Expertbulletin.com provides you step by step NCERT Solutions for Class 12 Maths Chapter 7 Integrals solved by Experts as per NCERT (CBSE) Book guidelines. All Relations and Functions Exercise Questions with Solutions to help you to score high marks in upcoming exams and other competitive exams. following topics and sub-topics are covered in the Chapter 7 Integrals
7.1 Introduction
7.2 Integration as an Inverse Process of Differentiation
7.2.1 Geometrical interpretation of indefinite integral
7.2.2 Some properties of indefinite integral
7.2.3 Comparison between differentiation and integration
7.3 Methods of Integration
7.3.1 Integration by substitution
7.3.2 Integration using trigonometric identities
7.4 Integrals of Some Particular Functions
7.5 Integration by Partial Fractions
7.6 Integration by Parts
7.6.1 Integral of the type
7.6.2 Integrals of some more types
7.7 Definite Integral
7.7.1 Definite integral as the limit of a sum
7.8 Fundamental Theorem of Calculus
7.8.1 Area function
7.8.2 First fundamental theorem of integral calculus
7.8.3 Second fundamental theorem of integral calculus
7.9 Evaluation of Definite Integrals by Substitution
7.10 Some Properties of Definite Integrals.

Integrals NCERT Solutions – Class 12 Maths

NCERT Solutions for Class 12 Maths Chapter 7

Exercise 7.1 : Solutions of Questions on Page Number : 299
Q1 :sin 2x
Answer :The anti derivative of sin 2x is a function of x whose derivative is sin 2x.
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 .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 + 2x2 = 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 x3 = t
∴ 3x2 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 x3 = t
⇒ 3x2 dx = dt


Q9 :
Answer :
Let tan x = t
∴ sec2x 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 x2 + 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 – x2) 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 x2, 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 x2 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 x2 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 (x3 + x + 1) by x2 – 1, we obtain

Let

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


Q13 :
Answer :

Equating the coefficient of x2, 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 x3, x2, x, and constant term, we obtain

On solving these equations, we obtain


Q16 :[Hint: multiply numerator and denominator by xn – 1 and put xn = t]
Answer :

Multiplying numerator and denominator by xn – 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 x3, x2, 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 x2 = t ⇒ 2x dx = dt


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



Q20 :
Answer :

Multiplying numerator and denominator by x3, we obtain

Let x4 = t ⇒ 4x3dx = dt


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


Q21 :[Hint: Put ex = t]
Answer :

Let ex = t ⇒ ex 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 x2, 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 x2 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 : x2 log x
Answer :
LetTaking log x as first function and x2 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 = I1 + I2 … (1)
Where,and
Taking log x as first function and x2 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 I1 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 = t2 ⇒ dx = 2tdt
When x = 0, and when x = 2, t = 2


Q8 :
Answer :

Let x + 2 = t2 ⇒ 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 sin2(-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 sin7(-x)=(sin(-x))7=(-sinx)7=-sin7x,therefore,sin2x 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
B. C. 
0
D. 
-2

Answer :

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

Hence the correct answer is C.
 


Exercise Miscellaneous : Solutions of Questions on Page Number : 352


Q1 :
Answer :


Equating the coefficients of x2, 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 x2, 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 x4 =t ⇒ 4x3 dx = dt


Q13 :
Answer :
Let ex = t ⇒ ex dx = dt


Q14 :
Answer :

Equating the coefficients of x3, x2, 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 x2, 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 x2, 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 – x2 = 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.