FUNCTIONS

5. BINOMIAL THEOREM

PREVIOUS EMACET BITS [EAMCET 2009]

( ) ( )( )1. The coefficient of x24 in the expansion of 1+ x2 12 1+ x12 1+ x24

1) 12 C6 2) 12 C6 + 2 3) 12 C6 + 4 4) 12 C6 + 6

Ans: 2

( ) ( )Sol: 1+ x2 12 1+ x12 + x24 + x36

( )= ⎡⎣12 C0 +12 C1x2 +12 C2x4 + ...... +12 C12x24 ⎤⎦ 1+ x12 + x24 + x36

Coefficient of x24 is 12 C0 +12 C6 +12 C12 =12 C6 + 2

2. If x is numerically so small so that x2 and higher powers of x can be neglected, then

⎜⎝⎛1 + 2x ⎞3/ 2 (32 www.NetBadi.in+5x)−1/ 5is approximately equal to [EAMCET 2009]
3 ⎟⎠

1) 32 + 31x 2) 31+ 32x 3) 31− 32x 4) 1− 2x
64 64 64 64

Ans: 1

Sol: ⎡⎢⎣1 + 3 ⎛ 2x ⎞⎤ ⎡⎣⎢32 ⎝⎜⎛1 + 5x ⎞ ⎤ −1/ 5
2 ⎝⎜ 3 ⎟⎠⎥⎦ 32 ⎟⎠⎦⎥

⎝⎜⎛1 5x ⎞−1/ 5
32 ⎟⎠
= [1 + x ] ( )32 −1/ 5 +

= 1 (1 + x ) ⎣⎢⎡1 − 1 × 5x ⎤ = 1 (1 + x ) ⎜⎛⎝1 − x ⎞
2 5 32 ⎦⎥ 2 32 ⎠⎟

= 1 ⎝⎛⎜1 + x − x ⎞ = 32 + 31x
2 32 ⎠⎟ 64

1+ x + x2 + x3 5 = 15 ak xk then 7
( ) ∑ ∑3.If [EAMCET 2008]
a2k = ..... 4) 1024

k=0 k=0

1) 128 2) 256 3) 512

Ans: 3

( )Sol: a0 + a1x + a2x2 + ...... + a15x15 = 1+ x + x2 + x3 5

Put x = 1

a0 + a1 + a2 + ...... + a15 = 415 ………..(1)

Put x = – 1

a0 − a1 + a2 − ...... − a15 = 0 …………(2)

Binomial Theorem

(1) + (2)

2 (a0 + a2 + a4 + ...... + a14 ) = 45

∴a0 + a2 + a4 + ...... + a14 = 512

4. If a = 5+ 5.7 + 5.7.9 + ....then a2 + 4a = [EAMCET 2008]
2!3 3!32 4!33

1) 21 2) 23 3) 25 4) 27

Ans: 2

Sol: a = 3.5 + 3.5.7 + ......
2!.32 3!.33

2 + a = 1+ 3 + 3.5 + 3.5.7 + .... this one comparing with
3 3.6 3.6.9

(1+ x)n = 1+ nx + n (n −1) x2 + ....

2!
www.NetBadi.in
nx = 1, nx (nx − x) = 3.5

2 3.6

1(1− x) = 3.5 ⇒ 1− x − 5 ⇒ x = −2
2 3.6 33

nx = 1 ⇒ n ⎛ − 2 ⎞ = 1 ⇒ n = − 3
⎝⎜ 3 ⎟⎠ 2

⎜⎛⎝1 − 2 ⎞−3/ 2 ⎛ 1 ⎞−3 / 2
3 ⎟⎠ ⎜⎝ 3 ⎠⎟
∴ 2 + a = (1+ x)n = ⇒

2 + a = 33/2 ⇒ (2 + a )2 = 33 ⇒ a2 + 4a = 23

( )5. n
If ak is the coefficient of xK in the expansion of 1+ x + x2
for k = 0, 1, 2, ……2n, then

a1 + 2a2 + 3a3 + ...... + 2na2n is equal to [EAMCET 2007]

1) −a0 2) 3n 3) n.3n+1 4) n.3n

Ans: 4

( )Sol: We have 1+ x + x2 n = a0 + a1x + a2x2 + a3x3 + ....a2n x2n on differentiating both sides, we get

( )n 1+ x + x2 n−1 (1+ 2x ) = a1 + 2a2x + 3a3x2 + ...... + 2na2nx2n−1

Put x = 1 [EAMCET 2007]
a1 + 2a2 + 3a3 + ........... + 2na2n = n.3n
6. The sum of the series 3 − 3.5 + 3.5.7 ....... =

4.8 4.8.12 4.8.12.16

2

1) 3 − 3 2) 2 − 3 3) 3 − 1 Binomial Theorem
24 34 24 4) 2 − 1

Ans: 2 34

Sol: 3 − 3.5 + 3.5.7 ....... + 3 − 3 [EAMCET 2006]
4.8 4.8.12 4.8.12.16 44 4) 43/ 2

= 3 + 3 − 3.5 + .....− 3 [EAMCET 2006]
4 4.8 4.8.12 4

= 1− 1 + 1.3 − 1.3.5 + ..... − 3
4 2.4.4 4.4.2.4.3 4

= ⎡ + 1⎝⎛⎜ − 1 ⎞ + 1(1 + 2) ⎛ − 1 ⎞2 + ⎤ − 3
⎢1 4 ⎠⎟ ⎝⎜ 4 ⎠⎟ .....⎥ 4
⎣⎢ 2!
⎥⎦

= ⎛⎝⎜1 − 1 ⎞−1/ 2 − 3 = 2−3
4 ⎟⎠ 4 34
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7. 1+ 2 + 2.5 + 2.5.8 + ....is equal to
4 4.8 4.8.12

1) 4−2/3 2) 3 16 3) 3 4

Ans: 2

Sol: Let s = 1+ 2 + 2.5 + .... on comparing with
4 4.8

(1+ x)n = 1+ nx + n (n −1) x2 + .....

2

We get , nx = 2 , nx (nx − x) = 2.5
42 4.8

2⎛ 2 − x ⎞ 2.5 −3
4 ⎝⎜ 4 ⎟⎠
= = ⇒ x =
2 4.8 4

n ⎛ − 3 ⎞ = 2 ⇒ n = − 2
⎜⎝ 4 ⎠⎟ 4 3

∴s = (1+ x )n = ⎝⎜⎛1 − 3 ⎞−2/3 = ⎛ 1 ⎞−2/3 = 3 16
4 ⎟⎠ ⎝⎜ 4 ⎠⎟

8. The correct matching of List- I from List – II is

List – I List – II

A) (1− )x −n (i) x
x +1

3

Binomial Theorem

B) (1+ x )−n (ii) 1− nx + n (n +1) x2.......if x < 1

2!

C) If x > 1 then 1+ 1 + 1 + ..... is (iii) 1+ nx + n (n +1) x2 + .......if x < 1
x x2
2!

D) If x >1 then 1− 2 + 3 − 4 + .... is (iv) x
x2 x4 x6 x −1

x4
x2 +1 2
( )(v)

x4
x2 −1 2
( )(vi)

ABCD AB C D
iv v
1) i iii iv vwww.NetBadi.in 2) ii iii i v

3) iii ii iv v 4) ii iii

Ans:

Sol: We know that (i) (1− x)−n = 1+ nx + n (n +1) x2 + ....if x < 1

2!

(ii) (1+ )x −n = 1− nx + n (n −1) x2 + ....if x < 1

2!

(iii) 1+ 1 + 1 + ......... = 1 1 = x
x x2 − x −1
1
x

1− 2 + 3 = x4
x2 x4 x2 +1 2
( )(iv)
.................

(1+ )x 15 = a0 + a1x + ..... + a15x15, 15 ar
If∑9. is equal to [EAMCET 2005]
then r 4) 135
r=1 a r −1

1) 110 2) 115 3) 120

Ans: 3

15 r. a r 15 n − (r −1)

= r.
∑ ∑Sol:
ar=1 r−1 r=1 r

15 15

= ∑ ⎡⎣(n +1) − r⎤⎦ = ∑(16 − r)
r=1 r=1

=15 + 14 + …….+ 1

= 15×16 = 120
2

4

10. If x < 1 , then the coefficient of x′ in the expansion of 1+ 2x Binomial Theorem
2 [EAMCET 2005]
(1− 2x)2 , is

1) r2r 2) (2r −1) 2r 3) r22r+1 4) (2r +1) 2r

Ans: 4

( )Sol: 1+ 2x = (1+ 2x)(1− 2x)−2
1− 2x2

= (1 + 2x ) ⎡ + 2 ( 2x ) + 2.3 ( 2x )2 + .... + 2.3.....r ( 2x )r−1 + 2.3.4.... ( r + 1) ( 2x )r ⎤
⎢1 1! 2! r ! ⎥
⎣⎢ (r −1)! ⎦⎥

The coefficient of xr = 2 r! .2r −1 + (r +1)! 2r

( r −1)! r!

= r.2r + (r +1).2r = 2r (2r +1)
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11. The coefficient of x3y4z5 in the expansion of ( xy + yz + zx )6 is [EAMCET 2005]

1) 70 2) 60 3) 50 4) none of these

Ans: 2
Sol: If the general term in the above expansion contains x3y4z5 then r + t = 3, r + s = 4 and s + t = 5

Also, r + s + t = 6

Solving these equation, we get r = 1, s = 3, t = 2

Coefficient x3y4z5 = 6! = 60
1!3!2!

12. The binomial coefficients which are in decreasing order are [EAMCET 2004]

1) 15 C5 ,15 C6 ,15 C7 2) 15 C10 ,15 C9 ,15 C8 3) 15 C6 ,15 C7 ,15 C8 4) 15 C7 ,15 C6 ,15 C5

Ans: 4

Sol: The series of binomial coefficient is

15 C0 ,15 C1,15 C2 ,.......15 C7 , 15 C8 ,15 C9 ........ 15 C14 ,15 C15

decrea sin g value greatest value decrea sin g value

From the above discussion, we can say that decreasing series is 15 C7 ,15 C6 ,15 C5

∴ Option (4) is correct

13. If the coefficient (2r + 1)th term and (r + 2)th term in the expansion of (1 + x)42 are equal, then r is

equal to : [EAMCET 2003]

1) 12 2) 14 3) 16 4) 18

Ans: 2
Sol: Given that coefficient of (2r + 1)th term = coefficient of (r + 2)th term

C43 =43 C r +1 ⇒ 43 = 2r +(r +1) (or) 2r = r +1
2r

5

Binomial Theorem

⇒ r = 14(or) r = 1

Thus r = 14 [EAMCET 2003]

( )14. The coefficient of x5 in the expansion of 1+ x2 5 (1+ x)4 is

1) 60 2) 50 3) 40 4) 56

Ans : 1

1+ x2 5 (1+ x )4 = ⎢⎡⎣1+5 C1x2 +5 C2x4 + ..... + 5⎤
( ) ( )Sol: x2 ⎥⎦
We have

⎡⎣1+4 C1x +4 C2x2 +4 C3x3 + x4 ⎦⎤

Coefficient to x5

=5 C1.4 C3 +5 C2.4 C1 = 20 + 40 = 60

15. If the coefficientwww.NetBadi.inofxintheexpansion of ⎛ x 2 + k ⎞5 is 270 then k is equal to [EAMCET 2002]
⎜⎝ x ⎠⎟

1) 1 2) 2 3) 3 4) 4

Ans: 3

Sol: General term in the expansion of ⎛ x 2 + k ⎞5 is
⎝⎜ x ⎠⎟

( )Tr+1 =5 Crx25−r ⎛ k ⎞r =5 Cr kr x10−3r
⎝⎜ x ⎠⎟

Let this term contains x then 10 − 3r = 1 ⇒ r = 2 then

Coefficient x =5 C3k3 = 10k3
10k3 = 270

k3 = 27 ∴ k = 3

( )16. n
The sum of the coefficients in the expansion of 1+ x + x2 [EAMCET 2002]
is

1) 2 2) 2n 3) 3n 4) 4n

Ans: 3

( )Sol: n
we have 1+ x + x2
, put x = 1

= (1+1+1)n = 3n

17. In the expansion of (1+ x)n the coefficients of pth and (p +1)th terms are respectively p and q, then

p + q is equal to [EAMCET 2002]

1) n 2) n + 1 3) n + 2 4) n + 3

Ans: 2

Sol: TP =n Cp−1 = P

6

Binomial Theorem

TP+1 =n Cp = q

∴p = Cn ⇒ p = p ⇒ p+q = n +1
p−1

q n Cp q n − p +1

18. 1+ 1 + 1.3 + 1.3.5 + ....... = [EAMCET 2001]
4 4.8 4.8.12

1) 2 2) 1 3) 3 4) 1
2 3

Ans: 1

Sol: This one comparing with (1+ x)n = 1+ nx + n (n −1) x2 + .....

