If $g(x)=\int_{\sin x}^{\sin (2x)} \sin^{-1}(t) dt$, then

(A) $g'\left(\frac{\pi}{2}\right)=-2\pi$

(B) $g'\left(-\frac{\pi}{2}\right)=2\pi$

(C) $g'\left(\frac{\pi}{2}\right)=2\pi$

(D) $g'\left(-\frac{\pi}{2}\right)=-2\pi$

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Definite Integrals

g (x) = \int_{s i n x}^{s i n (2 x)} sin^{- 1} (t) d t

, then

Using Leibniz rule for differentiation under the integral sign:

g^{'} (x) = sin^{- 1} (sin 2 x) \cdot 2 cos 2 x - sin^{- 1} (sin x) \cdot cos x

x = \frac{π}{2}

g^{'} (\frac{π}{2}) = sin^{- 1} (sin π) \cdot 2 cos π - sin^{- 1} (sin \frac{π}{2}) \cdot cos \frac{π}{2}

= sin^{- 1} (0) \cdot (- 2) - sin^{- 1} (1) \cdot 0 = 0

x = - \frac{π}{2}

g^{'} (- \frac{π}{2}) = sin^{- 1} (sin (- π)) \cdot 2 cos (- π) - sin^{- 1} (sin (- \frac{π}{2})) \cdot cos (- \frac{π}{2})

= sin^{- 1} (0) \cdot (- 2) - sin^{- 1} (- 1) \cdot 0 = 0

Neither value matches any option. This question was awarded bonus marks to all candidates.

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options