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 $f^{\prime }\left(\frac{\pi }{2}\right)$ is

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

Answer 1

The correct option is C $-\frac{\pi }{2}$

$\underset{\pi /4}{\overset{\beta }{\int }}f\left(x\right)dx=\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta$
Differentiating both sides w.r.t. $\beta ,$ we get
$\therefore f\left(\beta \right)=\beta \cos \beta +\sin \beta -\frac{\pi }{4}\sin \beta +\sqrt{2}$
$\Rightarrow f^{\prime }\left(\beta \right)=-\beta \sin \beta +\cos \beta +\cos \beta -\frac{\pi }{4}\cos \beta$
$\Rightarrow f^{\prime }\left(\frac{\pi }{2}\right)=-\frac{\pi }{2}$