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Page No 299:

Question 1:

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

Page No 299:

Question 2:

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 .

Page No 299:

Question 3:

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 .

Page No 299:

Question 4:

Answer:

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

It is known that,

Therefore, the anti derivative of .

Page No 299:

Question 5:

Answer:

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

It is known that,

Therefore, the anti derivative of is .

Page No 299:

Question 6:

Answer:

Page No 299:

Question 7:

Answer:

Page No 299:

Question 8:

Answer:

Page No 299:

Question 9:

Answer:

Page No 299:

Question 10:

Answer:

Page No 299:

Question 11:

Answer:

Page No 299:

Question 12:

Answer:

Page No 299:

Question 13:

Answer:

On dividing, we obtain

Page No 299:

Question 14:

Answer:

Page No 299:

Question 15:

Answer:

Page No 299:

Question 16:

Answer:

Page No 299:

Question 17:

Answer:

Page No 299:

Question 18:

Answer:

Page No 299:

Question 19:

Answer:

Page No 299:

Question 20:

Answer:

Page No 299:

Question 21:

The anti derivative of equals

(A) (B)

(C) (D)

Answer:

Hence, the correct answer is C.

Page No 299:

Question 22:

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.



Page No 304:

Question 1:

Answer:

Let = t

∴2x dx = dt

Page No 304:

Question 2:

Answer:

Let log |x| = t

Page No 304:

Question 3:

Answer:

Let 1 + log x = t

Page No 304:

Question 4:

sin x ⋅ sin (cos x)

Answer:

sin x ⋅ sin (cos x)

Let cos x = t

∴ −sin x dx = dt

Page No 304:

Question 5:

Answer:

Let

∴ 2adx = dt

Page No 304:

Question 6:

Answer:

Let ax + b = t

adx = dt

Page No 304:

Question 7:

Answer:

Let

dx = dt

Page No 304:

Question 8:

Answer:

Let 1 + 2x2 = t

∴ 4xdx = dt

Page No 304:

Question 9:

Answer:

Let

∴ (2x + 1)dx = dt

Page No 304:

Question 10:

Answer:

Let

Page No 304:

Question 11:

Answer:

Let I=xx+4 dxput x+4=tdx=dtNow, I=t-4tdt=t-4t-1/2dt=23t3/2-42t1/2+C=23.t.t1/2-8t1/2+C=23x+4x+4-8x+4+C=23xx+4+83x+4-8x+4+C=23xx+4-163x+4+C=23x+4x-8+C

                 

Page No 304:

Question 12:

Answer:

Let

Page No 304:

Question 13:

Answer:

Let

∴ 9x2dx = dt

Page No 304:

Question 14:

Answer:

Let log x = t

Page No 304:

Question 15:

Answer:

Let

∴ −8x dx = dt

Page No 304:

Question 16:

Answer:

Let

∴ 2dx = dt

Page No 304:

Question 17:

Answer:

Let

∴ 2xdx = dt



Page No 305:

Question 18:

Answer:

Let

Page No 305:

Question 19:

Answer:

Dividing numerator and denominator by ex, we obtain

Let

Page No 305:

Question 20:

Answer:

Let

Page No 305:

Question 21:

Answer:

Let 2x − 3 = t

∴ 2dx = dt

tan22x-3dx = sec22x-3 - 1dx=sec2t- 1dt2= 12sec2t dt - 1dt= 12tant - t + C= 12tan2x-3 - 2x-3 + C

Page No 305:

Question 22:

Answer:

Let 7 − 4x = t

∴ −4dx = dt

Page No 305:

Question 23:

Answer:

Let

Page No 305:

Question 24:

Answer:

Let

Page No 305:

Question 25:

Answer:

Let

Page No 305:

Question 26:

Answer:

Let

Page No 305:

Question 27:

Answer:

Let sin 2x = t

Page No 305:

Question 28:

Answer:

Let

∴ cos x dx = dt

Page No 305:

Question 29:

cot x log sin x

Answer:

Let log sin x = t

Page No 305:

Question 30:

Answer:

Let 1 + cos x = t

∴ −sin x dx = dt

Page No 305:

Question 31:

Answer:

Let 1 + cos x = t

∴ −sin x dx = dt

Page No 305:

Question 32:

Answer:

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

Page No 305:

Question 33:

Answer:

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

Page No 305:

Question 34:

Answer:

Page No 305:

Question 35:

Answer:

Let 1 + log x = t

Page No 305:

Question 36:

Answer:

Let

Page No 305:

Question 37:

Answer:

Let x4 = t

∴ 4x3 dx = dt

Let

From (1), we obtain

Page No 305:

Question 38:

equals

Answer:

Let

Hence, the correct answer is D.

Page No 305:

Question 39:

equals

A.

B.

C.

D.

Answer:

Hence, the correct answer is B.



Page No 307:

Question 1:

Answer:

Page No 307:

Question 2:

Answer:

It is known that,

Page No 307:

Question 3:

cos 2x cos 4x cos 6x

Answer:

It is known that,

Page No 307:

Question 4:

sin3 (2x + 1)

Answer:

Let

Page No 307:

Question 5:

sin3x cos3x

Answer:

Page No 307:

Question 6:

sin x sin 2x sin 3x

Answer:

It is known that,

Page No 307:

Question 7:

sin 4x sin 8x

Answer:

It is known that,

sin A . sin B = 12cosA-B-cosA+Bsin4x sin8x dx=12cos4x-8x-cos4x+8xdx=12cos-4x-cos12xdx=12cos4x-cos12xdx=12sin4x4-sin12x12+C

Page No 307:

Question 8:

Answer:

Page No 307:

Question 9:

Answer:

Page No 307:

Question 10:

sin4x

Answer:

Page No 307:

Question 11:

cos4 2x

Answer:

Page No 307:

Question 12:

Answer:

Page No 307:

Question 13:

Answer:

Page No 307:

Question 14:

Answer:

Page No 307:

Question 15:

Answer:

Page No 307:

Question 16:

tan4x

Answer:

From equation (1), we obtain

Page No 307:

Question 17:

Answer:

Page No 307:

Question 18:

Answer:

Page No 307:

Question 19:

Answer:

1sinxcos3x=sin2x+cos2xsinxcos3x=sinxcos3x+1sinxcosx
1sinxcos3x=tanxsec2x+1cos2xsinxcosxcos2x=tanxsec2x+sec2xtanx

Page No 307:

Question 20:

Answer:

Page No 307:

Question 21:

sin−1 (cos x)

Answer:

It is known that,

Substituting in equation (1), we obtain

Page No 307:

Question 22:

Answer:

Page No 307:

Question 23:

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.

Page No 307:

Question 24:

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.



Page No 315:

Question 1:

Answer:

Let x3 = t

∴ 3x2dx = dt

Page No 315:

Question 2:

Answer:

Let 2x = t

∴ 2dx = dt

Page No 315:

Question 3:

Answer:

Let 2 − x = t

⇒ −dx = dt

Page No 315:

Question 4:

Answer:

Let 5x = t

∴ 5dx = dt

Page No 315:

Question 5:

Answer:

Page No 315:

Question 6:

Answer:

Let x3 = t

∴ 3x2dx = dt

Page No 315:

Question 7:

Answer:

From (1), we obtain

Page No 315:

Question 8:

Answer:

Let x3 = t

⇒ 3x2dx = dt

Page No 315:

Question 9:

Answer:

Let tan x = t

∴ sec2x dx = dt



Page No 316:

Question 10:

Answer:

Page No 316:

Question 11:

19x2+6x+5

Answer:

19x2+6x+5dx=13x+12+22dx
Let (3x+1)=t
3 dx=dt
13x+12+22dx=131t2+22dt
                                 =13×2tan-1t2+C
                                =16tan-13x+12+C
 

Page No 316:

Question 12:

Answer:

Page No 316:

Question 13:

Answer:

Page No 316:

Question 14:

Answer:

Page No 316:

Question 15:

Answer:

Page No 316:

Question 16:

Answer:

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

4A = 4 ⇒ A = 1

A + B = 1 ⇒ B = 0

Let 2x2 + x − 3 = t

∴ (4x + 1) dx = dt

Page No 316:

Question 17:

Answer:

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

From (1), we obtain

From equation (2), we obtain

Page No 316:

Question 18:

Answer:

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

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

Page No 316:

Question 19:

Answer:

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

Page No 316:

Question 20:

Answer:

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

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

Page No 316:

Question 21:

Answer:

Let x2 + 2x +3 = t

⇒ (2x + 2) dx =dt

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

Page No 316:

Question 22:

Answer:

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

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

Page No 316:

Question 23:

Answer:

Equating the coefficients of x and constant term, we obtain

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

Page No 316:

Question 24:

equals

A. x tan−1 (x + 1) + C

B. tan− 1 (x + 1) + C

C. (x + 1) tan−1x + C

D. tan−1 x + C

Answer:

Hence, the correct answer is B.

Page No 316:

Question 25:

equals

A.

B.

C.

D.

Answer:

Hence, the correct answer is B.



Page No 322:

Question 1:

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

Page No 322:

Question 2:

Answer:

Let

Equating the coefficients of x and constant term, we obtain

A + B = 0

−3A + 3B = 1

On solving, we obtain

Page No 322:

Question 3:

Answer:

Let

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

A = 1, B = −5, and C = 4

Page No 322:

Question 4:

Answer:

Let

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

Page No 322:

Question 5:

Answer:

Let

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

A = −2 and B = 4

Page No 322:

Question 6:

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

Page No 322:

Question 7:

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

Page No 322:

Question 8:

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

Page No 322:

Question 9:

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

Page No 322:

Question 10:

Answer:

Let

Equating the coefficients of x2 and x, we obtain

Page No 322:

Question 11:

Answer:

Let

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

Page No 322:

Question 12:

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

Page No 322:

Question 13:

Answer:

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

AB = 0

BC = 0

A + C = 2

On solving these equations, we obtain

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

Page No 322:

Question 14:

Answer:

Equating the coefficient of x and constant term, we obtain

A = 3

2A + B = −1 ⇒ B = −7

Page No 322:

Question 15:

Answer:

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

On solving these equations, we obtain

Page No 322:

Question 16:

[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

Page No 322:

Question 17:

[Hint: Put sin x = t]

Answer:

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

A = 1 and B = −1



Page No 323:

Question 18:



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

Page No 323:

Question 19:

Answer:

Let x2 = t ⇒ 2x dx = dt

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

Page No 323:

Question 20:

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

Page No 323:

Question 21:

[Hint: Put ex = t]

Answer:

Let ex = t exdx = dt

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

A = −1 and B = 1

Page No 323:

Question 22:

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.

Page No 323:

Question 23:

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.



Page No 327:

Question 1:

x sin x

Answer:

Let I =

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

Page No 327:

Question 2:

Answer:

Let I =

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

Page No 327:

Question 3:

Answer:

Let

Taking x2 as first function and ex as second function and integrating by parts, we obtain

Again integrating by parts, we obtain

Page No 327:

Question 4:

x logx

Answer:

Let

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

Page No 327:

Question 5:

x log 2x

Answer:

Let

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

Page No 327:

Question 6:

x2 log x

Answer:

Let

Taking log x as first function and x2 as second function and integrating by parts, we obtain

Page No 327:

Question 7:

Answer:

Let

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

Page No 327:

Question 8:

Answer:

Let

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

Page No 327:

Question 9:

Answer:

Let

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

Page No 327:

Question 10:

Answer:

Let

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

Page No 327:

Question 11:

Answer:

Let

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

Page No 327:

Question 12:

Answer:

Let

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

Page No 327:

Question 13:

Answer:

Let

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

Page No 327:

Question 14:

Answer:

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

I=log x 2xdx-ddxlog x 2xdxdx=x22log x 2-2log x .1x.x22dx=x22log x 2-xlog x dx

Again integrating by parts, we obtain

I= x22logx 2-log x x dx-ddxlog x x dxdx=x22logx 2-x22log x -1x.x22dx=x22logx 2-x22log x +12x dx=x22logx 2-x22log x +x24+C

 

Page No 327:

Question 15:

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



Page No 328:

Question 16:

Answer:

Let

Let

It is known that,

Page No 328:

Question 17:

Answer:

Let

Let

It is known that,

Page No 328:

Question 18:

Answer:

Let

It is known that,

From equation (1), we obtain

Page No 328:

Question 19:

Answer:

Also, let

It is known that,

Page No 328:

Question 20:

Answer:

Let

It is known that,

Page No 328:

Question 21:

Answer:

Let

Integrating by parts, we obtain

Again integrating by parts, we obtain

Page No 328:

Question 22:

Answer:

Let

= 2θ

Integrating by parts, we obtain

Page No 328:

Question 23:

equals

Answer:

Let

Also, let

Hence, the correct answer is A.

Page No 328:

Question 24:

equals

Answer:

Let

Also, let

It is known that,

Hence, the correct answer is B.



Page No 330:

Question 1:

Answer:

 

Page No 330:

Question 2:

Answer:

Page No 330:

Question 3:

Answer:

Page No 330:

Question 4:

Answer:

Page No 330:

Question 5:

Answer:

Page No 330:

Question 6:

Answer:

Page No 330:

Question 7:

Answer:

Page No 330:

Question 8:

Answer:

Page No 330:

Question 9:

Answer:

Page No 330:

Question 10:

is equal to

A.

B.

C.

D.

Answer:

Hence, the correct answer is A.

Page No 330:

Question 11:

is equal to

A.

B.

C.

D.

Answer:

Hence, the correct answer is D.



Page No 334:

Question 1:

Answer:

It is known that,

Page No 334:

Question 2:

Answer:

It is known that,

Page No 334:

Question 3:

Answer:

It is known that,

Page No 334:

Question 4:

Answer:

It is known that,

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

Page No 334:

Question 5:

Answer:

It is known that,

Page No 334:

Question 6:

Answer:

It is known that,



Page No 338:

Question 1:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 2:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 3:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 4:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 5:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 6:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 7:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 8:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 9:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 10:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 11:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 12:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 13:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 14:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 15:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 16:

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

Page No 338:

Question 17:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 18:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 19:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 20:

Answer:

By second fundamental theorem of calculus, we obtain

Page No 338:

Question 21:

equals

A.

B.

C.

D.

Answer:

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

Page No 338:

Question 22:

equals

A.

B.

C.

D.

Answer:

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is C.



Page No 340:

Question 1:

Answer:

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

Page No 340:

Question 2:

Answer:

Also, let

Page No 340:

Question 3:

Answer:

Also, let x = tanθdx = 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

Page No 340:

Question 4:

Answer:

Let x + 2 = t2dx = 2tdt

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

Page No 340:

Question 5:

Answer:

Let cos x = t ⇒ −sinx dx = dt

When x = 0, t = 1 and when

Page No 340:

Question 6:

Answer:

Let dx = dt

Page No 340:

Question 7:

Answer:

Let x + 1 = t dx = dt

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

Page No 340:

Question 8:

Answer:

Let 2x = t ⇒ 2dx = dt

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

Page No 340:

Question 9:

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.

Page No 340:

Question 10:

If

A. cos x + x sin x

B. x sin x

C. x cos x

D. sin x + x cos x

Answer:

Integrating by parts, we obtain

Hence, the correct answer is B.



Page No 347:

Question 1:

Answer:

Adding (1) and (2), we obtain

Page No 347:

Question 2:

Answer:

Adding (1) and (2), we obtain

Page No 347:

Question 3:

Answer:

Adding (1) and (2), we obtain

Page No 347:

Question 4:

Answer:

Adding (1) and (2), we obtain

Page No 347:

Question 5:

Answer:

It can be seen that (x + 2) ≤ 0 on [−5, −2] and (x + 2) ≥ 0 on [−2, 5].

Page No 347:

Question 6:

Answer:

It can be seen that (x − 5) ≤ 0 on [2, 5] and (x − 5) ≥ 0 on [5, 8].

