MATHS COMBIND

Jpacademy 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 +β2 = -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

Jpacademy 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

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

Jpacademy 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

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

Jpacademy 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

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

Jpacademy 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

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

Jpacademy 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

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

Jpacademy 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

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

Jpacademy 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)

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

Jpacademy 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

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

Jpacademy 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

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

Jpacademy 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

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

Jpacademy 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)

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

Jpacademy Theory of Equations

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

Ans: 4

Sol. Let roots be α , β ,γ we have to find α 4 + β 4 + γ 4

Let f(x) = x3+x+1

f1(x) = 3x2+1

( )Now 1(x)
− f (x) = − 3x2 +1
f x3 + x +1

301

10 0 3 1

10 3 3 1
3 3 12

∴ α4 +β4 +γ4 = 2

25. If α , β ,γ are the roots of x3+4x+1=0 then the equation whose roots are α2 β2 , γ2 is
,
β +γ γ +α α +β

1) x3-4x-1=0 2) x3-4x+1=0 3) x3+4x-1=0 4) x3+4x+1=0

[EAMCET 2009]

Ans: 3

Sol. Let y = α 2 = α 2 = −α = −x [∵α + β + γ = 0]

β +γ α

∴ Required equation is (-x)3+4(-x)+1=0

⇒ x3 + 4x −1= 0
26. If f(x)=2x4-13x2+ax+b is divisible by x2-3x+2, then (a,b) =

1) (-a,-2) 2) (6,4) 3) (9,2) 4) (2,9)

Ans: 3
Sol. x2-3x+2 = (x-1)(x-2)

f(1)=0, f(2)=0

2-13+a+b=0 32-52+2a+b=0

a+b=11 2a+b=20

Solving (1) & (2) we get

(a,b) = (9,2)

7

Jpacademy NUMERICAL INTEGRATION

PREVIOUS EAMCET BITS

1. The lines x = π divides the area of the region bounded by y = sinx, y = cosx and x-axis
4

⎛ 0 ≤ x ≤ π⎞ into two regions of area A1 and A2. The A1 : A2: [EAMCET 2009]
⎜⎝ 2 ⎟⎠

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

Ans: 4

π/4

Sol. A1 = ∫ (cos x − sin x) dx
0

π/2

A2 = ∫ (sin x − cos x) dx
π/4

∴ A1 : A2 = 1:1

2. The velocity of a particle which starts from rest is given by the following table

t (in seconds) : 0 2 4 6 8 10

v (in m/sec) : 0 12 16 20 35 60

The total distance travelled (in meters) by the particle in 10 seconds, using Trapezoidal rule is

given by [EAMCET 2009]

1) 113 2) 226 3) 143 4) 246

Ans: 2

Sol. Distance travelled = h ⎣⎡( y0 + yn ) + 2( y1 + y2 + ...... + y n −1 )⎤⎦
2

3. The area (in square units) of the region bounded by the curve 2x = y2 – 1 and x = 0 is

[EAMCET 2008]

1) 1 2) 2 3) 1 4) 2
33
1
Ans: 2

1 y2 −1 dy 1
∫ ∫ ( )Sol.
Area of the region = − = − y2 −1 dy -1/2 0
−1 2
0

⎡ y3 ⎤1 2 -1
⎢ 3 y⎥ 3
= − ⎣ − = sq.units
⎦0

4. The area (in square units) of the region enclosed by the curves y = x2 and y = x3 is

[EAMCET 2007]

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

Ans: 1

Sol. x2 = x3

x2 − x3 = 0 ⇒ x2 ( x −1) = 0 ⇒ x = 0,1

Jpacademy Numerical Integration

∫ ( )∴ Area = 1 ⎡ x3 x4 ⎤1 1
0 x2 − x3 dx = ⎢ 3 − 4 ⎥ = 12 sq.units
⎣ ⎦0

6

5. Dividing the interval [0, 6] into 6 equal parts and by using trapezoidal rule the value of ∫ x3dx is
0

approximately. [EAMCET 2006]

1) 330 2) 331 3) 332 4) 333

Ans: 4

Sol. a = 0, b = 6; n = 6; yn = a+ (r – 1) h

h = b−a =1
n

∫6 x3dx = h ⎡⎣( y0 + y6 ) + 2( y1 + y2 + y3 + y4 +5 )⎤⎦ = 333
2
0

6. The area (in square units) bounded by the curves y2 = 4x and x2 = 4y in the plane is

[EAMCET 2005]

1) 8 2) 16 3) 32 4) 64
33 3 3

Ans: 2 Y
Sol. Area bounded by the curves y2 = 4ax and x2 = 4by is 16ab OX

3
where a = 1, b = 1

∴ Area = 16 sq.units [EAMCET 2004]
3

7. The area bounded by y = x2 + 2, x-axis, x = 1 an x = 2 is

1) 16 2) 17 3) 13 4) 20
3 3 3 3

Ans: 3

( )∫Sol. 2 x2 + 2 dx = 13
13

6 1
8. ∫2 x2 − x dx
If [2, 6] is divided into four intervals of equal length, then the approximate value of

by using Simpson’s rule is [EAMCET 2003]

1) 0.3222 2) 0.2333 3) 0.5222 4) 0.2555

Ans: 3

∫6 1

Sol. dx
2 x2 − x

(6 − 2) ⎡⎣( y0 + y4 ) + 4( y1 + y3 ) + 2y2 ⎤⎦ = 0.5222

3.4

9

∫9. The approximate value of x2dx by using trapezoidal rule with equal, intervals is[EAMCET 2002]
1

1) 248 2) 242.5 3) 242.8 4) 243

Jpacademy Numerical Integration

Ans: 1

Sol. h = b − a = 9 −1 = 2
n4

x 1 3579
y = x2 1 9 25 49 81

9 x 2dx = 2 ⎣⎡(1 + 81) + 2(9 + 25 + 49)⎦⎤ = (82 + 166) = 248
2
∫

1

10.

X1 2 3 4

Y 0.7111 0.7222 0.7333 0.7444

4

∫Using the above table and trapezoidal rule, the approximately value of ydx is [EAMCET 2001]
1

1) 0.1833 2) 1.1833 3) 2.1833 4) 3.1833

Ans:3

∫Sol. b h ⎣⎡( + ) + ( + + + )⎤⎦
Trapezoidal rule = 2 y0 yn 2 y1 y2 ..... y n −1
f (x)dx =

a

Where h = b − a
n

∴ 4 ydx = 1 (0.7111 + 0.7444) + 2 ( 0.7222 + 0.7333) = 2.1833

∫ 2

1

11. The area (in square units ) of the region bounded by the curve x2 = 4y, the line x = 2 and the X-

axis is [EAMCET 2000]

1) 1 2) 2 y
3 → x2= 4y

3) 4 4) 8 Ox
3 3

Ans: 2

∫ ∫Sol. 2 ydx = 2 x2 dx = 2 →x=0
0 04 3

12. The area (in square units) bounded by the curves y = x3, y = x2 and the ordinates x = 1, x = 2 is

[EAMCET 2000]

1) 17 2) 12 3) 2 4) 7
12 17 7 2

Ans: 1

∫ ( )Sol. 2 x3 − x2 dx = 17
1 12

UUUU

Jpacademy DEFINITE INTEGRATION

PREVIOUS EAMCET BITS

∫π 1 [EAMCET 2009]
[EAMCET 2008]
1. dx = [EAMCET 2008]
0 1+ sin x

1) 1 2) 2 3) – 1 4) – 2
4) π
Ans: 2
16
∫π 1 × 1− sin x dx 4) π

Sol: 0 1+ sin x 1− sin x

∫ ∫ ( )= π 1− sin x dx ⇒ π sec2 x − sec x tan x dx
0 cos2 x 0

= ( tan x − sex)π
0

= [tan π − sec π] − [tan 0° − sec 0°]

= (0 +1) − (0 −1) = 2

1 3) π
12
∫2. x3/2 1− xdx =
0

1) π 2) π
69

Ans: 4
Sol: Put x = sin2 θ then dx = 2sin θ cos θdθ

Also x = 0, 1 ⇒ 0, π/2

1 π/2

∫ ∫x3/2 1− xdx = sin3 θ 1− sin2 θ.2sin θ cos θdθ

00

π/ 2 1 3 1 π π

2 sin 4 s2
∫= θ co θdθ = 2 × × × × =
0 6 4 2 2 16

π/2

3. ∫ sin x dx =
−π/2

1) 0 2) 1 3) 2

Ans:3

π/2 0 π/2

Sol: ∫ sin x dx = ∫ sin (−x) dx + ∫ sin xdx
−π/2 −π/2 0

[ ] [ ]= cos x 0 + − cos x π/2
−π/2 0

=1−0−0+1= 2

Jpacademy Definite Integration
∫4. If f (x ) = t e− x dx then Lt f ( t ) is [EAMCET 2007]
−t 2 t→∞ 4) – 1

