JEE Main 2023 · 31 January · morning · Single correct · +4 / −1
Let y=f(x) represent a parabola with focus (-1/2, 0) and directrix y=-1/2. Then
S={x ∈ ℝ: tan ⁻¹(√f(x))+sin ⁻¹(√(f(x)+1))=π/2} :
- (a) is an empty set
- (b) contains exactly one element
- (c) contains exactly two elements
- (d) is an infinite set
Answer and solution
Answer (c) contains exactly two elements
Solution
Maths sir · 4 steps
Before solving anything, check the domains: the square root and sin⁻¹ squeeze f(x) to exactly 0.
- Find the parabola. Every point on a parabola is equally far from the focus and the directrix. So (x+1/2)² + y² = (y+1/2)².
(x+1/2)² = y + 1/4 ⇒ f(x) = x² + x
- Squeeze from both sides. √f(x) needs f(x) ≥ 0. sin⁻¹(√(f(x)+1)) needs √(f(x)+1) ≤ 1, so f(x) ≤ 0. Together these force f(x) = 0.
- Check it works. At f(x) = 0 the left side is tan⁻¹ 0 + sin⁻¹ 1 = 0 + π/2 = π/2. The equation holds.
- So. x² + x = 0 gives x = 0 or x = -1. So S = {-1, 0} has exactly two elements, option (c).
The idea to keep. When an inverse trig equation has square roots inside, write down every domain condition first. Often they pin the variable down completely and the identity work is not needed at all.
Answer from the NTA final answer key. Solution by PrepWiser.