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 $\mathrm{f}\left(\frac{\pi}{2}\right)$ is

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

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

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

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

JEE Main · 2005 · 08 May 2005 · Offline · Definite Integrals

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{π}{4}

and

x = β > \frac{π}{4}

(β sin β + \frac{π}{4} cos β + 2 β)

. Then

f (\frac{π}{2})

Answer: (D)

Differentiating the area relation

\int_{π /4}^{β} f (x) d x = β sin β + \frac{π}{4} cos β + 2 β

with respect to the upper limit

β

gives

f (β) = sin β + β cos β - \frac{π}{4} sin β + 2

. Setting

β = \frac{π}{2}

, where

cos \frac{π}{2} = 0

and

sin \frac{π}{2} = 1

, we obtain

f (\frac{π}{2}) = 1 - \frac{π}{4} + 2

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option