tan21 1−+ =

=

−

+−−

55

Tan

31

.21

1

31

21

Tan 11

⇒ Tan-1(1) =4π

Choice (3)

4. Given: Sin-1x + Sin-1y = 32π

We know that Sin-1x + Cos-1x = 2π

∴ Sin-1x + Sin-1y = 32π

can be written as

)yCosxCos(32

yCos2

xCos2

1111 −−−− +−π⇒π=−π+−π

= 32π

⇒ Cos-1x + Cos-1y = π − 32π

= 3π

Choice (1)

5.

Let Cos-1

178

= α ⇒ cosα = 178

∴ tanα = 8

15

∴ tan

−

178

Cos 1 = tanα =8

15 Choice (2)

6. Sin-1

+

−

71

Tan53 1 = Tan–1 1Tan

43 −+

71

=

=

−

+−−

2525

tan

283

1

71

43

tan 11 =4π

Choice (3)

7. We know that Sin-1 and32

3 π=

21

Sin2 1− = 26π

= 3π

∴ Sin-1

332

1Sin2

2

3 1 π+π=+ − =32π

Choice (4)

8. Given: cos [Cot-1{tan(Cos-1x)}]

=

−π −− xCos2

cotcotcos 11

⇒ cos

−π − xCos2

1 ⇒ sin(Cos-1x) = 2x1−

Choice (3) 9. Given: Sin–1x + Sin–1y + Sin–1z = 3π/2

Sin–1 x = Sin–1 y = Sin–1 z = 2π

x = sinπ/2 = 1; y = sinπ/2 = 1 z = sinπ/2 = 1;

Now, x100 + y200 + z300 + 3zyx

6301201101 +++

1 + 1 + 1 + 3111

6+++

= 3 + 1 = 4

Choice (4)

10. Cos-1

+

−

by

Cosax 1 = α

Cos–1

−α=

−

by

Cosax 1

∴

−α= −

by

Coscosax 1 = cosα.

b

ybsin

by 22 −

α+

ax

– cosα by

= sin α b

yb 22 −

−α=

α−2

22

2

b

y1sin

by

cosax

⇒ ( )α++α− 222

2

2

2sincos

b

ycos

abxy2

a

x = sin2α

∴ 2

2

2

2

b

ycos

abxy2

a

x +α− = sin2α

Choice (4)

11. Tan-1(2 − 3 ) =12π

Choice (1)

12. Sin-1

π4

5sin = Sin-1

π+π4

Sin =

π−−

4sinSin 1

=4π−

Choice (2)

y

b 22 yb −


13. Let

−

35

Cos21 1 = α ⇒ Cos2α =

35

Tan

−

35

cos21 1 = tanα =

35

1

35

1

2cos12cos1

+

−=

α+α−

= 53

53

+−

= 53

53

53

53

−−×

+−

= 59

53

−

−=

253 −

Choice (4)

14. Sin

−2

3 + 2Cos–1 (– 1/2) =

− Sin-1

−π+

−

21

Cos223 1 =

32

23

π−π+π−

⇒ 2π − π = π Choice (4)

15. sin2

+ −−

92

Tan41

Tan 11

=

++

+

−−

−−

92

Tan41

Tantan1

92

Tan41

Tantan2

112

11

∴ sin2α = α+α+

tan1cos2

=

−

++÷

−

+2

92

.41

1

92

41

1

92

.41

1

92

41

.2

=

+÷

2

2

34

171

3417.2

= 1 ×54

= 54

.

Alternate method:

92

Tan41

Tan 11 −− + = Tan-1

9.42.1

1

92

41

−

+

= Tan-1

21

Tan3417 1−=

sin

−

21

Tan2 1 = 2sin

−−

21

Tancos21

Tan 11

= 2.54

5

2.

5

1 = Choice (3)

16. Given: cos–1 x + 3sin–1 x = π

Cos-1x + Sin-1x + 2Sin-1x = π ⇒ π=+π − xSin22

1

⇒ Sin-1x = 4π

⇒ x =2

1 Choice (2)

17. Cos-1x + Cos-1

−+2

x332x 2

= Cos-1x + Cos-1

−π+π 2x13

sinx3

cos

= Cos-1x + Cos-1

θπ+θπSin

3SinCos

3Cos

Where, cosθ = x and sinθ = 2x1−

θ + Cos-1

θ−π3

Cos ⇒ θ + 3π − θ =

3π

Choice (1)

18. Given: Cot-1(secx − tanx) = Tan-1

− xtanxsec1

= Tan-1(secx + tanx)

⇒ Tan–1

+π=

−

+−

2x

4tanTan

2x

sin2x

cos

2x

sin2x

cos1

⇒ 2x

4+π

But it is given, Cot–1 (secx – tanx) = 4π

+ kx.

⇒ 42

x4

π=+π + kx ⇒ k =21

Choice (1)

19. Given:

3Sin-1

+−−

+−

2

21

2 x1

x1Cos4

x1

x23x1

x2Tan2

21 π=

−+ −

Put x = tanα

3Sin-1(sin2α) − 4(Cos-1(cos2α)) + 2Tan-1(tan2α) = 3π

⇒ 6α − 8α + 4α = 3π

2α =

3π

α = 6π⇒ x = tanα = tan

6π

= 3

1 Choice (4)

20. Let ba

cos21 1− = α ⇒ cos2α =

ba → (1)

Tan

−π+

+π −−

ba

Cos21

4Tan

ba

Cos21

411

Tan

α−π+

α+π4

Tan4

= ( ) ( )

α−α−+α+=

α+α−+

α−α+

2

22

tan1

tan1tan1tan1tan1

tan1tan1

⇒( )

α−α+

2

2

tan1

tan12=

α2cos2

= ab2

(from(1)) Choice (3)

21. Cos-1 x2 is defined only when 0 ≤ x2 ≤ 1 0 ≤ 2x ≤ 1

∴ The domain of cos–1 ( )x2 is 0 ≤ x ≤ 21

Choice (2)

22. Given: Sin-1x − Cos-1x = xCos22

1−−π

The minimum is 23

)(22

π−=π−π

The maximum is2

)0(22

π=−π

∴ The range is

ππ−2

,23

Choice (4)


