Question

Let $f\left(x\right)=$ ${\int }_0^x{\sin }^2$ $\left(\frac{t}{2}\right)dt.$ Then the value of $\underset{x\to 0}{\lim }$ $\frac{f\left(\pi +x\right)-f\left(\pi \right)}{x}$ is equal to

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

Answer 1

Answer: (D)

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