2!

nx = 1 , n (n −1) x2 = 1.3
4 2! 4.8
www.NetBadi.in
nx (nx − x) 1.3 1 ⎛ 1 − x ⎞ 1.3
4 ⎝⎜ 4 ⎠⎟
= ⇒ =
2 4.8 2 4.8

⇒ 1 − x = 3 ⇒ x = − 1 = nx = 1
44 24

n ⎛ − 1 ⎞ = 1 ⇒ n = − 1
⎜⎝ 2 ⎟⎠ 4 2

1+ 1 + 1.3 + .... = ⎛⎜⎝1 − 1 ⎞−1/ 2 = 2
4 4.8 2 ⎟⎠

19. The coefficient of x4 in the expansion of (1− 3x )2 is equal to [EAMCET 2001]
4) 4
1− 2x

1) 1 2) 2 3) 3

Ans: 4

( )Sol: we have (1− 3x )2 = 1+ 9x2 − 6x (1− 2x )−1
1− 2x

( )= 1+ 9x2 − 6x ⎡⎣1+ 2x + (2x )2 + (2x)3 + (2x )4 + .....⎦⎤

Coefficient of x4 = 16 + 9(4) − 6(8) = 4

20. If (1+ x )n = C0 + C1x + C2x2 + ...... + Cnxn then C0 + 2C1 + 3C2 + ...... + (n +1) Cn is equal to

[EAMCET 2001]

1) 2n + n.2n−1 2) 2n−1 + n.2n 3) 2n + (n +1) 2n 4) 2n−1 + (n −1) 2n

Ans:1

7

Binomial Theorem

Sol: ⎛ n +1 + 1 ⎞ .2n = (n + )2 .2n−1
⎝⎜ 2 ⎟⎠

= n.2n−1 + 2n (or)

Let S = C0 + 2C1 + 3C2 + ....... + (n +1)Cn …………..(1)

S = Cn + 2Cn−1 + 3Cn−2 + ....... + (n +1)C0

S =(n+1) C0 +n C1 + ....... + Cn ………………(2)

(1) + (2)

⇒ 2S = C0 (1+ n +1) + C1 (2 + n ) + ...... + Cn (n +1+1)

2S = (n + 2)(C0 + C1 + C2 + .... + Cn ) = (n + 2).2n

S = (n + )2 .2n−1 = n.2n−1 + 2n

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21. If C0 , C1, C2,...... are binomial coefficient, then C1 + C2 + C3 + ...... + Cr + ..... + Cn is equal to
[EAMCET 2000]

1) 2n 2) 2n−1 3) 2n −1 4) 22n

Ans: 3

Sol: we have (1+ x )n = 1+n C1x +n C2x2 + .....n Cn xn

Put x = 1

1+n C1 +n C2 + ..... +n Cn = 2n
C1 + C2 + ..... + Cn = 2n −1

22. The coefficient of x –n in (1+ x )n ⎛⎝⎜1 + 1 ⎞n [EAMCET 2000]
x ⎟⎠ 4)

1) 0 2) 1 3)2n

Ans: 2

Sol: (1+ x )n ⎛⎜⎝1 + 1 ⎞ = (1+ x )2n
x ⎟⎠
xn

= x−n ⎡⎣ 2n C0 +2n C1x + ... +2n Cn x2n ⎦⎤

Coefficient of x−n =2n C0 = 1

23. If the coefficient of rth term and (2+1)th term in the expansion of (1+x)3n are in the ratio 1 : 2 the r

is equal to [EAMCET 2000]

1) 6 (n +1) 2) 1 (3n +1) 3) 1 (n + 2) 4) 1 (3n + 2)

5 3 4 3

Ans: 2

Sol: coefficient of rth term = Tr = C3n
r −1

8

Binomial Theorem

Coefficient of (r+1)th term = Tr+1 = 3n Cr

Given Tr = 1⇒ C3n = 1⇒ r= 1 ⇒ r = 1 (3n +1)
Tr +1 2 r −1 2 3n − r +1 2
3
3n Cr

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9

7. EXPONENTIAL SERIES AND LOGARITHMIC SERIES

PREVIOUS EAMCET BITS

( )1.1 = a0 + a1x + a2x2 + ..... ⇒ 2a1 + 23 a3 + 25 a5 + ....... =
e3x ex + e5x [EAMCET 2009]
[EAMCET 2008]
1) e 2) e−1 3) 1 4) 0
[EAMCET 2007]
Ans:

Sol: e−2x + e2x

⎡ (2x)2 (2x)4 ⎤
= 2 ⎢1+ + + .......⎥
⎣⎢ 2! 4! ⎥⎦

⇒ a1 = 0, a3 = 0, a5 = 0..... and so on.

2. 1 + 1 + 1 + ...... =
1.3 2.5 3.7
www.NetBadi.in
1) 2 log 2 − 2 2) 2 − 2 loge2 3) 2 log 4 4) loge4
e e

Ans: 2

∑Sol:1 + 1 1 = ∞ 1
1.3 2.5 + 3.7 + ...... n =1 n +1)
(2n

∑ ∑=2 ∞ 1 = ∞ ⎡ 1 − 1⎤
n =1 2n +1) ⎣⎢ 2n 2n +1⎦⎥
(2n 2

n =1

= 2 ⎡ 1 − 1 + 1 − 1 + .....⎤⎦⎥
⎣⎢ 2 3 4 5

= 2 − 2 ⎢⎡⎣1− 1 + 1 − 1 + .....⎤⎦⎥ = 2 − 2 loge2
2 3 4

3. The coefficient of xk in the expansion of 1− 2x − x2 is
e−x

1) 1− k − k2 2) k2 +1 3) 1− k 4) 1
k! k! k! k!

Ans : 1

1− 2x − x2
e−x
( )Sol:
Now = 1− 2x − x2 ex

= ⎣⎡1 − 2x − x2 ⎦⎤ ⎢⎡1 + x + x2 + x3 + .... + xk + ....∞ ⎤
⎣ 2! 3! k! ⎥
⎦

= ⎜⎛1 + x + x2 + .... + xk + ....∞ ⎞ −2 ⎡ + x3 + ..... + ( xk + x k+1 ⎤
⎝ 2! k! ⎟ ⎢x 2! k! + .....∞⎥
⎠ ⎣ k −1)!
⎦

− ⎡ 2 + x3 + x4 + .... + xk + x k+1 + ⎤
⎢x 4! .....∞⎥
⎣ ( k − 2)! ( k −1)!
⎦

1

Exponential series and logarithmic series

∴ coefficient of xk is 1 − (k 2 1)! − (k 1 2)!
k! − −

1 − 2k − k (k −1) = 1− k − k2
k! k! k! k!

4. 1 − 1 + 1 − 1 + ... = [EAMCET 2007]
2 2.22 3.23 4.24

1) 1 2) log3 ⎛ 3⎞ 3) loge ⎛ 3⎞ 4) log e ⎛ 2 ⎞
4 ⎜⎝ 4 ⎟⎠ ⎝⎜ 2 ⎠⎟ ⎝⎜ 3 ⎠⎟

Ans: 2

Sol: x − x2 + x3 − x4 + ...... = log (1+ x )

234

⎛ 1 ⎞2 ⎛ 1 ⎞3 ⎛ 1 ⎞4
1 − ⎜⎝ 2 ⎠⎟ + ⎜⎝ 2 ⎟⎠ − ⎝⎜ 2 ⎟⎠
22 3 4 + ..... = log ⎝⎜⎛1 + 1 ⎞ = log3 ⎛ 3 ⎞
www.NetBadi.in 2 ⎟⎠ ⎜⎝ 2 ⎠⎟

5. The coefficient of xn in 1− 2x is [EAMCET 2006]
ex
4) (−1)n (1+ 4n)
1) 1+ 2n 2) (−1)n (1+ 2n) 3) (−1)n (1− 2n)
n! n!
n! n!

Ans :

Sol: 1− 2x = (1− 2x ) e−x
ex

(1 − 2x ) ⎡⎢1 − x + x2 + x3 + ........ + (−1)n xn + ⎤
⎣ 1! 2! 3! n! .....⎥
⎦

Coefficient of xn in 1− 2x
ex

= ( −1)n − 2 ( −1)n ⇒ (1)n (1− 2n)
(n −1)!
n! n!

6. If |x| < 1 and y = x − x2 + x3 − x4 + ...... then x is equal to [EAMCET 2006]
234

1) y + y2 + y3 + ...... 2) y − y2 + y3 − y4 + .....
23 234

3) y + y2 + y3 + ..... 4) y − y2 + y3 − y4 + .....
2! 3! 2! 3! 4!

Ans:

Sol: y = x − x2 + x3 − x4 + .....
234

y = loge (1+ x ) ⇒ 1+ x = ey

2

Exponential series and logarithmic series

⇒ 1+ x = 1+ y + y2 + .....
2!

∴ x = y + y2 + y3 + .....
2! 3!

∑7. ∝ 2n2 + n +1 = 2) 2e +1 3) 6e −1 [EAMCET 2005]
n=1 n ! 4) 6e +1
1) 2e −1

Ans: 3

∝ 2n2 +n +1 ∝ ⎡ 2n2 n+ 1⎤
∑ ∑Sol: = n =1 n! = n =1 + n! n !⎥⎦
Let s ⎢ n!
⎣

= ∝ ⎡ ( n 2 2)! + ( n 3 1)! + 1⎤
⎢ − − n !⎥⎦
∑ ⎣

n =1

= 2 ⎢⎡⎣1 + 1 + 1 www.NetBadi.in+1+.... ∝⎤⎦⎥ + 3 ⎢⎡⎣1 + 1 + 1 + .... ∝⎤⎦⎥ + ⎡1 + 1 + 1 + ... ∝⎦⎤⎥
1! 2! 3! 1! 2! ⎣⎢1! 2! 3!