1) 1 2) 1/2 3) 0 [EAMCET 2007]
4) – π
Ans:1
[EAMCET 2006]
∫ ∫Sol: f ( t ) = 0 ex dx + t e−x dx 4) 3π
−t 2 02
2
= 1 ⎡⎣ex ⎤⎦ 0 t + 1 ⎡⎣−e−x ⎤⎦ t
2 − 2 0

= 1 ⎣⎡1 − e−t ⎦⎤ − 1 ⎣⎡e−t −1⎦⎤ = 1 − 1 e−t − 1 e−t + 1
2 2 2 2 2 2

f(t) = 1− e−t

Lt f (t) = Lt ⎡⎣1− e−t ⎦⎤ =1− e−∞ =1− 0 =1

t→∞ x→∞

2π

∫5. sin6 x cos5 xdx =
0

1) 2π 2) π 3) 0
2

Ans: 3

2π

∫Sol: I = sin6 x cos5 xdx
0

Let f (x) = sin6 x cos5 x

f (2π − x) = f (x)

π

∫I = 2 sin6 x cos5 xdx
0

f (π − x) = sin6 (π − x) cos5 (π − x)

= − sin6 x cos5 x = −f (x)

∴I = 0

∫6. π/2 dx = 2) π 3) π
0 1+ tan3 x 2 4
1) π

Ans: 3

Let∫Sol:I= π/ 2 dx x
0 1+ tan3

Jpacademy∫π/ 2 cos3 x dx ……….(1) Definite Integration

= 0 sin3 x + cos3 x [EAMCET 2006]
4) e2 + 2
∫π/ 2 cos3 ⎛ π − x ⎞
⎝⎜ 2 ⎟⎠ 2e
= dx
⎛ π ⎞ ⎛ π ⎞ [EAMCET 2005]
0 sin 3 ⎜⎝ 2 − x ⎟⎠ + cos3 ⎝⎜ 2 − x ⎠⎟ 4) 150π

∫= π/2 sin3 x 3 x dx …………(2)
0 cos3 x + sin

(1) + (2)

π/2

2I = ∫ 1.dx
0

2I = π ⇒ I = π
24

∫7.1 cosh x dx =
−1 1+ e2x

1) 0 2) 1 3) e2 −1
2e

Ans: 3

∫ ( )1 ex + e−x ⎡⎢∵ cosh x = ex + e−x ⎤
⎣ 2 ⎥
Sol: dx ⎦
−1 2 1 + e2x

e2x +1 dx ⇒ 1 1 e−xdx
∫ ( ) ∫1

=
−1 2ex 1+ e2x 2 −1

∫= 1 1 ⇒ 1 ⎣⎡−e−x ⎤⎦1−1
2 2
e− x dx

−1

= − 1 ⎣⎡e−1 − e1 ⎦⎤ ⇒ e2 −1
2 2e

∫8. π/2 200sin x +100 cos x dx =

0 sin x + cos x

1) 50π 2) 25π 3) 75π

Ans: 3

∫Sol: π/2 a sin x + b cos x dx = (a + b) π
0 sin x + cox 4

∫π/2 200sin x +100 cos x dx = (200 +100) π = 75π
0 sin x + cos x 4

Jpacademy Definite Integration

∫9. π 1 θ sin θ θ dθ = [EAMCET 2005]
0 + cos2 [EAMCET 2004]

π2 π3 3) π2 π2
1) 2) 4)

2 3 4

Ans: 4

π θsin θ

Let I =
∫Sol: 0 1+ cos2 θ dθ

= π (π − θ)sin (π − θ) dθ ⇒ π (π − θ)sin θ dθ
1+ cos2 (π − θ)
∫ ∫ 1+ cos2 θ

0 0

π sin θ π/2 sin θ
0 + cos2 + cos2
∫ ∫I= π 1 θ dθ − I = 2I = π.2 1 θ dθ let cosθ = 5 , - sinθdθ = dt

0

0 −dt 1 1
1+ t2 1+ t2
∫ ∫I= π ⇒ π dt

1 0

= π ⎡⎣ tan −1 t ⎦⎤10 ⇒ I = π ⎡ π − 0⎦⎥⎤ ⇒ I = π2
⎣⎢ 4 4

∫10. π/2 log ⎛ 2 − sin θ ⎞ dθ =
−π/2 ⎝⎜ 2 + sin θ ⎟⎠

1) 0 2) 1 3) 2 4) – 1

Ans: 1

Sol: Let f (θ) = log ⎛ 2 − sin θ ⎞
⎜⎝ 2 + sin θ ⎠⎟

f (−θ) = log ⎛ 2 + sin θ ⎞ = − log ⎛ 2 − sin θ ⎞
⎜⎝ 2 − sin θ ⎠⎟ ⎜⎝ 2 + sin θ ⎠⎟

f (−θ) = −f (θ)

∴ It is an odd function

∫∴ I = π/2 log ⎛ 2 − sin θ ⎞ dθ = 0
−π/2 ⎜⎝ 2 + sin θ ⎟⎠

∫11. 2 2x − 2 dx = [EAMCET 2004]
0 2x − x2 4) 4

1) 0 2) 2 3) 3

Ans: 1

∫2 2x − 2

Sol: Let I = 0 2x − x2 dx

Jpacademy Definite Integration

Put 2x − x2 = t ⇒ (2 − 2x) dx = dt

∫0 dt = 0

− 0t

2 [EAMCET 2003]
4) 4
12. ∫ [x] dx =
−2

1) 1 2) 2 3) 3

Ans: 4

−1 0 1 2

Sol: ∫ [x] dx + ∫ [x] dx + ∫ [x] dx + ∫ [x] dx
−2 −1 0 1

=2+1+0+1=4

∫13.1 sin ⎛ 2 tan −1 1+ x ⎞ = [EAMCET 2003]
0 ⎜⎜⎝ 1− x ⎠⎟⎟dx

ππ π 4) π
1) 2) 3)
2
64

Ans: 2

Sol: Put x = cosθ L.L : 0 = cosθ ⇒ θ = π / 2
dx = - sinθdθ U.L : 1 = cosθ ⇒ θ = 0°

0 ⎡ 1+ cos θ ⎤
⎢2 1− cos θ ⎥
∫= sin tan −1 ⎦ ( − sin θ)dθ

π/2 ⎣

π/2 π/2

= ∫ sin (π − θ)sin θdθ = ∫ sin2 θdθ
00

= 1×π = π
22 4

∫14.3 3x +1 = [EAMCET 2003]
0 dx
x 2 + 9 ( )4) Log 2 2 + π
3
( ) ( )1) Log 2 2 + π 2) Log 2 2 + π ( )3) Log 2 2 + π
12 2 6

Ans: 1

=∫ ∫Sol:33 2x 9 dx + 3 x 1 9 dx
20 x2 + 0 2+

( )= 3 3 1 ⎡ −1 x ⎤3
2 ⎡⎣Log x2 +9 ⎤⎦ 0 + 3 ⎢⎣ tan 3 ⎥⎦0

= 3 [Log18 − Log9] + 1 ⎣⎡ tan −1 (1) − tan −1 (0)⎦⎤
2 3

Jpacademy Definite Integration
= 3 Log2 + π = Log23/2 + π
2 12 12 [EAMCET 2002]
4) log 1
( )= Log 2 2 + π
12 4

∫15. 3 dx =
2 x2 − x

1) log 2 2) log 4 3) log 8
3 3 3

Ans: 2 3) 3π
572
3 dx 3 dx
2 x2 − x 2 3) 5π
∫ ∫Sol: = x (x −1) 512

∫= 3 ⎛ 1 − 1 ⎞ dx = ⎡ ⎛ x −1⎞⎤3
2 ⎜⎝ x −1 x ⎟⎠ ⎣⎢log ⎜⎝ x ⎟⎠⎦⎥x=2

= log 2 − log 1 = log 4
32 3

π/2 [EAMCET 2002]

