# NCERT Solutions for Class 12 Maths Part-2 Chapter 7 Integrals

**Exercise 7.1 :** Solutions of Questions on Page Number** : 299**

**Q1 :sin 2x**

**Answer :**The anti derivative of sin 2*x* is a function of *x* whose derivative is sin 2*x*.

It is known that,

Therefore, the anti derivative of

**Q2 :Cos 3x**

**Answer : **The anti derivative of cos 3x is a function of x whose derivative is cos 3x.

It is known that,

Therefore, the anti derivative of .

**Q3 :e2x**

**Answer :**The anti derivative of e2x is the function of x whose derivative is e2x.

It is known that,

Therefore, the anti derivative of .

**Q4 :**

**Answer :**The anti derivative of is the function of x whose derivative is .

It is known that,

Therefore, the anti derivative of

**Q5 :**

**Answer :**

The anti derivative of is the function of x whose derivative is

It is known that,

Therefore, the anti derivative of is .

**Q6 :**

**Answer :**

**Q7 :**

**Answer :**

**Q8 :**

**Answer :**

**Q9 :**

**Answer :**

**Q10 :**

**Answer :**

**Q11 :**

**Answer :**

**Q12 :**

**Answer :**

**Q13 :**

**Answer :**

On dividing, we obtain

**Q14 :**

**Answer :**

**Q15 :**

**Answer :**

**Q16 :**

**Answer :**

**Q17 :**

**Answer :**

**Q18 :**

**Answer :**

**Q19 :**

**Answer :**

**Q20 :**

**Answer :**

**Q21 :The anti derivative of equals**

**(A) (B)**

**(C) (D)**

**Answer :**

Hence, the correct answer is C.

**Q22 :If such that f(2) = 0, then f(x) is**

**(A) (B)**

**(C) (D)**

**Answer :**It is given that,

∴Anti derivative of

∴

Also,

Hence, the correct answer is A.

