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

Answer 1

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

From the given condition $\underset{\frac{\pi }{4}}{\overset{\beta }{\int }}f\left(x\right)dx=\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta$
Differentiating with respect to $\beta ,$ we get
$f\left(\beta \right)=\beta \cos \beta +\sin \beta -\frac{\pi }{4}\sin \beta +\sqrt{2}\text{(Using Leibnitz Theorem)}$
$f\left(\frac{\pi }{2}\right)=\beta \cdot 0+(1-\frac{\pi }{4})\sin \frac{\pi }{2}+\sqrt{2}$
$\Rightarrow f\left(\frac{\pi }{2}\right)=1-\frac{\pi }{4}+\sqrt{2}$