∫16. sin4 x cos6 xdx =
−π/2

1) 3π 2) 3π 4) 3π
128 256 64

Ans: 2

π/2 π/2

∫ ∫Sol: sin4 x cos6 xdx = 2 sin4 x cos6 xdx
−π/2 0

= 2 ⎡3 . 1 . 5 . 3 . 1 . π ⎤ = 3π
⎣⎢10 8 6 4 2 2 ⎦⎥ 256

π/2 [EAMCET 2001]
4) 7π
∫17. sin8 x cos2 xdx =
0 512

1) π 2) 3π [EAMCET 2001]
512 512

Ans: 4

π/2

∫Sol: sin8 x cos2 xdx
0

= 7 .5. 3. 1 . 1 . π = 7π
10 8 6 4 2 2 512

1

18. ∫ (ax3 + bx) dx = 0 for ;
−1

Jpacademy 2) a > 0 and b > 0 only Definite Integration
1) any values of a and b [EAMCET 2000]
[EAMCET 2000]
3) a > 0 and b < 0 only 4) a < 0 and b < 0 only
[EAMCET 2000]
Ans: 1

Sol: ax3 + bx is odd function

1

∴ ∫ (ax3 + bx) dx = 0
−1

for any values of ‘a’ and ‘b’

⎛ 10 −2n ⎞⎛ 10 2 n +1 ⎞
⎜ xdx ⎟ + ⎜ ⎟
∑ ∫ ∑ ∫19. sin 27
sin 27 xdx

⎝ n=1 2n−1 ⎠ ⎝ n=1 2n ⎠

1) 272 2) – 54 3) 54 4) 0

Ans: 4
Sol: sin27x is odd function

∫20. 1 x dx =
0
(1− x )5/4

1) 16 2) 3 3) −3 4) −16
3 16 16 3

Ans: 4

1 x 1 1− x 1 1− x
0 0 0 x5/4
(1− x )5/4
∫ ∫ ∫Sol: dx = )5 / 4 dx ⇒ dx
⎡⎣(1 ⎤
− 1 − x ⎦

∫= 1 ⎛ 1 − 1 ⎞ dx = −16
0 ⎜⎝ x5/4 x1/ 4 ⎟⎠ 3

21. If f(x) is integrable on [0, a] then ∫a f (x) =
(a − x) dx
0 f (x)+f

1) 0 2) 1 3) a 4) a/2

Ans: 4

Sol: I = a f f (x) − x) dx ………………(1)
(x)+f (a
∫

0

⇒ I = a f ( f( a − x) ( x ) dx …………..(2)
x )+f
∫ a−

0

(1) + (2)

∫⇒ 2I = a I dx ⇒ I = a
02

Jpacademy Definite Integration

22. lim ⎡ 1 + 1 2 + ..... + 1 n ⎤ [EAMCET 2000]
⎢⎣ 2n +1 2n + 2n + ⎦⎥
n→∞

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

Ans: 3

Sol: lim ⎡ 1 + 1 2 + ..... + 1 n ⎤
⎣⎢ 2n +1 2n + 2n + ⎦⎥
n→∞

∑= n ⎛ 1 ⎞
lim r =1 ⎝⎜ 2n + r ⎠⎟

n→∞

∑n 1

= lim ⎛ r ⎞
n→∞ r =1 ⎜⎝ n ⎟⎠
n 2 +

∫= 1 dx = ⎡⎣log (2 + x )⎦⎤10 = loge ⎛ 3 ⎞
0 2+x ⎜⎝ 2 ⎟⎠

DDD

Jpacademy DIFFERENTIAL EQUATIONS

PREVIOUS EAMCET BITS

1. The solution of the differential equation dy = sin (x + y) tan ( x + y) −1 is [EAMCET 2009]

dx

1) cos ec( x + y) + tan (x + y) = x + c 2) x + cos ec (x + y) = c

3) x + tan (x + y) = c 4) x + sec (x + y) = c

Ans: 2

Sol: Let x + y = v

1+ dy = dv
dx dx

we have, dv −1 = sin v tan v −1
dx

dv = sin v tan v
dx

∫ ∫dv = sin2 v ⇒ cos v dt = dx
sin 2 v
dx cos2 v

− 1 =x+c
sin v

∴ cosec(x + y) + x = c

2. The differential equation of the family y = aex + bxex + cx2ex of curves, where a, b, c arbitrary

constants, is [EAMCET 2009]

1) y′′′ + 3y′′ + 3y′ = 0 2) y′′ + 3y′′ − 3y′ = 0

3) y′′′ − 3y′′ − 3y′ + y = 0 4) y′′′ − 3y′′ + 3y′ − y = 0

Ans: 4
Sol: y = aex + bxex + cx2ex

( )y = a + bx + cx2 ex

d.b.s.w.r.t ‘x’

( )y′ = a + bx + cx2 ex + ex (b + 2cx)

y′ = y + ex (b + 2cx)
y′′ = y′ + ex (2cx) + (b + 2cx) ex

y′′ = y′ + y′ − y + 2cxex

84

Jpacademy Differential Equations
y′′ = 2y′ − y + 2cxex [EAMCET 2008]
y′′′ = 2y′′ − y′ + 2cex

y′′′ = 2y′′ − y′ + [y′′ − 2y′ + y]

∴ y′′′ − 3y′′ + 3y′ − y = 0

3. The solution of the differential equation dy = x − 2y +1 is
dx 2x − 4y

1) (x − 2y)2 + 2x = c 2) (x − 2y)2 + x = c

3) x−y = log ⎛ cx ⎞ 4) (x − 2y) + x2 = c
⎜ y ⎟
⎝ ⎠

Ans: 1

Sol: Put x − 2y = z .

Then 1− 2 dy = dz ⇒ 2 dy =1− dz ⇒ dy = 1 ⎝⎛⎜1 − dz ⎞
dx dx dx dx dx 2 dx ⎠⎟

dy = x − 2y +1 ⇒ 1 ⎝⎛⎜1 − dz ⎞ = z +1 ⇒1− dz = 1+ 1
dx 2x − 4y 2 dx ⎠⎟ 2z dx z

⇒ dz = −1 ⇒ zdz = −dx ⇒ z2 = −x + c / 2 ⇒ z2 = −2x + c ⇒ (x − 2y)2 + 2x = c
dx z 2

4. The solution of the differential equation dy − y tan x = ex sec x is [EAMCET 2008]
dx

1) yex cos x + c 2) y cos x = ex + c 3) y = ex sin x + c 4) y sin x = ex + c

Ans: 2

Sol: I.F = e∫Pdx = e∫−tan xdx = cos x

∫ ∫The solution is y cos x = ex sec x cos xdx = exdx = ex + c ⇒ y cos x = ex + c

( )5. The solution of the differential equation xy2dy − x3 + y3 dx = 0 is [EAMCET 2008]

1) y3 = 3x3 + c 2) y3 = 3x3 log (cx ) 3) y3 = 3x3 + log (cx ) 4) y3 + 3x3 = log (cx)

Ans: 2

Sol: Put y = vx. Then dy = ν + dν
dx dx

( )xy2dy − x3 + y3 dx = 0 ⇒ dy = x3 + y3 ⇒ ν + x dv = x3 + x3v3 ⇒ v + x dv = 1 + v
dx xy2 dx x3ν2 dx v2

⇒ x dν = 1 ⇒ ν2dν ⇒ dx = ν3 = log x + log c
dx ν2 x3

85

Jpacademy Differential Equations

⇒ ν3 = 3log (cx) ⇒ y3 = 3log (cx)
x3

∴ y3 = 3x3 log (cx)

6. The differential equation obtained by eliminating the arbitrary constants a and b from

xy = aex + be−x is [EAMCET 2007]

1) x d2y + 2 dy − xy = 0 2) d2y + 2y dy − xy = 0
dx2 dx dx2 dx

3) x d2y + 2 dy − y = 0 4) d2y + dy − xy = 0
dx2 dx dx2 dx

Ans: 1
Sol: xy = aex + be−x

⇒ xy1 + y = aex − be−x [EAMCET 2007]
xy2 + y1 + y1 = aex + be−x
∴ xy2 + 2y1 − xy = 0

7. The solutions of ( x + y +1) dy = 1 is

dx

1) y = ( x + 2) + cex 2) y = − ( x + 2) + cex 3) x = − ( y + 2) + cey 4) x = ( y + 2)2 + cey