**Exercise 7.2 :** Solutions of Questions on Page Number** : 304**

**Q1 :**

**Answer :**

Let = t

∴2x dx = dt

**Q2 :**

**Answer :**

Let log |x| = t

∴

**Q3 :**

**Answer :**

Let 1 + log x = t

∴

**Q4 :sin x . sin (cos x)
**

**Answer :**sin x Ã¢â€¹â€¦ sin (cos x)

Let cos x = t

∴ – sin x dx = dt

**Q5 :**

**Answer :**

Let∴ 2adx = dt

**Q6 :**

**Answer :
**Let ax + b = t

⇒ adx = dt

**Q7 :**

**Answer :**Let

∴ dx = dt

**Q8 :**

**Answer :
**Let 1 + 2×2 = t

∴ 4xdx = dt

**Q9 :**

**Answer :
**Let

∴ (2x + 1)dx = dt

**Q10 :**

**Answer : **

Let

∴

**Q11 :**

**Answer :
**

**Q12 :**

**Answer :
**Let

∴

**Q13 :**

**Answer :
**Let

∴ 9×2 dx = dt

**Q14 :**

**Answer :
**Let log x = t

∴

**Q15 :**

**Answer :
**Let∴ – 8x dx = dt

**Q16 :**

**Answer :
**Let∴ 2dx = dt

**Q17 :**

**Answer :
**Let∴ 2xdx = dt

**Q18 :**

**Answer :
**Let

**Q19 :**

**Answer :**

Dividing numerator and denominator by ex, we obtain

Let

∴

**Q20 :**

**Answer :
**Let

∴

**Q21 :**

**Answer :**

**
**Let 2x – 3 = t

∴ 2dx = dt

**Q22 :**

**Answer :
**Let 7 – 4x = t

∴ – 4dx = dt

**Q23 :**

**Answer :
**Let

∴

**Q24 :**

**Answer :**

Let

∴

**Q25 :**

**Answer :
**Let

∴

**Q26 :**

**Answer :
**Let

∴

**Q27 :**

**Answer :
**Let sin 2x = t

∴

**Q28 :**

**Answer :
**Let∴ cos x dx = dt

**Q29 :****cot x log sin x**

**Answer :
**Let log sin x = t

**Q30 :**

**Answer :
**Let 1 + cos x = t

∴ – sin x dx = dt

**Q31 :**

**Answer :
**Let 1 + cos x = t

∴ – sin x dx = dt

**Q32 :**

**Answer :**

Let sin x + cos x = t ⇒ (cos x – sin x) dx = dt

**Q33 :**

**Answer :**

Put cos x – sin x = t ⇒ ( – sin x – cos x) dx = dt

**Q34 :**

**Answer :**

**Q35 :**

**Answer :
**Let 1 + log x = t

∴

**Q36 :**

**Answer :**

Let

∴

**Q37 :**

**Answer :
**Let x4 = t

∴ 4x3dx = dt

Let

∴

From (1), we obtain

**Q38 :****equals**

**Answer :
**Let

∴

Hence, the correct answer is D.

**Q39 :****equals**

**Answer :
**Let

∴

Hence, the correct answer is D.

**Exercise 7.3 : **Solutions of Questions on Page Number **: 307**

**Q1 :**

**Answer :**

**Q2 :**

**Answer :
**It is known that,

**Q3 : ****cos 2x cos 4x cos 6x**

**Answer :
**It is known that,

**Q4 : ****sin3 (2x + 1)**

**Answer :
**Let

**Q5 : ****sin3 x cos3 x**

**Answer :**

**Q6 : ****sin x sin 2x sin 3x**

**Answer :
**It is known that,

**Q7 : ****sin 4x sin 8x**

**Answer :
**It is known that,

**Q8 :**

**Answer :**

**Q9 :**

**Answer :**

**Q10 : ****sin4 x**

**Answer :**

**Q11 : ****cos4 2x**

**Answer :**

**Q12 :**

**Answer :**

**Q13 :**

**Answer :**

**Q14 :**

**Answer :**

**Q15 :**

**Answer :**

**Q16 : ****tan4x**

**Answer :**

From equation (1), we obtain

**Q17 :**

**Answer :**

**Q18 :**

**Answer :**

**Q19 :**

**Answer :
**

**Q20 :**

**Answer :**

**Q21 : ****sin-1 (cos x)**

**Answer :**

It is known that,

Substituting in equation (1), we obtain

**Q22 :**

**Answer :**

**Q23 :****is equal to**

**A. tan x + cot x + C
**

**B. tan x + cosec x + C**

**C. – tan x + cot x + C**

**D. tan x + sec x + C**

**Answer :**

Hence, the correct answer is A.

**Q24 :****equals**

**A. – cot (exx) + C
**

**B. tan (xex) + C**

**C. tan (ex) + C**

**D. cot (ex) + C**

**Answer :**

Let exx = t

Hence, the correct answer is B.

