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NCERT solution class 12 chapter 1 Integrals exercise 1.12 mathematics part 2

EXERCISE 1.12


Page No 352:

Question 1:

Answer:

Equating the coefficients of x2x, and constant term, we obtain

A + B − C = 0

B + = 0

A = 1

On solving these equations, we obtain

From equation (1), we obtain

Question 2:

Answer:

Question 3:

 [Hint: Put]

Answer:

Question 4:

Answer:

Question 5:

 

Answer:

On dividing, we obtain

Question 6:

Answer:

Equating the coefficients of x2x, and constant term, we obtain

A + B = 0

C = 5

9A + = 0

On solving these equations, we obtain

From equation (1), we obtain

Question 7:

Answer:

Let  a ⇒ dx = dt

Question 8:

Answer:

Question 9:

Answer:

Let sin x = t ⇒ cos x dx = dt

Question 10:

Answer:

Question 11:

Answer:

Question 12:

Answer:

Let x= t ⇒ 4x3 dx = dt

Question 13:

Answer:

Let ex = t ⇒ ex dx = dt

Question 14:

Answer:

Equating the coefficients of x3x2x, and constant term, we obtain

A + C = 0

B + D = 0

4A + C = 0

4D = 1

On solving these equations, we obtain

From equation (1), we obtain

Question 15:

Answer:

= cos3 x × sin x

Let cos x = t ⇒ −sin x dx = dt

Question 16:

Answer:

Question 17:

Answer:

Question 18:

Answer:

Question 19:

Answer:

Let I=∫sin-1x-cos-1xsin-1x+cos-1xdx

 

It is known that, sin-1x+cos-1x=π2

 

⇒I=∫π2-cos-1x-cos-1xπ2dx

 

=2π∫π2-2cos-1xdx

 

=2π.π2∫1.dx-4π∫cos-1xdx

 

=x-4π∫cos-1xdx           …(1)

 

Let I1=∫cos-1x dx

 

Also, let x=t⇒dx=2 t dt

 

⇒I1=2∫cos-1t.t dt

 

=2cos-1t.t22-∫-11-t2.t22dt

 

=t2cos-1t+∫t21-t2dt

 

=t2cos-1t-∫1-t2-11-t2dt

 

=t2cos-1t-∫1-t2dt+∫11-t2dt

 

=t2cos-1t-t21-t2-12sin-1t+sin-1t

 

=t2cos-1t-t21-t2+12sin-1t

From equation (1), we obtain

I=x-4πt2cos-1t-t21-t2+12sin-1t  =x-4πxcos-1x-x21-x+12sin-1x

=x-4πxπ2-sin-1x-x-x22+12sin-1x 

Question 20:

Answer:

Question 21:

Answer:

Question 22:

Answer:

Equating the coefficients of x2x,and constant term, we obtain

A + C = 1

3A + B + 2= 1

2A + 2B + C = 1

On solving these equations, we obtain

A = −2, B = 1, and C = 3

From equation (1), we obtain

Page No 353:

Question 23:

Answer:

Question 24:

Answer:

Integrating by parts, we obtain

Question 25:

Answer:

Question 26:

Answer:

When = 0, = 0 and 

Question 27:

Answer:

When and when

Question 28:

Answer:

When and when 

As , therefore, is an even function.

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

Question 29:

Answer:

Question 30:

Answer:

Question 31:

Answer:

From equation (1), we obtain

Question 32:

Answer:

Adding (1) and (2), we obtain

Question 33:

Answer:

From equations (1), (2), (3), and (4), we obtain

Question 34:

Answer:

Equating the coefficients of x2x, 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.

Question 35:

Answer:

Integrating by parts, we obtain

Hence, the given result is proved.

Question 36:

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.

Question 37:

Answer:

Hence, the given result is proved.

Question 38:

Answer:

Hence, the given result is proved.

Question 39:

Answer:

Integrating by parts, we obtain

Let 1 − x2 = t ⇒ −2x dx = dt

Hence, the given result is proved.

Question 40:

Evaluate as a limit of a sum.

Answer:

It is known that,

Question 41:

is equal to

A. 

B. 

C. 

D. 

Answer:

Hence, the correct answer is A.

Question 42:

is equal to

A. 

B. 

C. 

D. 

Answer:

Hence, the correct answer is B.

Page No 354:

Question 43:

If then is equal to

A. 

B. 

C. 

D. 

Answer:

Hence, the correct answer is D.

Question 44:

The value of is

A. 1

B. 0

C. − 1

D. 

Answer:

Adding (1) and (2), we obtain

Hence, the correct answer is B.


 

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