Ans: 3

Sol: Put x + y + 1 = z ⇒ dy = dz −1
dx dx

(x + y + 1) dy = 1⇒ z ⎛ dz −1⎞⎠⎟ =1
dx ⎜⎝ dx

⇒ dz =1+ 1 ⇒ ∫ z z dz = ∫ dx
dx z +1

⇒ z − log (z +1) = x + c

x + y +1 = x + log ( x + y + 2) + c

y = log (x + y + 2) + log c

⇒ x + y + 2 = cey

8. The solution of dy = y2 is [EAMCET 2007]
dx xy − x2 4) e−y/ x = ky

1) ey/ x = kx 2) ey/ x = ky 3) e−y/ x = kx

Ans: 2

86

Jpacademy Differential Equations

Sol: Put y = vx

dy = y2 ⇒ V + x dV = V2
dx xy − x2 dx V −1

⇒ ∫ V −1 dV = ∫ 1 dx
V x

⇒ V − log V = log x + log c

⇒ V = log (cVx)

∴ cy = ey/ x 3) ex+y + x + c = 0 [EAMCET 2007]
4) ex+y − x + c = 0
9. The solution of dy +1 = ex+y is
dx [EAMCET 2006]

1) e−(x+y) + x + c = 0 2) e−(x+y) − x + c = 0
Ans: 1
Sol: Put x + y = z ⇒ 1+ dy = dz

dx dx

∫ ∫dz = ez ⇒ e−zdz = dx

dx
−e−z = x + c ⇒ e−(x+y) + x + c = 0

( )10. The solution of x2 + y2 dx = 2xy dy is

( ) ( )1) c x2 − y2 = x 2) c x2 + y2 = x ( )3) c x2 − y2 = y ( )4) c x2 + y2 = y

Ans: 1

Sol: Homogeneous differential equation

Put y = V ⇒ V + V′x = 1+ V2
x 2V

⇒ ∫ 2V dV = ∫ dx ⇒ − log 1 − v2 = log x + log c
1− V2 x

( )Solving c x2 − y2 = x

( )11. The solution of 1+ x2 dy + 2xy − 4x2 = 0 is [EAMCET 2006]
dx

( )1) 3x 1+ x2 = 4x3 + c ( )2) 3y 1+ x2 = 4x3 + c

( )3) 3x 1− x2 = 4x3 + c ( )4) 3y 1− x2 = 4x3 + c

Ans: 2

Sol: dy + ⎛ 2x ⎞ y = 4x2
dx ⎝⎜ 1+ x2 ⎟⎠ 1+ x2

87

Jpacademy Differential Equations
Linear differential equation dy + P ( x) y = Q ( x)
[EAMCET 2006]
dx 4) 1 = cx − y log x

= e∫ pdx = 2x dx =1+ x2 y
[EAMCET 2005]
I.F e ∫ 1+ x 2
4) 2x + y + c
4x2
1+ x2 [EAMCET 2005]
( ) ( )∫y. 1+ x2 4) 3co s x + c
= 1+ x2 dx

( )y. 1+ x2 = 4x3 + c
3

( )3y 1+ x2 = 4x3 + c

12. The solution of dx + x = x2 is
dy y

1) 1 = cx − x log x 2) 1 = cy − y log y 3) 1 = cx + x log y
yx x

Ans: 2

Sol: ------

13. dx + dy = (x + y)(dx − dy) ⇒ log (x + y) =

1) x + y + c 2) x + 2y + c 3) x − y + c

Ans: 3
Sol: dx + dy = dx − dy

x+y

∫ d(x + y) = ∫ dx −∫ dy
(x + y)

log (x + y) = x − y + C

14. x2y − x3 dy = y4 cos x ⇒ x3y−3 =
dx

1) sinx 2) 2sin x + c 3) 3sin x + c

Ans: 3

Sol: x2y − x3 dy = y4 cos x
dx

x2ydx − x3dy = cos x
y4dx

= x2 dx − x3 dx = cos xdx
y3 y4

∫⇒ d ⎛ x3 ⎞ = 3sin x + C
⎜ y3 ⎟
⎝ ⎠

88

Jpacademy3x3 3x3 Differential Equations
∫ ∫⎛y3 dx − y4 dy ⎞ = 3cos xdx [EAMCET 2005]
⎜ ⎟
⎝ ⎠

⇒ x3 = 3sin x +C
y3

15. Observer the following statements :
I. dy + 2xydx = 2e−x2 ⇒ yex2 = 2x + C

II. ye2 2x = c ⇒ dx = 2e−x2 − 2xy dy

Which of the following is a correct statements

1) Both I and II are true 2) Neither I nor II is true

3) I is true , II is false 4) I is false, II is true

Ans: 3

Sol: dy + 2x.y = 2e−x2
dx

which is linear differential equation

I.F = e∫ 2xdx = ex2

∫y.ex2 = 2.e−x2 ex2 dx = 2x + C True. ∴ I is true, II is false

16. dy = y + x tan y / x ⇒ sin y = [EAMCET 2005]
dx x x [EAMCET 2004]

1) cx2 2) cx 3) cx3 4) cx4

Ans: 2

Sol: Put y = V ⇒ dy = V + x dv
x dx dx

V + x. dv = V + TanV
dx

∫ dv =∫ dv
tan V x

log sin V = log Cx

sin ⎛ y ⎞ = Cx
⎜⎝ x ⎠⎟

( )17. Integrating factor of x + 2y3 dy = y2 is
dx

⎛1⎞ −⎛⎜ 1 ⎞ 4) −1
⎜⎟ ⎟ y
2) e ⎝ y ⎠
1) e⎝ y ⎠ 3) y

Ans: 1

89

Jpacademy Differential Equations

Sol: dx = x + 2y
dy y2

dx + x ⎛ − 1 ⎞ = 2y
dy ⎜ y2 ⎟
⎝ ⎠

e∫− 1 dy 1
y2
I.F = = ey

18. y = Aex + Be2x + Ce3x satisfies the differential equation [EAMCET 2004]

1) y′′′ − 6y′′ + 11y′ − 6y = 0 2) y′′′ + 6y′′ + 11y′ − 6y = 0

3) y′′′ − 6y′′ −11y′ − 6y = 0 4) y′′′ − 6y′′ −11y′ + 6y = 0

Ans: 1

Sol: y′′′ − (1+ 2 + 3) y′′

+ (1.2 + 2.3 + 3.1) y′ −1.2.3y = 0

19. Observe the following statements : [EAMCET 2004]

A : Integrating factor of dy + y = x2 is ex
dx

R : Integrating factor of dy + P ( x) y = Q ( x ) is e∫P(x)dx

dx

Then the true statement among the following is

1) A is true, R false 2) A is false, R is true
3) A is true, R is true, R ⇒ A 4) A is false, R is false

Ans: 3

Sol: I.F of dy + y = x2 is e∫1dx = ex
dx

20. The differential equation of the family of parabola with focus at the origin and the X-axis as axis
is [EAMCET 2003]

1) y ⎛ dy ⎞2 + 4x ⎛ dy ⎞ = 4y 2) − y ⎛ dy ⎞2 = 2x dy −y
⎜⎝ dx ⎠⎟ ⎜⎝ dx ⎟⎠ ⎜⎝ dx ⎟⎠ dx

3) y ⎛ dy ⎞2 + y = 2xy dy 4) y ⎛ dy ⎞2 + 2xy ⎛ dy ⎞ + y = 0
⎝⎜ dx ⎟⎠ dx ⎜⎝ dx ⎠⎟ ⎜⎝ dx ⎟⎠

Ans: 2

Sol: Focus = (0, 0); directrix is x+ a = 0
Equation of the parabola is y2 = a(2x + a)

90

Jpacademy2ydy=2a ⇒ a = yy1 Differential Equations
dx [EAMCET 2003]
[EAMCET 2003]
y2 = 2xyy1 + y2y12

⇒ y2 = y ⎣⎡2xy1 + yy12 ⎦⎤

⇒ y = 2xy1 + yy12

⇒ −yy12 = 2xy1 − y

⇒ − y ⎛ dy ⎞2 = 2x ⎛ dy ⎞ − y
⎜⎝ dx ⎟⎠ ⎝⎜ dx ⎟⎠

21. Solution of dy = x log x2 + x is
dx sin y + y cos y

1) y sin y = x2 log x + c 2) y sin y = x2 + c

3) y sin y = x2 + log x + c 4) y sin y = x log x + c

Ans: 1

Sol: dy = x log x2 + x
dx sin y + y cos y

⇒ sin ydy + y cos ydy = x log x2dx + xdx

Integrating on both sides y sin y = x2 log x2 + C

( )22. The general solution of y2dx + x2 − xy + y2 dy = 0 is

1) tan −1 ⎛ x ⎞ + log y + c = 0 2) 2 tan−1 ⎛ x ⎞ + log y + c = 0
⎜ y ⎟ ⎜ y ⎟
⎝ ⎠ ⎝ ⎠

( )3) log y + x2 + y2 + log y + c = 0 4) 2 sinh−1 ⎛ x⎞ + log y + c = 0
⎜ ⎟
⎝ y ⎠

Ans: 1
Sol: dy = −y2

dx x2 − xy + y2
Put y = vx

91

Jpacademy Differential Equations

v + x dv = x2 −v2x2 = −v2
dx − vx2 + v2 x2 1− v + v2

( )dx = − 1 − v + v2 dv

x v 1+ v2

log xv = tan−1 v + c

log y = tan −1 ⎛ y ⎞ + c
⎜⎝ x ⎟⎠

⇒ log y + tan −1 ⎛ x ⎞ + c = 0
⎜ y ⎟
⎝ ⎠

23. Order of the differential equation of the family of all concentric circles centered at (h, k) is
[EAMCET 2002]

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

Ans: 1

Sol: ( x − h )2 + ( y − k)2 = r2

Centre (h, k) is fixed

Radius = r is a variable

Hence order is 1

24. The solution of dy + y = 1 is [EAMCET 2002]
dx 3 4) 3y = c + c−x / 3

1) y = 3 + cex / 3 2) y = 3 + ce−x / 3 3) 3y = c + ex / 3

Ans: 2

Sol: dy + y = 1 ⇒
dx 3

Integrating factor : I.F = e∫ 1 dx = ex /3
3

∫Solution yex / 3 = ex / 3dx

yex / 3 = 3ex / 3 + C
y = 3 + Ce−x / 3

25. y + x2 = dy has the solution [EAMCET 2002]
dx
2) y + x + 2x2 + 2 = cex
1) y + x2 + 2x + 2 = cex

3) y + x + x2 + 2 = ce2x 4) y2 + x + x2 + 2 = ce2x

Ans: 1
Sol: y + x2 = dy

dx

92

Jpacademy Differential Equations
dy − y = x2
dx [EAMCET 2002]
I.F = e∫ −1dx = e−x 4) y1/ 3 − x1/ 3 = c

∫Solution ye−x = x2e−xdx [EAMCET 2001]
( )ye−x = −e−x x2 + 2x + 2 + C

y + x2 + 2x + 2 = Cex

dy ⎛ y ⎞1/ 3
dx ⎜⎝ x ⎟⎠
26. The solution of = is

1) x2 / 3 + y2 / 3 = c 2) y2 / 3 − x2 / 3 = c 3) x1/ 3 + y1/ 3 = c

Ans: 2

Sol: dy = y1/ 3 ⇒ y −1/ 3dy = x −1/ 3dx
dx x1/ 3

∫ ∫y−1/ 3dy = x−1/ 3dx

3 ⎡⎣y2 / 3 − x2/3 ⎦⎤ = C1
2

⇒ y2/3 − x2/3 = C

27. The solution of xdx + ydy = x2 ydy − xy2dx is

( )1) x2 −1 = c 1 + y2 ( )2) x2 + 1 = c 1 − y2

( )3) x3 −1 = c 1+ y3 ( )4) x3 +1 = c 1− y3

Ans: 1
Sol: xdx + ydy = x2ydy − xy2dx

x (1+ y2 ) dx = −y(1− x2 ) dy

−x dx = y dy
1− x2 1+ y2

∫ x = ∫ 1 y dy
dx + y2

x2 −1

⇒ log (x2 −1) = log (1 + y2 ) + log c

( )∴x2 −1 = c 1 + y2

28. The solution of x2 + y2 dy = 4 is 3) x3 + y3 = 3x + c [EAMCET 2001]
dx 4) x3 + y3 = 12x + c

1) x2 + y2 = 12x + c 2) x2 + y2 = 3x + c
Ans: 4

93

Jpacademy Differential Equations

Sol: x2 + y2 dy = 4
dx

( )4 − x2 dx = y2dy
∫ (4 − x2 )dx = ∫ y2dy

x3 + y3 = 12x + c

29. The solution of dy + y = ex is [EAMCET 2001]
dx 4) 2ye2x = 2ex + c

1) 2y = e2x + c 2) 2yex + ex + c 3) 2yex = e2x + c

Ans: 3
Sol: dy + y = ex

dx
P = 1; Q = x
I.F is e∫pdx = ex

∫y.ex = ex .ex + c ⇒ 2yex = e2x + c

30. If c is a parameter, then the differential equation whose solution is y = c2 + c [EAMCET 2000]
x

1) y = x4 ⎛ dy ⎞ − x ⎛ dy ⎞2 2) y = x 4 ⎛ dy ⎞2 + x ⎛ dy ⎞
⎜⎝ dx ⎟⎠ ⎝⎜ dx ⎟⎠ ⎝⎜ dx ⎠⎟ ⎝⎜ dx ⎠⎟

3) y = x4 ⎛ dy ⎞2 − x ⎛ dy ⎞ 4) y = x 4 ⎛ d2y ⎞ − x ⎛ dy ⎞
⎝⎜ dx ⎠⎟ ⎝⎜ dx ⎠⎟ ⎜ dx 2 ⎟ ⎜⎝ dx ⎟⎠
⎝ ⎠

Ans: 3

Sol: y = c2 + C ⇒ dy = −C
x dx x2

⇒ C = −x2 dy
dx

y = ⎛ −x 2 dy ⎞2 + 1 ⎛ − x 2 dy ⎞
⎝⎜ dx ⎟⎠ x ⎜⎝ dx ⎟⎠

⇒∴ y = x4 ⎛ dy ⎞2 − x ⎛ dy ⎞
⎜⎝ dx ⎠⎟ ⎜⎝ dx ⎠⎟

31. The equation of curve passing through the origin and satisfying the differential equation

dy = ( x − y)2 is [EAMCET 2000]

dx

1) e2x (1 − x + y) = (1 + x − y) 2) e2x (1 + x − y) = (1 − x + y)

3) e2x (1 − x + y) = − (1 + x + y) 4) e2x (1 + x + y) = (1 − x + y)

94

Jpacademy Differential Equations

Ans: 1

Sol: dy = (x − y)2 Let x − y = t

dx

⇒ dy = 1 − dt
dx dx

1− dt = t2 ⇒ dx = dt
dx 1− t2

⇒ ∫ dx =∫ (1 dt )
− t2

x = 1 loge ⎛1+ t ⎞ ⇒ 2x = log ⎛1+ t ⎞
2 ⎝⎜ 1 − t ⎠⎟ ⎝⎜ 1 − t ⎠⎟

⇒ e2x = 1+ t
1− t

∴ e2x (1 − x + y) = (1 + x − y)

777

95

Jpacademy INDEFINITE INTEGRATION
PREVIOUS EAMCET BITS

1. ∫ ( x dx + 3 [EAMCET 2009]

+1) 4x

1) tan−1 4x + 3 + c 2) 3 tan−1 4x + 3 + c 3) 2 tan−1 4x + 3 + c 4) 4 tan−1 4x + 3 + c

Ans:

x = t2 −3 1 tdt
4 2
Sol: 4x + 3 = t2 = ∫ ⎛ t 2 − 3 ⎞
⎜ 1⎟
⎝ 4 + t
⎠

∫ ( )= 2
4dx = 2tdt t2 + dt = 2 tan−1 t + c = 2 tan −1 4x + 3 + c
t

dx = 1 tdt
2

2. ∫ ⎛ 2 − sin 2x ⎞ ex dx [EAMCET 2009]
⎜⎝ 1− cos 2x ⎟⎠

1) −ex cot x + c 2) ex cot x + c 3) 2ex cot x + c 4) −2ex cot x + c

Ans: 1

⎛ 2 − 2sin x cos x ⎞
⎝⎜ 2 sin 2 x ⎟⎠
( )∫ ∫Sol:
e x dx = cos ec2x − cot x exdx

= ∫ ex ⎣⎡(− cot x) + cos ec2x⎦⎤ dx

ex (− cot x) + c ⎡⎣∵ ∫ ex ⎣⎡f ( x) + f ′(x)⎦⎤ dx = exf (x) + c⎤⎦

∫3. If In = sinn xdx then nIn − (n −1) In−2 = [EAMCET 2009]