**Exercise 7.4 : **Solutions of Questions on Page Number** : 315**

**Q1 :**

**Answer :
**Let x

^{3}= t

∴ 3x

^{2}dx = dt

**Q2 :**

**Answer :
**Let 2x = t

∴ 2dx = dt

**Q3 :**

**Answer :
**Let 2 – x = t

⇒ – dx = dt

**Q4 :**

**Answer :
**Let 5x = t

∴ 5dx = dt

**Q5 :**

**Answer :**

**Q6 :**

**Answer :
**Let x3 = t

∴ 3×2 dx = dt

**Q7 :**

**Answer :**

From (1), we obtain

**Q8 :**

**Answer :
**Let x

^{3}= t

⇒ 3x

^{2}dx = dt

**Q9 :**

**Answer :
**Let tan x = t

∴ sec

^{2}x dx = dt

**Q10 :**

**Answer :**

**Q12 :**

**Answer :**

**Q13 :**

**Answer :**

**Q14 :**

**Answer :**

**Q15 :**

**Answer :**

**Q16 :**

**Answer :**

Equating the coefficients of x and constant term on both sides, we obtain

4A = 4 ⇒ A = 1

A + B = 1 ⇒ B = 0

Let 2×2 + x – 3 = t

∴ (4x + 1) dx = dt

**Q17 :**

**Answer :**

Equating the coefficients of x and constant term on both sides, we obtain

From (1), we obtain

From equation (2), we obtain

**Q18 :
Answer:**

Equating the coefficients of x and constant term on both sides, we obtain

Substituting equations (2) and (3) in (1), we obtain

**Q19 :**

**Answer :Answre **

Equating the coefficients of x and constant term, we obtain

2A = 6 ⇒ A = 3

– 9A + B = 7 ⇒ B = 34

∴ 6x + 7 = 3 (2x – 9) + 34

Substituting equations (2) and (3) in (1), we obtain

**Q20 :**

**Answer :**

Equating the coefficients of x and constant term on both sides, we obtain

Using equations (2) and (3) in (1), we obtain

**Q21 :**

**Answer :**

Let x^{2} + 2x +3 = t

⇒ (2x + 2) dx =dt

Using equations (2) and (3) in (1), we obtain

**Q22 :**

**Answer :**

Equating the coefficients of x and constant term on both sides, we obtain

Substituting (2) and (3) in (1), we obtain

**Q23 :**

**Answer :**

Equating the coefficients of x and constant term, we obtain

Using equations (2) and (3) in (1), we obtain

**Q24 :****equals**

**A. x tan – 1 (x + 1) + C
**

**B. tan – 1 (x + 1) + C**

**C. (x + 1) tan – 1 x + C**

**D. tan – 1x + C**

**Answer :**

Hence, the correct answer is B.

**Q25 :****equals**

**A.**

**B.**

**C.**

**D.**

**Answer :**

Hence, the correct answer is B.

**Exercise 7.5 :** Solutions of Questions on Page Number** : 322**

**Q1 :**

**Answer :
**Let

Equating the coefficients of x and constant term, we obtain

A + B = 1

2A + B = 0

On solving, we obtain

A = – 1 and B = 2

**Q2 :**

**Answer :
**Let

Equating the coefficients of x and constant term, we obtain

A + B = 0

– 3A + 3B = 1

On solving, we obtain

**Q3 :**

**Answer :
**Let

Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

A = 1, B = – 5, and C = 4

**Q4 :**

**Answer :
**Let

Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

**Q5 :**

**Answer :
**Let

Substituting x = – 1 and – 2 in equation (1),we obtain

A = – 2 and B = 4

**Q6 :**

**Answer :
**It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (1 – x

^{2}) by x(1 – 2x), we obtain

Let

Substituting x = 0 and in equation (1), we obtain

A = 2 and B = 3

Substituting in equation (1), we obtain

**Q7 :**

**Answer :
**Let

Equating the coefficients of x

^{2}, x, and constant term, we obtain

A + C = 0

– A + B = 1

– B + C = 0

On solving these equations, we obtain

From equation (1), we obtain

**Q8 :**

**Answer :
**Let

Substituting x = 1, we obtain

Equating the coefficients of x

^{2}and constant term, we obtain

A + C = 0

– 2A + 2B + C = 0

On solving, we obtain

**Q9 :**

**Answer :**

Let

Substituting x = 1 in equation (1), we obtain

B = 4

Equating the coefficients of x^{2} and x, we obtain

A + C = 0

B – 2C = 3

On solving, we obtain

**Q10 :**

**Answer :**

Let

Equating the coefficients of x2 and x, we obtain

**Q11 :**

**Answer :**

Let

Substituting x = – 1, – 2, and 2 respectively in equation (1), we obtain

**Q12 :**

**Answer :
**It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (x

^{3}+ x + 1) by x

^{2}– 1, we obtain

Let

Substituting x = 1 and – 1 in equation (1), we obtain

**Q13 :**

**Answer :**

Equating the coefficient of x^{2}, x, and constant term, we obtain

A – B = 0

B – C = 0

A + C = 2

On solving these equations, we obtain

A = 1, B = 1, and C = 1

**Q14 :**

**Answer :**

Equating the coefficient of x and constant term, we obtain

A = 3

2A + B = – 1 ⇒ B = – 7

**Q15 :**

**Answer :**

Equating the coefficient of x^{3}, x^{2}, x, and constant term, we obtain

On solving these equations, we obtain

**Q16 :****[Hint: multiply numerator and denominator by x ^{n – 1} and put x^{n} = t]**

**Answer :**

Multiplying numerator and denominator by x

^{n – 1}, we obtain

Substituting t = 0, – 1 in equation (1), we obtain

A = 1 and B = – 1

**Q17 :****[Hint: Put sin x = t]**

**Answer :**

Substituting t = 2 and then t = 1 in equation (1), we obtain

A = 1 and B = – 1

**Q18 :**

**Answer :**

Equating the coefficients of x^{3}, x^{2}, x, and constant term, we obtain

A + C = 0

B + D = 4

4A + 3C = 0

4B + 3D = 10

On solving these equations, we obtain

A = 0, B = – 2, C = 0, and D = 6

**Q19 :**

**Answer :**

Let x^{2} = t ⇒ 2x dx = dt

Substituting t = – 3 and t = – 1 in equation (1), we obtain

**Q20 :**

**Answer :**

Multiplying numerator and denominator by x^{3}, we obtain

Let x4 = t ⇒ 4x3dx = dt

Substituting t = 0 and 1 in (1), we obtain

A = – 1 and B = 1

**Q21 :****[Hint: Put e ^{x} = t]**

**Answer :**

Let e

^{x}= t ⇒ e

^{x}dx = dt

Substituting t = 1 and t = 0 in equation (1), we obtain

A = – 1 and B = 1

**Q22 :**

**A.**

**B.**

**C.**

**D.**

**Answer :**

Substituting x = 1 and 2 in (1), we obtain

A = – 1 and B = 2

Hence, the correct answer is B.

**Q23 :**

**A.**

**B.**

**C.**

**D.**

**Answer :**

Equating the coefficients of x^{2}, x, and constant term, we obtain

A + B = 0

C = 0

A = 1

On solving these equations, we obtain

A = 1, B = – 1, and C = 0

Hence, the correct answer is A.

**Exercise 7.6 :** Solutions of Questions on Page Number** : 327**

**Q1 : ****x sin x
**

**Answer :**

Let I =

Taking x as first function and sin x as second function and integrating by parts, we obtain

**Q2 :**

**Answer :
**Let I =

Taking x as first function and sin 3x as second function and integrating by parts, we obtain

**Q3 :**

**Answer :
**Let

Taking x

^{2}as first function and ex as second function and integrating by parts, we obtain

Again integrating by parts, we obtain

**Q4 : ****x logx
**

**Answer :**

Let

Taking log x as first function and x as second function and integrating by parts, we obtain

**Q5 :****x log 2x
**

**Answer :**

LetTaking log 2x as first function and x as second function and integrating by parts, we obtain

**Q6 : ****x ^{2} log x
**

**Answer :**

LetTaking log x as first function and x

^{2}as second function and integrating by parts, we obtain

**Q7 :**

**Answer :
**Let

Taking as first function and x as second function and integrating by parts, we obtain

**Q8 :
**

**Answer :**

Let

Taking as first function and x as second function and integrating by parts, we obtain

**Q9 :
**

**Answer :**

Let

Taking cos – 1 x as first function and x as second function and integrating by parts, we obtain

**Q10 :
**

**Answer :**

Let

Taking as first function and 1 as second function and integrating by parts, we obtain

**Q11 :
**

**Answer :**

Let

Taking as first function and as second function and integrating by parts, we obtain

**Q12 :
**

**Answer :**

Let

Taking x as first function and sec2x as second function and integrating by parts, we obtain

**Q13 :
**

**Answer :**

Let

Taking as first function and 1 as second function and integrating by parts, we obtain

**Q14 :
**

**Answer :**

Taking as first function and x as second function and integrating by parts, we obtain

**Q15 :
**

**Answer :**

Let

Let I = I

_{1}+ I

_{2}… (1)

Where,and

Taking log x as first function and x

^{2}as second function and integrating by parts, we obtain

Taking log x as first function and 1 as second function and integrating by parts, we obtain

Using equations (2) and (3) in (1), we obtain

**Q16 :
**

**Answer :**

Let

Let

⇒

∴

It is known that,

**Q17 :
**

**Answer :**

Let

Let

⇒

It is known that,

**Q18 :
**

**Answer :**

Let⇒

It is known that,

From equation (1), we obtain

**Q19 :
**

**Answer :**

Also, let ⇒

It is known that,

**Q20 :
**

**Answer :**

Let

⇒

It is known that,

**Q21:
**

**Answer :**

Let

Integrating by parts, we obtain

Again integrating by parts, we obtain

**Q22 :
**

**Answer :**

Let ⇒

= 2θ

⇒

Integrating by parts, we obtain

**Q23 :****equals**

**Answer :**

Let

Also, let ⇒

Hence, the correct answer is A.

