MATHS COMBIND

Jpacademy Limits

Sol. lim log (1+ x )
3x
x→0 e

−1

log (e1+ x ) e
e
lim x = log = log3e

x→0 ⎛ 3x −1 ⎞ log3e
⎜ ⎟
⎝ x ⎠

œœœ

6

Jpacademy
MAXIMA AND MINIMA
[EAMCET 2009]
PREVIOUS EAMCET BITS

1. The function f ( x ) = x3 + ax2 + bx + c, a2 ≤ 3b has

1) one maximum value 2) one minimum value
3) no extreme value 4) one maximum and one minimum value
Ans: 3

Sol. f (x ) = x3 + ax2 + bx + c

f ′( x) = 3x2 + 2ax + b

( )∴ Δ = 4a2 −12b = 4 a2 − 3b < 0

∴ f ′(x) never zero.

2. The maximum value of log x , 0 < x < ∞ is [EAMCET 2009]
x

1) ∞ 2) e 3) 1 4) e–1

Ans: 4

Sol. f (x) = log x

x

x ⎛ 1 ⎞ − log x (1) 1− log x
⎜⎝ x ⎠⎟ x
∴ f′(x) = = 0 ⇒ f′(x) = = 0
x2

⇒x=e

∴ Max. f = 1
e

3. If m and M respectively denote the minimum and maximum of f (x) = ( x −1)2 + 3 for x ∈[−3,1]

then the order pair (m, M) [EAMCET 2008]
4) (–19, –3)
1) (-3, 19) 2) (3, 19) 3) (–19, 3)

Ans: 2

Sol. f (x) = (x −1)2 + 3 ⇒ f ′(x) = 2(x −1) ⇒ f ′′(x) = 2 > 0

f ′( x) = 0 ⇒ x = 1 . Minimum value = f(1) = 3

−3 ≤ x ≤ 1 ⇒ −4 ≤ x −1 ≤ 0 ⇒ ( x −1)2 + 3 ≤ 19

Minimum value = 3 ∴ (m, M) = (3, 19)

4. The condition for f ( x ) = x3 + px2 + qx + r ( x ∈ \) to have no extreme value, is[EAMCET 2007]

1) p2 < 3q 2) 2p2 < q 3) p2 < 1/ 4 q 4) p2 > 3q

Ans: 1

Sol. f(x) has no extreme value then f ′( x ) = 0 have no real roots.

f ′( x) = 3x2 + 2px + q

4p2 −12q < 0 ⇒ p2 < 3q

1

Jpacademy Maxima and Minima

5. Observe the statements given below: [EAMCET 2007]

Assertion (A) : f ′(x ) = xe−x has the maximum at x = 1

Reason (R) : f ′(1) = 0 and f ′′(1) < 0

Which of the following is correct ?
1) Both A and R are true and R is the correct reason for A
2) Both A and R are true , but R is not the correct reason for A
3) A is true, R is false
4) A is false, R is true
Ans: 1

Sol. f (x ) = xe−x ⇒ f ′( x) = (1− x) e−x

f′(x) = 0 ⇒ x =1

f ′′(x) = (2 − x) e−x ⇒ f ′′(1) = − 1 < 0

e

∴ f(x) has maximum at x = 1

6. In the interval (- 3, 3) the function f (x ) = x + 3 , x ≠ 0 is

3x

1) increasing 2) decreasing

3) neither increasing nor decreasing 4) partly increasing and partly decreasing

Ans:

Sol. f (x) = x + 3 , x ≠ 0

3x

f′(x) = 1 − 3 = 0
3 x2

f′(x) = x2 −9
3x 2

x2 − 9 < 0, ∀x ∈ (−3,3)

i.e., f ′(x) < 0 ∀x ∈(−3,3)

∴ decreasing

7. The perimeter of a sector is a constant. If its area is to be maximum, the sectorial angel is

[EAMCET 2006]

πc πc 3) 4c 4) 2c
1) 2)

6 4

Ans: 4

Sol. A + 2r = k (say)

θl
r

2

Jpacademy Maxima and Minima
but A = rθ
rθ + 2r = k [EAMCET 2005]

r= k [EAMCET 2003]
θ+2 4) 9

Area A = 1 r2θ = k2θ 4
[EAMCET 2002]
2 2(θ + 2)2
4) 1 < x < 2
A1 = k2 ⎪⎧ 2(θ + 2)θ − (θ + 2)2 ⎪⎫ = 0
2 (θ + 2)4 ⎬
⎨ ⎪⎭
⎪⎩

⇒ θ = 2c

8. Observe the following statements :

A : f ′( x) = 2x3 − 9x2 +12x − 3 is increasing in [−∞,1] ∪[2, ∞]

R : f ′(x) < 0 for x ∈(1, 2)

Then which of the following is true?
1) Both A and R are true and R is the correct reason for A
2) Both A and R are true but R is not the correct reason for A
3) A is true, R is false
4) A is false, R is true
Ans: 4

Sol. A : f ′(x ) = 6x2 −18x +12 > 0
⇒ x ∈(−∞,1) ∪ (2,∞)

∴A is false

B : f ′(x) < 0,∀x ∈(1, 2) True

9. The minimum value of 2x2 + x – 1 is

1) 1 2) 3 3) − 9
42 8

Ans: 3

Sol. Min. value = 4ac − b2 = −9
4a 8

10. If log (1+ x) − 2x is increasing then …..

2+x

1) 0 < x < ∞ 2) – ∞ < x < 0 3) – ∞ < x < ∞

Ans: 1

Sol. f (x) = log (1+ x) − 2x

2+x

f ′( x) = 1 − (2 + x)2 − 2x (1) > 0
1+ x (2+ x
)2

3

Jpacademy Maxima and Minima
⇒ x2 > 0 ∴x>0
[EAMCET 2001]
(1+ x)(2 + x)2
4) −1(α − β)2
⇒ 0<x < ∞
4
11. The minimum value of ( x − α)(x − β) is

1) 0 2) αβ 3) 1 (α − β)2

Ans: 4 4

Sol. x2 − (α + β) x + αβ

The min. value of ax2 + bx + c is 4ac − b2
4a

4αβ − (α + β)2 − (α − β)2

∴=
44

12. The maximum value of xy subject to x + y = 7 is [EAMCET 2001]
4) 55/4
1) 12 2) 10 3) 49/4

Ans: 3

Sol. Given x + y = 7

xy = x(7 – x)

Let f (x) = x (7 − x) ⇒ f ′(x) = 7 − 2x = 0

⇒ x = 7;y= 7 [EAMCET 2000]
22

13. The minimum value of xx is

1) e 2) ee 3) e−1/ e 4) e1/ e

Ans:

Sol. y = xx ⇒ log y = x log x ⇒ dy = xx (1+ logx)
dx

(1+ log x ) = 0 ⇒ x = e−1

∴ Min.value = e−1/e

”””

4

Jpacademy PARTIAL DIFFERENTIATION

PREVIOUS EAMCET BITS

1. z = tan ( y + ax ) + y − ax ⇒ zxx − a2zyy = [EAMCET 2009]

1) 0 2) 2 3) zx + zy 4) zxzy

Ans: 1

Sol. Z = f ( y + ax) + g ( y − ax)

⇒ Zxx − a2Zyy = 0

⎛ x 4 + y4 − 8x2y2 ⎞ x ∂z + y ∂z
⎜ x2 + y2 ⎟ ∂x ∂y
2. If z = sec−1 ⎠ then = [EAMCET 2008]

⎝

1) cot z 2) 2 cotz 3) 2tanz 4) 2 secz
Ans: 2

Sol. z= z = sec−1 ⎛ x4 + y4 − 8x2y2 ⎞ ⇒ sec z is a homogeneous function of degree 2
⎜ x2 + y2 ⎟
⎝ ⎠

⇒ x ∂ [sec z] + y ∂ [sec z] = 2sec z ⇒ x sec z tan z ∂z + y sec z tan z ∂z = 2sec z
∂x ∂y ∂x ∂y

⇒ x ∂z + y ∂z = 2 cot z
∂x ∂y

3. If z = log ( tan x + tan y), then (sin 2x ) ∂z + (sin 2y) ∂z = [EAMCET 2007]

∂x ∂y

1) 1 2) – 2 3) 3 4) 4
Ans: 2
Sol. ∂z = sec2 x
∂x tan x + tan y

∂z = sec2 y
∂y tan x + tan y

sin 2x ∂z + sin 2y ∂z = 2
∂x ∂y

4. u = sin −1 ⎛ x2 + y2 ⎞ ⇒ x ∂u + y ∂u = [EAMCET 2006]
⎜ x + y ⎟ ∂x ∂y
⎝ ⎠

1) sin u 2) tan u 3) cos u 4) cot u

Ans: 2

Sol. sin u = x2 + y2 = f ( x, y)

x+y

f(x, y) is a Homogeneous equation of ‘1’ degree
∴ by Euler’s theorem

1

Jpacademy Partial Differentiation

x ∂f + y ∂f = n.f . = f (x, y)

∂x ∂y

x ∂u + y. ∂u = n. f (u)
∂x ∂y f ′(u)

⇒ sin u = tan u
cos u

5 z = cos−1 ( x + y) + sec−1 (y+ 2x) ⇒ ∂2z + 2 ∂2z = [EAMCET 2005]
∂x 2 ∂y2 ∂2z
4)
3∂ 2 z −∂ 2 z −3∂ 2 z ∂x∂y
1) 2) 3)

∂x∂y ∂x∂y ∂x∂y

Ans: 1

Sol. differentiating partially

( )( ) ( )6.
f x, y = 2 x − y 2 − x4 − y4 ⇒ f xx f yy − f 2 (0,0) [EAMCET 2004]
xy

1) 32 2) 16 3) 0 4) – 1

Ans: 3

Sol. f (x, y) = 2( x − y)2 − x4 − y4

( )fxx = 4 −12x2 = 4

(0,0)

( )fyy = −4 −12y2

(0,0)

( )fyy = −4

(0,0)

( )∴ − f 2 = 4× 4 − 42 = 0
f xx .f yy yy
(0,0)

7. If u ( x, y) = y log x + x log y, then uxuy − ux log x − uy log y + log x log y = ... [EAMCET 2003]

1) 0 2) – 1 3) 1 4) 2

Ans: 3

Sol. u ( x, y) = y log x + x log y

ux = y + log y and uy = log x + x
x y

uxuy − ux log x − uy log y + log x log y

= (ux − log y)(uy − log x)

= y.x =1
xy

8. If z = y ⎡ x + cos ⎝⎛⎜1 + y ⎞⎤ , then x ∂z = [EAMCET 2002]
x ⎢⎣sin y x ⎟⎠⎥⎦ ∂x 4) 2y ∂z

1) y ∂z 2) −y ∂z 3) 2y ∂z ∂x
∂y ∂y ∂y

Ans: 2

2

Jpacademy Partial Differentiation

Sol. z= y ⎡ x + cos ⎝⎛⎜1 + y ⎞⎤ then
x ⎢sin y x ⎟⎠⎥⎦
⎣

x ∂z + y ∂z = 0 ⇒ x ∂z = −y ∂z
∂x ∂y dx dy

9. If z = sec ( y − ax) + tan ( y + ax), then ∂2z − a2 ∂2z = [EAMCET 2002]
∂x 2 ∂y2

1) z 2) 2z 3) 0 4) – z
Ans: 3

Sol. z = sec( y − ax) + tan ( y + ax)

z = f (ax + y) + φ( y − ax)

∂2z + a2 ∂2z
∂x 2 ∂y2

∴ ∂2z + a2 ∂2z ⇒ ∂2z − a2 ∂2z =0
∂x 2 ∂y2 ∂x 2 ∂y2

10. If u = e−x2 −y2 , then [EAMCET 2001]
4) x2uy + y2ux = 0
1) xux = yuy 2) yux = xuy 3) yux + xuy = 0

Ans: 2

Sol. u = e−x2 −y2

ux = −2xe−x2 −y2 ; uy = −2ye−x2 −y2

∴ yux = xuy

11. If u = xy2 tan −1 ⎛ y ⎞ , then x ∂u + y ∂u = [EAMCET 2001]
⎝⎜ x ⎠⎟ ∂x ∂y

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

Ans: 3

Sol. Given u = xy2 . tan −1 ⎛ y⎞
⎜⎝ x ⎠⎟

x ∂ u + y ∂u = nu = 3u (∵n = 3)

∂ x ∂y

( )12. −1 ⎛ y ⎞ ∂2y ∂2u
If u = loge x2 + y2 + tan ⎜⎝ x ⎠⎟ , then ∂x 2 + ∂y2 = [EAMCET 2000]
4) u
1) 0 2) 2u 3) 1/u
[EAMCET 2000]
Ans: 1

Sol. ∂2u + ∂2u = 0
∂x 2 ∂y2

13. ⎛ x+ y ⎞ x ∂u + y ∂u =
If u = cos−1 ⎜⎜⎝ x+ y ⎟⎟⎠ , then ∂x ∂y

3

Jpacademy 3) − 1 cot u Partial Differentiation
1) 1 cot u 2) 2 cot u 2 4) 3 cot u
2

Ans: 3

Sol. n = 1 ⇒ x ∂u + y ∂u = −n cot u
2 ∂x ∂y

= − 1 cot u
2

‘‘‘‘

4

Jpacademy RATE OF MEASURE

PREVIOUS EAMCET BITS
1. A stone thrown upwards, has its equation of motion s = 490t –4.9t2. Then the maximum height

reached by it is [EAMCET 2005]

1) 24500 2) 12500 3) 12250 4) 25400

Ans: 3

Sol. S = 490t − 4.9t2

v = ds = 490 − 9.8t
dt

at maximum height v = 0

∴ t = 50

∴ S = 12250
2. A particle moves along the curve y =x2 + 2x. Then the point on the curve such that x and y

coordinates of the particle change with the same rate: [EAMCET 2004]

1) (1, 3) 2)  1 , 5  3)  − 1 , − 3  4) (−1, −1)
Ans: 3  2 2   2 4 

Sol. dy = ( 2x + 2) dx
dt dt

⇒ x = −1 ∵ dy = dx 
2 dt dt 

y = −3
4

∴( x, y) =  −1 , −3 
 2 4 

69. A point is moving on y = 4 – 2x2. The x-coordinate of the point is decreasing at the rate of

5 units per second. Then the rate at which ‘y’-coordinate of the point is changing when the point

is (1, 2) is [EAMCET 2004]

1) 5 units/sec 2) 10 units/sec 3) 15 units /sec 4) 20 units/ sec

Ans: 4

Sol. y = 4 − 2x2 ⇒ dy = −4x dx
dt dt

= 4(1)(−5) = 20 units / sec

3. Gas is being pumped into a spherical balloon at the rate of 30 ft3/ min. Then the rate at which the

radius increases when it reaches the value 15 ft is [EAMCET 2003]

1) 1 ft / min 2) 1 ft / min 3) 1 ft / min 4) 1 ft / min
30π 15π 20 25

Ans: 1

Sol. v = 4 πr3 ⇒ dv = 4πr 2 dr
3 dt dt

30 = 4π (15)2 dr ⇒ dr = 1 ft / min
dt dt 30π

1

Jpacademy Rate of measure

4. The family of curves, in which the substangent at any point to any curve is double the abscissa, is

given by 2) y = cx2 3) x2 = cy2 [EAMCET 2001]
1) x = cy2 4) y = cx

Ans: 1

Sol. Let x = cy2 ⇒ dy = 1
dx 2cy

⇒ Subtangent = y = 2cy2 = 2x
1/ 2cy

5. The distance moved by the particle in time t is given by x = t3 −12t2 + 6t + 8 . At the instant,

when its acceleration is zero, the velocity is [EAMCET 2000]

1) 42 2) – 42 3) 48 4) – 48

Ans: 2

Sol. S = t3 −12t2 + 6t + 8

V = ds = 3t 2 − 24t + 6
dt

a = dv = 6t − 24 = 0 ⇒ t = 4
dt

∴ t = 4 ⇒ υ = −42

²²²

2

Jpacademy TANGENT AND NORMALS

PREVIOUS EAMCET BITS

1. The equation to the normal to the curve y4 = ax3 at (a, a) is [EAMCET 2008]
4) 4x − 3y = 0
1) x + 2y = 3a 2) 3x − 4y + a = 0 3) 4x + 3y = 7a
[EAMCET 2008]
Ans: 3 4) 5/8