1) sinn−1 x cos x 2) cosn−1 x sin x 3) – sinn−1 x cos x 4) − cosn−1 x sin x

Ans: 1

∫Sol: sinn−1 x sin x
In = dx

uv

In = sinn−1 x (− cos x) − ∫ (n −1)(sin )x n−2 (cos x)(− cos x) dx

∫ ( )In = − sinn−1 x cos x + (n −1) (sin x )n−2 1− sin2 x dx

∫ ∫In = − sinn−1 x cos x + (n −1) sinn−2 xdx − (n −1) sinn xdx

In = − sinn−1 x cos x + (n )−1 In−2 − (n −1) In ⇒ nIn − (n )−1 In−2 = sinn−1 x cos x

Jpacademy Indefinite Integration

4. If ∫ ex ⎛ 1− sin x ⎞ dx = f (x)+c then f (x) = [EAMCET 2008]
⎜⎝ 1− cos x ⎠⎟

1) ex cot ⎛ x⎞ 2) e−x cot ⎛ x ⎞ 3) −ex cot ⎛ x⎞ 4) −e−x cot ⎛ x⎞
⎝⎜ 2 ⎠⎟ ⎝⎜ 2 ⎠⎟ ⎝⎜ 2 ⎟⎠ ⎝⎜ 2 ⎟⎠

Ans:3

⎛ 1− 2sin x cos x ⎞ ⎡1 x x ⎤
⎜ 2 x 2 ⎟ ⎣⎢ 2 2 2 ⎥⎦
ex ⎜ ⎟dx ex cos ec2
⎜ 2 sin 2 ⎟
∫ ∫Sol: = − cot dx

⎝ 2⎠

= ex ⎛ − cot x ⎞ + c ⎡⎣∵∫ ex ⎡⎣f ( x) + f ′(x)⎤⎦dx = exf (x) + c⎦⎤
⎜⎝ 2 ⎠⎟

∴f ( x) = −ex cot x

2

∫5. If In = xnecxdx for n ≥ 1 then c.In + n.In−1 = [EAMCET 2008]
4) xn + ecx
1) xnecx 2) xn 3) ecx
Ans: 1 [EAMCET 2008]

∫Sol: xn ecx 4) tan (xex )
In = . dx

uv

∫ ∫xn n −1. ⎛ ecx ⎞
ecxdx − n.x ⎜ c ⎟dx
⎠
⎝

In = xn ⎛ ecx ⎞ − n I n −1 ⇒ CIn = x n ecx − nIn−1
⎜ ⎟ c
⎝ c ⎠

C.In + nIn−1 = xnecx

6. If ∫ ex (1+ x).sec2 (xex ) dx = f (x) + c then f (x) =

1) (cos xex ) 2) sin (xex ) 3) 2 tan−1 x

Ans: ( )∫ ∫ex (x +1)sec2 xex dx = sec2 tan
Sol: Let xex = t ( )= tan t + c ⇒ tan xex + c

( )x.ex + ex.1 dx = dt

ex ( x +1) dx = dt ∴f (x) = tan (xex )

7. I∫ ex −1 dx =f (x)+c then f(x) is equal to [EAMCET 2007]
ex +1

Jpacademy Indefinite Integration
( ) ( )1) 2log ex +1 3) (2 log ex +1) − x
2) log e2x −1 ( )4) log e2x +1

Ans: 3

∫ ∫ ( )Sol:
= ex + ex − ex −1 dx ⇒ 2ex − ex +1
ex +1 ex +1 dx

2ex ex +1 dx ex
ex + ex +1 ex +
∫ ∫ ∫ ∫= 1 dx − = 2 1 dx − 1dx

= 2 log ex +1 − x + c

∴f (x) = 2log (ex +1) − x

∫8. tan −1 ⎛ 1− x ⎞ = [EAMCET 2007]
⎜⎝⎜ 1+ x ⎠⎟⎟ dx

( )1) 1 x cos−1 x − 1− x2 + c ( )2) 1 x cos−1 x + 1− x2 + c
2 2

( )3) 1 x sin−1 x − 1− x2 + c ( )4) 1 x sin−1 x + 1− x2 + c
2 2
Ans:
Sol: Put x = cos2θ

∫= −1 ⎛ 1 − cos 2θ ⎞ 1 ⎢⎡cos−1 −1 ⎤
⎝⎜⎜ 1 + cos 2θ ⎟⎠⎟ 2 ⎣ 1− x2 .xdx ⎥
tan dx ∫= x.x −
⎦

( )∫= tan−1 tan 2 θ dx 1 ⎡⎢cos−1 1 −2x ⎤
2 ⎣ 2
∫= x.x − 1− x2 dx ⎥
⎦

= ∫ θdx = ∫ 1 cos−1 xdx = 1 ⎢⎣⎡cos−1 x.x − 1 × 2 1 − x 2 ⎤ + c
2 2 ⎦⎥
2

∫= 1 c os−1 x.1.dx = 1 ⎡⎣cos−1 x.x − 1 − x 2 ⎤ + c
2 2 ⎦

9. ∫ sin x + 8cos x dx [EAMCET 2007]
4sin x + 6 cos x

1) x + 1 log 4sin x + 6 cos x + c 2) 2x + log 2sin x + 3cos x + c
2 4) 1 log 4sin x + 6 cos x + c

3) x + 2 log 2sin x + 3cos x + c 2
Ans: 1

∫Sol: a cos x + b sin x dx = ⎛ ac + bd ⎞ x + ⎛ ad − bc ⎞ log (cos x + d sin x) + c
c cos x + d sin x ⎜⎝ c2 + d2 ⎟⎠ ⎜⎝ c2 + d2 ⎠⎟

Jpacademy Indefinite Integration

∫ 8cos x +1.sin x dx = ⎛ 48 + 4 ⎞ x + ⎛ 32 − 6 ⎞ log 6 cos x + 4sin x +c
6 cos x + 4sin x ⎝⎜ 36 +16 ⎟⎠ ⎝⎜ 36 +16 ⎟⎠

= x + 1 log 6 cos x + 4sin x + c
2

10. If ∫ x dx = g(x)+c then g(x) is equal to [EAMCET 2006]
a3 − x3

1) 2 cos−1 x 2) 2 sin −1 ⎛ x3 ⎞ 3) 2 sin −1 ⎛ x3 ⎞ 4) 2 co s−1 ⎛ x ⎞
3 3 ⎜ a3 ⎟ 3 ⎜⎝⎜ a3 ⎟⎠⎟ 3 ⎜⎝ a ⎠⎟
⎝ ⎠

Ans:

∫ ∫Sol: x dx = 1 x / a dx Let x = sin2/3 θ
a3 ⎛⎜1 − x3 ⎞ a ⎛ x ⎞3 a
⎝ a3 ⎟ − ⎝⎜ a ⎠⎟
⎠ 1

∫= 1 (sin θ)1/ 3 × 2a × cos θ dθ dx = a. 2 (sin 2 −1 cos θdθ
3
a 1− sin2 θ (sin θ)1/ 3 3 θ)3

= 2 ∫1.dθ ⇒ 2 θ + c dx = 2a × cos θ dθ
3 3 3
(sin )θ 1/3

= 2 sin −1 ⎛ x3 ⎞ + c
3 ⎝⎜⎜ a3 ⎟⎠⎟

∴g (x) = 2 sin −1 ⎛ x3 ⎞ + c
3 ⎜⎝⎜ a3 ⎟⎠⎟

11. If ∫ x2 dx =f (x)+c then f(x) is equal to [EAMCET 2006]
+ 2x + 2
4) 3 tan−1 ( x +1)
1) tan−1 ( x +1) 2) 2 tan−1 ( x +1) 3) − tan−1 ( x +1)

Ans: 1

Sol: ∫ dx = tan −1 ( x + 1) c

( x + 1)2 + 1

∴f (x) = tan−1 (x +1)

12. Observe the following statements : [EAMCET 2006]

∫A : ⎛ x2 +1⎞ x2 −1 = x2 −1 + c
⎜ ⎟
⎝ x2 ⎠ e x dx ex

∫R : f 1 (x ).ef (x)dx = f ( x) + c , then which of the following is true ?

