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NDA 2020 Maths · Mathematics · Derivatives · hard

If y=sin ^-1 (2x√(1-x^2) for |x|<(1)/(√2), then dy/dx equals

Single correct option

If y=sin⁡−1 ⁣(2x1−x2)y=\sin^{-1}\!\left(2x\sqrt{1-x^2}\right) for ∣x∣<12|x|<\dfrac1{\sqrt2}, then dydx\dfrac{dy}{dx} equals
  1. A. 11−x2\dfrac{1}{\sqrt{1-x^2}}
  2. B. 21−x2\dfrac{2}{\sqrt{1-x^2}}
  3. C. 2x1−x2\dfrac{2x}{\sqrt{1-x^2}}
  4. D. 21−x22\sqrt{1-x^2}
Show answer and solution

Answer: (B) 21−x2\dfrac{2}{\sqrt{1-x^2}}

Put x=sin⁡θx=\sin\theta with ∣θ∣<π4|\theta|<\dfrac\pi4. Then 2x1−x2=sin⁡2θ2x\sqrt{1-x^2}=\sin2\theta and y=sin⁡−1(sin⁡2θ)=2θ=2sin⁡−1xy=\sin^{-1}(\sin2\theta)=2\theta=2\sin^{-1}x because 2θ∈(−π2,π2)2\theta\in\left(-\dfrac\pi2,\dfrac\pi2\right). Therefore dydx=21−x2\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-x^2}}. The option with 11 in the numerator ignores the doubling, and the domain restriction is what makes y=2sin⁡−1xy=2\sin^{-1}x valid; outside it the result changes.
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Original question written in the NDA 2020 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake