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NCERT Solutions for class 12 Maths | Chapter 7 - Integrals

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Free download in PDF Class 12 Maths Chapter 7 Integrals Multiple Choice Questions and Answers for Board, JEE, NEET, AIIMS, JIPMER, IIT-JEE, AIEE and other competitive exams. MCQ Questions for Class 12 Maths with Answers were prepared based on the latest exam pattern. These short solved questions or quizzes are provided by Gkseries. These will help the students for preparation of their examination./p>

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(1) If dx = log |x – 2| + k log f(x) + c, then
[A] f(x) = |x² + 2x + 2|
[B] f(x) = x² + 2x + 2
[C] k = –1/2
[D] All of these
Answer: All of these
(2) ∫ cos-1(1/x)dx equals
[A] x sec-1 x + log |x ++C
[B] x sec-1 x – log |x ++C
[C] -x sec-1 x – log |x ++C
[D] None of these
Answer: x sec-1 x – log |x ++C

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(3) dx is equal to
[A] tan x + cos x + c
[B] tan x + cosec x + c
[C] tan x + cot x + c
[D] tan x+ sec x + c
Answer: tan x + cot x + c
(4) |x|dx =
[A] 0
[B] 2
[C] 1
[D] 4
Answer: 4
(5) Evaluate: ∫(2tan x – 3cot x)2 dx
[A] -4tan x – 9cot x – 25x + C
[B] 4tan x – 9cot x – 25x + C
[C] -4tan x + 9 cot x + 25x + C
[D] 4tan x + 9cot x + 25x + C
Answer: 4tan x – 9cot x – 25x + C
(6)  If ∫ sec²(7 – 4x)dx = a tan (7 – 4x) + C, then value of a is
[A] 7
[B] -4
[C] 3
[D] −1/4
Answer: −1/4
(7) ∫cot²x dx equals to
[A] cot x – x + C
[B] cot x + x + C
[C] -cot x + x + C
[D] -cot x – x + C
Answer: -cot x – x + C
(8) ∫1.dx =
[A] x + k
[B] 1 + k
[C] x2/2 + k
[D] log x + k
Answer: x + k
(9) =
[A] tan x/2 + k
[B] 1/2 tan x/2 + k
[C] 2 tan x/2 + k
[D] tan² x/2 + k
Answer: tan x/2 + k
(10)
[A] 1/3(1+log)³ + c
[B] 1/2(1+log)² + c
[C] log (log 1 + x) + 2
[D] None of these
Answer: 1/3(1+log)³ + c
(11) dx is equal to
[A] log |1 + cos x | + c
[B] log | x + sin x | + c
[C] x – tan + c
[D] x. tan x/2 + c
Answer: x. tan x/2 + c
(12)
[A] tan x/2 + k
[B] 1/2 tan x/2 + k
[C] 2 tan x/2 + k
[D] tan² x/2 + k
Answer: tan x/2 + k
(13)
[A] 1
[B] π2/64
[C] π2192
[D] None of these
Answer: π2192
(14) dx is equal to
[A]
[B]
[C]
[D]
Answer:
(15) What is the value of dx?
[A] 1
[B] π/2
[C] 0
[D] –π/2
Answer: 0
(16) dx =
[A] 1 – log 2
[B] 2
[C] 1 + log 2
[D] log 2
Answer: 1 – log 2
(17) Value of
[A] sin-1 (x – 1) + c
[B] sin-1 (1 + x) + c
[C] sin-1 (1 + x²) + c
[D]
Answer: sin-1 (x – 1) + c
(18) If = a log |1 + x²| + b tan-1 x + 1/5 log |x + 2| + C, then
[A] a = −1/10, b = −2/5
[B] a = 1/10, b = −2/5
[C] a = −1/10, b = 2/5
[D] a = 1/10, b = 2/5
Answer: a = −1/10, b = −2/5
(19) Evaluate: dx
[A] -sin-1 (x−4/5) + C
[B] sin-1 (x+4/5) + C
[C] sin-1 (x−4/5) + C
[D] None of these
Answer: sin-1 (x−4/5) + C
(20)
[A]
[B]
[C]
[D]
Answer:

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