Question

Let $f\left(x\right)$ be a non-negative continuous function such that the area bounded by the curve $y=f\left(x\right)$, $x$-axis and the ordinates $x=\frac{\pi }{4}$ and $x=\beta >\frac{\pi }{4}$ is $(\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta )$. Then, $f\left(\frac{\pi }{2}\right)$ is

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

Answer 1

The correct option is A $(1-\frac{\pi }{4}+\sqrt{2})$

Given, ${\int }_{\pi /4}^{\beta }f\left(x\right)dx=\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta$
On differentiating with respect to $\beta$ on both sides, we get
$f\left(\beta \right)=\sin \beta +\beta \cos \beta -\frac{\pi }{4}\sin \beta +\sqrt{2}$ (by Leibnitz rule)
Put $\beta =\frac{\pi }{2}$
Then, $f\left(\frac{\pi }{2}\right)=\sin \frac{\pi }{2}+\frac{\pi }{2}\cos \frac{\pi }{2}-\frac{\pi }{4}\sin \frac{\pi }{2}+\sqrt{2}$
$=1+0-\frac{\pi }{4}+\sqrt{2}$
$=1-\frac{\pi }{4}+\sqrt{2}$