JEE Advanced 2017 · Paper 2 · Mathematics Q8

MathematicsMultiple correct+4 / -2Question 44 of 54

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

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

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

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

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

Answer and worked solution

Official answer: Bonus: full marks to everyone

Official key: marks to all.

Differentiate (Leibniz rule).
$g'(x)=2\cos2x\,\sin^{-1}(\sin2x)-\cos x\,\sin^{-1}(\sin x).$

At $x=\pm\frac\pi2$ . $\sin2x=0$ , so the first term is 0. $\cos x=0$ , so the second term is 0. Hence $g'(\pm\frac\pi2)=0$ .

None of the four options is true, which is why marks were given to all.

JEE Advanced 2017 · Paper 2, Mathematics question 8