Chapter Exam

Chapter 2: Inverse Trigonometric Functions — Online MCQ Test

Q: What is the range of the function sin⁻¹(x)?

Answer: B. [-π/2, π/2] — The principal range of sin⁻¹(x) is [-π/2, π/2] as defined in NCERT Grade 12.

Q: The domain of cos⁻¹(x) is ______.

Answer: B. [-1, 1] — The domain of cos⁻¹(x) is [-1, 1] because cosine values lie between -1 and 1.

Q: Simplify: sin⁻¹(sin(π/3)).

Answer: A. π/3 — Since π/3 ∈ [-π/2, π/2], sin⁻¹(sin(π/3)) = π/3 directly.

Q: Simplify: cos(tan⁻¹(x)).

Answer: A. 1/√(1+x²) — If tan⁻¹(x) = θ, then tan(θ) = x. Using cos(θ) = 1/√(1+tan²(θ)), we get cos(tan⁻¹(x)) = 1/√(1+x²).

Q: Simplify: tan⁻¹(1/2) + tan⁻¹(1/3).

Answer: A. π/4 — Using tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1-ab)) when ab < 1: tan⁻¹((1/2+1/3)/(1-1/6)) = tan⁻¹(5/6 ÷ 5/6) = tan⁻¹(1) = π/4.

MATHS · Grade 12 · CBSE(NCERT)

Practice Chapter 2: Inverse Trigonometric Functions with a free chapter-wise online MCQ test for CBSE(NCERT) Grade 12 MATHS. This chapter covers: CBSC MATHS Chapter 2: Inverse Trigonometric Functions. AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter 2: Inverse Trigonometric Functions — Important Questions & Answers (FAQ)

Frequently asked questions from CBSE(NCERT) Grade 12 MATHS — Chapter 2: Inverse Trigonometric Functions, with answers and explanations. These are sample questions; the exam has its own separate question set.

What is the range of the function sin⁻¹(x)?

A. [0, π]
B. [-π/2, π/2] ✓
C. (-π/2, π/2)
D. [0, π/2]

Answer: B. [-π/2, π/2]
The principal range of sin⁻¹(x) is [-π/2, π/2] as defined in NCERT Grade 12.

The domain of cos⁻¹(x) is ______.

A. (-∞, ∞)
B. [-1, 1] ✓
C. (0, 1)
D. [0, π]

Answer: B. [-1, 1]
The domain of cos⁻¹(x) is [-1, 1] because cosine values lie between -1 and 1.

Simplify: sin⁻¹(sin(π/3)).

A. π/3 ✓
B. 2π/3
C. π/2
D. -π/3

Answer: A. π/3
Since π/3 ∈ [-π/2, π/2], sin⁻¹(sin(π/3)) = π/3 directly.

Simplify: cos(tan⁻¹(x)).

A. 1/√(1+x²) ✓
B. x/√(1+x²)
C. √(1+x²)
D. (1+x²)

Answer: A. 1/√(1+x²)
If tan⁻¹(x) = θ, then tan(θ) = x. Using cos(θ) = 1/√(1+tan²(θ)), we get cos(tan⁻¹(x)) = 1/√(1+x²).

Simplify: tan⁻¹(1/2) + tan⁻¹(1/3).

A. π/4 ✓
B. π/6
C. π/3
D. π/2

Answer: A. π/4
Using tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1-ab)) when ab < 1: tan⁻¹((1/2+1/3)/(1-1/6)) = tan⁻¹(5/6 ÷ 5/6) = tan⁻¹(1) = π/4.

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