= 2e + 3e + e −1= 6e – 1

∑8. If a < 1, b = ∝ ak , then a is equal to [EAMCET 2005]

n=1 k

∑∝ (−1)k bk ∑ ( )∝ −1 bk−1 k ∑∝ (−1)k bk ∑ ( )∝ −1 bk−1 k

1) 2) 3) k=1 (k −1)! 4) k=1 (k +1)!
n=1 k k=1 k !

Ans: 2

∑Sol: b = ∝ ak = a + a2 + a3 + .... ∝
k=1 k! 1 2 3
b = − loge (1− a )

e−b = 1− a ⇒ a = 1− e−b = 1 − ⎡ − b + b2 − b3 + .... ∝⎥⎤
⎢1 1! 2! 3! ⎦
⎣

∑∴a = ∝ (−1)k−1 bk
k=1 k !

( )∝ −1 k−1 bk [EAMCET 2004]

∑9. The value of the series ∴a =
k=1 k !

( ) ( )1) cosh x logae ( )3) sin h ( )4)
2) coth x logae x log a tan h x log a
e e

Ans: 3.

e − ex logea − x logae ⎡⎢∵ ex − e−x x3 x5 ⎤
⎣ 2 3! 5! ....⎥
Sol: 2 = x+ + +
⎦

= sinh ⎡⎣ x log a ⎦⎤
e

3

Exponential series and logarithmic series

10. Coefficient of x10 in the expansion of (2 + 3x ) e−x is [EAMCET 2004]

1) − 26 2) − 28 3) − 30 4) − 32

(10)! (10)! (10)! (10)!

Ans: 2

Sol: (2 + 3x) e−x = (2 + 3x ) ⎡ − x + x2 − x3 ...... x9 + x10 ⎤
⎢1 1! 2! 3! 9! 10!.....⎥⎦
⎣

∴ Coefficient of x10 in the above series

= 2 − 3 = 1 (2 − 30) ⇒ − 28
10! 9! 10! 10!

11. 1 + 1+ 2 + 1+ 2 + 3 + .... is equal to [EAMCET 2003]
2! 3! 4!

1) e 2) e 3) e 4) e
23 4 5
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Ans: 1

∑Sol: = ∝ 1+ 2 + 3 + ..... + n

n =1 (n +1)!

= ∝ 2(n n (n +1) = 1 ∝ (n 1
+1) n (n −1)! 2
∑ ∑ −1)!

n =1 n =1

= 1 ⎡⎣⎢1 + 1+ 1 + 1 + .....⎥⎦⎤ = 1 e = e
2 1! 2! 3! 2 2

( )12. 2 2 2
If 0 < y < 21/3 and x y3 −1 = 1, then x + 3x3 + 5x5 + ...... is equal to [EAMCET 2003]

1) log ⎛ y3 ⎞ 2) log ⎛ y3 ⎞ 3) log ⎛ 2y3 ⎞ 4) log ⎛ y3 ⎞
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ 2 − y3 ⎠ ⎝ 1− y3 ⎠ ⎝ 1− y3 ⎠ ⎝ 1 − 2y3 ⎠

Ans: 1

Sol: y3 −1 = 1
x

x + x3 + x5 + ...... = 1 log ⎛ 1 + x ⎞
3 5 2 ⎝⎜ 1 − x ⎠⎟

⎡ ⎛ 1 ⎞3 ⎛ 1 ⎞5 ⎤
⎢ ⎝⎜ x ⎠⎟ ⎝⎜ x ⎠⎟ ⎥
= 2 ⎢ 1 + 3 + 5 + .....⎥ ⇒ 2× 1 log ⎛ 1+ 1/ x ⎞
⎥ 2 ⎜⎝ 1− 1/ x ⎟⎠
⎢x
⎢⎥
⎣⎦

= log ⎡1 + y3 −1⎤ = log ⎡ 2 y3 ⎤
⎣⎢1 − y3 + 1⎦⎥ ⎢ − y3 ⎥
⎣ ⎦

x2 2 + x3 3 + ......(a > 0, x ∈ R ) is equal to
2! 3!
( ) ( )13.1+ a + a a
x log e log e log e [EAMCET 2002]

4

Exponential series and logarithmic series

1) a 2) ax 3) a logex 4) x

Ans: 2 [EAMCET 2002]
4) e2 − e
Sol: 1+ x + x2 + x3 + ........ = ex
1! 2! 3!

( )1+x log a + x log a 2
e e
+ ........ = exlogea = elogaex = a x
1! 2!

14. 1+ 1+ 2 + 1+ 2 + 22 + ..... is equal to
2! 3!

1) e2 + e 2) e2 3) e2 −1

Ans: 4

(2n −1)
∑ ∑Sol: ∝ 1+ 2 + 22 + ..... + 2n−1 = ∝ 1 2 −1
n=1 n ! n=1 n !
www.NetBadi.in
∑ ∑ ∑= ∝ 2n −1 = ∝ 2n − ∝ 1
n=1 n ! n=1 n ! n=1 n !

( )= e2 −1 − (e −1) = e2 − e

15. 2 + 2 + 4 + 2 + 4 + 6 + ...... [EAMCET 2001]
2! 3! 4!

1) e 2) e−1 3) e−2 4) e−3

Ans: 1

∝ 2 + 4 + 6 + ......... + 2n ∝ 2(1+ 2 + 3 + .....n)
n =1 n =1 (n +1)!
(n +1)!
∑ ∑Sol:= ⇒

= ∝ ( n n(n +1) 1)! ⇒ ∝ ( n 1 1) !
+1) n (n − −
∑ ∑

n =1 n =1

= 1+ 1 + 1 + 1 + ....... = e
1! 2! 3!

( )16. |x| < 1, the coefficient x3 in the expansion of log 1+ x + x2 in ascending powers of x is

[EAMCET 2001]

1) 2 2) 4 3) − 2 4) − 4
33 3 3

Ans: 3

( ) ( )Sol: ⎡ 1+ x + x2 (1− x) ⎤ ⎡1− x3 ⎤
We have log 1+ x + x2 = log ⎢ = log ⎥
⎥ ⎢ 1− x ⎦
⎣⎢ 1− x ⎦⎥ ⎣

= log ⎣⎡1 − x3 ⎤⎦ − log[1− x] = − ⎡ x 3 + x6 + x9 + ⎤ + ⎡ + x2 + x3 + ⎤
⎢ 2 3 ......⎥ ⎢x 2 3 ......⎥
⎣ ⎦ ⎣ ⎦

5

Exponential series and logarithmic series

Coefficient of x3 is −1+ 1 = − 2
33

www.NetBadi.in

6

3. ALGEBRA OF MATRICES [EAMCET 2009]
PREVIOUS EAMCET BITS

35x
1 If one roots of 7 x 7 = 0 is – 10, then the other roots are

x53

1) 3, 7 2) 4, 7 3) 3, 9 4) 3, 4

Ans: 1

Sol: 3(3x − 35) − 5(21− 7x) + x (35 − x2 ) = 0

⇒ x3 − 79x + 210 = 0
Verify S1 ; S1 = 0
−10 + α + β = 0

α + β = 10

www.NetBadi.in
∴ Roots are 3, 7
2. If x, y, z are all positive and are the pth, qth and rth terms of a geometric progression respectively,

log x p 1

then the value of the determent log y q 1 = [EAMCET 2009]

log z r 1

1) log(xyz) 2) (p −1)(p −1)(r −1) 3) pqr 4) 0

Ans: 4
Sol: Let x = AR p−1, y = ARq−1, Z = AR r−1

∴Logx = LogA + (p −1) LogR
Logy = LogA + (q −1) LogR
Logz = LogA + (r −1) LogR

log A + (p −1) log R p 1
log A + (q −1) log R q 1 = 0
log A + (r −1) log R r 1

1 −1 x [EAMCET 2009]
3. If 1 x 1 has no inverse, then the real value of x is

x −1 1

1) 2 2) 3 3) 0 4) 1

Ans: 4

Sol: Det A = 0 ⇒ x = 1

4. If A = ⎡1 −2⎤ and f (t) = t2 − 3t + 7 then f ( A) + ⎡3 6⎤ = [EAMCET 2008]
⎢⎣4 5 ⎦⎥ ⎢⎣−12 −9⎥⎦

1

Matrices

⎡1 0⎤ ⎡0 0⎤ ⎡0 1⎤ ⎡1 1⎤
1) ⎣⎢0 1⎦⎥ 2) ⎢⎣0 0⎥⎦ 3) ⎢⎣1 0⎥⎦ 4) ⎣⎢0 0⎥⎦

Ans: 2

Sol: A2 = ⎡1 −2⎤ ⎡1 −2⎤ = ⎡−7 −12⎤
⎢⎣4 5 ⎥⎦ ⎣⎢4 5 ⎦⎥ ⎢⎣ 24 17 ⎦⎥

f (A) = ⎡3 6⎤ = A2 − 3A + 7I + ⎡3 6⎤
⎣⎢−12 −9⎦⎥ ⎣⎢−12 −9⎦⎥

= ⎡−7 −12⎤ − 3 ⎡1 −2⎤ + 7 ⎡1 0⎤ + ⎡3 6⎤ = ⎡0 0⎤
⎢⎣ 24 17 ⎥⎦ ⎢⎣4 5 ⎥⎦ ⎢⎣0 1⎥⎦ ⎢⎣−12 −9⎥⎦ ⎢⎣0 0⎥⎦

⎡ 7 −3 3⎤ [EAMCET 2008]
5. The inverse of the matrix ⎢⎢−1 1 0⎥⎥ ⎡1 3 3⎤
4) ⎢⎢1 4 3⎥⎥
⎣⎢−1 0 1⎦⎥ ⎢⎣1 3 4⎦⎥

⎡1 1 1⎤ www.NetBadi.in ⎡1 3 1⎤ ⎡1 1 1⎤ [EAMCET 2008]
1) ⎢⎢3 4 3⎥⎥ 2) ⎢⎢4 3 8⎥⎥ 3) ⎢⎢3 3 4⎥⎥ 4) (a + b + c)3

⎢⎣3 3 4⎦⎥ ⎣⎢3 4 1⎥⎦ ⎣⎢3 4 3⎥⎦ [EAMCET 2007]
4) 3
Ans:

Sol: AA−1 = I

⎡ 7 −3 3⎤ ⎡1 3 3⎤ ⎡1 0 0⎤
⎢⎢−1 1 0⎥⎥ ⎢⎢1 4 3⎥⎥ = ⎢⎢0 1 0⎥⎥
⎣⎢−1 0 1⎥⎦ ⎣⎢1 3 4⎦⎥ ⎣⎢0 0 1⎥⎦

a − b − c 2a 2a

6. 2b b − c − a 2b =

2c 2c c − a − b

1) 0 2) a + b + c 3) (a + b + c)2

Ans: 4

Sol: Put a = 1, b = 1, c = 1

−1 2 2

2 −1 2 = −1(−3) − 2(−6) + 2(6) = 27

2 2 −1

(a + b + c)3 = (1+1+1)3 = 27

⎡1 2 x ⎤
If ⎢⎢4 −1 ⎥
7. 7 ⎥ is a singular matrix, then x =

⎢⎣2 4 −6⎦⎥

1) 0 2) 1 3) – 3

Ans: 3

Sol: det A = 0

2

Matrices

1(6 − 28) − 2(−24 −14) + x (16 + 2) = 0

−22 + 76 +18x = 0
⇒ x = −3

⎛4 0 0⎞
⎜ ⎟
8. If A is a square matrix such that A (Adj A) = ⎝⎜⎜ 0 4 0 ⎠⎟⎟ then det (Adj A) = [EAMCET 2007]
0 0 4