23. Given: Tan-1

=

−−

+ −−

a4

Tana2

aTana2

a 11

⇒ a4

a2

aa2

a1

a2

aa2

a=

−

++

+−+ ⇒

a4

a

4a1

a4

22

=

−+

a2 + a4 − 4 = a2 ⇒ a4 = 4 ⇒ a = ± 2 Choice (1)

24. Given: Tan–1 (1 – x) + Tan–1 (1 – x) = Tan–1 (3x) + tan–1 x

Tan–1

−+=

−+−−++ −

x.x31xx3

tan)x1)(x1(1)x1()x1( 1

Tan–1

−=

−2

12 x31

x4Tan

x

2

22 x31

x4

x

2

−= ⇒ 2 – 6x2 = 4x3 ⇒ 2x3 +3x2 =1

x = ½ satisfies the above equation. Choice (2) 25. Tan-1x + Tan-1y = π − Tan-1z.

xy1yx

−+

= Tan(π − Tan-1z) = −z

⇒ x + y = − z + xyz ⇒ x + y + z = xyz Choice (4) 26. Put x = tanθ,

⇒ tan

θ+θ−+

θ+θ −−

2

21

21

tan1

tan1Cos

21

tan1

tan2Sin

21

= tan

+ −− xTan.221

xTan2.21 11

⇒ tan

−−

21

x1

x2Tan =

2x1

x2

− Choice (3)

27. Tan–1

+−

+

+−+

+− −−

32

231

21

121

1

1

cc1

ccTan

cc1cc

Tanxycyxc

+ …+ Tan–1

nc

1

⇒ Tan–1

+

−

1

1

c1

.yx

1

c1

yx

+

−+ −

21

211

cc1

1

c1

c1

Tan + … +

Tan–1

nc1

⇒ Tan–1

−

+

−

−−−

2

1

1

1

1

1

c

1Tan

c

1Tan

c

1Tan

y

x

++ −

n

1

c1

Tan... ⇒ Tan–1

=

−

xy

Cotyx 1

Choice (2)

28. Given: 3Cot–1 )x(Cot32

1 1−−

−= Cot–1(3)

⇒ 3Tan–1(2 − 3 ) = Tan–1

+

−

31

Tanx1 1

⇒ 3.

−

+=π −

31

.x1

1

31

x1

Tan12

1 ⇒ tan

31

.x1

1

31

x1

4 −

+=π

⇒ 1x3x3

−+

= 1 ⇒ 3 + x = 3x − 1

⇒ 2x = 4 ⇒ x = 2 Choice (4)

29. Given: α = Cot-1

−−+

−++22

22

x1x1

x1x1

⇒ tanα = 22

22

x1x1

x1x1

−++

−−+

Put x2 = cos2θ ⇒ Tanα = θ−+θ+θ−−θ+

2cos12cos1

2cos12cos1

⇒ tanα = )sin(cos2

)sin(cos2

θ+θθ−θ

= tan

θ−π4

⇒ tanα = tan

θ−π4

or α = 4π − θ

2α = 2π −2θ

sin2α = sin

θ−π2

2= cos2θ

sin2α = x2 Choice (1) 30. f(x) = cot–1x + cos-1 x – cos –1 (cosx) = cos-1 x = cot–1 x – x The range of x is [–1, 1]

If x = 1, cos–1 (1) + cot–1 (– 1) = 0 + 4

4−π

If x = – 1, cos–1 (– 1) + cot–1 (– 1) + 1

= π + 4

3π + 1 =

447 +π

∴ The range of the function is

+π−π4

47,

44

Choice (3)

Exercise – 4


1. By sine rule, we have Bsin

bAsin

a =

°

=°

⇒

30sinb

60sin6

⇒ b = 6 × 323

221 =×

Given: ABC a right triangle ⇒ c2 = a2 + b2

c2 = 62 + 12 ⇒ c = 4 3 Choice (2)

2. From sine rule, we have Csin

83sin

6 =°

⇒ sinC = 32

⇒ C = Sin-1

32

⇒ sin–1 x = sin–

32

∴ x = 32

9x2 – 3x – 2 = 0232

394

.9 =−

−

Choice (1) 3. From Cosine rule we have, a2 = b2 + c2 – 2b c cosA.

cosA = 9 + 16 – 2 x 3 x 4 x 87

⇒ a2 = 4 ∴ a = 2. Choice (2)


4. Given ∆ ABC is isosceless and ∠ B = 120° ⇒ ∠ A = ∠ C = 30°

by Sine rule, Bsin

bAsin

a =°

=° 120sin

32

30sin

a

⇒ a= 2

The area of ∆ ABC = 21

ab sinC = 21

x 2 x 2 3 sin 30°.

= 3 sq. units. Choice (3)

5. From the given sides 24 is the least.

∴ The angle opposite to 24 is the smallest.

Let c = 24 , a = 6+ 12 and b = 48

cosC = ab2

cba 222 −+

)48()126(.2

)24()48()126( 222

+−++=

⇒ cosC = 23

)33(3.16

)33(24 =+

+ ⇒ ∠ C =

6π

Choice (3)

6. Given: 2acos2

2C

+ 2ccos2

2A

= 3b

⇒ a(1 + cosC) + c(1 + cosA) = 3b ⇒ a + c + acosC + ccosA = 3b ⇒ a + c + b = 3b (from projection rule) ⇒ a + c = 2b ⇒ a, b and c are in A.P. Choice (1)

7. Given 4sin2

2A

. sin2 2C

= sin2 2B

⇒ 4. ac

)cs()as(ab

)bs()as(.

bc)cs()bs( −−=−−−−

⇒ 2

2

b

)bs(4 − = 1 ⇒ 2(s – b) = b

2s = 3b ⇒ a + b + c = 3b ⇒ a + c = 2b. ∴ a, b, c are in A.P. Choice (3) 8. Given: In ∆ABC, ∠ C = 90° ⇒ B = 90° – A

From Napier’s rule, tan2

Acot

cb

cb

2

CB

+−

=

−

⇒ tan2

Acot

cb

cb

2

90A90

+−

=

°−−°

⇒ tan2

2A

= bcbc

+−

Choice (4)

9. We know that r = 2215

230

s==

∆

from 13th problem, R = 24

33

∴ The ratio of inradius and circum-radius is

r : R = 2 2 : 24

33 = 16 : 33. Choice (4)

10. The ratio of inradius and circum-radius of an equilateral triangle is 1 : 2.

∴ r = 21

Rr

2R =⇒

⇒ r : R = 1 : 2. Choice (3)

11. acosA + bcosB + ccosC = 2RsinAcosA + 2RsinBcosB + 2RsinCcosC = R[sin2A + sin2B + sin2C] = R 4. sinAsinBsinC

= 4R.R2

R2

R4

R2

abcR2c

.R2b

.R2a

22

∆=∆==

Choice (2) 12. We know, that

31

18321

61

91

181

r1

r1

r1

r1

321=++=++=++=

∴ The area of the triangle = 6x9x18x3rrrr 321 =

= 54 units. Choice (3) 13. Given: h1, h2 and h3 are the altitudes of ∆ABC

⇒ ∆ = 21

h1a ⇒ h1 = a

2∆

Similarly, h2 = b

2∆ and h3 =

c2∆

c2

Ccos

b2

Bcos

a2

Acos

h

Ccos

h

Bcos

h

Acos

321 ∆+

∆+

∆=++

= ∆21

[2RsinAcosA + 2RsinBcosB + 2RsinCcosC]

=∆2

R [sin2A+sin2B+sin2C] =

∆2R

[4sinAsinBsinC]

= ∆2

R [4sinAsinBsinC] =

R1

R2C

.R2b

.R2a4

.2R =∆

(Q ∆ = 2R2sinAsinBsinC) Choice (4)

14. Given: (a – b)2 cos2 2C

+ (a + b)2 sin2 2C

= (a – b)2

−++

+

2

Ccos1)ba(

2

Ccos1 2

= 21

{(a – b)2 + (a + b)2 + [(a – b)2 – (a + b)2]cosC}

= 21

{2(a2 + b2) – 4abcosC}

= 22

(a2 + b2 – 2abcosC)

= c2. Choice (2) 15. Given: r1 : r2 : r3 = 4 : 8 : 12

⇒ 121

:81

:41cs

:bs

:as =

∆−

∆−

∆−

⇒ s – a : s – b : s – c = 6 : 3 : 2 ⇒ s – a = 6k, s – b = 3k and s – c = 2k a = s – b + s – c =3k + 2k = 5k. b = s – a + s – c = 6k + 2k = 8k c = s – a + s – b = 6k + 3k = 9k ∴ a : b : c = 5k : 8k : 9k = 5 : 8 : 9. Choice (2) 16. Given: The ratio of the angles of the triangle 3 : 4 : 5. Then the angles of triangle are 45°, 60°, 75°.

From the Sine rule, we have oo 60sin

b

45sin

a=

The ratio of the two smaller sides is

23

b

2

1a

=

3

2

3.2

2

b

a==

a : b = 2 : 3 . Choice (3)


C

B

θ φ

A

D

17. Given: 2R + r = r1 ⇒ 2R = r1 – r

⇒ 2R = 4Rsin2A

cos2B

cos2C

– 4R sin2A

sin2B

sin2C

⇒ 2R = 4Rsin 2A

−2

Csin

2

Bsin

2

Ccos

2

Bcos

21

= sin2A

cos

2B

+

2C

= sin2A

sin2A

.

sin2 2A

= 21

⇒ 2A

= 45° ⇒ ∠ A = 90°.

∴ ABC is a right triangle. Choice (3) 18. Given: A1, A2, A3 and A are the areas of ex-circles and

incircle of ∆ABC respectively ⇒ A1 = πr1

2, A2 = πr22, A3 = πr3

2 and A = πr2. Where r1, r2, r3 and r are the radii of ex-circles and incircle of ∆ABC.

⇒ A

1

r

1r1

r1

r11

A

1

A

1

A

12

321321

=π

=

++π

=++

Choice (4)

19. Given: in ∆ ABC

−

−

3

1

2

1

rr

1rr

1 = 2

⇒

−−−

−−−

ascs

1asbs

1 = 2

⇒ as

bsas−

+−−

−+−−

as)csas(

= 2 ⇒

−−−2)as(

)ac()ab(= 2

⇒ (b – a) (c – a) = 2

a2

cba2

−++

⇒ 2(bc – ab – ac + a2) = (b + c – a)2 ⇒ a2 = b2 + c2 ∴ ABC is a right triangle. Choice (3)

20. cosA + cosB + cosC = 2cos2

BA +cos

2BA −

+cosC

= 2cos2

C180 −°cos

2BA −

+ cosC

= 2sin2C

cos2

BA − + 1 – 2sin2

2C

= 2sin2C

−−

2

Csin

2

BAcos + 1

= 2sin2C

+−

−2

BAcos

2

BAcos +1

= 2sin2C

2

Bsin

2

Asin2 + 1 = 1+2sin

2A

sin2B

sin2C

= 1 + RR2

sin2A

sin2B

sin2C

= 1 + Rr

.

Choice (4) 21. Given: Σ a2 (cos2B – cos2C) = Σa2(1 – sin2B – 1 + sin2C)

⇒ Σa2(sin2C – sin2B) = Σa2

−

2

2

2

2

R4

b

R4

c

= 2R4

1Σa2(c2 – b2)

∴ 2R4

1 [ ])ab(c)ca(b)bc(a 222222222 −+−+− = 0

Choice (4)

22. Given: r1, r2 and r3 are in H.P

⇒2r2

= ∆−+

∆−=

∆−

⇒+ csas)bs(2r1

r1

31

⇒ 2s – 2b = 2s – a – c ⇒ 2b = a + c ⇒ a, b, c are in A.P. Choice (3)

23. Given: In ∆ ABC, 2222 bc

a

ac

b

−+

−= 0

b(c2 – b2) + a(c2 – a2) = 0. bc2 – b3 + ac2 – a3 = 0. c2(b + a) = b3 + a3 c2 (b + a) = (b + a) (b2 + a2 – ba) ⇒ a2 + b2 – ab = c2

⇒ a2 + b2 – 2.abcos60° = c2 ⇒ ∠ C = 3π

Choice (2) 24. Given: r + r1 + r2 – r3 = 0 ⇒ r1 + r2 = r3 – r

⇒ ∆

−−

∆=

−+

− s1

cs1

bs1

as1

⇒ )cs(scss

)bs)(as(bas2

−+−=

−−−−

⇒ s(s – c) = (s – a)(s – b) or )bs()as(

)cs(s−−

− = 1

⇒ cot2 2C

= 1 ⇒ 2C

= 4π

⇒ C = 2π

Choice (1).

