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JEE Main 2016 · Mathematics · Differentiation and Application of Derivatives · hard

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

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If y=sin⁡−1 ⁣(2x1−x2)y=\sin^{-1}\!\left(2x\sqrt{1-x^2}\right) for ∣x∣<12|x|<\tfrac1{\sqrt2}, then dydx\dfrac{dy}{dx} at x=12x=\tfrac12 equals
  1. A. 43\tfrac{4}{\sqrt3}
  2. B. 23\tfrac{2}{\sqrt3}
  3. C. 13\tfrac{1}{\sqrt3}
  4. D. 22
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Answer: (A) 43\tfrac{4}{\sqrt3}

Put x=sin⁡θx=\sin\theta, so y=sin⁡−1(sin⁡2θ)=2θ=2sin⁡−1xy=\sin^{-1}(\sin2\theta)=2\theta=2\sin^{-1}x in the given range. Then y′=21−x2y'=\dfrac{2}{\sqrt{1-x^2}}, which at x=12x=\tfrac12 is 23/2=43\dfrac{2}{\sqrt{3}/2}=\dfrac{4}{\sqrt3}.
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Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake