NCERT Solutions for Class 12 Maths Chapter 7 – Integrals

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,

img 620f5949f2096

Therefore, the anti derivative ofimg 620f594a413b7

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,

img 620f594a5a911

Therefore, the anti derivative of img 620f594a9fe6e.

Question 3:

e2x

Answer:

The anti derivative of e2is the function of x whose derivative is e2x.

It is known that,

img 620f594ab9d87

Therefore, the anti derivative of img 620f594ad4a74.

Question 4:

img 620f594aeb1cf

Answer:

The anti derivative of

img 620f594aeb1cfis the function of whose derivative is img 620f594aeb1cf.

It is known that,

img 620f594b0d82b

Therefore, the anti derivative of img 620f594b5224e.

Question 5:

img 620f594b9596d

Answer:

The anti derivative of img 620f594baa644 is the function of x whose derivative is img 620f594baa644.

It is known that,

img 620f594bc1899

Therefore, the anti derivative of img 620f594baa644 is img 620f594c0fe1c.

Question 6:

img 620f594c29786

Answer:

img 620f594c29786

img 620f594c40b3e

Question 7:

img 620f594c5740a

Answer:

img 620f594c5740a

img 620f594c70c11

Question 8:

img 620f594c87358

Answer:

img 620f594c87358

img 620f594ca2f70

Question 9:

img 620f594ce4260

Answer:

img 620f594ce4260

img 620f594d06b45

Question 10:

img 620f594d21251

Answer:

img 620f594d21251

img 620f594d3a31b

Question 11:

img 620f594d7f1c4

Answer:

img 620f594d7f1c4

img 620f594d926cc

Question 12:

img 620f594dd42bb

Answer:

img 620f594dd42bb

img 620f594deba85

Question 13:

img 620f594e434e0

Answer:

img 620f594e434e0

On dividing, we obtain

img 620f594e5c7d9

Question 14:

img 620f594e7361c

Answer:

img 620f594e7361c

img 620f594e8a709

Question 15:

img 620f594ecc338

Answer:

img 620f594ecc338

img 620f594f0ce6d

Question 16:

img 620f594f545c1

Answer:

img 620f594f545c1

img 620f594f69fcc

Question 17:

img 620f594fa8f93

Answer:

img 620f594fa8f93

img 620f594fec292

Question 18:

img 620f59503c0b0

Answer:

img 620f59503c0b0

img 620f595080b96

Question 19:

img 620f5950c2aac

Answer:

img 620f5950db880

img 620f5951010c6

Question 20:

img 620f595145cfe

Answer:

img 620f595145cfe

img 620f59515cd45

Question 21:

The anti derivative of img 620f59519e3c5equals

(A) img 620f5951b7982 (B) img 620f5951cdb67

(C) img 620f5951e779b (D) img 620f59520ca26

img 620f595225eb6

Answer:

img 620f59523fd2d

Hence, the correct answer is C.

Question 22:

If img 620f59527f3dcsuch that f(2) = 0, then f(x) is

(A) img 620f595295e68 (B) img 620f5952b2714

(C) img 620f5952c9fef (D) img 620f5952e052c

Answer:

It is given that,

img 620f59527f3dc

∴Anti derivative of img 620f595306c60

img 620f5953207c5

Also,

img 620f59536735d

Hence, the correct answer is A.

Page No 304:

Question 1:

img 620f5953aa2f2

Answer:

Let img 620f5953c1ef7t

∴2x dx = dt

img 620f5953dfa3e

img 620f59540576d

Question 2:

img 620f59541f072

Answer:

Let log |x| = t

∴ img 620f595436734

img 620f59544a383

Question 3:

img 620f59548e1fd

Answer:

img 620f5954a7931

Let 1 + log t

∴ img 620f5954e7f93

img 620f59550d353

img 620f59554d797

Question 4:

sin x ⋅ sin (cos x)

Answer:

sin x ⋅ sin (cos x)

Let cos x = t

∴ −sin x dx = dt

img 620f595564df3

Question 5:

img 620f5955a6d17

Answer:

img 620f5955bd2cd

Let img 620f5956350c0

∴ 2adx = dt

img 620f59564bd0a

Question 6:

img 620f5956901c3

Answer:

Let ax + b = t

⇒ adx = dt

img 620f5956aa0fe

img 620f5956c15f9

img 620f59570eed5

Question 7:

img 620f59572304f

Answer:

Let img 620f59573662c

∴ dx = dt

img 620f59574f924

Question 8:

img 620f59579338c

Answer:

Let 1 + 2x2 = t

∴ 4xdx = dt

img 620f5957a9964

Question 9:

img 620f5957efc00

Answer:

Let img 620f595815a12

∴ (2x + 1)dx = dt

img 620f59582ed38

img 620f595873986

Question 10:

img 620f5958b506c

Answer:

img 620f5958cc82f

Let img 620f5958e3cc8

img 620f59590ad52

img 620f595921ceb

img 620f595967004

Question 11:

img 620f59597e4b6

Answer:

Let I=∫xx+4 dxput x+4=t⇒dx=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

 

Question 12:

img 620f595999aff

Answer:

Let img 620f5959b0f8f

∴ img 620f5959c82a4

img 620f5959e31ae

Question 13:

img 620f595a65819

Answer:

Let img 620f595a7a30e

∴ 9x2 dx = dt

img 620f595a913f9

Question 14:

img 620f595ad3fc0

Answer:

Let log x = t

∴ img 620f595aed26a

img 620f595b133f7

img 620f595b56346

Question 15:

img 620f595b71537

Answer:

Let img 620f595b8abd5

∴ −8x dx = dt

img 620f595ba14f1

Question 16:

img 620f595be6ac3

Answer:

Let img 620f595c09a21

∴ 2dx = dt

img 620f595c1fdc4

Question 17:

img 620f595c64d6f

Answer:

Let img 620f595c7f1da

∴ 2xdx = dt

img 620f595c96838

img 620f595ca9a05

Page No 305:

Question 18:

img 620f595cec03b

Answer:

Let img 620f595d0be88

∴ img 620f595d25498

img 620f595d3f589

img 620f595d55c1c

Question 19:

img 620f595d7240e

Answer:

img 620f595d7240e

Dividing numerator and denominator by ex, we obtain

img 620f595d889c9

Let img 620f595da182d

∴ img 620f595dbbabb

img 620f595dd5dee

img 620f595e24aed

Question 20:

img 620f595e3f116

Answer:

Let img 620f595e58ae1

∴ img 620f595e6eec2

img 620f595e894cf

img 620f595ea3672

Question 21:

img 620f595ee8311

Answer:

img 620f595ee8311img 620f595f16bc6

Let 2x − 3 = t

∴ 2dx = dt

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

Question 22:

img 620f595f311da

Answer:

Let 7 − 4x = t

∴ −4dx = dt

img 620f595f4a83a

Question 23:

img 620f595f9550e

Answer:

Let img 620f595fa88b7

∴ img 620f595fbf8fd

img 620f595fd6284

img 620f595fed2c9

Question 24:

img 620f59600faf7

Answer:

img 620f5960265e6

Let img 620f596069c22

∴ img 620f596083701

img 620f5960c863b

Question 25:

img 620f59614a11a

Answer:

img 620f5961608a2

Let img 620f5961a0279

∴ img 620f5961b6a09

img 620f5961ccf7e

Question 26:

img 620f59621bc76

Answer:

Let img 620f596232a86

∴ img 620f59624d322

img 620f596266ef2

Question 27:

img 620f5962a8c64

Answer:

Let sin 2x = t

∴ img 620f5962c3506

img 620f5962db6be

Question 28:

img 620f59635a926

Answer:

Let img 620f5963715cb

∴ cos x dx = dt

img 620f59638c97d

Question 29:

cot x log sin x

Answer:

Let log sin x = t

img 620f5963ee94d

img 620f596414aed

Question 30:

img 620f596459294

Answer:

Let 1 + cos x = t

∴ −sin x dx = dt

img 620f596473e74

Question 31:

img 620f5964b2ecf

Answer:

Let 1 + cos x = t

∴ −sin x dx = dt

img 620f5964ccdd5

Question 32:

img 620f59651fc9c

Answer:

img 620f596536cc3

Let sin x + cos x = t ⇒ (cos x − sin xdx = dt

img 620f5965aa97a

Question 33:

img 620f5965ebd96

Answer:

img 620f59660fade

Put cos x − sin x = t ⇒ (−sin x − cos xdx = dt

img 620f5966848dd

Question 34:

img 620f5966c9576

Answer:

img 620f5966e7f70

Question 35:

img 620f59676da55

Answer:

Let 1 + log x = t

∴ img 620f596784d44

img 620f59679f25d

Question 36:

img 620f5967e015c

Answer:

img 620f596802cfc

Let img 620f596871e40

∴ img 620f59688bd20

img 620f5968a68d4

Question 37:

img 620f596924e0b

Answer:

Let x4 = t

∴ 4x3 dx = dt

img 620f59693f560

Let img 620f5969df635

img 620f596a0527e

From (1), we obtain

img 620f596a1b236

img 620f596a77959

Question 38:

img 620f596a8f6a3equals

img 620f596aa5860

Answer:

Let img 620f596b16b26

∴ img 620f596b2d6af

img 620f596b6bf51

img 620f596baddc2

Hence, the correct answer is D.

Question 39:

img 620f596bc6be1equals

A. img 620f596be524d

B. img 620f596c0c178

C. img 620f596c20088

D. img 620f596c36e47

Answer:

img 620f596c50be8

Hence, the correct answer is B.

Page No 307:

Question 1:

img 620f596c98599

Answer:

img 620f596cb1d89

Question 2:

img 620f596d27f63

Answer:

It is known that, img 620f596d41b2d

img 620f596d87815

Question 3:

cos 2x cos 4x cos 6x

Answer:

It is known that,img 620f596e01cd9

img 620f596e426cc

Question 4:

sin3 (2x + 1)

Answer:

Let img 620f596eb0aeb

img 620f596ec795d

img 620f596f42d4e

Question 5:

sin3 x cos3 x

Answer:

img 620f59babff37

img 620f59bb1b4f3

Question 6:

sin x sin 2x sin 3x

Answer:

It is known that, img 620f59bb61f47

img 620f59bbab0b2

Question 7:

sin 4x sin 8x

Answer:

It is known that,

sin A . sin B = 12cosA-B-cosA+B∴∫sin4x sin8x dx=∫12cos4x-8x-cos4x+8xdx=12∫cos-4x-cos12xdx=12∫cos4x-cos12xdx=12sin4x4-sin12x12+C

Question 8:

img 620f59bc318d9

Answer:

img 620f59bc488db

Question 9:

img 620f59bc6c2be

Answer:

img 620f59bc8465c

Question 10:

sin4 x

Answer:

img 620f59bd0969b

img 620f59bd4c8d3

Question 11:

cos4 2x

Answer:

img 620f59bdc3732

Question 12:

img 620f59be42416

Answer:

img 620f59be5cac8

Question 13:

img 620f59becedd1

Answer:

img 620f59beeb0cd

Question 14:

img 620f59bf71a23

Answer:

img 620f59bf89787

Question 15:

img 620f59c00bc39

Answer:

img 620f59c01fecb

Question 16:

tan4x

Answer:

img 620f59c098ed8

From equation (1), we obtain

img 620f59c11ce6b

Question 17:

img 620f59c15ee0d

Answer:

img 620f59c173351

Question 18:

img 620f59c1e7499

Answer:

img 620f59c207594

Question 19:

img 620f59c27c1fc

Answer:

1sinxcos3x=sin2x+cos2xsinxcos3x=sinxcos3x+1sinxcosx

 

⇒1sinxcos3x=tanxsec2x+1cos2xsinxcosxcos2x=tanxsec2x+sec2xtanx

img 620f59c296751

Question 20:

img 620f59c2b3798

Answer:

img 620f59c2cb165

Question 21:

sin−1 (cos x)

Answer:

img 620f59c347ed8

img 620f59c362c73

It is known that,

img 620f59c3d9d4e

Substituting in equation (1), we obtain

img 620f59c429f7b

Question 22:

img 620f59c46cfcd

Answer:

img 620f59c4844dc

Question 23:

img 620f59c50a1be 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:

img 620f59c521991

Hence, the correct answer is A.