25. 2C

cot)bs)(as(

)cs(s

css

bsasrrrr 2

3

21 =−−

−=

−∆∆

−∆

−∆

= Choice (2)

26. We know that,

r = s∆

, r1 = as−

∆, r2 =

bs−∆

and r3 = cs−

∆

Now,

∆−−∆+

∆−−∆=

−+−

scsc1

sbsb1

c

rr

brr 32

⇒

−+−∆+

−+−∆

)cs(scss

c)bs(sbss

b

⇒ )as()as(

x)cs)(bs(s

a)cs()bs(

bscss −

−−−

∆=

−−−+−∆

= 1

2 ra

as

a)as(a =

−∆

=∆

−∆ Choice (2)

27. Given: AB = AC

Let D be the mid point of AC

⇒ AD = DC = 21

AC

Q ∠ CAB = ∠ ABC = 45° In ∆BCD,

cotθ = CDCD2

CDBC = = 2 → (1) and

In ∆ ADB, cotφ = cot(45 – θ)


A B

K

53°

B C

A

θ

B C

A

60° D 40 m

45

C D

A

45°

20

B

x

⇒ cotφ = 1221

1cotcot1

−+=

−θθ+

= 3 (from (1) cotθ = 2)

∴ The required ratio is 2 : 3. Choice (1)

28. b(rr2 +r1r3) =

−−∆+

−∆

)cs()as()bs(sb

22

−−−−+++−∆)cs()as()bs(ssbsac)ca(ss

b22

2

⇒

∆+++−∆

2

22 ac)cba(ss2b

= b{2s2 – s(2s) + ac} = b(ac) = abc. Choice (4) 29. Given: a, b and c are in A. P and a + c = 2b.

∴ Cot A/2 = ∆− )as(s

Cot C/2 = ∆− )cs(s

Cot 2A

cot 2C

= ∆−

∆− )cs(s)as(s

= )bs(

s)cs()bs()as(s2 −∆

−−− =

bss−

= b2cba

cbab2s2

s2−++

++=−

= bb3

bbb2 =+

= 3.

Choice (1) 30. Given: A, B, C are in A.P and A + C = 2B We know that A + B + C = 180° 2B + B = 180° B = 60°

Given: 13

6cb

−= =

13

6CsinBsin

−=

Sin C = ( )

6

Bsin13 − =

6

13 −sin 60°

= 23

.6

13 −

Sin C = 2,,2

13 −

∴ C = 15° ∴ A = 105°. Choice (4)

Exercise – 5


Let K be the position of the kite and BK is the length of the string.

∴ In ∆KAB, sin53o = KBKA

⇒ KA = KB sin53° = m202554 =× . Choice (2)

2.

Let AB be the pole and BC be the length of its shadow.

Given: BC = .AB3

1

In ∆ABC, tanθ = 3BCAB =

∴θ = 60°. Choice (1) 3.

Let BD be the ladder AB = 16 ft Let BC = x Since ∠ ADC = 45°, DC = AC = 16 + x Now, BC2 + DC2 = 202 ⇒ x2 + (16 + x)2 = 202 2x2 + 32x − 144 = 0 ⇒ x2 + 16x − 72 = 0

2

344162

54416x

±−=±−=

3428x +−= ⇒∴ The height of the pole (AC) = 16 + x

= ( 3428 + ) ft Choice (1)

4. Let AB be the height of the chimney Since ∠ ACB = 45°, AB = CB = CD + DB

AB = 40 + DB ∴ DB = AB − 40

∆ABD, tan 60° = 40AB

ABDBAB

−=

⇒ 3 (AB – 40) = AB ⇒ ( ) 34013AB =−

∴ ( ) ( )m3320

23340

AB +=+= .

Choice (2)


D

E

C A

1⋅6

B

12

53°

C D A 60°

B

45°

C D A

β

B

θ

60 m

B 37° 53°

53° C A D

37°

A

C

10 m 17 m 8 m

B D

5.

Let AB be the man and CD be the tower. AC = 12 m and AC = BE

In ∆BED, tan 53° = BEDE

⇒ DE = BE tan 53° = .m1634

12 =×

∴ The height of the tower is 16 + 1⋅6 = 17⋅6 m. Choice (3) 6.

Let AB be the tower and C and D denote the heads of the two men. Then AC = 10 m and AD = 17 m

∴ CD = CB + BD = 2222 ABADABAC −+−

= 2222 817810 −+− = 6 + 15 = 21 m Choice (4)

7. Let AB be the chimney

CD = 36 m Since ∠ BCA = 45°, AB = CA = CD + DA ∴ DA = AB − CD = AB − 36

In ∆BAD, tan 60° = 36AB

ABDAAB

−=

⇒ AB = ( ) ( ) .m3318

23336 +=+

Choice (1) 8.

Let AB be the tower

Given: tanθ = 54

, tanβ = 25

and CD = 32 m

In ∆ABC, tanθ = 54

ACAB =

⇒ 54

DA32AB

54

DACDAB =

+⇒=

+

∴ ( )

4128AB5

DA−=

In ∆DAB, tanβ = ( ) 25

4128AB5

AB25

DAAB =

−⇒=

⇒ m17640

AB = Choice (2)

9.

Let A be the position of the aeroplane and DC be the building.

Since ∠ ACB = 45°, BC = AB = DE

CD = BE = ( )336 − m

In ∆AED, tan 30° = ( )

AB336AB

DEAE −−=

⇒ ( ) 3336AB3AB ×−−=

∴ AB = ( )

( )( )

1813

1318

13

3336 =−

−=−

−m.

Choice (4) 10. Let AB be the cliff. Let C and D denote the position of the two boats.

In ∆BAC, tan 53° = CAAB

∴ CA = AB cot 53° = AB cot (90° − 37°) = AB tan 37°

In ∆BAD, tan 37° = ADAB

⇒ AD = AB cot37o

∴ CD = CA + AD = AB (tan 37° + cot 37°)

Since, sin 37° = ,53

tan 37° = 43

and cot 37° = 34

∴ CD = m12525534

43

60 =×=

+ . Choice (1)

45° 30°

A

E

B C

D

45°

30°


A

D lamp

C B 12 18 E θ

6

B

C

A P α θ

B

A

D

C M 45° 60°

A

B D C 45° 60°

A

B

D

C

45° 37°

E

C1

C

A

B

160 60°

B1

D

Level of the lake

30°

•

11.