**Q24 :****equals**

**Answer :**

**
**Let

Also, let ⇒

It is known that,

Hence, the correct answer is B.

**Exercise 7.7 :** Solutions of Questions on Page Number** : 330**

**Q1 :**

**Answer :**

**Q2 :**

**Answer :**

**Q3 :**

**Answer :**

**Q4 :**

**Answer :**

**Q5 :**

**Answer :**

**Q6 :**

**Answer :**

**Q7 :**

**Answer :**

**Q8 :**

**Answer :**

**Q9 :**

**Answer :**

**Q10 :****is equal to**

**A.
**

**B.**

**C.**

**D.**

**Answer :**

Hence, the correct answer is A.

**Q11 :****is equal to
**

**A.**

**B.**

**C.**

**D.**

**Answer :**

Hence, the correct answer is D.

**Exercise 7.8 :** Solutions of Questions on Page Number **: 334**

**Q1 :**

**Answer :
**It is known that,

**Q2 :**

**Answer :**

It is known that,

**Q3 :**

**Answer :
**It is known that,

**Q4 :**

**Answer :**

It is known that,

From equations (2) and (3), we obtain

**Q5 :**

**Answer :**

It is known that,

**Q6 :**

**Answer :
**It is known that,

**Exercise 7.9 :** Solutions of Questions on Page Number **: 338**

**Q1 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q2 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q3 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q4 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q5 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q6 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q7 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q8 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q9 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q10 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q11 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q12 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q13 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q14 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q15 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q16 :**

**Answer :
**Let

Equating the coefficients of x and constant term, we obtain

A = 10 and B = – 25

Substituting the value of I

_{1}in (1), we obtain

**Q17 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q18 :**

**Answer :**

By second fundamental theorem of calculus, we obtain

**Q19 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q20 :**

**Answer :**

**
**By second fundamental theorem of calculus, we obtain

**Q21 : ****equals
**

**A.**

**B.**

**C.**

**D.**

**Answer :**

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

**Q22 : ****equals
**

**A.**

**B.**

**C.**

**D.**

**Answer :**

**By second fundamental theorem of calculus, we obtain**

Hence, the correct answer is C.

**Exercise 7.10 :** Solutions of Questions on Page Number **: 340**

**Q1 :**

**Answer :**

When x = 0, t = 1 and when x = 1, t = 2

**Q2 :**

**Answer :**

When x = 0, t = 1 and when x = 1, t = 2

**Q3 :**

**Answer :**

Also, let

**Q4 :**

**Answer :**

Also, let

**Q5 :**

**Answer :**

Also, let x = tanθ ⇒ dx = sec2θ dθ

When x = 0, θ = 0 and when x = 1,

Takingθas first function and sec2θ as second function and integrating by parts, we obtain

**Q6 :**

**Answer :**

Also, let x = tanθ ⇒ dx = sec2θ dθ

When x = 0, θ = 0 and when x = 1,

Taking θ as first function and sec2θ as second function and integrating by parts, we obtain