Question 24:

img 620f59c59419f equals

A. âˆ’ cot (exx) + C

B. tan (xex) + C

C. tan (ex) + C

D. cot (ex) + C

Answer:

img 620f59c5ad4a9

Let exx = t

img 620f59c5c4eff

Hence, the correct answer is B.

Page No 315:

Question 1:

img 620f59c617f76

Answer:

Let x3 = t

∴ 3x2 dx = dt

img 620f59c62efcc

Question 2:

img 620f59c670ea3

Answer:

Let 2x = t

∴ 2dx = dt

img 620f59c68af1d

Question 3:

img 620f59c7082ae

Answer:

Let 2 − t

⇒ −dx = dt

img 620f59c722a3a

Question 4:

img 620f59c79496c

Answer:

Let 5x = t

∴ 5dx = dt

img 620f59c7abad1

Question 5:

img 620f59c7f261f

Answer:

img 620f59c8185ff

Question 6:

img 620f59c85f358

Answer:

Let x3 = t

∴ 3x2 dx = dt

img 620f59c87b665

Question 7:

img 620f59c8c2045

Answer:

img 620f59c8dbf01

From (1), we obtain

img 620f59c95d9ba

Question 8:

img 620f59c9d28d4

Answer:

Let x3 = t

⇒ 3x2 dx = dt

img 620f59c9e941d

Question 9:

img 620f59ca3ba08

Answer:

Let tan x = t

∴ sec2x dx = dt

img 620f59ca56eb5

Page No 316:

Question 10:

img 620f59caca6e5

Answer:

img 620f59cae1ed1

Question 11:

19×2+6x+5

Answer:

∫19×2+6x+5dx=∫13x+12+22dx

 

Let (3x+1)=t

3 dx=dt

 

⇒∫13x+12+22dx=13∫1t2+22dt

 

=13×2tan-1t2+C

 

=16tan-13x+12+C

 

Question 12:

img 620f59cb6205c

Answer:

img 620f59cb7c755

Question 13:

img 620f59cbf02c8

Answer:

img 620f59cc163b6

Question 14:

img 620f59cc8fb9c

Answer:

img 620f59cca6368

Question 15:

img 620f59cd27f64

Answer:

img 620f59cd41b24

Question 16:

img 620f59cdbbc14

Answer:

img 620f59cdd2984

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

img 620f59ce24d19

Question 17:

img 620f59ce69810

Answer:

img 620f59ce82fd0

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

img 620f59cec8476

From (1), we obtain

img 620f59cede86f

From equation (2), we obtain

img 620f59cf630ec

Question 18:

img 620f59cfa56ea

Answer:

img 620f59cfbba46

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

Related:  NCERT Solutions for Class 12 English Chapter 1 My Mother at Sixty-Six

img 620f59d00b820

img 620f59d0824c6

img 620f59d0c5abc

img 620f59d145ddb

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

img 620f59d1b8ec6

Question 19:

img 620f59d235bd2

Answer:

img 620f59d24b63a

img 620f59d2907bc

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

img 620f59d2cebe2

img 620f59d354720

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

img 620f59d3c89d4

Question 20:

img 620f59d4455d7

Answer:

img 620f59d45eca6

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

img 620f59d4a1540

img 620f59d5213ea

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

img 620f59d59a1ff

Question 21:

img 620f59d613070

Answer:

img 620f59d629c6f

Let x2 + 2x +3 = t

⇒ (2x + 2) dx =dt

img 620f59d69d64f

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

img 620f59d71ecdeimg 620f59d735795

Question 22:

img 620f59d7a2f6a

Answer:

img 620f59d7b67d5

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

img 620f59d804120

img 620f59d87834b

img 620f59d8e8605

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

img 620f59d967423

Question 23:

img 620f59d9d9bdc

Answer:

img 620f59d9f3543

Equating the coefficients of x and constant term, we obtain

img 620f59da41d73

img 620f59dacd225

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

img 620f59db4cb25

Question 24:

img 620f59dbbbb42equals

A. x tan−1 (x + 1) + C

B. tan− 1 (x + 1) + C

C. (x + 1) tan−1 x + C

D. tan−1 x + C

Answer:

img 620f59dbd2258

Hence, the correct answer is B.

Question 25:

img 620f59dc23813equals

A. img 620f59dc3d10b

B. img 620f59dc5728c

C. img 620f59dc6dac3

D. img 620f59dc8672c

Answer:

img 620f59dc9ff58

Hence, the correct answer is B.

Page No 322:

Question 1:

img 620f59dd27a05

Answer:

Let img 620f59dd3f111

img 620f59dd812f1

Equating the coefficients of x and constant term, we obtain

A + = 1

2A + B = 0

On solving, we obtain

A = −1 and B = 2

img 620f59ddc0887

Question 2:

img 620f59de40661

Answer:

Let img 620f59de590e8

img 620f59de9b7f2

Equating the coefficients of x and constant term, we obtain

A + B = 0

−3A + 3B = 1

On solving, we obtain

img 620f59deb43ab

img 620f59decdc77

Question 3:

img 620f59df4db59

Answer:

Let img 620f59df62702

img 620f59dfd3354

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

A = 1, B = −5, and C = 4

img 620f59e04ffcf

Question 4:

img 620f59e0bfa4e

Answer:

Let img 620f59e0dcfb6

img 620f59e1516cf

Substituting x = 1, 2, and 3 respectively in equation (1), we obtain img 620f59e1c108a

img 620f59e20f764

Question 5:

img 620f59e27b224

Answer:

Let img 620f59e2970bd

img 620f59e315da8

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

A = −2 and B = 4

img 620f59e3df2f2

Question 6:

img 620f59e45bceb

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

img 620f59e479b83

Let img 620f59e4cdeca

img 620f59e51dae9

Substituting x = 0 and img 620f59e58d0a8 in equation (1), we obtain

= 2 and B = 3

img 620f59e5a3fc6

Substituting in equation (1), we obtain

img 620f59e5e822b

Question 7:

img 620f59e6664fb

Answer:

Let img 620f59e67d78b

img 620f59e6c1ca3

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

A + C = 0

A + B = 1

B + C = 0

On solving these equations, we obtain

img 620f59e70d03b

From equation (1), we obtain

img 620f59e74eb19

Question 8:

img 620f59e7c5977

Answer:

Let img 620f59e7d9dde

img 620f59e82992a

Substituting x = 1, we obtain

img 620f59e86f03c

Equating the coefficients of x2 and constant term, we obtain

A + C = 0

−2A + 2B + C = 0

On solving, we obtain

img 620f59e88307b

img 620f59e89cad1

Question 9:

img 620f59e91bb0d

Answer:

img 620f59e93352a

Let img 620f59e9773ad

img 620f59e9b8366

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

img 620f59ea331b6

img 620f59ea4d2f4

Question 10:

img 620f59eac152a

Answer:

img 620f59ead7a49

Let img 620f59eb30942

img 620f59eba1807

Equating the coefficients of x2 and x, we obtain

img 620f59ec1fee4

img 620f59ec66eb3

Question 11:

img 620f59ecdb7dc

Answer:

img 620f59ed01486

Let img 620f59ed4559f

img 620f59edb7faf

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

img 620f59ee34051

img 620f59ee77c23

Question 12:

img 620f59eee7b65

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

img 620f59ef0d2da

Let img 620f59ef2054c

img 620f59ef62c12

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

img 620f59efa3945

img 620f59efbb117

Question 13:

img 620f59f039748

Answer:

img 620f59f04fadc

Equating the coefficient of x2x, 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

img 620f59f09486f

Question 14:

img 620f59f1159fc

Answer:

img 620f59f12c10a

Equating the coefficient of x and constant term, we obtain

A = 3

2A + = −1 ⇒ B = −7

img 620f59f16d9bf

Question 15:

img 620f59f1b4b20

Answer:

img 620f59f1cdff1

img 620f59f250037

Equating the coefficient of x3x2, x, and constant term, we obtain

img 620f59f2bd439

On solving these equations, we obtain

img 620f59f2d738e

img 620f59f329d88

Question 16:

img 620f59f39b655 [Hint: multiply numerator and denominator by xn − 1 and put xn = t]

Answer:

img 620f59f39b655

Multiplying numerator and denominator by x− 1, we obtain

img 620f59f3b1785

img 620f59f42e17d

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

A = 1 and B = −1

img 620f59f49daba

img 620f59f4b8705

Question 17:

img 620f59f53b87a [Hint: Put sin x = t]

Answer:

img 620f59f552785

img 620f59f59844d

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

A = 1 and B = −1

img 620f59f615523

img 620f59f656996

Page No 323:

Question 18:

img 620f59f6c906b

Answer:

img 620f59f6e007c

img 620f59f72e00b

Equating the coefficients of x3x2x, 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

img 620f59f7a1294

img 620f59f7e690e

Question 19:

img 620f59f865d3b

Answer:

img 620f59f865d3b

Let x2 = t ⇒ 2x dx = dt

img 620f59f87f09a

img 620f59f8ed756

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

img 620f59f965a30

img 620f59f97c50c

img 620f59f9c2c37

Question 20:

img 620f59fa405e0

Answer:

img 620f59fa405e0

Multiplying numerator and denominator by x3, we obtain

img 620f59fa58ffb

Let x4 = t ⇒ 4x3dx = dt

img 620f59fa98c76

img 620f59fada59c

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

A = −1 and B = 1

img 620f59fb2bc4d

img 620f59fb43b19

Question 21:

img 620f59fb871b0 [Hint: Put ex = t]

Answer:

img 620f59fb871b0

Let ex = ⇒ ex dx = dt

img 620f59fba033c

img 620f59fbde3da

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

A = −1 and B = 1

img 620f59fc5e3d9

img 620f59fc72806

Question 22:

img 620f59fcb791d

A. img 620f59fcd2265

B. img 620f59fceb626

C. img 620f59fd0df9d

D. img 620f59fd24191

Answer:

img 620f59fd3a3a6

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

A = −1 and B = 2

img 620f59fda8870

Hence, the correct answer is B.

Question 23:

img 620f5a3a40f5d

A. img 620f5a3a5b4ad

B. img 620f5a3a9caa6

C. img 620f5a3addc6b

D. img 620f5a3b2ed7a

Answer:

img 620f5a3b70510

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

A + B = 0

C = 0

A = 1

On solving these equations, we obtain

= 1, B = −1, and C = 0

img 620f5a3bb54d1

Hence, the correct answer is A.

Page No 327:

Question 1:

x sin x

Answer:

Let I = img 620f5a3c06af1

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

img 620f5a3c1fbe8

Question 2:

img 620f5a3c61e87

Answer:

Let I = img 620f5a3c768eb

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

img 620f5a3c9023d

Question 3:

img 620f5a3cd2d65

Answer:

Let img 620f5a3cebf40

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

img 620f5a3d0e7d5

Again integrating by parts, we obtain

img 620f5a3d4fe44

Question 4:

x logx

Answer:

Let img 620f5a3d90340

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

img 620f5a3da9e00

Question 5:

x log 2x

Answer:

Let img 620f5a3dec993

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

img 620f5a3e11c58

Question 6:

xlog x

Answer:

Let img 620f5a3e544b0

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

Related:  NCERT Solutions for Class 12 Maths Chapter 12 – Linear Programming

img 620f5a3e6e451

Question 7:

img 620f5a3eb4363

Answer:

Let img 620f5a3eccf81

Taking img 620f5a3ee7641as first function and x as second function and integrating by parts, we obtain

img 620f5a3f0a375

Question 8:

img 620f5a3f7f2bc

Answer:

Let img 620f5a3f95f03

Taking img 620f5a3fb0814 as first function and x as second function and integrating by parts, we obtain

img 620f5a3fc73da

Question 9:

img 620f5a77ac017

Answer:

Let img 620f5a77c64bd

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

img 620f5a77dd021

Question 10:

img 620f5a785f8f9

Answer:

Let img 620f5a7876161

Taking img 620f5a785f8f9 as first function and 1 as second function and integrating by parts, we obtain

img 620f5a788cb27

Question 11:

img 620f5a790e92d

Answer:

Let img 620f5a7924530

img 620f5a793a423

Taking img 620f5a797eb79 as first function and img 620f5a7994f3b as second function and integrating by parts, we obtain

img 620f5a79ab7e3

Question 12:

img 620f5a7a2590f

Answer:

Let img 620f5a7a3f6ae

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

img 620f5a7a5af5f

Question 13:

img 620f5a7a9fa5e

Answer:

Let img 620f5a7ab637d

Taking img 620f5a7a9fa5e as first function and 1 as second function and integrating by parts, we obtain

img 620f5a7acf628

Question 14:

img 620f5a7b1e6d7

Answer:

img 620f5a7b31675

Taking img 620f5a7b4a938 as first function and x as second function and integrating by parts, we obtain

I=log x 2∫xdx-∫ddxlog x 2∫xdxdx=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 +12∫x dx=x22logx 2-x22log x +x24+C

Question 15:

img 620f5a7b64aa5

Answer:

Let img 620f5a7b7e833

Let I = I1 + I2 … (1)

Where, img 620f5a7bc2ecaand img 620f5a7bd70d3

img 620f5a7befcc3

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

img 620f5a7c15959

img 620f5a7bd70d3

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

img 620f5a7c84f3a

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

img 620f5a7cf13a1

Page No 328:

Question 16:

img 620f5a7d40f5d

Answer:

Let img 620f5a7d5a764

Letimg 620f5a7d9c5f8

⇒ img 620f5a7db5a14

∴ img 620f5a7dcc3a8

It is known that, img 620f5a7e19539

img 620f5a7e5e5fa

Question 17:

img 620f5a7e75f0b

Answer:

Let img 620f5a7e8c71c

img 620f5a7ecfdab

Let img 620f5a7f21c2f ⇒ img 620f5a7f388cd

img 620f5a7f4ebec

It is known that, img 620f5a7f9372e

img 620f5a7fd8b1e

Question 18:

img 620f5a8028d62

Answer:

img 620f5a803fb81

Let img 620f5a80bb137 ⇒ img 620f5a80cede7

It is known that, img 620f5a80e85d0

From equation (1), we obtain

img 620f5a8137b02

Question 19:

img 620f5a817b866

Answer:

img 620f5a8195cad

Also, let img 620f5a81db523 ⇒ img 620f5a8203559

It is known that, img 620f5a82198e8

img 620f5a825ea43

Question 20:

img 620f5a82728d2

Answer:

img 620f5a828be72

Let img 620f5a82caa8d ⇒ img 620f5a82e41f3

It is known that, img 620f5a8306919

img 620f5a834aae0

Question 21:

img 620f5a8390152

Answer:

Letimg 620f5a83aa657

Integrating by parts, we obtain

img 620f5a83eea9a

Again integrating by parts, we obtain

img 620f5a843fdfa

Question 22:

img 620f5a84b54e5

Answer:

Letimg 620f5a84ceeb2 ⇒ img 620f5a84e734f

img 620f5a850cc71 = 2θ

⇒ img 620f5a857f1ccimg 620f5a85c53c0

Integrating by parts, we obtain

img 620f5a85df122

Question 23:

img 620f5a862eceb equals

img 620f5a8645681

Answer:

Let img 620f5a8688768

Also, let img 620f5a86a09cf ⇒ img 620f5a86b4a91

img 620f5a86cdfd4

Hence, the correct answer is A.

Question 24:

img 620f5a86e117d equals

img 620f5a8709122

Answer:

img 620f5a86e117d

Let img 620f5a87755b2

Also, let img 620f5a87e5308 ⇒ img 620f5a880aab1

It is known that, img 620f5a88244cf

img 620f5a88712a6

Hence, the correct answer is B.

Page No 330:

Question 1:

img 620f5a888a6a3

Answer:

img 620f5a88a112f

img 620f5a88bcc9fimg 620f5a88d8a4b img 620f5a88f262b

img 620f5a89154d1

 

Question 2:

img 620f5a8932943

Answer:

img 620f5a8949cae

Question 3:

img 620f5a89bf066

Answer:

img 620f5a89d55f3

img 620f5a8a274c7

Question 4:

img 620f5a8a9ac92

Answer:

img 620f5a8ab16f4

img 620f5a8b058f4

Question 5:

img 620f5a8b775ff

Answer:

img 620f5a8b8af45

img 620f5a8bcdeab

Question 6:

img 620f5a8c45bc6

Answer:

img 620f5a8c5f626

img 620f5a8ca66cd

Question 7:

img 620f5a8d1e5ea

Answer:

img 620f5a8d3a70c

img 620f5a8d7a19a

Question 8:

img 620f5a8dea414

Answer:

img 620f5a8e105c3

img 620f5a8e557e7

Question 9:

img 620f5a8ec7ebe

Answer:

img 620f5a8edf412

img 620f5a8f59281

Question 10:

img 620f5a8f9e643is equal to

A. img 620f5a8fb4da3

B. img 620f5a90002bd

C. img 620f5a9016d88

D. img 620f5a902d3c1

Answer:

img 620f5a90723f1

Hence, the correct answer is A.

Question 11:

img 620f5a90e2776is equal to

A. img 620f5a9104cb5

B. img 620f5a9175d78

C. img 620f5a91e70be

D. img 620f5a926627b

Answer:

img 620f5a92d8fab

img 620f5a93288c6

Hence, the correct answer is D.

Page No 334:

Question 1:

img 620f5a939b403

Answer:

It is known that,

img 620f5a93b4b89

img 620f5a9433661

Question 2:

img 620f5a94aabcf

Answer:

img 620f5a94be72e

It is known that,

img 620f5a94d82de

img 620f5a95570ac

Question 3:

img 620f5a95cf60e

Answer:

It is known that,

img 620f5a95e65f4

img 620f5a96612ad

Question 4:

img 620f5adc0522b

Answer:

img 620f5adc1a682

It is known that,

img 620f5adc8c5ae

img 620f5add0ab16

img 620f5add809e9

img 620f5ade03379

img 620f5ade737b8

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

img 620f5adee99c6

Question 5:

img 620f5adf38d31

Answer:

img 620f5adf5342c

It is known that,

img 620f5adf9864c

img 620f5ae0184aa

Question 6:

img 620f5ae08f6f1

Answer:

It is known that,

img 620f5ae0a995c

img 620f5ae12abf4

Page No 338:

Question 1:

img 620f5ae1a338e

Answer:

img 620f5ae1ba0e3

By second fundamental theorem of calculus, we obtain

img 620f5ae20876a

Question 2:

img 620f5ae220f36

Answer:

img 620f5ae23ae35

By second fundamental theorem of calculus, we obtain

img 620f5ae254d99

Question 3:

img 620f5ae29b3b8

Answer:

img 620f5ae2e2a6d

By second fundamental theorem of calculus, we obtain

img 620f5ae360cf6

Question 4:

img 620f5ae3d8513

Answer:

img 620f5ae3f1dab

By second fundamental theorem of calculus, we obtain

img 620f5ae443197

Question 5:

img 620f5ae489912

Answer:

img 620f5ae4a26fd

By second fundamental theorem of calculus, we obtain

img 620f5ae4e4d52

Question 6:

img 620f5ae532e16

Answer:

img 620f5ae546771

By second fundamental theorem of calculus, we obtain

img 620f5ae55f902

Question 7:

img 620f5ae579073

Answer:

img 620f5ae58f855

By second fundamental theorem of calculus, we obtain

img 620f5ae5d3a08

Question 8:

img 620f5ae623617

Answer:

img 620f5ae63b93f

By second fundamental theorem of calculus, we obtain

img 620f5ae67e219

Question 9:

img 620f5ae6c35a0

Answer:

img 620f5ae6dc83b

By second fundamental theorem of calculus, we obtain

img 620f5ae72db87

Question 10:

img 620f5ae7445d0

Answer:

img 620f5ae75dfcb

By second fundamental theorem of calculus, we obtain

img 620f5ae7a416c

Question 11:

img 620f5ae7bb886

Answer:

img 620f5ae7d1ea0

By second fundamental theorem of calculus, we obtain

img 620f5ae81d3d8

Question 12:

img 620f5ae860158

Answer:

img 620f5ae87a300

By second fundamental theorem of calculus, we obtain

img 620f5ae8e90c0

Question 13:

img 620f5ae93adce

Answer:

img 620f5ae954b8f

By second fundamental theorem of calculus, we obtain

img 620f5ae9c7058

Question 14:

img 620f5aea1ac2b

Answer:

img 620f5aea324ac

By second fundamental theorem of calculus, we obtain

img 620f5aeaa979f

Question 15:

img 620f5aeb2bbc4

Answer:

img 620f5aeb50840

By second fundamental theorem of calculus, we obtain

img 620f5aeb97fb5

Question 16:

img 620f5aebaf47a

Answer:

Let img 620f5aebc5d34

img 620f5aebdd0b0

img 620f5aec7f24d

Equating the coefficients of x and constant term, we obtain

A = 10 and B = −25

img 620f5aecc38e0

img 620f5aed126c3

img 620f5aed84066

Substituting the value of I1 in (1), we obtain

img 620f5aee05048

Question 17:

img 620f5aee4792c

Answer:

img 620f5aee62474

By second fundamental theorem of calculus, we obtain

img 620f5aeed1906

Question 18:

img 620f5aef52447

Answer:

img 620f5aef69799

By second fundamental theorem of calculus, we obtain

img 620f5aefaf242

Question 19:

img 620f5aefc69ef

Answer:

img 620f5aefdb0b0

By second fundamental theorem of calculus, we obtain

img 620f5af02e2b6

Question 20:

img 620f5af0a1958

Answer:

img 620f5af0b86fc

By second fundamental theorem of calculus, we obtain

img 620f5af138c50

Question 21:

img 620f5af17d6eaequals

A. img 620f5af19357f

B. img 620f5af1a67bd

C. img 620f5af1ba1b1

D. img 620f5af1d3c1f

Answer:

img 620f5af1edda3

By second fundamental theorem of calculus, we obtain

img 620f5af23be46

Hence, the correct answer is D.

Question 22:

img 620f5af2834e2equals

A. img 620f5af29d591

B. img 620f5af2b3453

C. img 620f5af2cedff

D. img 620f5af2e5b16

Answer:

img 620f5af3063c7

By second fundamental theorem of calculus, we obtain

img 620f5af34a43d

Hence, the correct answer is C.

Page No 340:

Question 1:

img 620f5af3902e3

Answer:

img 620f5af3aa4f9

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

img 620f5af3eb73c

Question 2:

img 620f5af436ee2

Answer:

img 620f5af44b03d

Also, let img 620f5af4bd51a

img 620f5af50849e

Question 3:

img 620f5af57afb5

Answer:

img 620f5af5954b1

Also, let x = tanθ ⇒ dx = sec2θ dθ

When x = 0, θ = 0 and when = 1, img 620f5af5d7dd3

img 620f5af5ef3c8

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

img 620f5af640157

Question 4:

img 620f5af6b9125

Answer:

img 620f5af702e46

Let + 2 = t2 ⇒ dx = 2tdt

When x = 0, img 620f5af71996b and when = 2, = 2

img 620f5af730675

Question 5:

img 620f5af7a7d78

Answer:

img 620f5af7a7d78

Let cos x = t ⇒ −sinx dx = dt

When x = 0, = 1 and whenimg 620f5af7c1004

img 620f5af7dc7d4

Question 6:

img 620f5af82dc47

Answer:

img 620f5af84468e

Let img 620f5af88b3cf⇒ dx = dt

img 620f5af8a538a

Question 7:

img 620f5af92720e

Answer:

img 620f5af940946

Let x + 1 = ⇒ dx = dt

When x = −1, = 0 and when x = 1, = 2

img 620f5af9b0d49

Question 8:

img 620f5afa019db

Answer:

img 620f5afa019db

Let 2x = t ⇒ 2dx = dt

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

img 620f5afa15976

Question 9:

The value of the integral img 620f5b7685bdc is

A. 6

B. 

C. 3

D. 4

Answer:

img 620f5b76a1afe

img 620f5b76e431b

Let cotθ = t ⇒ −cosec2θ dθdt

img 620f5b7763c09

Hence, the correct answer is A.

Question 10:

If img 620f5b77db5d8

A. cos x + x sin x

B. x sin x

C. x cos x

D. sin x cos x

Related:  NCERT Solutions for Class 11 Maths Chapter 6 - Linear Inequalities

Answer:

img 620f5b7828c8a

Integrating by parts, we obtain

img 620f5b783ff00

Hence, the correct answer is B.

Page No 347:

Question 1:

img 620f5b7883af9

Answer:

img 620f5b789dbfb

Adding (1) and (2), we obtain

img 620f5b791adfb

Question 2:

img 620f5b795b8a7

Answer:

img 620f5b797596f

Adding (1) and (2), we obtain

img 620f5b79ebbc7

Question 3:

img 620f5b7a37b92

Answer:

img 620f5b7a4edd5

Adding (1) and (2), we obtain

img 620f5b7ac5290

Question 4:

img 620f5b7b17b1c

Answer:

img 620f5b7b2ef14

Adding (1) and (2), we obtain

img 620f5b7b9f458

Question 5:

img 620f5b7be1b2d

Answer:

img 620f5b7c07794

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

img 620f5b7c1f180

Question 6:

img 620f5b7c8e642

Answer:

img 620f5b7ca7b9d

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

img 620f5b7cc134b

Question 7:

img 620f5b7d3aa36

Answer:

img 620f5b7d5105b

Question 8:

img 620f5b7dc90b2

Answer:

img 620f5b7de319e

Question 9:

img 620f5b7e6bc0c

Answer:

img 620f5b7e82cb8

Question 10:

img 620f5b7f0a477

Answer:

img 620f5b7f4e60e

Adding (1) and (2), we obtain

img 620f5b7fc5950

Question 11:

img 620f5b80129a3

Answer:

img 620f5b8028ef0

As sin(−x) = (sin (−x))2 = (−sin x)2 = sin2x, therefore, sin2is an even function.

It is known that if f(x) is an even function, then img 620f5b803f2cf

img 620f5b807e5f9

Question 12:

img 620f5b80c5ac7

Answer:

img 620f5b80dcc4f

img 620f5b81543d0

Adding (1) and (2), we obtain

img 620f5b81c150c

Question 13:

img 620f5b82107b0

Answer:

img 620f5b822872e

As sin(−x) = (sin (−x))7 = (−sin x)7 = −sin7x, therefore, sin2is an odd function.

It is known that, if f(x) is an odd function, then img 620f5b826ae4c

img 620f5b8281b93

Question 14:

img 620f5b829b211

Answer:

img 620f5b82ae45a

It is known that,

img 620f5b82efca5

img 620f5b8342ab9

img 620f5b8359b94

Question 15:

img 620f5b83cb2ee

Answer:

img 620f5b83e19cb

img 620f5b842ecc5

Adding (1) and (2), we obtain

img 620f5b849c9f2

Question 16:

img 620f5b84e1dfb

Answer:

img 620f5b850d81d

img 620f5b85541c7

Adding (1) and (2), we obtain

img 620f5b85c332b

sin (π − x) = sin x

img 620f5b863fd68

img 620f5b86b1690

Adding (4) and (5), we obtain

img 620f5b872e386

Let 2x = t ⇒ 2dx = dt

When x = 0, = 0 and when

x=π2, t=π∴

I=12∫0πlog sin tdt-π2log 2

⇒I=I2-π2log 2       [from 3]

⇒I2=-π2log 2

⇒I=-πlog 2

Question 17:

img 620f5b879fff0

Answer:

img 620f5b87ba121

It is known that, img 620f5b8807a72

img 620f5b884734d

Adding (1) and (2), we obtain

img 620f5b888ba14

Question 18:

img 620f5b88d1854

Answer:

img 620f5b88eb2ca

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

img 620f5b890a953

Question 19:

Show that img 620f5b898f884if f and g are defined as img 620f5b89d63e9and img 620f5b89ec94e

Answer:

img 620f5b8a0e779

img 620f5b8a5ec8e

Adding (1) and (2), we obtain

img 620f5b8acfbb4

Question 20:

The value of img 620f5b8b4bde3is

A. 

B. 2

C. π

D. 1

Answer:

img 620f5b8b90a42

It is known that if f(x) is an even function, then img 620f5b8c188c0 and

if f(x) is an odd function, then img 620f5b8c5c7d4

img 620f5b8c73183

Hence, the correct answer is C.

Question 21:

The value of img 620f5b8c899a6is

A. 2

B. img 620f5b8ca35aa

C. 0

D. img 620f5b8cb9aef

Answer:

img 620f5b8cd30b6

img 620f5b8d4b792

Adding (1) and (2), we obtain

img 620f5b8dc25e9

Hence, the correct answer is C.

Page No 352:

Question 1:

img 620f5bc57a9c4

Answer:

img 620f5bc59a001

img 620f5bc5e0b05

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

A + B − C = 0

B + = 0

A = 1

On solving these equations, we obtain

img 620f5bc66285a

From equation (1), we obtain

img 620f5bc6ad5ce

Question 2:

img 620f5bc731f3e

Answer:

img 620f5bc74b5f0

Question 3:

img 620f5bc7c4e1f [Hint: Putimg 620f5bc7e2ccf]

Answer:

img 620f5bc809ac0

Question 4:

img 620f5bc882588

Answer:

img 620f5bc89e3da

Question 5:

img 620f5bc933ad5 img 620f5bc953cb6

Answer:

img 620f5bc99b4a5

On dividing, we obtain

img 620f5bc9e4ff9

Question 6:

img 620f5bca67a1d

Answer:

img 620f5bca9309b

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

A + B = 0

C = 5

9A + = 0

On solving these equations, we obtain

img 620f5bcb16dbe

From equation (1), we obtain

img 620f5bcb5e273

Question 7:

img 620f5bcbd4642

Answer:

img 620f5bcbd4642

Let  a ⇒ dx = dt

img 620f5bcbedaf2

Question 8:

img 620f5bcc72b4b

Answer:

img 620f5bcc93096

Question 9:

img 620f5bccdf4a5

Answer:

img 620f5bccdf4a5

Let sin x = t ⇒ cos x dx = dt

img 620f5bcd0a4e1

Question 10:

img 620f5bcd52285

Answer:

img 620f5bcd7094d

Question 11:

img 620f5bcdea733

Answer:

img 620f5bce154d7

Question 12:

img 620f5bce9effe

Answer:

img 620f5bce9effe

Let x= t ⇒ 4x3 dx = dt

img 620f5bcebdcd8

Question 13:

img 620f5bcf1598a

Answer:

img 620f5bcf1598a

Let ex = t ⇒ ex dx = dt

img 620f5bcf3150a

Question 14:

img 620f5bcfab2df

Answer:

img 620f5bcfca012

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

img 620f5bd04f742

From equation (1), we obtain

img 620f5bd0970b6

Question 15:

img 620f5bd119dd1

Answer:

img 620f5bd119dd1= cos3 x × sin x

Let cos x = t ⇒ −sin x dx = dt

img 620f5bd137916

Question 16:

img 620f5bd1829ea

Answer:

img 620f5bd19f342

Question 17:

img 620f5bd2375ad

Answer:

img 620f5bd27c4b1

Question 18:

img 620f5bd300a53

Answer:

img 620f5bd31fa63

Question 19:

img 620f5bd3a7c38

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 img 620f5bd3eb5a1

Question 20:

img 620f5bd419aa1

Answer:

img 620f5bd437e3e

img 620f5bd4aea06

img 620f5bd506a81

Question 21:

img 620f5bd57e1fe

Answer:

img 620f5bd59c0eb

Question 22:

img 620f5bd61ef29

Answer:

img 620f5bd639250

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

img 620f5bd6b3bec

Page No 353:

Question 23:

img 620f5bd736cbc

Answer:

img 620f5bd751f2d

Question 24:

img 620f5bd7c54fe

Answer:

img 620f5bd81c3d0

Integrating by parts, we obtain

img 620f5bd898106

img 620f5bd8dc757

Question 25:

img 620f5bd92e9e4

Answer:

img 620f5bd94baea

img 620f5bd9c38fe

img 620f5bda13cc6

Question 26:

img 620f5bda2f59c

Answer:

img 620f5bda4d9b9

When = 0, = 0 and img 620f5bda970d5

img 620f5bdab8236

Question 27:

img 620f5bdb0cdb4

Answer:

img 620f5bdb2a178

When img 620f5bdba4d57and whenimg 620f5bdbc5b3e

img 620f5bdbe419a

Question 28:

img 620f5bdc39eaa

Answer:

img 620f5bdc55791

Whenimg 620f5bdcd313c and when img 620f5bdcf1756

img 620f5bdd1d278

As img 620f5bdd3c9ba, therefore, img 620f5bdd58be2is an even function.

It is known that if f(x) is an even function, then img 620f5bdd74adb

img 620f5bddba443

Question 29:

img 620f5bddd8d52

Answer:

img 620f5bddf36c2

Question 30:

img 620f5bde77560

Answer:

img 620f5bde9b3f1

Question 31:

img 620f5bdf4aafa

Answer:

img 620f5bdf93ed9

From equation (1), we obtain

img 620f5be01b5cc

Question 32:

img 620f5c1a2c95d

Answer:

img 620f5c1a4cf0e

img 620f5c1ac4e43

Adding (1) and (2), we obtain

img 620f5c1b43c9b

Question 33:

img 620f5c1bb8d59

Answer:

img 620f5c1c10018

img 620f5c1cce555

img 620f5c1dc442e

img 620f5c1e4f7a4

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

img 620f5c1edb852

Question 34:

img 620f5c1f06369

Answer:

img 620f5c1f4f3e2

img 620f5c1f6d7f0

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

img 620f5c1fb5777

img 620f5c203ce9b

Hence, the given result is proved.

Question 35:

img 620f5c2058e7f

Answer:

img 620f5c20767d2

Integrating by parts, we obtain

img 620f5c2094fe6

Hence, the given result is proved.

Question 36:

img 620f5c20de44e

Answer:

img 620f5c2104469

Therefore, f (x) is an odd function.

It is known that if f(x) is an odd function, then img 620f5c21787dc

img 620f5c219398d

Hence, the given result is proved.

Question 37:

img 620f5c21dc7cb

Answer:

img 620f5c220717a

Hence, the given result is proved.

Question 38:

img 620f5c225439c

Answer:

img 620f5c227031f

img 620f5c22e7e97

Hence, the given result is proved.

Question 39:

img 620f5c2346c87

Answer:

img 620f5c23654fd

Integrating by parts, we obtain

img 620f5c2386d32

Let 1 − x2 = t ⇒ −2x dx = dt

img 620f5c23cdf2e

Hence, the given result is proved.

Question 40:

Evaluate img 620f5c2425288as a limit of a sum.

Answer:

img 620f5c244296c

It is known that,

img 620f5c245d9c0

img 620f5c24d477e

img 620f5c25564f3

img 620f5c259a516

img 620f5c25e2d81

img 620f5c26375d6

Question 41:

img 620f5c26aece2is equal to

A. img 620f5c26cac4c

B. img 620f5c26e940e

C. img 620f5c271368e

D. img 620f5c2741db0

Answer:

img 620f5c275f89d

Hence, the correct answer is A.

Question 42:

img 620f5c27ab45fis equal to

A. img 620f5c27ca79f

B. img 620f5c27e9d93

C. img 620f5c2811722

D. img 620f5c282ffc3

Answer:

img 620f5c284beb6

Hence, the correct answer is B.

Page No 354:

Question 43:

If img 620f5c28eafd9then img 620f5c290ccc0is equal to

A. img 620f5c2926657

B. img 620f5c293fdef

C. img 620f5c29598f5

D. img 620f5c297361b

Answer:

img 620f5c298df05

img 620f5c2a087f8

Hence, the correct answer is D.

Question 44:

The value of img 620f5c2a7a386is

A. 1

B. 0

C. − 1

D. img 620f5c2a91458

Answer:

img 620f5c2aab7a2

img 620f5c2b2945b

Adding (1) and (2), we obtain

img 620f5c2b9c2ea

Hence, the correct answer is B.

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