Page No 347:

Question 7:

Answer:

Page No 347:

Question 8:

Answer:

Page No 347:

Question 9:

Answer:

Page No 347:

Question 10:

Answer:

Adding (1) and (2), we obtain

Page No 347:

Question 11:

Answer:

As sin2 (−x) = (sin (−x))2 = (−sin x)2 = sin2x, therefore, sin2x is an even function.

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

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Question 12:

Answer:

Adding (1) and (2), we obtain

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Question 13:

Answer:

As sin7 (−x) = (sin (−x))7 = (−sin x)7 = −sin7x, therefore, sin2x is an odd function.

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

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Question 14:

Answer:

It is known that,

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Question 15:

Answer:

Adding (1) and (2), we obtain

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Question 16:

Answer:

Adding (1) and (2), we obtain

sin (π − x) = sin x

Adding (4) and (5), we obtain

Let 2x = t ⇒ 2dx = dt

When x = 0, t = 0
and when x=π2, t=π
I=120πlog sin tdt-π2log 2
I=I2-π2log 2       [from 3]
I2=-π2log 2
I=-πlog 2

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Question 17:

Answer:

It is known that,

Adding (1) and (2), we obtain

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Question 18:

Answer:

It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4

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Question 19:

Show that if f and g are defined as and

Answer:

Adding (1) and (2), we obtain

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Question 20:

The value of is

A. 0

B. 2

C. π

D. 1

Answer:

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

if f(x) is an odd function, then

Hence, the correct answer is C.

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Question 21:

The value of is

A. 2

B.

C. 0

D.

Answer:

Adding (1) and (2), we obtain

Hence, the correct answer is C.



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Question 1:

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

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Question 2:

Answer:

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Question 3:

[Hint: Put]

Answer:

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Question 4:

Answer:

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Question 5:

Answer:

On dividing, we obtain

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Question 6:

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

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Question 7:

Answer:

Let x a = t dx = dt

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Question 8:

Answer:

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Question 9:

Answer:

Let sin x = t ⇒ cos x dx = dt

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Question 10:

Answer:

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Question 11:

Answer:

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Question 12:

Answer:

Let x4 = t ⇒ 4x3dx = dt

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Question 13:

Answer:

Let ex = texdx = dt

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Question 14:

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

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Question 15:

Answer:

= cos3x × sin x

Let cos x = t ⇒ −sin x dx = dt

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Question 16:

Answer:

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Question 17:

Answer:

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Question 18:

Answer:

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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
     =2ππ2-2cos-1xdx
     =2π.π21.dx-4πcos-1xdx
     =x-4πcos-1xdx           ...(1)
Let I1=cos-1x dx          
Also, let x=tdx=2 t dt
I1=2cos-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
 

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Question 20:

Answer:

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Question 21:

Answer:

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Question 22:

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



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Question 23:

Answer:

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Question 24:

Answer:

Integrating by parts, we obtain

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Question 25:

Answer:

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Question 26:

Answer:

When x = 0, t = 0 and

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Question 27:

Answer:

When and when

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Question 28:

Answer:

When and when

As , therefore, is an even function.

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

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Question 29:

Answer:

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Question 30:

Answer:

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Question 31:

Answer:

From equation (1), we obtain

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Question 32:

Answer:

Adding (1) and (2), we obtain

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Question 33:

Answer:

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

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Question 34:

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.

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Question 35:

Answer:

Integrating by parts, we obtain

Hence, the given result is proved.

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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.

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Question 37:

Answer:

Hence, the given result is proved.

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Question 38:

Answer:

Hence, the given result is proved.

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Question 39:

Answer:

Integrating by parts, we obtain

Let 1 − x2 = t ⇒ −2x dx = dt

Hence, the given result is proved.

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Question 40:

Evaluate as a limit of a sum.

Answer:

It is known that,

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Question 41:

is equal to

A.

B.

C.

D.

Answer:

Hence, the correct answer is A.

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Question 42:

is equal to

A.

B.

C.

D.

Answer:

Hence, the correct answer is B.



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Question 43:

If then is equal to

A.

B.

C.

D.

Answer:

Hence, the correct answer is D.

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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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