Question

Let $f \left(x\right)=$ $\displaystyle \int_0^x \sin^2$ $\left(\frac{t}{2}\right) dt .$ Then the value of $\displaystyle \lim_{x \to 0} $ $\frac{f \left(\pi+x\right)-f \left(\pi\right)}{x} $ is equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. 1
• E. 0

Answer 1

Answer: (D)

Given, $f(x)=\int\limits_{1}^{x} \sin ^{2}\left(\frac{t}{2}\right) d t$
On differentiating both sides, we get
$f^{'}(x)=\sin ^{2}\left(\frac{x}{2}\right)\left[\frac{d}{d x}(x)-\frac{d}{d x}(1)\right]$ (by Leibnitz's rule)
$\Rightarrow f^{'}(x) =\sin ^{2} \frac{x}{2} $
$\therefore \displaystyle\lim _{x \rightarrow 0} \frac{f(\pi+x)-f(\pi)}{x} =f^{'}(\pi) $
$ =\sin ^{2} \frac{\pi}{2}=1 $