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NCERT Solutions for class 12 Maths | Chapter 7 - Integrals

(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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