Question

If $f\left(x\right)$ is continuous function such that the area bounded by the curve $y=f\left(x\right),$ the $x$-axis and the lines $x=a$and $x=0$ is $\frac{\left(a^2\right)}{2}+\frac{a}{2}\sin a+\left(\frac{\pi }{2}\right)\cos a.$

Then the value of $f\left(\frac{\mathrm{\pi }}{2}\right)$ is.

• A. $\frac{1}{2}$
• B. $\frac{a}{2}$
• C. $\frac{a^2}{2}$
• D. $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is A

$\frac{1}{2}$

Explanation for correct option:

Find the value of $f(\pi /2)$:

Given that,

${\int }_0^{a}f\left(x\right)dx=\frac{a^2}{2}+\frac{a}{2}\sin a+\frac{\mathrm{\pi }}{2}\cos a$

Differentiating the above equation with respect to $a$, we get

$f\left(a\right)=a+\frac{1}{2}(\sin a+a\cos a)-\frac{\mathrm{\pi }}{2}\sin a\left[\because \frac{d\left(uv\right)}{dx}=vu\text{'}+v\text{'}u\right]$

Now putting $a=\frac{\mathrm{\pi }}{2}$

$$\begin{gathered} f\left(\frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{\pi }}{2}+\frac{1}{2}-\frac{\mathrm{\pi }}{2} \\ =\frac{1}{2} \end{gathered}$$
Hence, the correct option is A.