JEE Main 2021 · 26 August · morning · Single correct · +4 / −1
Let f(x) = cos ( 2tan ⁻¹sin ( cot ⁻¹√((1 - x)/x) ) ), 0 < x < 1. Then :
- (a) (1 - x)²f'(x) - 2(f(x))² = 0
- (b) (1 + x)²f'(x) + 2(f(x))² = 0
- (c) (1 - x)²f'(x) + 2(f(x))² = 0
- (d) (1 + x)²f'(x) - 2(f(x))² = 0
Answer and solution
Answer (c) (1 - x)²f'(x) + 2(f(x))² = 0
Solution
Maths sir · 4 steps
Peel the inverse functions one layer at a time and f(x) collapses to (1-x)/(1+x).
- Innermost layer. Let θ = cot⁻¹√((1-x)/x). For 0 < x < 1, θ lies in (0, π/2), so sin θ = 1/√(1+cot² θ) = 1/√(1 + (1-x)/x) = √x.
- Outer layer. Now let φ = tan⁻¹√x, so tan² φ = x. Use the identity cos 2φ = (1-tan² φ)/(1+tan² φ).
f(x) = (1-x)/(1+x), f'(x) = (-(1+x)-(1-x))/(1+x)² = -2/(1+x)²
- Bring in f squared. Multiply f'(x) by (1-x)²: (1-x)² f'(x) = -2(1-x)²/(1+x)² = -2(f(x))².
- So. (1-x)² f'(x) + 2(f(x))² = 0, which is option (c).
The idea to keep. For nested inverse trig expressions, name each inner angle, turn it into a right-triangle ratio, and use an identity like cos 2φ = (1-tan² φ)/(1+tan² φ) to reach an algebraic function.
Answer from the NTA final answer key. Solution by PrepWiser.