Class 12 Maths Chapter 2 Inverse Trigonometric Functions MCQs are available here with the correct answers and solutions. MCQs of Class 12 Maths Chapter 2 covers all the concepts of the NCERT curriculum, such as the restrictions on domains and ranges of trigonometric functions, which ensure the existence of their inverses and observe their behaviour through graphical representations. These MCQs will help Class 12 students to prepare for the board exam 2022-2023. All these MCQs are given here based on the latest CBSE guidelines for Class 12 students.
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MCQs for Chapter 2 Inverse Trigonometric Functions
The expert teachers solved all the MCQs of chapter 2, Inverse trigonometric functions of Class 12 Maths. Each of these multiple-choice questions contains a correct answer and an explanation. Students must solve these multiple-choice questions to understand how to apply all the concepts and formulas of Class 12 Maths.
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Practising MCQs of Chapter 2 Class 12 Maths will help boost your confidence in recognising the methods of solving any problem that appears in the CBSE 2022-2023 board exam.
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MCQs for Chapter 2 Inverse Trigonometric Functions with Answers
1. The principal value of tan-1(tan 3π/5) is
(a) 2π/5
(b) -2π/5
(c) 3π/5
(d) -3π/5
Correct option: (b) -2π/5
Solution:
tan-1 (tan 3π/5)
This can be written as:
tan-1 (tan 3π/5) = tan-1 (tan[π – 2π/5])
= tan-1 (- tan 2π/5) {since tan(π – x) = -tan x}
= –tan-1 (tan 2π/2)
= –2π/5
2. sin[π/3 – sin-1(-½)] is equal to:
(a) ½
(b) ⅓
(c) -1
(d) 1
Correct option: (d) 1
Solution:
sin[π/3 – sin-1(-½)]
= sin[π/3 – sin-1[sin (-π/6))]
sin[π/3 – (-π/6)]
= sin(π/3 + π/6)
= sin π/2
= 1
3. The domain of sin–1(2x) is
(a) [0, 1]
(b) [– 1, 1]
(c) [-1/2, 1/2]
(d) [–2, 2]
Correct option: (c) [-1/2, 1/2]
Solution:
Let sin–1(2x) = θ.
Thus, 2x = sin θ.
As we know, – 1 ≤ sin θ ≤ 1
We can write this as– 1 ≤ 2x ≤ 1, which gives -1/2 ≤ x ≤ 1/2.
Therefore, the domain of sin-1(2x) is [-½, ½].
4. If sin–1x + sin–1y = π/2, then value of cos–1x + cos–1y is
(a) π/2
(b) π
(c) 0
(d) 2π/3
Correct option: (a) π/2
Solution:
Given,
sin–1 x + sin–1 y = π/2
[(π/2) – cos-1x] + [(π/2) – cos-1y] = π/2(π/2) + (π/2) – (π/2) = cos-1x + cos-1y
Therefore, cos–1x + cos–1y = π/2.
5. Which of the following is the principal value branch of cos–1x?
(a) [–π/2, π/2]
(b) (0, π)
(c) [0, π]
(d) (0, π) – {π/2}
Correct option: (c) [0, π]
Solution:
The principal value branch of cos–1x is [0, π].
6. The value of the expression sin [cot–1 (cos (tan–1 1))] is
(a) 0
(b) 1
(c) 1/√3
(d) √(2/3)
Correct option: (d) √(2/3)
Solution:
sin [cot–1 (cos (tan–1 1))]
= sin[cot-1 {cos (tan-1 (tan π/4))}] {since tan π/4 = 1}
= sin[cot-1 (cos π/4)]
= sin[cot-1(1/√2)]
= sin [sin-1(√(⅔))] {by Pythagoras theorem}
= √(⅔)
7. The domain of y = cos–1 (x2 – 4) is
(a) [3, 5]
(b) [0, π]
(c) [-√5, -√3] ∩ [-√5, √3]
(d) [-√5, -√3] ∪ [√3, √5]
Correct option: (d)
Solution:
Given,
y = cos–1 (x2 – 4 )
⇒ cos y = x2 – 4
As we know, –1 ≤ cos y ≤ 1
So, – 1 ≤ x2 – 4 ≤ 1
Adding 4 on both sides, we get;
⇒ 3 ≤ x2 ≤ 5
Taking square root on both sides, we get;
⇒ √3 ≤ x ≤ √5
⇒ x∈ [-√5, -√3] ∪ [√3, √5]
8. If α ≤ 2 sin–1x + cos–1x ≤ β, then
(a) α = -π/2, β = π/2
(b) α = 0, β = π
(c) α = -π/2, β = 3π/2
(d) α = 0, β = 2π
Correct option: (b) α = 0, β = π
Solution:
Given,
α ≤ 2 sin–1x + cos–1x ≤ β
We know that,
-π/2 ≤ sin–1 x ≤ π/2
⇒ (-π/2) + (π/2) ≤ sin–1x + (π/2) ≤ (π/2) + (π/2)
⇒ 0 ≤ sin–1x + (sin–1x + cos–1x) ≤ π
⇒ 0 ≤ 2 sin–1x + cos–1x ≤ π
By comparing with α ≤ 2 sin–1x + cos–1x ≤ β, we get α = 0, β = π.
9. The value of sin (2 tan–1 (.75)) is equal to
(a) .75
(b) 1.5
(c) .96
(d) sin 1.5
Correct option: (c) .96
Solution:
sin (2tan–1 (.75))
Let, tan–1 (.75) = θ
tan θ = 0.75
tan θ = 3/4
Thus by Pythagoras theorem, we get;
sin θ = 3/5 and cos θ = 4/5.
Now,
sin (2tan–1 (.75)) = sin 2θ {as tan-1(.75) = θ}
= 2 sin θ cos θ
= 2 × (3/5) × (4/5)
= 24/25
= 0.96
Therefore, sin (2tan–1 (.75)) = .96.
10. sin(tan-1 x), where |x| < 1, is equal to:
(a) x/√(1 – x2)
(b) 1/√(1 – x2)
(c) 1/√(1 + x2)
(d) x/√(1 + x2)
Correct option: (d) x/√(1 + x2)
Solution:
Let tan-1x = θ.
So, tan θ = x = x/1
From this, we can write the sin θ and cos θ values as:
sin θ = x/√(1 + x2)
cos θ = 1/√(1 + x2)
Now,
sin(tan-1 x) = sin θ = x/√(1 + x2).
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