Chapter 2: Inverse Trigonometric Functions — Online MCQ Test
MATHS · Grade 12 · CBSE(NCERT)
Practice Chapter 2: Inverse Trigonometric Functions with a free chapter-wise online MCQ test.
This chapter covers: CBSC MATHS Chapter 2: Inverse Trigonometric Functions.
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Chapter 2: Inverse Trigonometric Functions — Important Questions & Answers
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 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.
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.
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).
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.
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.