**Q7 :**

**Answer :**

Let x + 2 = t^{2} ⇒ dx = 2tdt

When x = 0, and when x = 2, t = 2

**Q8 :**

**Answer :**

Let x + 2 = t^{2} ⇒ dx = 2tdt

When x = 0, and when x = 2, t = 2

**Q9 :**

**Answer :**

Let cos x = t ⇒ – sinx dx = dt

When x = 0, t = 1 and when

**Q10 :**

**Answer :**

Let cos x = t ⇒ – sinx dx = dt

When x = 0, t = 1 and when

**Q11 :**

**Answer :**

**
**Let ⇒ dx = dt

**Q12 :**

**Answer :**

**
**Let ⇒ dx = dt

**Q13 :**

**Answer :**

Let x + 1 = t ⇒ dx = dt

When x = – 1, t = 0 and when x = 1, t = 2

**Q14 :**

**Answer :**

Let x + 1 = t ⇒ dx = dt

When x = – 1, t = 0 and when x = 1, t = 2

**Q15 :**

**Answer :**

Let 2x = t ⇒ 2dx = dt

When x = 1, t = 2 and when x = 2, t = 4

**Q16 :**

**Answer :**

** **Let 2x = t ⇒ 2dx = dt

When x = 1, t = 2 and when x = 2, t = 4

**Q17 : ****The value of the integral is
**

**A. 6**

**B. 0**

**C. 3**

**D. 4**

**Answer :**

Let cotθ = t ⇒ – cosec2θ dθ= dt

Hence, the correct answer is A.

**Q18 : ****The value of the integral is
**

**A. 6**

**B. 0**

**C. 3**

**D. 4**

**Answer :**

Let cotθ = t ⇒ – cosec2θ dθ= dt

Hence, the correct answer is A.

**Q19 : ****If**

**A. cos x + x sin x
**

**B. x sinx**

**C. x cos x**

**D. sin x + x cos x**

**Answer :**

Integrating by parts, we obtain

Hence, the correct answer is B.

**Q20 : ****If**

**A. cos x + x sin x
**

**B. x sinx**

**C. x cos x**

**D. sin x + x cos x**

**Answer :**

Integrating by parts, we obtain

Hence, the correct answer is B.

**Exercise 7.11 :** Solutions of Questions on Page Number** : 347**

**Q1 :**

**Answer :**

Adding (1) and (2), we obtain

**Q2 :**

**Answer :**

**
**Adding (1) and (2), we obtain

**Q3 :**

**Answer :**

**
**Adding (1) and (2), we obtain

**Q4 :**

**Answer :**

Adding (1) and (2), we obtain

**Q5 :****Answer :**

Adding (1) and (2), we obtain

**Q6 :**

**Answer :
**Adding (1) and (2), we obtain

**Q7 :****Answer :**

Adding (1) and (2), we obtain

**Q8 :** **Answer :**

**Q9 :**

**Answer :**

**Q10 :
**

**Answer :**

Adding (1) and (2), We obtain

**Q11 :** **Answer :
**As sin

^{2}(-x)=(sin(-x)) = (-sinx)

^{2}= sin2x, therefore, sin2x is an even function.

**It is known that if f(x) is an even function,then**

**Q12 :**

**Answer :**

**Q13 :**

**Answer :
**As sin

^{7}(-x)=(sin(-x))

^{7}=(-sinx)

^{7}=-sin

^{7}x,therefore,sin

^{2}x is an even function.

It is known that if f(x) is an even function,then

**Q14 :** **
Answer :
**It is known that,

**Q15 :**

**Answer :
**

Adding (1) and (2), we obtain,

**Q16 :**

**Answer :
**Adding (1) and (2), we obtain,

sin(Π-x)=sinx

Adding (4) and (5), we obtain,

**Q17 :**

**Answer :
**It is known that,

Adding (1) and (2),we obtain,

**Q18 :**

**Answer :
**It can be seen that,(x-a) ≤ 0 when 0 ≤ x ≤1 and(x-1) ≥0 when 1 ≤ x ≤ 4.

**Q19 : Show that if f and g defined asand
**

**Answer :**

Adding (1) and (2), we obtain

**Q20 : The value of
A**.0

**B**.2

**C.**Π

**D.**

**Answer :**

It is known that if f(x) is an even function,then

and f(x) is an odd function,then

Hence,the correct answer is C.

**Q21 : The value of
**

**A.**2

**0**

B. C.

B. C.

**-2**

D.

D.

**Adding (1) and (2),we obtain,**

Answer :

Answer :

Hence the correct answer is C.