Sol. y4 = ax3 ⇒ 4y3 dy = 3ax2 ⇒ 3ax 2 ⇒ m = ⎛ dy ⎞ = 3a 3 = 3
dx 4y3 ⎝⎜ dx ⎠⎟(a,a) 4a 3 4

Equation of the normal is y − a = − 4 (x − a ) ⇒ 4x + 3y = 7a

3
2. The length of the substangent at (2, 2) to the curve x5 = 2y4 is

1) 5/2 2) 8/5 3) 2/5

Ans: 2

Sol. x5 = 2y4 ⇒ 5x4 = 8y3 dy ⇒ dy = 5x4 = 5(2)4 = 5
dx dx 8y3 8 ( 2 )3 4

Length of the subtangent 2 = 2 = 8
m 5/4 5

3. The length of tangent, subtangent normal and subnormal for the curve y = x2 + x – 1 at (1, 1) are

A, B, C and D respectively, then their increasing order is [EAMCET 2007]

1) B, D, A, C 2) B, A, C, D 3) A, B, C, D 4) B, A, D, C

Ans:4

Sol. m = ⎛ dy ⎞ = 3
⎜⎝ dx ⎟⎠(1,1)

A = length of tangent = 10
3

B = length of subtangent = 1
3

C = length of normal = 10

D = length of subnormal = 3

∴B<A<D<C

4. If θ is the angle between the curves xy = 2 and x2 + 4y = 0 , then tanθ = [EAMCET 2006]

1) 1 2) – 1 3) 2 4) 3
Ans: 4

Sol. y = 2 , y = −x2
x 4

x3 = −8 (−2, −1)

x = −2

y = −1

m1 = dy = −2 = −1
dx x2 2

1

Jpacademy Tangent and Normals

m2 = dy = −2x = −x =1
dx 4 2

tan θ = m1 + m2 = −1 −1 3 =3
2= 2

1 + m1m2 1 − 1 1
22

5. Match the points on the curve 2y2 = x + 1 with the slopes of normals at those points and choose

the correct answer. [EAMCET 2004]

List –I List – II

I) (7, 2) a) −4 2

II) ⎛ 0, 1⎞ b) – 8
⎜⎝ 2 ⎟⎠

III) (1, –1) c) 4
d) 0
(IV) 3, 2 )

e) −2 2

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

Ans: 2

Sol. 2y2 = x +1 ⇒ −dx = −4y = slope of the normal
dy

− dx at (7, 2) = −8

dy

− dx at ⎛ 0, 1 ⎞ = −2 2
dy ⎝⎜ 2 ⎟⎠

− dx at (1, −1) = 4 ;

dy

( )− dx at 3, 2 = −4 2
dy

6. The angle between the curves y = sin x and y = cos x is [EAMCET 2003]

( )1) tan−1 2 2 ( )2) tan−1 3 2 ( )3) tan−1 3 3 ( )4) tan−1 5 2

Ans: 1

Sol. y = sin x; y = cos x ⇒ x = π
4

m1 = cos π = 1
4 2

m2 = − sin π = −1
4 2

( )θ ⎛ m1 − m2 ⎞
= tan −1 ⎜ 1+ m1m2 ⎟ = tan −1 2 2
⎝ ⎠

2

Jpacademy Tangent and Normals

7. The two curves x = y2, xy = a3 cut orthogonally at a point, then a2 = [EAMCET 2002]

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

Ans: 2

Sol. x = y2 , xy = a3

2ym1 = 1 m2 = −a3 ⇒ m1 = 1
x2 2y

m1m2 = −1 ⇒ 1 a3 = +1
2y . x2

xy = 1 ⇒ x = 1
2x2y 2

( )a6 = x2y2 = x3 = 1 ⇒ a2 3 = ⎛ 1 ⎞3 ∴ a2 = 1
8 ⎝⎜ 2 ⎠⎟ 2

8. The equation of the tangent to the curve 6y = 7 – x3 at (1, 1) is [EAMCET 2001]
4) x + y + 2 = 0
1) 2x + y = 3 2) x + 2y = 3 3) x + y = - 1
[EAMCET 2000]
Ans: 2