1) 4 2) 16 3) 64 4) 256

Ans: 2

Sol: A (AdjA) = A I

⎛4 0 0⎞ ⎛| A | 0 0 ⎞
⎜ ⎟ ⎜ ⎟
⎝⎜⎜ 0 4 0 ⎟⎟⎠ = ⎜⎜⎝ 0 |A| 0 | ⎟⎠⎟
0 0 4 0 0 A
|

|A| = 4 www.NetBadi.in

det (AdjA) = (det A)n−1

= (4)3−1

= 16

9. The number of nontrivial solutions of the system x − y + z = 0, x + 2 y − z = 0, 2x + y + 3z = 0 is

[EAMCET 2007]

1) 0 2) 1 3) 2 4)3

Ans: 1

⎡1 −1 1 ⎤ ⎡x⎤ ⎡0⎤
Sol: Write given system of equation is matrix form AX = B ⇒ ⎢⎢1 −1⎥⎥ ⎢⎢ y ⎥ ⎢⎢0⎥⎥
2 ⎥ =

⎢⎣2 1 3 ⎥⎦ ⎣⎢z ⎥⎦ ⎣⎢0⎦⎥

⎡1 −1 1 ⎤
Now | A |= ⎢⎢1 2 −1⎥⎥ = 9

⎢⎣2 1 3 ⎦⎥

Since | A |≠ 0 . So given system of equation is has only trivial solution, so there is no non-trivial
solution.

⎡1 2 2⎤ [EAMCET 2006]
10. A = ⎢⎢2 1 2⎥⎥ then A3 − 4A2 − 6A is equal to

⎢⎣2 2 1⎥⎦

1) 0 2) A 3) – A 4) I
Ans: 3

⎡1 2 2⎤ ⎡1 2 2⎤ ⎡9 8 8⎤
Sol: A2 = ⎢⎢2 1 2⎥⎥ ⎢⎢2 1 2⎥⎥ = ⎢⎢8 9 8⎥⎥

⎣⎢2 2 1⎦⎥ ⎣⎢2 2 1⎥⎦ ⎣⎢8 8 9⎥⎦

3

Matrices

⎡9 8 8⎤ ⎡1 2 2⎤ ⎡41 42 42⎤ [EAMCET 2006]
A3 A2.A = ⎢⎢8 9 8⎥⎥ ⎢⎢2 1 2⎥⎥ = ⎢⎢42 41 42⎥⎥

⎢⎣8 8 9⎥⎦ ⎢⎣2 2 1⎥⎦ ⎢⎣42 42 41⎦⎥

⎡41 42 42⎤ ⎡9 8 8⎤ ⎡1 2 2⎤
A3 − 4A2 − 6A = ⎢⎢42 41 42⎥⎥ − 4 ⎢⎢8 9 8⎥⎥ − 6 ⎢⎢2 1 2⎥⎥

⎢⎣42 42 41⎥⎦ ⎢⎣8 8 9⎦⎥ ⎣⎢2 2 1⎦⎥
⎡−1 −2 −2⎤
= ⎢⎢−2 −1 −2⎥⎥ = −A
⎢⎣−2 −2 −1⎦⎥
log e log e2 log e3
11. log e2 log e3 log e4 =
log e3 log e4 log e5
www.NetBadi.in
1) 0 2) 1 3) 4 loge 4) 5 loge
Ans: 1

log e 2 log e 3log e 1 2 3
Sol: 2 log e 3log e 4 log e = 2 3 4

3log e 4 log e 5log e 3 4 5

C2 – C1 and C3 – C2
1 11

= 2 1 1 =0
3 11

12. If A is an invertible matrix of order n, then the determinant of adjA is equal to [EAMCET 2006]

1) A n 2) A n+1 3) A n−1 4) A n+2

Ans: 3
Sol: Since A is invertiable matrix of order n, then the determinant of adjA = A n−1

13. m[−3 4] + n [4 −3] = [10 −11] ⇒ 3m + 7n = [EAMCET 2005]
4) 1
1) 3 2) 5 3) 10
Ans: 1
Sol: −3m + 4n = 10 …………..(1)
4m − 3n = −11 ………….(2)
Solve (1) & (2) we get
m = – 2, n = 1
∴3m + 7n = −6 + 7 = 1

4

Matrices

⎡ 1 0 2 ⎤ ⎡ 5 a −2⎤
Adj ⎢⎢−1 −2⎥⎥ ⎢ ⎥
14. 1 = ⎢ 1 1 0 ⎥ ⇒ [a b] = [EAMCET 2005]
4) [4 –1]
⎣⎢ 0 2 1 ⎦⎥ ⎣⎢−2 −2 b ⎦⎥

1) [ -4 1] 2) [–4 –1] 3) [4 1]

Ans: 3

⎡ 1 0 2 ⎤ ⎡ 5 a −2⎤
Sol: Give that adj⎢⎢−1 −2⎥⎥ ⎢ ⎥
1 = ⎢ 1 1 0 ⎥

⎢⎣ 0 2 1 ⎦⎥ ⎢⎣−2 −2 b ⎦⎥

⎡ 1 0 2 ⎤ ⎡ 5 1 −2⎤
Cofactors of ⎢⎢−1 1 −2⎥⎥ ⎢ 1 −2⎥⎥
= ⎢ 4

⎢⎣ 0 2 1 ⎦⎥ ⎣⎢−2 0 1 ⎥⎦

⎡ 1 0 2 ⎤ ⎡ 5 4 −2⎤ ⎡ 5 a −2⎤
Adj ⎢⎢−1 −2⎥⎥ ⎢ ⎥ ⎢ ⎥
0 www.NetBadi.in=⎢ 1 1 0 ⎥ = ⎢ 1 1 0 ⎥

⎢⎣ 0 2 1 ⎥⎦ ⎢⎣−2 −2 1 ⎦⎥ ⎢⎣−2 −2 b ⎥⎦

⇒ a = 4, b = 1

[a b] = [4 1]

15 A = ⎡−1 0⎤ ⇒ A3 − A2 = [EAMCET 2005]
⎣⎢ 0 2⎦⎥

1) 2A 2) 2I 3) A 4) I

Ans: 1

Sol: A2 = ⎡−1 0⎤ ⎡−1 0⎤ = ⎡1 0⎤
⎢ 2⎥⎦ ⎢ 2⎦⎥ ⎢⎣0 4⎥⎦
⎣ 0 ⎣ 0

A3 = ⎡1 0⎤ ⎡−1 0⎤ = ⎡−1 0⎤
⎢⎣0 4⎥⎦ ⎢ 2⎥⎦ ⎢ 8⎦⎥
⎣ 0 ⎣ 0

A3 − A2 = ⎡−2 0⎤ = 2A
⎢ 4⎥⎦
⎣ 0

⎡1 −1 0⎤
16. Match the following elements of ⎢⎢0 4 2⎥⎥ with their cofactors and choose the correct answer

⎢⎣3 −4 6⎦⎥
[EAMCET 2004]

Element Cofactor

I) – 1 a) –2

II) 1 b) 32

III) 3 c) 4

IV) 6 d) 6

e) – 6

1) b, d, a, c 2) b, d, c, a 3) d, b, a, c 4) d, a, b, c

5

Ans: 3 Matrices

Sol: Cofactor of −1 = (−1)1+2 0 2 [EAMCET 2004]
= +6 4) 0
36
[EAMCET 2004]
Cofactor of 1 = ( )−1 1+1 4 2 4) 3
= 32
−4 6 [EAMCET 2003]
4) a + b + c
Cofactor of 3 = ( )−1 3+1 −1 0 = −2

42

Cofactor of 6 = ( )−1 3+3 1 −1 = 4

04

⎡1990 1991 1992⎤
17. det ⎢⎢1991 1992 1993⎥⎥ =

⎣⎢1992 1993 1994⎥⎦

1) 1992 www.NetBadi.in2) 1993 3) 1994

Ans: 4

Sol: C2 − C1 and C3 − C2

1990 1 1

1991 1 1 = 0

1992 1 1

⎡ 1 −1 1 ⎤
⎢ 1 −1⎥⎥ is
18. The rank of ⎢ 1

⎢⎣−1 1 1 ⎦⎥

1) 0 2) 1 3) 2

Ans: 4

Sol: det A = 1(1+1) +1(1−1) +1(1+1)

=4
det A ≠ 0

∴ Rank of A = 3

pa qb rc abc

19. If p + q + r = 0 and qc ra pb = k c a b then k =

rb pc qa b c a

1) 0 2) abc 3) pqr

Ans: 3

Sol: Put p = 1, q = 1, r = – 2, a = 1, b = 1, c = 2

1 1 −4 1 1 2

2 −2 +1 = K 2 1 1

−2 2 1 121

6

Matrices

⇒ −8 = K (4) ⇒ K = −2

∴K = pqr

cos (A + B) − sin (A + B) cos 2B

20. If sin A cos A sin B = 0 then B = [EAMCET 2003]

− cos A sin A cos B

1) (2n +1) π 2) nπ 3) (2n +1) π 4)2nπ

2

Ans: 1

Sol: cos2 (A + B) + sin2 (A + B) + cos 2B = 0

1+ cos 2B = 0
2 cos2 B = 0

cos B = 0 ⇒ B = (2n +1) π , n ∈ Z

2
21. The number of solutions of the system equations 2x + y − z = 7, x − 3y + 3z = 1, x + 4y − 3z = 5 is

1) 3 2) 2 3) 1 4) 0 [EAMCET 2003]
Ans: 3
Sol: Given 2x + y − z = 7 ………….(1)
www.NetBadi.in
x − 3y + 2z = 1…………(2)

x + 4y − 3z = 5 ………….(3)

From (1) & (2) we get
5x − y = 15 ………...…..(4)

From (1) & (3) we get
5x − y = 16 ……………..(5)

(4) & (5) are parallel

∴ solution does not exist.

