• question_answer Let $f(x)$ be a non-negative continous function such that the area bounded by the curve $y=f(x)$, x-axis and the ordinates $x=\frac{\pi }{4}$, $x=\beta >\frac{\pi }{4}$ is $\left( \beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta \right)$. Then $f\ \left( \frac{\pi }{2} \right)$ is                                               [AIEEE 2005] 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)            $\left( \frac{\pi }{4}-\sqrt{2}+1 \right)$

Given that, $\int_{\pi /4}^{\beta }{f\ (x)dx}$$=\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta$                    Differentiating w.r.t. b, we get                     $\therefore$ $f(\beta )=\sin \beta +\beta \cos \beta -\frac{\pi }{4}\sin \beta +\sqrt{2}$,                    Hence, $f\ \left( \frac{\pi }{2} \right)=\left( 1-\frac{\pi }{4}+\sqrt{2} \right)$.