Objective Type Question and Answers On Class 12 Maths Chapter 7 Integrals

(1) dx is equal to

[A] 1

[B] 2

[C] 3

[D] 4

Answer: 1

(2) Evaluate: dx

[A] 1/6 tan^-1 (2 tan x) + C

[B] tan^-1 (2 tan x) + C

[C] 1/6 tan^-1(2tanx/3) + C

[D] None of these

Answer: 1/6 tan^-1(2tanx/3) + C

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(3) Evaluate: ∫ sec²(7 – 4x)dx

[A] –1/4 tan(7 – 4x) + C

[B] 1/4 tan(7 – 4x) + C

[C] 1/4 tan(7 + 4x) + C

[D] –1/4 tan(7x – 4) + C

Answer: –1/4 tan(7 – 4x) + C

(4)

[A]

[B]

[C]

[D] None of these

Answer:

(5) Evaluate: dx

[A]

[B]

[C]

[D] None of these

Answer:

(6) ∫ cos(log_e.x)dx is equal to

[A] 1/2 x[cos (log_ex) + sin(log_ex)]

[B] x[cos (log_ex) + sin(log_ex)]

[C] 1/2 x[cos (log_ex) – sin(log_ex)]

[D] x[cos (log_ex) – sin(log_ex)]

Answer: x[cos (log_ex) + sin(log_ex)]

(7) ∫ sin^-1 xdx is equal to

[A] cos^-1 x + C

[B] x sin^-1x ++ C

[C]

[D]

Answer: x sin^-1x ++ C

(8) If

[A] a = 1/3, b = 1

[B] a = −1/3, b = 1

[C] a = −1/3, b = -1

[D] a = 1/3, b = -1

Answer: a = 1/3, b = -1

(9) dx is equal to

[A] 2√2

[B] –1/3 cos x + c

[C] 0

[D] 2(√2 – 1)

Answer: 0

(10) dx is equal to

[A] 10^x – x¹⁰ + C

[B] 10^x + x¹⁰ + C

[C] (10^x – x¹⁰)^-1 + C

[D] log_e(10^x + x¹⁰) + C

Answer: log_e(10^x + x¹⁰) + C

(11)

[A] √x + k

[B] 2√x + k

[C] x + k

[D]

Answer: 2√x + k

(12)

[A]

[B] log | sin x + cos x | + c

[C] log | sin x – cos x | + c

[D]

Answer: log | sin x + cos x | + c

(13) d/dx ∫f(x)dx is equal to

[A] f'(x)

[B] f(x)

[C] f'(x’)

[D] f(x) + c

Answer: f(x)

(14) What is the value of

[A] π/2

[B] π/4

[C] π/8

[D] None of these

Answer: π/4

(15) is equal to

[A] 1/10010

[B] 1/10100

[C] 1/1010

[D] 1/110100

Answer: 1/10100

(16) ∫log₁₀ xdx =

[A] log_e 10.x loge (x/e) + c

[B] log₁₀ e.x loge (x/e) + c

[C] (x – 1) log_e x + c

[D] 1/x + c

Answer: log₁₀ e.x loge (x/e) + c

(17)

[A] tan^-1 (2x) + c

[B] cot^-1 (2x) + c

[C] cos^-1 (2x) + c

[D] sin^-1 (2x) + c

Answer: sin^-1 (2x) + c

(18)

[A]

[B]

[C]

[D]

Answer:

(19)

dx is equal to

[A]

[B]

[C]

[D]

Answer:

(20)

[A] log |1 + cotx/2| + C

[B] log |1 – tanx/2| + C

[C] log |1 – cotx/2| + C

[D] log |1 + tanx/2| + C

Answer: log |1 – cotx/2| + C

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