DC' = DC = the height of the cloud B'D = AB = 160 m In ∆BB' C Tan 30°

= ( )160CD3BBBB

CB 11

1−=⇒ -- (1)

In ∆BB1C1, tan 60° = ( )160CD3

160CD

BB

DBDC1

11

−+=+

⇒ 3CD − 3 × 160 = CD + 160 ⇒ 2CD = 640 m. ∴ CD = 320 m. Choice (2) 12.

BE = shadow of the man. AB = 18 ft.

In ∆ABE, tanθ = ECDC

EBAB = ⇒ ( )1218

DC186

+=

∴ .ft1018

306DC =×= Choice (1)

13.

AC = CB

In ∆PAC, tanα = n21

AP2AB

APAC ==

In ∆PAB, tan (α + θ) = n1

APAB =

⇒ n1

tan.tan1tantan =

θα−θ+α

⇒ n1

tann21

1

tann21

=θ−

θ+

⇒ θ−=θ+ tann2

1n1

tann21

2 ⇒ θ

+= tann2

11

n21

2

∴ cotθ = n

1n2 2 + Choice (1)

14.

Let M be the mid point of AC and AB = m3112

In ∆MAB, tan 60° = AMAB

∴ m1123

3112AM ==

since ∠ DMC = 45° , MC = CD = 112 m. (QMC = AM) Choice (4) 15.

Let AB be the tower. CD = 90 m

Since ∠ ACB = 45°, AB = CB = CD + DB = 90 + DB ∴ DB = AB − 90

Now , tan 60° = 90AB

ABDBAB

−=

⇒ 3 (AB – 90) = AB ⇒ ( ) 390AB13 =−

⇒ ( )

( ) ( )m334513

13390AB +=

−+= Choice (2)

16.

Let AB be the monument and DC be the building.

In ∆BED, tan 37° = ED24

EDEB = ⇒

ED24

43 =

∴ ED = 32m


← 3 m → B D C

A

θ 90−θ

A B

1 km

C D O 45°

53°

A

C B 100

3

100

θ

α θ C A

B

D α

h

E

A C 45°

P

53°

D B

Since ∠ ADE = 45°, AE = ED = 32 m The height of the monument is AB = AE + EB = 32 + 24 = 56 m. Choice (3)

17.

Let AB be the tower. Let ∠ ACB = θ.

Then, ∠ ADC = 90° − θ (∵they are complementary)

∆ABD, tan (90° − θ) = DBAB

⇒ cotθ = 12AB

∴ A B = 12cotθ --- (1)

In ∆ABC, tanθ = θ=⇒= tan3AB3

ABBCAB

--- (2)

∴ From (1) and (2), AB2 = 12 × 3 cotθ⋅ tanθ = 36 ∴ AB = 6 m. Choice (1) 18.

Let A be the position of the aeroplane at the given instant of time. Let v km/sec be the speed of the aeroplane. Then DC = AB = 12v km ∠ BOC = 45°, OC = BC = 1km ⇒ OD + DC = 1 OD = 1 − 12v

Now, tan 53° = v121

1ODAD

−=

⇒ tan (90° − 37°) = v121

1−

⇒ v121

137tan1

−=

°

Given: sin 37° = 0⋅6 = 53 ∴ tan 37° =

43

⇒ v211

134

−=

⇒ 4 − 48v = 3

∴ sec/km481

v = . Choice (2)

19.

Let AB be the tower

In ∆ACD, tanα = CAh

∴ CA = h cotα

In ∆CAB, tanθ = CAAB

∴ AB = CA tanθ = h tanθ cotα. Choice (3) 20.

Let AB be the pillar. CB = 100

In ∆ABC, tanθ = 3

1100

3100CBAB ==

∴θ = 30°. Choice (4) 21.

26 26

Let PB be the ladder.

Then, PB = PD = 26 m

In ∆PAB, Cos45° = 26

AP

⇒ 26

AP

2

1 = AP = 6 m.

In ∆PCD, cos53o = PDPC

⇒ 26

PC53 = (∴ cot 53o =

53

)

⇒ PC = m26.35

218 =

∴ The distance between the two walls of the room is

AP + PC = (6 + 3.6 m2 ) Choice (3)


A D C

B

45° 37° 80

A

h

B

β α

C

D

O

C1

x

h

x + h

R

O B AB

60° 30° x

22.

Let AB be the lighthouse and C and D denote the positions of the two ships.

CD = 80 m Since, ∠ ADB = 45°, AB = AD ∴ AC = AD = 80 = AB + 80

Now, tan 37° = ACAB

⇒ 80AB

AB43

+= (∴ sin 37° = 0⋅6 =

53

)

⇒ 3AB + 3 × 80 = 4AB ∴ AB = 240 m. Choice (2) 23.

Let C be the position of the cloud and OC be the surface level of the lake.

Let DC = x, Then OC1 = x + h.

In ∆ADC, tanα = ADCD

⇒ AD = αtan

x= x cotα ------ (1)

In ∆ADC1, tanβ = AD

DC1

⇒ AD = (x + 2h) cosβ ------ (2) ∴ From (1) and (2), x cosα = (x + 2h) cotβ

⇒ x = β−α

βcotcot

coth2

∴ The height of the cloud above the surface of the lake is x + h

= hcoscos

cosh2 +β−α

β =

( )β−αα+β

coscoscoscosh

=

ββ−

αα

αα+

ββ

sincos

sincos

sincos

sincos

h

=( )

βα−βαβα+βα

cossinsincossincoscossinh

=( )

( )α−ββ+α

sinsinh

units Choice (4)

24. Given: cosθ = 0⋅6 = 53

⇒ sinθ = 54

Let AC be the tree and let it break at the point B.

In ∆OAB, sinθ = OBAB

⇒ x

x1254 −=

1x

1254 −== ∴ .m

320

x =

Choice (1) 25.

Let OR be the rock.

