TIFR-2015-Maths-A-13

Question

For a real number $t >0$, let $\sqrt{t}$ denote the positive square root of $t$. For a real number $x > 0$, let $F(x)= \int_{x^{2}}^{4x^{2}} \sin \sqrt{t}$  $dt$. If $F'$ is the derivative of $F$, then

• A. $F'(\frac{\pi}{2}) = 0$
• B. $F'(\frac{\pi}{2}) = \pi$
• C. $F'(\frac{\pi}{2}) = - \pi$
• D. $F'(\frac{\pi}{2}) = 2\pi$

Answer 1

Ans is c...

Answer 2

By Leibniz rule,

$F’(x)= \sin \sqrt{4x^2} \frac{d}{dx}(4x^2)\; – \; \sin \sqrt{x^2} \frac{d}{dx}(x^2)$

$F’(x) = 8x\sin 2x\; – 2x \sin x$

$F’(\frac{\pi}{2}) = -\pi$