Question

Let $f\left(x\right)={\int }_1^x{\sin }^2\left(\frac{t}{2}\right)dt$, then the value of $\underset{x\to 0}{lim}\frac{f(\pi +x)-f\left(\pi \right)}{x}$ is equal to

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

Answer 1

The correct option is D $1$

Given, $f\left(x\right)={\int }_1^x{\sin }^2\left(\frac{t}{2}\right)dt$
On differentiating both sides, we get
$f^{\prime }\left(x\right)={\sin }^2\left(\frac{x}{2}\right)\left[\frac{d}{dx}\right(x)-\frac{d}{dx}(1\left)\right]$ (By Leibnitz's rule)
$\Rightarrow f^{\prime }\left(x\right)={\sin }^2\frac{x}{2}$
Therefore, $\underset{x\to 0}{lim}\frac{f(\pi +x)-f\left(\pi \right)}{x}=f^{\prime }\left(\pi \right)$
${\sin }^2=\frac{\pi }{2}=1$