Sol. 6y = 7 − x3 ⇒ dy at (1,1) = − 1

dx 2

Equation of the tangent is y – 1 = −1(x −1)

2
⇒ x + 2y −3 = 0

9. The angle between the curves y2 =4x, x2 = 4y at (4, 4) is

1) tan −1 ⎛ 1 ⎞ 2) tan −1 ⎛ 3⎞ 3) π 4) π
⎝⎜ 2 ⎟⎠ ⎝⎜ 4 ⎟⎠ 2 4

Ans: 2

Sol. y2 = 4x ⇒ dy at (4, 4) = 1 = m1
dx 2

x2 = 4y ⇒ dy at (4, 4) = 2 = m2
dx

T an θ = m1 − m2 ⇒ θ = tan −1 ⎛ 3 ⎞
1+ m1m2 ⎝⎜ 4 ⎠⎟

10. Area of the triangle formed by the normal to the curve x = esin yat (1, 0) with the coordinate axes is

1) 1 2) 1 3) 3 [EAMCET 2000]
42 4 4) 1

Ans: 2
Sol. x = esin y ⇒ log x = sin y ⇒ dy = 1

dx x cos y

dy at (1, 0) = 1 = slope of the tangent

dx

∴ Equation of the normal is x + y – 1 = 0

∴ Area of the triangle with normal and coordinate axes = 1/ 2
”””

3

Jpacademy 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 + 5x )−1/ 5 is 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)

Jpacademy 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!

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

Jpacademy 3) 3 − 1 Binomial Theorem
1) 3 − 3 2) 2 − 3 24 4) 2 − 1
24 34
34
Ans: 2
[EAMCET 2006]
Sol: 3 − 3.5 + 3.5.7 ....... + 3 − 3 4) 43/ 2
4.8 4.8.12 4.8.12.16 44
[EAMCET 2006]
= 3 + 3 − 3.5 + .....− 3
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

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

Jpacademy 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 v 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

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

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

Jpacademy 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 coefficient of x in the expansion 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

Jpacademy 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

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

Jpacademy 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

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

Jpacademy 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

9

Jpacademy
7. EXPONENTIAL SERIES AND LOGARITHMIC SERIES

PREVIOUS EAMCET BITS

( )1.1
e3x ex + e5x = a0 + a1x + a2x2 + ..... ⇒ 2a1 + 23 a3 + 25 a5 + ....... = [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

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

Jpacademy 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 ⎞
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

Jpacademy 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 + 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

Jpacademy 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

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+ x a + a a
log e log e log e [EAMCET 2002]

4

Jpacademy 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 !

∑ ∑ ∑= ∝ 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

Jpacademy Exponential series and logarithmic series

Coefficient of x3 is −1+ 1 = − 2
33

6

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

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

∴ 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

Jpacademy 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⎤ ⎡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

Jpacademy 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

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

Jpacademy 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

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

Jpacademy 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 = ⎢ 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

Jpacademy Matrices
Ans: 3
[EAMCET 2004]
Sol: Cofactor of −1 = (−1)1+2 0 2 4) 0
= +6
36 [EAMCET 2004]
4) 3
Cofactor of 1 = ( )−1 1+1 4 2
= 32 [EAMCET 2003]
−4 6 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 2) 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

Jpacademy 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)

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

Jpacademy 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

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

Jpacademy 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 [EAMCET 2000]
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 =

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

Jpacademy Matrices
adjA = A n−1
[EAMCET 2000]
= dn−1 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

b2 − ac = 0 [∵a ≠ 6]

b2 = ac
b2.b = abc
∴abc = b3

10

Jpacademy 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

= − 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

Jpacademy Partial fractions
a = −2 [EAMCET 2006]
b1
[EAMCET 2005]
∴a : b = −2 :1 [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

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

Jpacademy 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 = 1

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

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

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

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

2+k =5 ∴k = 3

4

Jpacademy 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 [EAMCET-2000]
= 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 =

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

Jpacademy 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

= n!r r )! = 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

Jpacademy 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

= 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

Jpacademy 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

= 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

Jpacademy 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!

⇒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

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

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

6

Jpacademy
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

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