In ∆BPR, tan60o = OBOR

⇒ OR = 3 OB ------ (1)

In ∆AOR, tan 30o = OAOR

⇒BAOB

OR

3

1+

=

⇒ 3 ( 3 OR) = OB + BA (∵ OR = 3 OB)

⇒ 2.OB = BA ⇒ OB = 2

AB

As it takes 15 minutes for the boat to cover the distance AB, the boat will reach the shore in 7 min and 30 seconds. Choice (3)

Exercise – 6

Solutions for questions 1 to 40: 1. i343 = (i2)171 . i = – 1 (i) = – i Choice (2)

2. 2

i3i1i1

i1i21 +=

−−

++

We know, the multiplicative inverse of x + iy is

iyx

y

yx

x2222 +

−+

.

∴ The multiplicative inverse of 5

i3is

2i3 −+

Choice (3)

3. Given: i45i23

i45i23

−−−

++

= x + iy

( )

1625i2815i2815

)i45)(i45()i45)(i23()i45(i23

+−−−−+=

−++−−−+

= 41

i4−

But it is given that, 41

i4− = x+ iy; ⇒ x = 0, y =

414−

,

∴ (x, y) = (0, – 4/41) Choice (4)

C1

x

B

x

O θ

A

12 − x


4. Given: x + iy = θ−θ+ sinicos1

1

= 2/cos2/sini22/cos2

12 θθ−θ

= )2/sini2/(cos2/cos2

1θ−θθ

= x)2/sini2/(cos

2cos2

1

θ−θθ

θ+θθ+θ

2/sini2/cos2/sini2/cos

= 2/cos2

2/sini2/cosθ

θ+θ

x + iy = 1/2 + i/2 tanθ/2 ∴ (y, x) = (1/2 tan 2θ

, 1/2)

Choice (1)

5. Given: ∑ +=

+11

1n

1nn )ii( = i + [i2 + i3 +….+ i11] 2+ i12

= i + 2i2 + 2i3 + 2i4 +….+2i11 + i12 = I + 2i2 (1 + i + i2 +…+i9) + (i2)6

= i – 2

−−

i1i1 10

+ (– 1)6 = i – 2i1)2(

− + 1

= – (1 + i) Choice (3) 6. We know that 1 + ω + ω2 = 0. Using this, we get (1 – ω + ω2)7 + (1 – ω2 + ω)7 = – 27 (ω7 + ω14) = (– 2)7 ((ω3)2 ω + (ω3)4 . ω2) = –27 (ω + ω2) , (Qω3 = 1)

= – 27 (– 1) = 27 Choice (2) 7. The centre and radii of the circles are C1 = (2, 0) and r1 = 1, c2 = (5, 0), r2 = 2 The distance between the centres is 3 and also the sum

of the radii is 3

∴ The two circles touch externally. Choice (1)

8. Given: n2

i1i1

+−

= ( ) ( ) n22n2

2i1

)i1)(i1()i1(i1

−=

−+−−

n2

2i211

−−

= (– i)2n. if n = 2 then (– i)2n = (– i)4 ∴ The least positive integer = 2. Choice (4) 9. Let i68 − = x – iy

⇒ 8 – 6i = x2 – y2 – 2xyi Comparing real and imaginary parts on both sides we get = x2 – y2 = 8 ----------- (1)

2xy = 6,

(x2 + y2) = 22222 yx4)yx( +−

⇒ x2 +y2 = 3682 + = 10___________(2) solving (1) and (2) for x2 and y2 we have x2 = 9 and y2 = 1 ∴ x – iy = ± (3 – i) Choice (1) 10. Let x + iy = i125 +−

(x + iy)2 = – 5 + 12i ⇒ x2 – y2 = – 5 and 2xy = 12 x2 + y2 = 13 ___________(2) solving (1) and (2), x2 = 4 and y2 = 9. ∴ The square root is ± (2 + 3i). Choice (2)

11. θ−θ+

sini21sini23

= θ+θ+

θ−θ+

sini21sini21

xsini21sini23

= θ+

θ+θ−2

2

sin41

sini8sin43 =

θ+θ+

θ+θ−

22

2

sin41

sini8

sin41

sin43.

(a) If the above complex number is purely real, then

imaginary part is 0, i.e. θ+

θ2sin41

sin8=0 ⇒ sinθ = 0

∴ θ = nπ where n ∈ z. Choice (1) (b) If the above complex number is purely imaginary,

then the real part is 0, i.e. θ+θ−

2

2

sin41

sin43=0

= 3 – 4 sin2 θ = 0 ⇒ sin2θ = 3/4 θ = nπ ± π/3. Choice (2)

12. i4i3i1 +

+−

= i33i11

+−

= i3

i113++−

xi3i3

−−

=5

i1851 +

Let 5

i1851 + = r (cos θ + i sin θ)

22 yxr += = 135

1851

22

=

+

; and

θ = tan–1

xy

= tan-1(18).

The mod aptitude form of

∴ i4i3i1 +

+−

= 13 [cos (tan-1 (18)] + i sin (tan-1 (18))

Choice (3)

13. Let z = x+ iy, z1 = x1 + iy1 and z2 = x2 + iy2 Given: z – z1 = z – z2 ⇒ (x – x1) + (y – y1)i = (x – x2 ) + (y – y2)i

⇒ 21

21 )yy()xx( −+−

= 22

22 )yy()xx( −+−

Taking the square on both sides we have (x – x1)

2 + (y – y1)2 = (x – x2)

2 + (y – y2)2

= 2x(x1 – x2) + 2y (y1 – y2) + 2

1

2

1

2

2

2

2 yxyx −−+ =0

It is a first degree equation. ∴ The locus of z is a straight line. Alternate method: Geometrically, |z1 – z2| represents the distance between

the two complex numbers z1 and z2 z – z1 = z – z2 ⇒ z is equidistant from z1 and z2 ⇒ z is the perpendicular bisecter of the line joining z1 and z2 Hence, the locus of z is a straight line. Choice (4) 14. Given: x = 2 – 3i Now x2 – 4x + 10 = (x – 2)2 + 6 = (2 – 3i – 2)2 + 6 = – 9 + 6 = – 3 Alternate method: Given: x = 2 – 3i (x – 2) = – 3i Taking the square on both sides, (x – 2)2 = (– 3i)2 x2 – 4x + 13 = 0 : x2 – 4x + 10 = – 3. Choice (3)


