Let f(x) be a non-negative continous function such that the area … - Vidyadip

MCQ

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

A
$\left( {1 - \frac{\pi }{4} - \sqrt 2 } \right)$
✓
$\left( {1 - \frac{\pi }{4} + \sqrt 2 } \right)$
C
$\left( {\frac{\pi }{4} + \sqrt 2 - 1} \right)$
D
$\left( {\frac{\pi }{4} - \sqrt 2 + 1} \right)$

✓

Answer

Correct option: B.

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

b
(b) Given that, $\int_{\pi /4}^\beta {f\;(x)dx} $

$ = \beta \sin \beta + \frac{\pi }{4}\cos \beta + \sqrt 2 \beta $

Differentiating w.r.t. $\beta$, we get

$\therefore $ $f(\beta ) = \sin \beta + \beta \cos \beta - \frac{\pi }{4}\sin \beta + \sqrt 2 $,

Hence, $f\;\left( {\frac{\pi }{2}} \right) = \left( {1 - \frac{\pi }{4} + \sqrt 2 } \right)$.

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