22. If A, B are square matrices of order 3, A is non-singular and AB = 0, then B is a

[EAMCET 2002]

1) Null matrix 2) Non –singular matrix

3) Singular matrix 4) Unit matrix

Ans: 1 and 3

Sol: Given AB = 0

⇒ B = 0 [∵A is non-singular] (or)

|AB| = 0
A B = 0 ⇒ B = 0 ⎡⎣∵A is non-singular A ≠ 0⎤⎦
∴ B is null matrix (or) singular matrix

7

Matrices

⎡1 0 1⎤ 3) 4 [EAMCET 2002]
23. If A = ⎢⎢2 1 0⎥⎥ then, det A = 4) 5

⎢⎣3 2 1⎦⎥
1) 2 2) 3
Ans: 1

Sol: detA = 1(1− 0) +1(4 − 3)

=2
24. If x2 + y2 + z2 ≠ 0, x = cy + bz, y = ax + cx and z = bx + ay, then a2 + b2 + c2 + 2abc =

1) 2 2) a + b + c 3) 1 [EAMCET 2002]
Ans: 3 4) ab + bc + ca
Sol: Given x − cy − bz = 0

cx − y + az = 0
bx + ay − z = 0
www.NetBadi.in
1 −c −b
⇒ c −1 a = 0

b a −1

⇒ 1(1− a2 ) + c(−c − ab) − b (ac + b) = 0

∴a2 + b2 + c2 + 2abc = 1

⎡2 2⎤ ⎡0 −1⎤ −1
⎢⎣−3 2⎥⎦ ⎣⎢1 0 ⎦⎥ , then
( )25. =
If A = , B = B−1A−1 [EAMCET 2001]
4)
1) 2) 3)

( ) ( ) ( )Sol: B−1A−1 −1 = A−1 −1 B−1 −1 = AB

= ⎡2 2⎤ ⎡0 −1⎤
⎣⎢−3 2⎦⎥ ⎣⎢1 ⎥
0 ⎦

= ⎡0 + 2 −2 + 0⎤ ⎡2 −2⎤
⎢⎣0 + 2 3 + 0 ⎥⎦ = ⎢⎣2 ⎥
3 ⎦

( )26. A square matrix aij in which aij = 0 for i ≠ j and aij = k (constant) for i = j is

[EAMCET 2001]

1) Unit matrix 2) Scalar matrix 3) Null matrix 4) Diagonal matrix

Ans: 2

⎡k 0 0⎤ ⎡∴a ij = 0 for i≠ j⎤
⎢ k 0⎥⎥ ⎢⎢⎣aij = k for i =j ⎥
Sol: ⎢ 0 0 k ⎥⎦ ⎦⎥

⎢⎣0

It is a scalar matrix

8

Matrices

27. If A = ⎡0 2⎤ , KA = ⎡0 3a ⎤ , then the values of k, a, b, are respectively [EAMCET 2001]
⎣⎢3 −4⎥⎦ ⎢⎣2b 24⎦⎥

1) −6, −12, −18 2) −6, 4,9 3) −6, −4,9 4) −6,12,18

Ans : 3

Sol: KA = ⎡0 3a ⎤
⎢⎣2b 24⎥⎦

⎡0 2k ⎤ = ⎡0 3a ⎤
⎣⎢3k −4k ⎥⎦ ⎢⎣2b 24⎥⎦

−4k = 24 2k = 3a 3k = 2b
k = −6 −12 = 3a −18 = 2b

a = −4 b = −9

( )28. If A and B are two square matrices such that = −A−1BA then A2 + B2 =
www.NetBadi.in [EAMCET 2000]

1) 0 2) A2 + B2 3) A2 + 2AB + B2 4) A + B

Ans: 2
Sol: Given B = A−1BA

(A + B)2 = A2 + AB + BA + B2

( )= A2 + A −A−1BA + BA + B2

= A2 − AA−1BA + BA + B2

= A2 − IBA + BA + B2
= A2 − BA + BA + B2
= A2 + B2

29 If A = ⎡1 3⎤ , then the determinant A2 – 2A is [EAMCET 2000]
⎢⎣2 1⎥⎦

1) 5 2) 25 3) –5 4) –25

Sol: A2 = ⎡1 3⎤ ⎡1 3⎤
⎣⎢2 1⎦⎥ ⎢⎣2 1⎦⎥

A2 = ⎡7 6⎤
⎣⎢4 7⎥⎦

A2 − 2A = ⎡7 6⎤ − ⎡2 6⎤ = ⎡5 0⎤
⎣⎢4 7⎦⎥ ⎢⎣6 2⎦⎥ ⎣⎢−2 5⎦⎥

A2 − 2A = 25 − 0 = 25

30. If ‘d’ is the determinant of a square matrix A of order n, then the determinant of its adjoint is

1) dn 2) dn−1 3) dn−2 4) d [EAMCET 2000]

Ans: 2

Sol: |A| = d

9

adjA = A n−1 Matrices

= dn−1 [EAMCET 2000]
4) ab – b –c
a 2b 2c

31. If a ≠ 6 , b, c satisfy 3 b c = 0 , then abc =

1) a + b + c 4a b 3) b3
2) 0

Ans:

Sol: a (b2 − ac) − 2b (3b − 4c) + 2c(3a − 4b) = 0

ab2 − a2c − 6b2 + 8bc + 6ac − 8bc = 0
ab2 − a2c − 6b2 + 6ac = 0

b2 (a − 6) − ac(a − 6) = 0

(b2 − ac)(a − 6) = 0
www.NetBadi.in
b2 − ac = 0 [∵a ≠ 6]

b2 = ac
b2.b = abc
∴abc = b3

10

6. PARTIAL FRACTIONS

PREVIOUS EAMCET BITS

1. For x < 1, the constant term in the expansion of 1 is [EAMCET 2009]

(x −1)2 (x − 2)

1) 2 2) 1 3) 0 4) − 1
2
Ans :4 [EAMCET 2008]

Sol: 1 = 1 x⎤
2 ⎥⎦
(x −1)2 (x − 2) (1− x)2 (−2) ⎢⎣⎡1−

( )= − 1 ⎜⎛⎝1 − x ⎞−1
2 2 ⎟⎠
1− x2 www.NetBadi.in

= − 1 ⎡⎣1 + 2x + 3x 2 + ....⎤⎦ ⎡ x + ⎛ x ⎞2 + ⎤
2 ⎢1 + 2 ⎜⎝ 2 ⎠⎟ ...⎥
⎣⎢
⎥⎦

∴ constant = − 1
2

2. If x2 + x +1 = A + B 1 + ( C then A – B =
x2 + 2x +1 x+
x + 1)2

1) 4c 2) 4c + 1 3) 3c 4) 2c

Ans: 4

Sol: x2 + x +1 = A ( x +1)2 + B( x +1) C

Put x = - 1 comparing coefficient of x2

c=1 A = 1 , put x = 0

A+B+C=1
1 + B +1 = 1 ⇒ B = – 1

A − B = 1− (−1) = 2 ⇒ 2C

3. If ( x − 3x − b ) = 2 + 1 , then a : b is equal to [EAMCET 2007]
x−a x−b
a)(x

1) 1 : 2 2) −2:1 3) 1 : 3 4) 3 : 1

Ans: 2

Sol: 3x = 2(x − b) + (x − a )

Put x = 0
0 = −2b − a ; a = −2b

1

a = −2 Partial fractions
b1 [EAMCET 2006]

∴a : b = −2 :1 [EAMCET 2005]
[EAMCET 2004]
If 3x + 2 = A + Bx + C then A + C − B is equal to
x +1 2x2 + 3
(x +1) 2x2
( )4. +3

1) 0 2) 2 3) 3 4) 5
Ans: 2

Sol: 3x + 2 = A (2x2 + 3) + (Bx + C)(x +1)

Put x = – 1

−1 = 5A ⇒ A = − 1
5

Put x = 0 ⇒ 2 = 3A + C
www.NetBadi.in
2 = − 3 + C ⇒ C = 13
55

Comparing coefficient of x2

2A + B = 0 ⇒ − 2 + B = 0 ⇒ B = 2
55

A + C − B = − 1 + 13 − 2 = 2
555

5. If x3 =A+ B+ C+ D, then A is equal to
2x −1 x+2 x−3
(2x −1)(x + 2)(x − 3)

1) 1 2) − 1 3) − 8 4) 27
2 50 25 25

Ans: 1

Sol: A = Coefficient of x3 in Nr
Coefficient x3in Dr

A=1
2

6. If x +1 = A+B, then 16A + 9B is equal to
2x −1 3x +1
(2x −1)(3x +1)

1) 4 2) 5 3) 6 4) 8
Ans: 3

Sol: x +1 = A (3x +1) + B(2x −1)

Put x = 1 ⇒ 3 = A ⎛ 5 ⎞ ⇒ A = 3
2 2 ⎜⎝ 2 ⎠⎟ 5

2

Partial fractions

Put x = −1 ⇒ 2 = B ⎝⎛⎜ −5 ⎞ ⇒ B=− 2
3 3 3 ⎟⎠ 5

∴16A + 9B = 16 ⎛ 3 ⎞ + 9 ⎛ −2 ⎞ = 6
⎜⎝ 5 ⎟⎠ ⎝⎜ 5 ⎟⎠

7. Let a, b and c be such that 1 =a +b+c then a+b+c
1− x 1− 2x 1− 3x 135
(1− x)(1− 2x)(1− 3x)

is equal to [EAMCET 2003]

1) 1 2) 1 3) 1 4) 1
15 6 5 3

Ans: 1

Sol: 1= a (1− 2x)(1− 3x) + b (1− x)(1− 3x) + c(1− x)(1− 2x)

Put x = 1

1 = a (−1)(−2) ⇒ a = 1www.NetBadi.in

2

Put x=1 ⇒ 1 = b ⎛ 1 ⎞ ⎛ − 1 ⎞ ⇒ b = −4
2 ⎜⎝ 2 ⎟⎠ ⎝⎜ 2 ⎟⎠

Put x = 1 ⇒1= c ⎛ 2 ⎞⎛ 1 ⎞ ⇒ c = 9
3 ⎝⎜ 3 ⎟⎠⎝⎜ 3 ⎟⎠ 2

Now a + b + c = 1 − 4 + 9 = 1
1 3 5 2 3 2.5 15

8. If 1− x + 6x2 = A + B + C , then A is equal to [EAMCET 2002]
x − x3 x 1− x 1+ x 4) 4

1) 1 2) 2 3) 3

Ans: 1

( )Sol: Given 1− x + 6x2 = A 1− x2 + Bx (1+ x) + Cx (1− x)

Put , x = 0, then A = 1

9. If x−4 = 2 − 1 , then k = [EAMCET 2001]
x2 − 5x − 2k x−2 x+k 4) 3

1) – 3 2) – 2 3) 2

Ans: 1

Sol: x2 x−4 2k = 2(x + k) −(x − 2)
− 5x − (x − 2)(x + k)

x2 x−4 2k = x2 x + 2k + 2 2k
− 5x −
+(k − 2)x −

Comparing coefficient of x in Dr

3

k − 2 = −5 ∴k = −3 Partial fractions
[EAMCET 2000]
If x2 +5 = 1 + k 4) 4
x2 + 2 2 x2 + 2 x2 + 2
( ) ( )10. , then k =

1) 1 2) 2 3) 3
Ans: 3

( )Sol: x2 + 5 = x2 + 2 + k

2+k =5 ∴k = 3

www.NetBadi.in

4

4. PERMUTATIONS AND COMBINATIONS

PREVIOUS EAMCET BITS

1. The number of ways in which 13 gold coins can be distributed among three persons such that

each one gets at least two gold coins is [EAMCET-2000]

1) 36 2) 24 3) 12 4) 6

Ans : 1

Sol : Required number of ways
= coefficient. x13 in (x2+x3+….)3
= coefficient. x7 in (1+x+x2+….)3
= coefficient. x7 in (1+x)-3
= 9C7= 9C2 = 36

2. If C (2n, 3) : C (n, 2) = 12 : 1, then n =
www.NetBadi.in [EAMCET-2000]

1) 4 2) 5 3) 6 4) 8

Ans: 2
Sol : 2nC3 : nC2 = 12 :1

2nC3 = 12nC2.nc2

⇒ 2n(2n −1)(2n − 2) = 12. n(n −1)

62

⇒ 2n −1 = 9 ⇒ n = 5

3. The number of quadratic expressions with the coefficients drawn from the set{0, 1, 2, 3} is

1) 27 2) 36 3) 48 4) 64 [EAMCET-2000]

Ans : 3
Sol : ax2+bx+c = 0

a can be filled in 3 ways

b can be filled in 4 ways

c can be filled in 4 ways

Required no. of ways = 3 x 4 x 4 = 48

4. The number of ways in which 5 boys are 4 girls sit around a circular table so that no two girls sit

together is [EAMCET-2001]

Ans : 1

1) 5! 4! 2) 5! 3! 3) 5! 4) 4!

Sol : First we arrange 5 boys around a circle in (5-1)! = 4! Ways then we have 5 gaps between them
then arrange 4 girls in 5 gaps arrangement of 4 girls in 5 gaps
Arrangement of 4 girls in 5 gaps = 5P4 = 5!

∴ Required no. of ways = 5! 4!

1

Permutation and combination

5. Using the digits 0, 2, 4, 6, 8 not more than once in any number, the number of 5 digited numbers

that can be formed is [EAMCET-2001]

1) 16 2) 24 3) 120 4) 96

Ans: 4

Sol : Required no. of ways = 5! – 4! = 120-24 = 96

6. If n and r are integers such that 1 < r < n then n . C (n-1, r-1) = [EAMCET-2002]

1) C (n, r) 2) n . C (n, r) 3) r C (n, r) 4) (n - 1) . C (n, r)

Ans: 3

Sol : n.c(n −1, r −1) = n.(n −1)cr−1

= n. (r (n −1)! r )! × r
−1)!(n − r

= www.NetBadi.inn!rr )!=r.n cr= r. c(n,r)

r!(n −

7. The least value of the natural number 'n' satisfying c(n,5) + c(n,6) > c(n+1,5) [EAMCET 2002]

1) 10 2) 12 3) 13 4) 11

Ans: 1

Sol : Given n c5 +n c6 >(n+1) c5

(n+1) c6 >(n+1) c5

(n +1)! > (n +1)!
6!(n − 5)! 5!(n − 4)!

⇒ n > 10

∴ The least value of ‘n’ is 10

8. The no. of ways such that 8 beads of different colour be strung in a neckles is...[EAMCET-2002]

1) 2520 2) 2880 3) 4320 4) 5040

Ans: 1

Sol : Required number of ways = (8 −1)! = 2520

2

9. The number of 5 digited numbers which are not divisible by 5 and which contains of 5 odd digits
is [EAMCET-2002]

1) 96 2) 120 3) 24 4) 32

Ans: 1

Sol : The 5 odd digits be 1,3,5,7,9

Required = 5! – 4!

= 120 – 124

= 96

2

Permutation and combination

10. Let l1 and l2 be two lines intersecting at P. if A1, B1, C1 are points on l1, and A2, B2, C2, D2, E2 are

points on l2 and if none of those coincides with P, then the number of triangles formed by these

eight points. [EAMCET-2003]

1) 56 2) 55 3) 46 4) 45

Ans: 4

Sol : If triangle is including point P the other points must be one from l1 and other point from l2,
Number of triangles formed with P.

n(E1 ) =3 c1 ×5 c1 = 15

When p is not included

Number of triangles formed

( )n E2 =3 c2 ×5 c1 +3 c1 ×5 c2

= 15 + 15

= 30
∴ Total number of triangles = n(E1) +n(E2)

= 15 + 30
www.NetBadi.in
= 45

11. The number of positive odd divisors of 216 is [EAMCET-2004]

1) 4 2) 6 3) 8 4) 12

Ans: 1
Sol: The factors of 216 = 23. 33

The odd divisors are the multiplied 3.
∴The number of positive odd divisors

=3+1=4

12. A three digit number n is such that the last two digits of it are equal and different from the first.

The number of such n’s is [EAMCET 2005]

1) 64 2) 72 3)81 4)900

Ans: 3

Sol: If the last two digits are equal to then the first digit may 1 to 9

If the last two digits are equal to 1 to 9 then the first digit may be selected in 8 ways.
∴The required number = 9 + 9 × 8

= 81

13. If N denotes Set of all positive integers and if and if is defined by the sum of positive divisors of .

then where is a positive integer is [EAMCET-2005]

1) 2k+1 −1 ( )2) 2 2k+1 −1 ( )3) 3 2k+1 −1 ( )4) 4 2k+1 −1

ns: 3
Sol: Given f(x) = the sum of positive divisors of n

3

Permutation and combination

( ) ( )f 2k.3 = 3 1+ 2 + 22 + 33 + ... + 2k

( )=
3⎜⎝⎜⎛ 1 2k+1 −1 ⎠⎞⎟⎟
2 −1

( )= 3 2k+1 −1

14. The number of natural numbers less than 1000, in which no two digits are repeated is

[EAMCET 2006]

1) 738 2) 792 3) 837 4) 720
Ans: 1

Sol : The number of 1 digit numbers = 9

The number of 2 digit numbers = 9 x 9 = 81

The number of 3 digit numbers = 9 × 9 p2 = 648
∴ The number of Required numbers
www.NetBadi.in
= 9 + 81+ 648 = 738

15. The number of ways of arranging 8 men and 4 women around a circular table such that no two

women can sit together, is [EAMCET-2007]

Ans:

1) 8! 2) 4! 3) 8! 4! 4) 7!.8 P4

Ans: 4

Sol: Number of ways of arranging 8 men around a circle = (8-1)! = 7!

Then we have 8 gaps between them
Number of ways of arranging 4 women in 8 gaps = 8p4
∴Required number of ways = 7!. 8p4

16. If a polygon of n sides has 275 diagonals, then n = [EAMCET-2007]

1) 25 2) 35 3) 20 4) 15

Ans: 1

Sol: Number of diagonals of a polygon of n sides = 275

n(n − 3) = 275

2

n(n − 3) = 550

n(n − 3) = 25× 22

∴ n = 25

17. 9 balls are to be placed in 9 boxes, and 5 of the balls can not fill into 3 small boxes. The numbers

of ways of arranging one ball in each of the boxes is [EAMCET-2008]

4

Permutation and combination

1) 18720 2) 18270 3) 17280 4) 12780

Ans: 3
Sol : 5 balls can be placed in 6 boxes (other than the 3 small boxes) in 6p5 ways

The remaining 4 balls can be placed in the remaining 4 boxes in 4! ways.
∴The required number of arrangements = 6 p5 × 4!

18. If n pr = 30240 and n cr = 252 then the ordered pair (n,r) =

1) (12,6) 2) (10,5) 3) (9,4) 4) (16,7)

Ans: 2

Sol : n pr = 30240
n cr 252
⇒ r! =120

⇒ r! = 5! www.NetBadi.in

⇒r =5

n p5 = 30240 = 10 p5 ⇒ n = 10
∴(n,r) = (10,5)

19. The number of subsets of {1,2,3,…9} containing at least one odd number is [EAMCET-2009]

1) 324 2) 396 3) 496 4) 512

Ans: 3
Sol : No of subsets = 29 – 24

= 512 – 16

= 496

20. ‘P’ points are chosen each of the three coplanar lines. The maximum number of triangles formed

with vertices at these points is [EAMCET-2009]

1) p3+3p2 2) (1 p3 + p) 3) p2 (5 p − 3) 4) p2 (4 p − 3)
2
2

Ans: 4

Sol : Let the lines be L1, L2, L3

( )Max no of triangles = 3 c2 ×p c2 × p c1 + p c1 3

= 6× p( p −1) × p + p3

2

= p2 (3 p − 3 + p)

= p2(4p-3)

21. A binary sequence is an array of 0’s and 1’s the number of n-digit binary sequences which

contain even number of 0’s is [EAMCET-2009]

5

1) 2n-1 2) 2 n-1 3) 2 n-1 -1 Permutation and combination
3) 2n

Ans : 1
Sol : If n is even, no of n-digit binary sequences = 2n-1.

www.NetBadi.in

6

QUADRATIC EXPRESSIONS
PREVIOUS EAMCET BITS

1. The roots of (x - a)(x - a -1) + (x - a -1)(x - a - 2) + (x - a)(x - a - 2) = 0 , a ∈ IR always :

1) equal 2) imaginary 3) real and distinct 4) rational and equal
[EAMCET 2009]
Ans: 3

Sol : Put a = 0

x (x -1) + (x -1)(x - 2) + (x)(x - 2) = 0

3x2 - 6x + 2 = 0 ⇒ 3x2 - 6x + 2 = 0

Δ = (6)2 - 4(3)(2)

Δ≠0

∴ Roots are real and distinct.

2. Let f (x) = x2 + ax + b , where a, b ∈ IR . If f (x) = 0 has all its roots imaginary, then the roots of
www.NetBadi.in
f (x) + f 1 (x) + f 11 (x) = 0 are : [EAMCET 2009]

1) real and distinct 2) imaginary 3) equal 4) rational and equal
Ans: 2 ⇒ Δ < 0 a2 - 4b < 0

Sol. Given roots of f (x) are imaginary

f 1 (x) = 2x + a

f 11 (x) = 2

f (x) + f 1 (x) + f 11 (x) = 0

x2 + ax + b + 2x + a + 2 = 0

⇒ x2 + (a + 2)x + (a + b + 2) = 0

∴Δ = (a + 2)2 - 4(1)(a + b + 2)

= a2 + 4a + 4 - 4a - 4b - 8

= a2 - 4b - 4 < 0

∴ Roots are imaginary.

