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 $x=\frac{\pi }{4}$, $x=\beta >\frac{\pi }{4}$ is $(\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta -\frac{\pi }{2})$. 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 B $(1-\frac{\pi }{4}+\sqrt{2})$

Given that,

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

Differentiating w.r.t x and we get

$f\left(\beta \right)=\beta \cos \beta +\sin \beta -\frac{\pi }{4}\sin \beta +\sqrt{2}-0$

$f\left(\frac{\pi }{2}\right)=(1-\frac{\pi }{4})\sin \frac{\pi }{2}+\sqrt{2}$

$=(1-\frac{\pi }{4})\times 1+\sqrt{2}$

$=1-\frac{\pi }{4}+\sqrt{2}$

Hence, this is the answer.