2 6i

z

2π

15. When n is not a multiple of 3, then ωn = ω or ω2 if n is not a multiple of 3 and ωn = ω then

ωn + ω2n = ωn + (ωn)2 = ω + ω2 = – 1 and if ωn = ω2, then ω2 + (ω2)2 = ω2 + ω = – 1 = ωn + ω2n = – 1 Choice (4) 16. Given: (a – b) (aω – bω2) (aω2 – bω) = (a – b) (a2ω3 – abω2 – abω4 + b2ω3) = (a – b) {a2 – ab (ω2 + ω) + b2} (Qω3 = 1) = (a – b) (a2 + ab + b2) Q ( 1 + ω + ω2 = 0 = a3 – b3

⇒ ω2 + ω = –1) Choice (4) 17. Let z = x + iy z – 2 = (x – 2) + iy z – 6i = x + i(y – 6)

)6y(ix

iy)2x(i6z

2z−++−=

−−

= [ ] ( )[ ] [ ])6y(ix)6y(ix

)6y(ixiy)2x(−−−+−−+−

= 22 )6y(x

))6y)(2x(xy(i)6y(y)2x(x

−+−−−+−+−

= 2222 6y(x

))6y)(2x(xy(i

)6y(x

))6y(y)2x(x

−+−−−+

−+−+−

∴ Amplitude = tan–1

−+−−−−

)6y(y)2x(x)6y)(2x(xy

(∵ The amplitude of x + iy is tan–1 y/x)

The amplitude of i6z

2z−−

is given as π/2

tan-1

−+−−−−

)6y(y)2x(x)6y)(2x(xy

= π / 2

It is true only when the denominator is zero. = x (x – 2) + y (y – 6) = 0 ⇒ x2 +y2 – 2x – 6y = 0

Alternate method:

Q arg

−−

i6z2z

= π /2,

So z lies on the circle and (2, 0), (0, 6) are the ends of the diameter.

∴ The equation of the circle is (x – 2) (x – 0) + (y – 0) (y – 6) = 0

x2 + y2 – 2x – 6y = 0 Choice (3) 18. The points on the argand plane are A (1, 0),

B ( )2

1,

2

1, C (0, 1)

AB =

22

2

1

2

12

+

−

BC =

22

2

12

2

1

−+

AC = 11+ Clearly, we notice that AB = BC The given points form an isosceles triangle. Choice (2)

19. Let z = x + iy; iz = – y + ix z + iz = x – y + i(y + x) The points on argand plane are z = (x, y), iz = (–y, x), z + iz = (x – y, x + y) The area of the triangle is

22 xxyyxy21

yxxyxyxyyx

21 −+−−=

−−+−−−+

22 yx21 + =

2z

21

Alternate method: Given: z, iz, z + iz are vertices of triangle Let z = x +iy iz = – y + ix z, iz are perpendicular to each other.

The Area of the triangle = 21

z iz = 1/2 z 2

Choice (4) 20. Given: az1 + bz2 + cz3 = 0 and a + b + c = 0

⇒ z3 = cbzaz 21

−+

and c = – (a + b)

z3 = babzaz 21

++

Clearly z3 divides the line joining z1 and z2 in the ratio a : b ⇒ z1, z2, z3 are collinear

∴ The three points lie on a line Choice (3) 21. Let z1 = x1 + iy1 and z2 = x2 + iy2 Given z1 + z2 = z1 + z2

⇒ (x1 + x2)2 + (y1 + y2)

2

= x 22

22

21

21 yxy +++ + 2 2

222

21

21 yxyx ++

⇒ x1 y2 – x2y1 = 0

⇒ 2

2

1

1

xy

xy = ⇒ tan–1

1

1

xy

= tan–1

2

2

xy

⇒ arg z1 = argz2

⇒ argz1 – arg z2 = 0 Choice (1) 22. Given: (–1 + i 3 )30 + (–1 – i 3 )30

The mod amplitude form of –1 + i 3 is

2 (cos +π32

i sin 32π

)

(– 1 + i 3 )30

= 30

32

sini32

cos2

π+π

= 230 (cos 32π

x 30 + i sin 32π

x 30)

= 230 (cos 20π + isin 20π) Similarly, ( – 1 – i√3)30 = 230 (cos 20 π – i sin 20 π)

(–1 + i 3 )30 + ( –1 + i 3 )30

= 230 (cos 20 π + i sin 20 π + cos 20 π – i sin 20 π) = 230 (2 cos 20 π) = 231 Alternate method: We know that

+−2

3i1 = ω,

−−2

3i1 = ω2

= (– 1 + 3 )30 + (– 1 – i 3 )30


=

3030

2i31

22

3i12

−−+

+−

= 230. ω30 + 230(ω2)30 = 230 (ω30 + ω60)

= 230 (1 + 1) = 231 (∵ ω3 = 1 ⇒ ω30 = 1) Choice (3)

23. Given: (1 + cosθ + I sin θ)n + (1 + cos θ – I sin θ)n = (2 cos2 θ /2 + i 2 sinθ/2cosθ/2)n + (2cos2 θ/2 – i 2 sin θ/2cos θ/2)n

= 2n cosn θ/2 (cos θ/2 + i sin θ/2)n + 2n cosn θ/2 n

2sini

2cos

θ−θ = 2n cosn θ/2

θ−θ+θ+θ2

nsini2

ncos

2n

sini2

ncos

= 2n cosn

2θ

2 cos 2

nθ = 2n+1 cosn

2θ

.cos 2nθ

Choice (2)

24. Given: x + x1

= 2 cos θ ⇒ x2 – 2 cos θ x + 1 = 0

⇒ x = cos θ ± i sin θ ∴ x10 = cos 10θ + i sin 10θ and

10x

1= cos 10θ – i sin 10θ

∴ x10 + 10x

1 = 2 cos 10θ Choice (4)

25. Given: α and β are the roots of x2 – 2x + 4 = 0

⇒ α = 1 + 3 i and β = 1 – i √3

α = 2( cos π/3 + i sin π/3) and β = 2 (cos π /3 – i sin 3π

).