3. Let α and β be the roots of the quadratic equations ax2 + bx + c = 0 observe the lists given

below. [EAMCET 2008]

LIST - I LIST – II

( ) ( )A) ac2 1/3 + a2c 1/3 + b = 0

i) α = β ⇒ B) 2b2 = 9ac

ii) α = 2β ⇒ C) b2 = 6ac

iii) α = 3β ⇒ D) 3b2 = 16ac

iv) α = β2 ⇒ E) b2 = 4ac

( ) ( )F) ac2 1/3 + a2c 1/3 = b

Quadratic Expressions

The correct match to List – I from List – Ii is

i ii iii iv i ii iii iv i ii iii iv i ii iii iv
4) E B D A
1) E B D F 2) E B A D 3) E D B F

Ans: 4

Sol : If the roots are in the ratio m : n then b2 = (m + n )2

ac mn

i) α =β ⇒ α = 1 ⇒ b2 = (1 + 1)2 ⇒ b2 = 4ac
β 1 ac
1x1

ii) α = 2β ⇒ α = 2 ⇒ b2 = (2 +1)2 ⇒ 2b2 = 9ac
β 1 ac
2x1

iii) α = 3β ⇒ α = 3 ⇒ b2 = (3 +1)2 ⇒ 3b2 = 16ac
β 1 ac
3x1

iv) α = β2 ⇒ β +β2www.NetBadi.in= -b/a, β3 = c
a

⇒ ⎛ c ⎞1/ 3 + ⎛ c ⎞2/3 = -b / a
⎝⎜ a ⎠⎟ ⎜⎝ a ⎠⎟

( ) ( )⇒ a2c 1/3 + ac2 1/3 = -b

4. If α + β = -2 and α3 + β3 = -56 then the quadratic equation whose roots are α and β is

1) x2 + 2x - 16 = 0 2) x2 + 2x - 15 = 0 3) x2 + 2x - 12 = 0 4) x2 + 2x - 8 = 0
[EAMCET 2008]
Ans: 4
Sol : α3 + β3 = -56

(α + β)3 - 3αβ (α + β) = -56

(-2)3 + 3αβ (-2) = -56

-8 + 6αβ = -56
αβ = -8
∴ Required equation is

x2 - (α + β) x + αβ = 0

x2 + 2x - 8 = 0

5. If α and β are the roots of the equation ax2 + bx + c = 0 and px2 + qx + r = 0 has roots

1− α and 1− β then r = [EAMCET 2007]
αβ

1) a + 2b 2) a + b + c 3) ab + bc + ca 4) abc

Ans: 2

Sol. The equation heaving roots 1 , 1 is cx2 + bx + a = 0 .
αβ

2

Quadratic Expressions

The equation having roots 1 - 1, 1 -1 is
α α

c(x +1)2 + b(x +1) + a = 0

⇒ cx2 + (2c + b) x + (c + b + a) = 0

px2 + qx + r = 0

∴r=a+b+c

6. The set of values of x for which the inequalities x2 − 3x −10 < 0, 10x − x2 −16 > 0 hold

simultaneously is [EAMCET 2007)

1) (−2,5) 2) (2,8) 3) (−2,8) 4) (2,5)

Ans: 4

Sol. x2 - 3x -10 < 0 ⇒ (x - 5)(x + 2) > 0

⇒ -2 < x < 5(or) x > 3
www.NetBadi.in
10x - x2 - 16 > 0 ⇒ x2 - 10x + 16 < 0

⇒ (x- 2) (x-8) < 0

⇒ 2<x<8
∴ common set = (2,5)

7. If 9x2 + 6x + 1 < (2 - x) then [EAMCET 2006]

1) x ∈ ⎛ - 3 , 1 ⎞ 2) x ∈ ⎛ -3 , 1 ⎤ 3) x ∈ ⎡ -3 , 1 ⎞ 4) x < 1/4
⎜⎝ 2 4 ⎠⎟ ⎝⎜ 2 4 ⎦⎥ ⎢⎣ 2 4 ⎠⎟

Ans: 1

Sol : 9x2 + 6x + 1 < (2 - x)

⇒ (3x + 1)2 < (2 - x)

Taking +ve sign

3x + 1 < (2 - x)

⇒ x <1/ 4

Taking –ve sign

-3x - 1 < 2 - x

x > −3
2

x ∈⎜⎛⎝ -3 , 1 ⎞
2 4 ⎠⎟

8. If x is real, then the minimum value of x2 − x +1 is [EAMCET 2005]
x2 + x +1 4)1

1) 1/3 2) 3 3) 1/2

Ans: 1

3

Quadratic Expressions

Sol. Let y = x2 - x +1
x2 + x +1

⇒ (y -1) x2 + (y +1) x + (y -1) = 1

x is real ⇒ (y + 1)2 - 4(y - 1)2 ≥ 0

⇒ 3y2 -10y + 3 ≤ 0

⇒ (3y -1)(y - 3) ≤ 0

⇒ 1 ≤ y ≤ 3
3

∴ Minimum value = 1/3

9. E1: a + b + c = 0 if 1 is a root of ax2 + bx + c = 0

E2: b2-a2=2ac if sin cos are the roots of ax2 + bx + c = 0 [EAMCET 2005]

1) E1 is true, E2 is true 2) E1 is true, E2 is false
www.NetBadi.in
3) E1 is false, E2 is true 4) E1 is false, E2 is false

Ans: 1

Sol. E1 : 1 is a root of ax2 + bx + c = 0

⇒ a(1)2 + b(1) + c = 0

⇒ a+b+c = 0

E2 : Sinθ + Cosθ = - b , SinθCosθ = c/a
a

(Sinθ + Cosθ )2 = ⎛ -b ⎞2
⎜⎝ a ⎠⎟

1+ 2SinθCosθ = b2 ⇒ 1+ 2 ⎛ c ⎞ = b2
a2 ⎜⎝ a ⎟⎠ a2

⇒ b2 - a2 = 2ac

∴E 1 is true, E 2 is true.

10. The set of all solutions of the inequation x2 − 2x + 5 ≤ 0 is [EAMCET 2004]

1) R − (−∞, −5) 2) R − (5, ∞) 3) φ 4) R − (−∞, −4)

Ans: 3

Sol. x2 - 2x + 5 = 0 ⇒ x = 2 ± 4 - 20 = 1± 2i
2

⇒ x2 - 2x + 5 > 0∀x ∈ R

∴ x2 - 2x + 5 ≤ 0 has no real solution.

11 If x-2 is a common factor of the expressions x2 + ax + b and x2 + cx + d , then b − d =
c−a
[EAMCET 2004]

1) -2 2) -1 3) 1 4) 2

Ans:4

4

Quadratic Expressions

Sol. x - 2 is a common factor of x2 + ax + b and x2 + cx + d
⇒ 4 + 2a + b = 0, 4 + 2c + d = 0
⇒ 2a + b = 2c + d

⇒ b −d = 2(c-a)

⇒ b-d = 2
c-a

12. The solution set contained in R of the inequation 3x + 31−x − 4 < 0 is [EAMCET 2003]

1) (1, 3) 2) (0, 1) 3) (1, 2) 4) (0, 2)

Ans: 2

Sol. 3x + 3 - 4 < 0
3x

32x + 3 - 4.3x < 0

(3x -1)(3x - 3) < 0
www.NetBadi.in
1 < 3x < 3

∴ The solution set is (0,1)

13. The minimum value of 2x2 +x-1 is [EAMCET 2003]

1) 1/4 2) 3/2 3) -9/8 4) 9/4

Ans. 3

Sol. Minimum value of ax2 + bx + c is 4ac - b2
4a

∴ Minimum value of 2x2 + x - 1 is 4 ( 2 ) ( -1) - 1 = - 9
4(2) 8

14. If the equations x2 + ax + b = 0 and x2 + bx + a = 0 (a ≠ b) have a common root then a+b=

[EAMCET 2002]

1) -1 2) 2 3) 3 4) 4

Ans: 1

Sol. Let 'α' be a common root of x2 + ax + b = 0 and x2 + bx + a = 0

⇒ α2 + aα + b = 0 -(1)

α2 + bα + a = 0 - (2) -(1)

Solve (1) & (2)

(1) – (2) ⇒ (a - b)α + (b - a) = 0

⇒α =1 [EAMCET 2002]
1+a+b =0 4) x2 − 5x + k = 0
∴ a + b = -1

15. If ‘3’ is a root of x2 + kx − 24 = 0 it is also root of
1) x2 + 5x + k = 0 2) x2 + kx + 24 = 0 3) x2 − kx + 6 = 0
Ans: 2

5

Quadratic Expressions

Sol. Put x = 3

9 + 3k - 24 = 0 ⇒ K = 5

Put K=5 in options.

∴ 3 is also a root of x2 - kx + 6 = 0

x2 - 5x + 6 = 0

16. If α , β are the roots of x2 + bx + c = 0 and α + h, β + h are the roots of x2 + qx + r = 0 then
h = [EAMCET 2001]

1) b+q 2) b-q 3) ½ (b + q) 4) ½ (b − q)

Ans: 4
Sol. α,β are the roots of x2 + bx + c = 0

⇒ α + β = -b

α + h,β + h are the roots of x2 + qx + r = 0

⇒ (α + h) + (β + h) = -q
www.NetBadi.in
α + β + 2h = -q

-b + 2h = -q

⇒ h = 1 (b - q)
2

( )17. If 203−2x2 = 40 5 3x2 −2 then x = [EAMCET 2001]

1) ± 13 2) ± 12 3) ± 4 4) ± 5
2 13 5 4

Ans: 2

( )Sol. 20 3-2x2 = ⎣⎡20x2 5 ⎦⎤3x2-2

= ⎣⎡20 20 ⎦⎤3x2 -2

( )=⎡ 20 ⎤3/2 3x2 -2
⎣ ⎦

( ) ( )20 3-2x2 = 9x2 -6

20 2

3 - 2x2 = 9x2 - 6
2

6 - 4x2 = 9x2 - 6
13x2 = 12

x=± 12
13

18. If α , β are the roots of 9x2 + 6x +1 = 0 then the equation with the roots 1/α , 1/ β is

[EAMCET 2000]

1) 2x2 + 3x +18 = 0 2) x2 + 6x − 9 = 0 3) x2 + 6x + 9 = 0 4) x2 − 6x + 9 = 0

6

Quadratic Expressions

Ans: 3

Sol f ⎛ 1 ⎞ = 0
⎝⎜ x ⎠⎟

⇒ 9 ⎛ 1 ⎞2 + 6 ⎛ 1 ⎞ +1 = 0
⎝⎜ x ⎟⎠ ⎜⎝ x ⎠⎟

x2 + 6x + 9 = 0

19. The equation formed by decreasing each root of ax2 + bx + c = 0 by 1 2x2 + 8x + 2 = 0 is then

[EAMCET 2000]

1) a = -b 2) b = -c 3) c = -a 4) b=a+c

Ans: 2

Sol. The equation formed by decreasing each root of ax2 + bx + c = 0 by 1 is 2x2 + 8x + 2 = 0

⇒ The equation formed by increasing each root of 2x2 + 8x + 2 = 0 is ax2 + bx + c = 0

∴ax2 + bx + c = 2(x - 1)2 + 8(x - 1) + 2
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ax2 + bx + c = 2x2 + 4x - 4

∴ a = b = c ⇒ 2a = b = -c
2 4 -4

20. If (3 + i ) is a root of the equation x2 + ax + b = 0 then a = [EAMCET 2000]

1) 3 2) -3 3) 6 4) -6

Ans: 4

Sol. If 3 + i is one root of the given equation then the other root be 3 – i

Sum of the roots of x2 + ax + b = 0 is –a

⇒ 3 + i + 3 - i = -a

∴a = -6

7

2. THEORY OF EQUATIONS

PREVIOUS EAMCET Bits

EAMCET-2001

1. Each of the roots of the equation x3 −6x2 + 6x −5 = 0 are increased by k so that the new
transformed equation does not contain term. Then k =

−1 −1 3. -1 4. -2
1. 3 2. 2

Ans: 4

Sol. The transformed equation is (x − k )3 − 6(x − k )2 + 6(x − k )− 5 = 0

Coefficient of x2 is 0 ⇒ − 3k − 6 = 0 ⇒ k = −2

2. The roots of the equation x3 −14x2 + 56x − 64 = 0 are in ......... progression.

1. Arithmetico-geometric 2. Harmonic 3. Arithmetic 4. Geometric

Ans : 4 www.NetBadi.in

Sol. By verification x = 2 is a factor of given equation.

