JEE Main 2023 · 13 April · evening · Single correct · +4 / −1
The range of f(x)=4 sin ⁻¹(x²/(x²+1)) is
- (a) [0,2 π]
- (b) [0,2 π)
- (c) [0, π)
- (d) [0, π]
Answer and solution
Answer (b) [0,2 π)
Solution
Maths sir · 4 steps
Find the range of the inside fraction first, then push it through sin⁻¹ and multiply by 4.
- Rewrite the inside. Let t = x²/(x²+1). Writing it as t = 1 - 1/(x²+1) makes its behaviour easy to see.
- Range of t. Since x² + 1 ≥ 1, the fraction 1/(x²+1) lies in (0, 1]. It equals 1 at x = 0 but never reaches 0. So t ∈ [0, 1): t = 0 is attained, and t = 1 is not.
- Apply inverse sine. sin⁻¹ is increasing on [0, 1]. So sin⁻¹ t runs from sin⁻¹ 0 = 0 up to, but not including, sin⁻¹ 1 = π/2.
0 ≤ sin⁻¹ t < π/2 ⇒ 0 ≤ 4 sin⁻¹ t < 2π
- So. The range of f is [0, 2π), which is option (b).
The idea to keep. For a composite like k sin⁻¹(g(x)), get the exact range of g(x) first, noting which endpoints are actually reached. A value that is only approached stays an open end all the way through.
Answer from the NTA final answer key. Solution by PrepWiser.