MCQ
Let $f (x)$ be a non-negative continuous function such that the area bounded by the curve $y= f (x), X$-axis and the ordinates $x=\frac{\pi}{4}$ and $x=\beta>\frac{\pi}{4}$ is $\left(\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta\right)$. Then $f\left(\frac{\pi}{2}\right)$ is
  • $1-\frac{\pi}{4}+\sqrt{2}$
  • B
    $1-\frac{\pi}{4}-\sqrt{2}$
  • C
    $\frac{\pi}{4}-\sqrt{2}+1$
  • D
    $\frac{\pi}{4}+\sqrt{2}-1$

Answer

Correct option: A.
$1-\frac{\pi}{4}+\sqrt{2}$
(A)
According to the given condition,
$\int_{\frac{\pi}{4}}^\beta f (x) d x=\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta$
Differentiating w.r.t. $\beta$, we get
$f(\beta)=\sin \beta+\beta \cos \beta-\frac{\pi}{4} \sin \beta+\sqrt{2}$
$\therefore f\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}+\frac{\pi}{2} \cos \frac{\pi}{2}-\frac{\pi}{4} \sin \frac{\pi}{2}+\sqrt{2}$
$=1-\frac{\pi}{4}+\sqrt{2}$

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