2 1 −14 56 −64
0 2 −24 64

1 −12 32 0

(x-2) (x2-12x+32) = 0
x2-12x+32=0

x = 4,8
∴Roots are 2,4,8
∴ These are in G.P.
3. If there is a multiple root of order 3 for the equation x4 − 2x3 + 2x −1= 0 , then the other root is

1. -1 2. 0 3. 1 4. 2

Ans: 1

Let f(x) = x4-2x3+2x-1 ⇒ f (1) = 0

⇒ f ′(x ) = 4x3 − 6x2 + 2 ⇒ f ′(1) = 0

⇒ f11(x) = 12x2-12x ⇒ f 11(1) = 0

Roots of given equation are 1,1,1
Let the other root be α

S1 = 2
1+1+1+ α
α = -1
∴ Other root is -1

4. The equation whose roots are the negatives of the roots of the equation
x7 +3x5 + x3 − x2 + 7x + 2 = 0

1. x7 + 3x5 + x3 + x2 − 7x + 2 = 0 2. x7 + 3x5 + x3 + x2 + 7x − 2 = 0

1

Theory of Equations

3. x7 + 3x5 + x3 − x2 − 7x − 2 = 0 4. x7 + 3x5 + x3 − x2 + 7x − 2 = 0

Ans: 2

Sol. f(-x) = 0
(-x)7 + 3(-x)5 +(-x)3 -(-x)2 + 7(-x) + 2 = 0
-x7-3x5-x3-x2-7x+2=0
x7+3x5+x3+x2+7x-2=0

5. The biquadratic equation, two of whose roots are 1 + i, 1− 2 is

1. x4 − 4x3 + 5x2 − 2x − 2 = 0 2. x4 − 4x3 −5x2 + 2x + 2 = 0
3. x4 + 4x3 −5x2 + 2x − 2 = 0 4. x4 + 4x3 + 5x2 − 2x + 2 = 0

Ans: 1

Sol. The roots of required equation are

1+i, 1-i, 1− 2 , 1+ 2

Here S1 = 1+i+1-i+1− 2 +1+ 2 =4 (sum of the roots)
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S4 = (1+i) (1-i) (1− 2 )(1+ 2 ) (product of the roots)
= (1-i2) (1- 2)

= -2

Now verify options.

6. To remove the 2nd term of the equation x4 −8x3 + x2 − x + 3 = 0 diminished the root of the

equation by [ EAMCET-2002 ]

1. 1 2. 2 3. 3 4. 4

Ans: 2

Sol. h = − a1 = − (− 8) = 2
na0 4(1)

7. The maximum possible number of real roots of the equation x5 −6x2 − 4x + 5 = 0 is

1. 3 2. 4 3. 5 4. 0

Ans: 1
Sol. Let f(x) = x5 – 6 x2 – 4x + 5 = 0, f(–x) = –x5 – 6x2 + 4x + 5 = 0

Number of positive real roots = Number of changes of signs in f(x)

=2

No. of negative roots = No. of changes of signs in f(-x)

=1

∴ No. of real roots = No. of positive roots + No. of negative roots =

= 2 +1 = 3

8. If α, β, γ are the roots of the equation x3 + ax2 + bx + c = 0 then α−1 + β−1 + γ−1 =

a −b c b

1. c 2. c 3. a 4. a

Ans: 2

Sol. α −1 + β −1 + γ −1 = 1 + 1 + 1
αβγ

2

Theory of Equations

= βγ + αγ + αβ = S2 = b
αβγ S3 − c

1+ 3i

9. If 2 is a root of the equation x4 − x3 + x −1= 0 then its real roots are

1. 1,1 2. -1, -1 3. 1, 2 4. 1, -1

Ans: 4

1+ 3i

Sol. If 2 is a roots of the given equation then the other root be roots are 1− 3i
2

Let the remaining roots be α , β

Now sum of the roots of given equation = S1 = 1

1+ 3i +1− 3i +α + β =1
22

1+α + β =1
α+β =0
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By verification roots are 1,-1

10. If α, β, γ are the roots of 2x3 − 2x −1= 0 then

1. -1 2. 1 3. 2 4. 3

Ans: 2

Sol. (Σαβ )2 = (S2 )2

= ⎜⎛ − 2 ⎟⎞2 = 1
⎝2⎠

11. If α,β, γ are the roots of the equation x3 + 4x +1 = 0 then (α + β)−1 + (β + γ )−1 + (γ + α)−1 =

EAMCET - 2003

1) 2 2) 3 3) 4 4) 5

Ans: 3
Sol. α + β + γ = 0

(α + β )−1 + (β + γ )−1 + (γ + α )−1 = (− γ )−1 + (−α )−1 + (− β )−1

= −1−1−1
γαβ

= − ⎜⎜⎝⎛ 1 + 1 + 1 ⎠⎞⎟⎟
α β γ

= − ⎜⎜⎛⎝ αβ + βγ + γα ⎠⎟⎟⎞ = − ⎛⎜ 4 ⎞⎟ = 4
αβγ ⎝ −1⎠

12. Let α ≠ 0 and P(x) be a polynomial of degree greater than 2. If P(x) leaves remainders
α and − α when divided respectively by x + α and x − α then the remainder when P(x) is divided by

x2 − α2 is

3

Theory of Equations

1) 2x 2) -2x 3) x 4) –x

Ans: 4 and R(a) = -a
pa+q = -a------(2)
Sol. Let the remainder be R(x), then

R(x) = p(x)+q

Given R(-a) = a

- pa + q = a -------(1)

Solving (1) & (2), we get

p = -1, q = 0

∴R(x) = −x

13. If the sum of two of the roots of x3 + px2 + qx + r = 0 is zero then pq =

1) -r 2) r 3) 2r 4) -2r

Ans: 2
Sol. Let the roots be α , β ,γ

Given α + β = 0 www.NetBadi.in

α +β +γ =−p⇒γ =−p

γ = − p is a root of x3 + px2 + qx + r = 0

⇒ (− p)3 + p(− p)2 + q(− p)+ r = 0

∴ pq = r

14. If the roots of the equation 4x3 −12x2 +11x + k = 0 are in A.P. Then K = [EAMCET-2004]

1) -3 2) 1 3) 2 4) 3

Ans: 1

Sol. Let the roots be a-d, a, a+d

(a-d) + a + (a+d) = − ⎜⎛ −12 ⎞⎟
⎝4⎠

3a = 3 ⇒ a = 1
a = 1 is a root of 4x3-12x2+11x+k = 0

⇒ 4(1)3-12(1)2+11(1)+k=0

⇒ 3+k = 0 ∴ k = -3

15. α,β, γ are the roots of the equation x3 −10x2 + 7x + 8 = 0 Match the following

1) α + β + γ − 43

a) 4

2) α2 + β2 + γ2 −7

b) 8

1 +1+1 c) 86
d) 0
3) α β γ

α+β + γ

4) βγ γα αβ

e) 10

1) e, c, a, b 2) d, c, a, b 3) e, c, b, a 4) e, b, c, a

4

Theory of Equations

Ans: 3
Sol. x3 −10x2 + 7x + 8 = 0

Now α + β + γ = 10

α 2 + β 2 + γ 2 = (α + β + γ )2 − 2(αβ + βγ + γα )

= (10)2 – 2(7)
= 86
1 + 1 + 1 = βγ + γα + αβ = 7
α β γ αβγ −8

α + β + α = α2 + β2 + γ2 = 86 = −43
βγ γα αβ αβγ −8 4

16. If f(x) is a polynomial of degree n with rational coefficients and 1 + 2i, 2 − 3 and 5 are three
roots of f(x)=0, then the least value of n is
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1) 5 2) 4 3) 3 4) 6

Ans: 1

Sol. Since 1+2i, 2 − 3 and 5 are the some roots of polynomial f(x) of degree n. As we know this

conjugate are also the roots of the polynomial is 1-2i, 2 + 3

∴ The least value of n is 5. [EAMCET-2005]
17. The roots of the equation x3 − 3x − 2 = 0 are

1) -1, -1, 2 2) -1, 1, -2 3) -1, 2, -3 4) -1, -1, -2

Ans 1

Sol. Verify S1
Here S1 = 0
By verification the roots are -1,-1,2

18. If α,β, γ are the roots of x3 + 2x2 − 3x −1 = 0 then α−2 + β−2 + γ−2 =

1) 12 2) 13 3) 14 4) 15

Ans: 2

α−2 + β−2 + γ−2 = 1 + 1 + 1
Sol. α2 β2 γ2

α 2β 2 + β 2γ 2 + γ 2α 2
= α 2β 2γ 2

αβγ = −2

αβ + βγ + γα = −3

αβγ = 1

(αβ + βγ + γα )2 = α 2β 2 + β 2γ 2 + γ 2α 2 + 2αβγ (α + β + γ )

9 = α 2β 2 + β 2γ 2 + γ 2α 2 + 2(1)(− 2)

α 2β 2 + β 2γ 2 + γ 2α 2 = 13

5

Theory of Equations

α −2 + β −2 + γ 2 = 13 = 13
1

19 The difference between two roots of the equation x3-13x2+15x+189=0 is 2. Then the roots of the

equation are [EAMCET : 2006]

1) -3,5,9 2) -3,-7,-9 3) 3,-5,7 4) -3,7, 9

Ans: 4

Sol. Verify S1
20. If α, β ,γ are the roots of the equation x3 − 6x2 +11x + 6 = 0 then Σα 2β + Σαβ 2 is equal to

1) 80 2) 84 3) 90 4) -84

Ans: 2

Sol. Σα 2β + Σαβ 2 = S1S2 − 3S3

= (6) (11) – 3(-6)

= 84

21. If 1, 2, 3 and 4 are the roots of the equation x4 + ax3 + bx2 + cx + d = 0, then a + 2b + c = (E-2007)
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1) -25 2) 0 3) 10 4) 24

Ans: 3

Sol. (x-1)(x-2)(x-3)(x-4) = x4+ax3+bx2+cx+d

( )( )⇒ x2 − 3x + 2 x2 − 7x +12 = x4 + ax3 + bx2 + cx + d

⇒ x4 −10x3 + 35x2 − 50x + 24 = x4 + ax3 + bx2 + cx + d
Now a = -10, b = 35, c = -50, d = 24
a +2b+c=-10+2(35)-50

= 10
22. If α , β ,γ are the roots of x3 − 2x2 + 3x − 4 = 0 then the value of α 2β 2 + β 2γ 2 + γ 2α 2 is

1) -7 2) -5 3) -3 4) 0

Ans: 1
Sol. α + β + γ = 2 , αβ + βγ + γα = 3 , αβγ = −4

α 2β 2 + β 2γ 2 + γ 2α 2 = (αβ + βγ + γα )2 − 2αβγ (α + β + γ )

= (3)2 − 2(4)(2) = -7

EAMCET 2008
23. The cubic equation whose roots are thrice to each of the roots of x3-2x2-4x+1=0 is

1) x3-6x2+36x+27=0 2) x3+6x2+36x+27=0 3) x3-6x2-36x+27=0 4) x3+6x2-36x+27=0
Ans: 4

Sol. x = 3α ⇒ x ⇒ f ⎜⎛ x ⎟⎞ = 0
3 ⎝3⎠

⎜⎛ x ⎞⎟3 + 2⎛⎜ x ⎞⎟2 − 4⎜⎛ x ⎞⎟ +1 = 0
⎝3⎠ ⎝3⎠ ⎝3⎠
⇒ x3 + 6x2 − 36x + 27 = 0
24. The sum of fourth powers of the roots of the equation x3 + x +1 = 0 is

6