αn = 2n (cos3

nπ+ i sin

3nπ

) and βn = 2n (cos3

nπ – i sin

3nπ

)

α n + βn = 2n (2 cos n π/3) = 2n+1 cos n π/3 Choice (4) 26. Let a = cos α + i sin α, b = cos β + i sin β and c = γ + i

sin γ a + b + c = cos α + cos β + cos γ + i (sin α + sin β + sin γ )

= 0 + i(0) a + b + c = 0 we know that if (a + b + c) = 0, then a3 + b3 + c3 = 3abc ⇒ (cos α + i sin α)3 + (cos β + i sin β)3 + (cos γ + i sin γ )3

= 3 (cos α + i sin α) ( cos β + i sin β) ( cos γ + i sin γ) Cos 3 α + i sin 3 α) + (cos 3 β + i sin 3 β) + (cos 3γ + i sin 3 γ)

= 3[cos(α + β + γ) + i sin (α + β + γ)] Comparing the real parts on both sides, we get, cos 3 α + cos 3 β + cos 3 γ = 3 cos (α + β + γ). Choice (1)

27. 2/1

qp

=

2/1

y2siniy2cosx2sinix2cos

++

= (cos 2 (x – y) + i sin 2 (x – y))1/2 = cos (x – y) + i sin (x – y).

2/1

pq

=

21

x2sinix2cosy2siniy2cos

++

= cos (y – x) + isin (y – x) = cos (x – y) – isin (x – y)

∴ pq

qp + = 2cos(x – y) Choice (3)

28.

50

23i3

+ =

50

2i

23

3

+

= 325 (cosπ/6 + i sin π/6)50

= 325

π+π3

25sini

325

cos

∴ 325 (cos 3

25π + i sin

3

25π) = 324 (x – iy)

⇒ 3 ( cos π/3 + i sin π/3) = x – iy

⇒

+

2i3

21

3 = x – iy

⇒ x = 3/2 and y = 2

33−

∴ (y, x) = (2

33−,

23

) Choice (2)

29. Given: 1, α1, α2, …….αn-1 are the nth roots of unity ⇒ xn – 1 = (x – 1) (x – ∝ 1) (x – ∝ 2) …… (x – ∝ n-1)

x1x1 n

−−

= (x – ∝ 1) (x – ∝ 2) ….(x – ∝ n-1)

(x – ∝ 1) (x – ∝ 2) …..(x – ∝ n-1) = (1 + x + x2 +…+ xn–1) Put x = 1 ∴ (1 – ∝ 1) (1 – ∝ 2) ….. (1 – ∝ n–1) = n Choice (3) 30. x4 – 1 = 0 ⇒ x = (1)1/4 = ( cos 0 + i sin 0)1/4

= cos 4k2 π

+ i sin 4k2 π

for k = 0, 1, 2, 3

When k = 0, 1, 2, 3 we get x = 1, i, – i, – 1 ∴ The roots are ± 1, ± i. Choice (4) 31. (cos π/3 + i sin π/ 3)3/4

= {cos (2 k π + π/3) + i sin (2 k π + π/3)}3/4

= cos ((6k +1) 4

π) + i sin ( 6k + 1) π/4 where k = 0, 1, 2, 3

The product of these values is

cos

π+π+π+π4

194

134

74

+ i sin

π+π+π+π4

194

134

74

= cos 10 π + i sin 10 π = 1

Choice (1)

32. We know that ω = 2

i31+− and ω2 =

2i31−−

6

3i1

3i1

−

+ +

6

3i1

3i1

+−

= ( )( )6

62

ω−ω−

+( )

62

6

)( ω−ω−

= ω6 + 6

1

ω = (ω3)2 + ( )23

1

ω = 1 + 1 = 2.

Choice (3)

33. We know that i = cos 2π

+ i sin 2π

i = 2/ie π

(i)i = [ ] i2/ie π = 2eπ−

. Choice (4)


34. We know that i = cos 2π

+ isin 2π

= 2i

eπ

∴ log I = log 2i

eπ

= 2iπ

log(log i) = log (i π/2) = log i + log π/2 = i π/2 + log π/2 Choice (2) 35. The thirteenth term of the series is t13 = (cos θ + i sin θ )12 = (cos 12 θ + i sin 12θ).

When θ = π/3, t13 = cos 12 ×3

π + i sin 12.

3

π = 1 + 0 = 1

Choice (4) 36. z = z +1 z – 0 = z – (– 1) z is equidistant from 0 and – 1 ∴ It lies on the perpendicular bisector of line joining

points (1, 0) and (– 1, 0). The roots of z are collinear. Choice (4)

37. =π∞

=n

1nx x1 x2 x3 ……

= cos

+π+π+π....

222 32 + i sin

+π+π+π....

222 32

32 222

π+π+π+ …… is a G.P of infinite terms

∴ Its sum is π { ....2

1

2

121

32+++ }

= π

− 2/112/1

= π

∴ x 1 x2 x3 …. = cos π + i sin π = – 1 Choice (2)

38. ∑∞

=0n

ni = 1 + i + i2 + …. = i1

1−

= i1

i1x

i1

1

++

−= i

2

1

2

1

2

i1 +=+

Given, ∑∞

=0n

ni = a + ib ⇒ 2i

21 + = a + ib

a = 1/2 b = 1/2 = (b, a) =

21

,21

Choice (2)

39. We know that cos nθ = nC0 cosnθ – nC2 cos n-2

θ . sin2θ + nC4 cosn-4 θ. sin4 θ – ……………

∴ cos 4 θ = 4C0 cos4θ – 4C2 cos 4-2 θ. sin2 θ + 4c4 cos 4-4 θ . sin4 θ.

∴ cos 4θ = cos4 θ – 6 cos2 θ (1 – cos2 θ) + (1 – cos2 θ)2 = cos4θ – 6 cos2θ + 6 cos4 θ + 1 + cos4 θ – 2 cos2 θ = 8 cos4 θ – 8 cos2 θ +1 Choice (3)

40. We have (2 i sin θ)n = (x – n)x1

where x = cos θ + i sin θ

(2 isinθ)6 = (x – 6)x1

= x6 – 6x4 + 15x2 – 20

+ 642 x

1

x

6

x

15 +−

– 26 sin6 θ =

+6

6

x

1x – 6 (x4 + )

x

14

+ 15 (x2 + )x

12

– 20

= 2 cos 6 θ – 12 cos 4θ + 30 cos 2θ – 20

(∵ xn + nx

1 = 2 cos nθ)

∴ sin6 θ = 32

1−(cos 6θ – 6 cos 4θ + 15 cos 2θ – 10)

Choice (4)

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