Question

If$f\left(x\right)$ is a non - negative continuous function such that the area bounded by the curve $y=f\left(x\right),$ x-axis and the coordinates $x=\frac{\pi }{4}$ and$x=\beta >(\pi /4)$ is $(\beta \sin \beta +\frac{\pi }{4}c\mathrm{os}\beta +\sqrt{2}\beta )$ then $f\left(\frac{\pi }{2}\right)=?$

• A. $\left(1–\frac{\pi }{4}+\sqrt{2}\right)$
• B. $\left(1–\frac{\pi }{4}–\sqrt{2}\right)$
• C. $\left(\frac{\pi }{4}–\sqrt{2}+1\right)$
• D. $\left(\frac{\pi }{4}+\sqrt{2}–1\right)$

Answer 1

The correct option is A

$\left(1–\frac{\pi }{4}+\sqrt{2}\right)$

Explanation for correct answer:

Find the value of $f\left(\frac{\pi }{2}\right)$:

${\int }_{\frac{\mathrm{\pi }}{4}}^{\beta }f\left(x\right)dx=\beta \sin \beta +\frac{\mathrm{\pi }}{4}\cos \beta +\sqrt{2}\beta$

Differentiating both sides with respect to $\beta$

$f\left(\beta \right)=\sin \beta +\beta \cos \beta -\frac{\mathrm{\pi }}{4}\sin \beta +\sqrt{2}$ [By Leibnitz rule]

So,

$$\begin{gathered} f\left(\frac{\mathrm{\pi }}{2}\right)=1+0-\frac{\mathrm{\pi }}{4}+\sqrt{2} \\ =1-\frac{\mathrm{\pi }}{4}+\sqrt{2} \end{gathered}$$

Hence, the correct option is A.