1) Both A and R are true and R is the correct reason of A

Jpacademy Indefinite Integration

2) Both A and R are true and R is the correct reason of A

3) A is true, R is false 4) A is false, R is true

Ans: 3

∫Sol: R : f 1 ( x ).ef (x)dx ∫A : ⎜⎝⎛1 + 1 ⎞ ⎛ x − 1 ⎞
x2 ⎠⎟ x
e⎝⎜ ⎠⎟dx

Let f(x) = t Let x − 1 = t
x

∫ f 1 (x) dn = dt ⎝⎜⎛1 + 1 ⎞ dx = dt
x2 ⎠⎟

∫ etdt = et + c ∫ etdt = et +c = x−1 +c

ex

= ef (x) + c x2 −1

= e x +c

Here A is true and R is false

13. If ∫ sin x x) dx = f (x) + c [EAMCET 2005]

cos x (1+ cos

1) log 1+ cos x 2) log c os x 3) log sinx 4) log 1+ sinx
cos x 1+ cos x 1+ sin x sin x

Ans: 1

Sol: Let cosx = t

– sinxdx = dt

∫ t −dt ) = ∫ ⎛ −1 + t 1 ⎞ dt
⎝⎜ t + 1 ⎟⎠
(1+ t

= − log t + log | t +1| +c

= log t +1 + c = log 1+ cos x + c
t cos x

∴ f ( x) = log 1+ cos x + c

cos x

∫ ( ) ( )14.
x49 tan−1 x50 dx = K ⎣⎡tan−1 x50 ⎦⎤2 + c , then K = [EAMCET 2005]
1 + x100 4) − 1

1) 1 2) − 1 3) 1 100
50 50 100

Ans:

( )Sol: Let tan−1 x50 = t

Jpacademy 1 2 × 50.x49dx = dt Indefinite Integration
( )1+ x 50 [EAMCET 2005]

∫= x49dx = dt = t dt = 1 × t2 +c ⇒ 1 ⎡⎣ tan −1 50⎤⎦ 2 +c 4) – x tan−1 x
1+ x100 50 50 50 2 100
[EAMCET 2004]
∫15. If sin −1 ⎛ 2x ⎞ dx = f ( x ) − log 1+ x2 +c then f(x) =
⎝⎜ 1+ x2 ⎠⎟ ( )4) 2 tan−1 x +100

1) 2x tan−1 x 2) −2x tan−1 x 3) x tan−1 x [EAMCET 2004]
4) x −1
Ans: 1
1+ x
sin −1 ⎛ 2x ⎞ 2 tan−1 xdx
⎝⎜ 1+ x2 ⎠⎟
∫ ∫Sol: dx =

∫= 2 tan−1 x.1.dx

⎡ −1 1 ⎤
∫ ∫= ⎢⎣ 1+ x2 ⎥⎦
2 tan x. 1.dx − .xdx

= 2x tan−1 x − log 1+ x2 + c

∴f ( x ) = 2x tan−1 x

16. ∫ ( x + dx x + 99 = f (x)+c⇒ f (x) =

100)

1) 2 ( x +100)1/2 2) 3( x +100)1/2 ( )3) 2 tan−1 x + 99

Ans: 3
Sol: Let x + 99 = t2

dx = 2t dt

x +100 = (x + 99) +1 ⇒ t2 +1

dx x + 99 = ∫ 2tdt = 2 tan−1 ( t ) + c
t2 +1
100) t
( )∫( x +

( )= 2 tan−1 x + 99 + c

( )∴f (x) = 2 tan−1 x + 99

17. ∫ 1 3− x2 x2 .exdx = exf ( x ) + c ⇒ f ( x ) =
− 2x +

1) 1+ x 2) 1− x 3) 1+ x
1− x 1+ x x −1

Ans:1

Jpacademy Indefinite Integration
( )∫ ∫Sol:
=− x2 −3 .e x dx ⇒ − x2 −1 − 2 [EAMCET 2004]
4) 2 cot x
( x −1)2 ( x −1)2 exdx
[EAMCET 2003]
= −∫ ⎡ ( x +1)(x − 1) − 2 ⎤ ex dx 4
⎢ ( x −1)2 ⎥ 4) 2 (1+ x)3/2 + c
⎣⎢ ( x −1)2 ⎦⎥
[EAMCET 2003]
∫= ⎡ x +1 − ( 2 ⎤ .ex .dx ⇒ −ex ⎡ x +1⎤ + c 4) xex+x−1 + c
⎢ x −1 ⎥ ⎢⎣ x −1⎦⎥
⎣⎢ x + 1)2 ⎦⎥

= e x ⎛ x +1 ⎞ + c ⇒ f ( x ) = 1 + x
⎜⎝ 1− x ⎠⎟ 1 − x

18. ∫ sin cot x x dx = −f ( x ) + c ⇒ f ( x ) =
x cos

1) 2 tan x 2) – 2 tan x 3) – 2 cot x

Ans: 4

cot x cos ec2 x cot x.cos ec2x dx
x cos cos ec 2 x cot x
∫ ∫Sol: x × dx =
sin

= −∫ cos ec2x dx = −2 cot x + c
cot x

∴f (x) = 2 cot x

∫19. 1+ x + x + x2 dx =
x + 1+ x

1) 1 1+ x + c 2) 2 (1+ x )3/2 + c 3) 1+ x + c
2 3) −xex+x−1 + c
3

Ans:

Sol: = ∫ ( )2 dx
1+ x + x. 1+ x

x + 1+ x

=∫ 1+ x ⎡⎣ 1+ x + x ⎦⎤dx
x + 1+ x

∫= 1+ x dx = 2(1+ )x 3/2 + c
3

∫ ( )20. 1+ x − x−1 ex+x−1dx =

1) (1+ )x ex+x−1 + c 2) ( x )−1 ex+x−1 + c

Ans:

Jpacademy Indefinite Integration

⎝⎜⎛1 − 1 ⎟⎞⎠e x + 1 ⎡⎢⎣⎛⎜⎝1 − 1 ⎞ x + 1 x+ 1 ⎤
∫ ∫Sol: x + x x dx = x ⎠⎟ e x + ⎥dx
xe x ⎦

x+1

Let xe x = t

⎡ x+1 ⎜⎝⎛1 − 1 ⎞ + x+1 ⎤ dx = dt
⎢x.e x x2 ⎟⎠ ⎥
⎣ ex ⎦

x+ 1 ⎛ x − 1 + 1⎟⎠⎞ dx = dt
⎝⎜ x
ex

= ∫1.dt = t + c

x+1

xe x + c

21. ∫ 1− dx sin x = [EAMCET 2002]
cos x −

1) log 1+ cot x + c 2) log 1− tan x + c 3) log 1− cot x + c 4) log 1+ tan x + c
22 2 2

Ans: 3

Sol: ∫ 1 − dx sin x
cos x −

= ∫ 2sin2 dx =∫ dx
x / 2 − 2sin x / 2 cos x / 2
2 sin 2 x ⎣⎢⎡1 − cot x⎤
2 2 ⎥⎦

= ∫ 1 cos ec2 x dx = log 1− cot x +c
2 2 2
x
1− cot
2

22. ∫ 7 + dx x = [EAMCET 2002]
5 cos

1) 1 tan −1 ⎛ 1 tan x ⎞ + c 2) 1 tan −1 ⎛ 1 tan x ⎞ + c
3 ⎜⎝ 3 2 ⎠⎟ 6 ⎜⎝ 6 2 ⎟⎠

3) 1 tan −1 ⎛ tan x ⎞ + c 4) 1 tan −1 ⎛ tan x ⎞ + c
7 ⎝⎜ 2 ⎟⎠ 4 ⎝⎜ 2 ⎠⎟

Ans: 2

Sol: ∫ 7 + dx x
5 cos

Jpacademy 1− tan2 x Indefinite Integration
2 [EAMCET 2002]
use cos x = [EAMCET 2001]
1+ tan2 x
2

∫ ( )sec2 . x dx2 x = 1 ∫ sec2 x dx
tan 2 2 2
12 + 2
6 2 + tan2 x
22

Put tanx/2 = t

= 1 tan −1 ⎛ 1 tan x ⎞ + C
6 ⎜⎝ 6 2 ⎠⎟

∫23. 3x dx =

9x −1

1) 1 log 3x + 9x −1 + c 2) 1 log 3x − 9x −1 + c
log3 log3

3) 1 log 3x − 9x −1 + c 4) 1 log 9x + 9x −1 + c
log9 log3

Ans: 1

∫Sol: 3x dx
9x −1

3x = t ⇒ 3x log 3dx = dt

∫1 dt = 1 3) log 3x + 9x −1 + C
t2 −1
log 3 (log

24. ∫ dx =

x (x +9)

( )1) 2 tan−1 x +C 2) 2 tan −1 ⎛ x ⎞ + C ( )3) tan−1 x + C 4) tan −1 ⎛ x ⎞ + C
3 3 ⎝⎜⎜ 3 ⎟⎠⎟ ⎜⎜⎝ 3 ⎟⎠⎟