**Exercise Miscellaneous :** Solutions of Questions on Page Number** : 352**

**Q1 :**

**Answer :
**Equating the coefficients of x

^{2}, x, and constant term, we obtain

– A + B – C = 0

B + C = 0

A = 1

On solving these equations, we obtain

From equation (1), we obtain

**Q2 :**

**Answer :
**

**Q3 :**

** [Hint: Put]**

**Answer :
**

**Q4 :**

**Answer :
**

**Q5 :**

**Answer :
**

On dividing, we obtain

**Q6 :**

**Answer :
**

Equating the coefficients of x

^{2}, x, and constant term, we obtain

A + B = 0

B + C = 5

9A + C = 0

On solving these equations, we obtain

From equation (1), we obtain

**Q7 :**

**Answer :
**Let x – a = t ⇒ dx = dt

**Q8 :
**

**Answer :**

**Q9 :**

**Answer :
**Let sin x = t ⇒ cos x dx = dt

**Q10 :**

**Answer :
**

**Q11 :**

**Answer :
**

**Q12 :**

**Answer :
**Let x

^{4}=t ⇒ 4x

^{3}dx = dt

**Q13 :**

**Answer :
**Let ex = t ⇒ e

^{x}dx = dt

**Q14 :**

**Answer :
**Equating the coefficients of x

^{3}, x

^{2}, x, and constant term, we obtain

A + C = 0

B + D = 0

4A + C = 0

4B + D = 1

On solving these equations, we obtain

From equation (1), we obtain

**Q15 :
**

**Answer :**

= cos3 x × sin x

Let cos x = t ⇒ – sin x dx =dt

**Q16 :**

**Answer :
**

**Q17 :**

**Answer :**

**Q18 :**

**Answer :
**

**Q19 :**

**Answer :
**

**Q20 :**

**Answer :
**

**Q21 :
**

**Answer :**

**Q22 :**

**Answer :
**Equating the coefficients of x

^{2}, x,and constant term, we obtain

A + C = 1

3A + B + 2C = 1

2A + 2B + C = 1

On solving these equations, we obtain

A = – 2, B = 1, and C = 3

From equation (1), we obtain

**Q23 :**

**Answer :
**

**Q24 :**

**Answer :
**

Integrating by parts, we obtain

**Q25 :**

**Answer :
**

**Q26 :**

**Answer :
**

When x = 0, t = 0 and

**Q27 :**

**Answer :
**When and when

**Q28 :**

**Answer :
**

When and when

As ,

therefore, is an even function.

It is known that if f(x) is an even function, then

**Q29 :**

**Answer :**

**Q30 :**

**Answer :
**

**Q31 :**

**Answer :
**

From equation (1), we obtain

**Q32 :**

**Answer :
**

Adding (1) and (2), we obtain

**Q33 :**

**Answer :
a
**From equation (1),(2),(3) and (4), we obtain

**Q34 :**

**Answer :
**Equating the coefficients of x

^{2}, x, and constant term, we obtain

A + C = 0

A + B = 0

B = 1

On solving these equations, we obtain

A = – 1, C = 1, and B = 1

Hence, the given result is proved.

**Q35 :** **Answer :
**Integrating by parts, we obtain

Hence, the given result is proved.

**Q36 :
**

**Answer :**

Therefore, f (x) is an odd function.

It is known that if f(x) is an odd function, then

Hence, the given result is proved.

**Q37 :**

**Answer :
**

Hence, the given result is proved.

**Q38 :****Answer :
**

Hence, the given result is proved.

**Q39 :**

**Answer :
**Integrating by parts, we obtain

Let 1 – x^{2} = t ⇒ – 2x dx = dt

Hence, the given result is proved.

**Q40 :**

** Evaluate as a limit of a sum.**

**Answer :
**It is known that,

**Q41 :**

** is equal to**

**A.**

**B.**

**C.**

**D.**

**Answer :
**

Hence, the correct answer is A.

**Q42 :**

is equal to

**A.**

**B.**

**C.**

**D.**

**Answer :
**

Hence, the correct answer is B.

**Q43 :** **If then is equal to**

**A.**

**B.**

**C.**

**D.**

**Answer :**

Hence, the cor

Correct answer is D.

**Q44 :**

** The value of is**

**A. 1**

**B. 0**

**C. – 1**

**D.**

**Answer :
**Adding (1) and (2), we obtain

Hence, the correct answer is B.