Ans: 2

Sol: ∫ dx

x (x +9)

Let x = t2 , dx = 2tdt

⇒ ∫ t ( 2tdt

)t2 + 9

∫= 2 dt = 2 tan −1 ⎛ t ⎞ + C
t2 + 32 3 ⎜⎝ 3 ⎟⎠

Jpacademy Indefinite Integration

= 2 tan −1 ⎛ x ⎞ + C
3 ⎜⎝⎜ 3 ⎟⎠⎟

25. ∫ ( x +1)2 exdx = [EAMCET 2001]

1) xex + c 2) x2ex + c 3) ( x +1) ex + c ( )4) x2 +1 ex + c

Ans: 4

Sol: ∫ ( x +1)2 exdx
∫ (x2 +1+ 2x).exdx

Let f (x) = x2 +1 ⇒ f ′(x) = 2x

∫ ⎣⎡f (x) + f ′(x)⎤⎦ exdx = f (x).ex + c
( )= x2 +1 ex + c

∫26. dx = [EAMCET 2001]
a2 sin2 x + b2 cos2 x

1) 1 tan −1 ⎛ a tan x ⎞ + c 2) tan −1 ⎛ a tan x ⎞ + c
ab ⎝⎜ b ⎟⎠ ⎝⎜ b ⎟⎠

3) 1 tan −1 ⎛ b tan x ⎞ + c 4) tan −1 ⎛ b tan x ⎞ + c
ab ⎝⎜ a ⎟⎠ ⎜⎝ a ⎠⎟

Ans: 1

dx = sec2 xdx sec2 x
a2 sin2 x + b2 cos2 x a2 tan2 x + b2 dx
∫ ∫ ∫Sol: ⇒
a2 sin2 x sec2 x + b2 cos2 x sec2 x

= 1 tan −1 ⎛ a tan x ⎞ + C
ab ⎝⎜ b ⎠⎟

( )∫27. ex 1− cot x + cot2 x dx = [EAMCET 2000]

1) ex cot x + c 2) −ex cot x + c 3) ex cos ec + c 4) −ex cos ec + c

Ans: 2

( )∫Sol: ex 1− cot x + cot2 x dx

( )∫= ex − cot x + cos ecx2 dx

Let f (x) = − cot x ⇒ f ′(x) = cos ec2x

⇒ ∫ ex ⎣⎡f (x) + f ′(x)⎦⎤ dx = exf (x) + c

= −ex cot x + c

Jpacademy Indefinite Integration

∫28. sin 6 x dx = [EAMCET 2000]
cos8 x

1) tan 7x + c 2) tan7 x + c 3) tan 7x + c 4) sec7 x + c
7 7

Ans: 2

sin 6 x tan 6 x sec2 tan7 x ⎡ (x)n ′(x) dx ⎣⎡f ( x )⎦⎤n+1 ⎤
cos8 x ⎢∵ ⎥
∫ ∫ ∫Sol: dx = = + c f f =
xdx 7 ⎣⎢ n +1 ⎦⎥

777

Jpacademy SUCCESSIVE – DIFFERENTIAL

PREVIOUS EAMCET BITS

( ) ( )1. y = easin−1 x ⇒ 1− x2 yn+2 − 2n +1 xyn+1 = [EAMCET 2009]

( )1) n2 + n2 yn ( )2) n2 − a2 yn ( )3) n2 + a2 yn ( )4) − n2 − a2 yn

Ans: 3

Sol: y = ea sin −1 ,⇒ y1 = y x a
x 1− x2

( )⇒ 1− x2 y12 = a2y2

( )⇒ 2 1− x2 y1y2 − 2xy12 = 2a2yy1

( )1− x2 y2 − xy1 = a2y ………….(1)

( ) ( )Diff. (1) ‘n’ times using Leibnitz theorem 1− x2 yn+2 − (2n +1) xyn+1 = n2 + a2 yn

2. If y = sin (logc x ) then x2 d2y + x dy = [EAMCET 2008]
dx 2 dx

1) sin (loge x) 2) cos (loge x) 3) y2 4) −y

Ans: 4

Sol: y = sin (log x) ⇒ dy = cos (log x) 1 ⇒ x dy = cos (log x)

dx x dx

⇒ x d2y + dy = − sin (log x) 1
dx 2 dx x

⇒ x2 d2y + x dy = −y
dx 2 dx

( )3. x = cos θ, y = sin 5θ ⇒ 1− x2 d2y − x dy = [EAMCET 2007]
dx2 dx 4) –25 y

1) −5y 2) 5y 3) 25 y

Ans: 4

Sol: dy = −5cos 5θ = −5 1− sin2 5θ
dx sin θ 1− cos2 θ

= −5 1− y2 = y1
1− x2

( ) ( ) ( )1− x2 y12 = 25 1− y2 ⇒ 1− x2 y2 − xy1 = −25y

49

Jpacademy Successive – Differential
4. f (x ) = ex sin x ⇒ f (6) ( x) = [EAMCET 2006]

1) e6x sin 6x 2) −8ex cos x 3) 8ex sin x 4) 8ex cos x
Ans: 2

Sol: f ( x ) = eax sin bx

( ) ( )f n ( x ) = a2 + b2 n/2 .eax sin bx + n tan−1 b / a

a = 1, b = 1, n = 6

6

ex sin
( ) ( )f 6 (x) =
(1 + 1) x + 6 tan−1 (1)

= 8ex sin ⎛ 3π + x ⎞ = −8ex cos x
⎜⎝ 2 ⎟⎠

( )5. d2y =
y = sin−1 x ⇒ 1− x2 dx 2 [EAMCET 2004]

1) −x dy 2) 0 3) x dy 4) x ⎛ dy ⎞2
dx dx ⎜⎝ dx ⎟⎠

Ans: 3

Sol: y = sin−1 x ⇒ y1 = 1
1− x2

( )⇒ 1− x2 y12 = 1

( )⇒ 1− x2 2y1y2 − 2xy12 = 0

( )∴ 1− x2 y2 = xy1

( )6. dn
If In = dx n xn log x , then In − nIn−1 = .... [EAMCET 2003]
4) (n –1)!
1) n 2) n – 1 3) n!

Ans: 4

( )Sol: dn
In = dx n xn log x

y = xn log x ⇒ y1 = xn ⎛ 1 ⎞ + nx n −1 log x
⎝⎜ x ⎟⎠

( y1 )n−1 = nIn−1 + (n −1)!
⇒ In − nIn−1 = (n −1)!

7. If y = aex + be−x + c , where a, b, c are parameters, then y′′′ = [EAMCET 2002]

1) y 2) y′ 3) 0 4) y′′

Jpacademy Successive – Differential

Ans: 2
Sol. y = aex + be−x + c

y′ = aex − be−x ;
y′′ = aex + be−x
y′′′ = aex − be−x
y′′′ = y′

8. If y = a cos (log x) + b sin (log x) , where a, b are parameters, then x2y′′ + xy′ = [EAMCET 2002]

1) y 2) – y 3) 2y 4) –2y
Ans: 2

Sol: y = a cos (log x) + b sin (log x)

xy′ = −a sin (log x) + b cos (log x)

xy′′ + y′ = −aco s (log x) − b sin (log x)

x

⇒ x2y′′ + xy′ = −y

9. If yk is the kth derivative of y with respect to x, y = cos(sinx) then y1 sin x + y2 cos x =
[EAMCET 2001]

1) y sin3 x 2) −y sin3 x 3) y cos3 x 4) −y cos3 x

Ans: 4
Sol: Given y = cos(sinx)

⇒ y1 = − sin (sin x ).cos x

y2 = sin (sin x).sin x − cos2 x.cos (sin x)

∴ y1 sin x + y2 cos x = − sin (sin x ).sin x cos x
+ sin (sin x )sin x.cos x − cos3 x.cos (sin x) = −y cos3 x

( )10.dn
dx n ex sin x = [EAMCET 2000]

1) 2n /2.ex cos ( x + nπ / 4) 2) 2n /2.ex cos ( x − nπ / 4)

3) 2n /2.ex sin ( x + nπ / 4) 4) 2n /2.ex sin ( x − nπ / 4)

Ans: 3

( )Sol: y = eax sin (bx) ⇒ yn = n ⎛ tan −1 b ⎞
a2 + b2 .eax ⎝⎜ a ⎟⎠
sin bx + n

where a = 1, b = 1