JEE MAIN - Mathematics (2005 - No. 23)
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 $$x = {\pi \over 4}$$ and $$x = \beta > {\pi \over 4}$$ is $$\left( {\beta \sin \beta + {\pi \over 4}\cos \beta + \sqrt 2 \beta } \right).$$ Then $$f\left( {{\pi \over 2}} \right)$$ is
$$\left( {{\pi \over 4} + \sqrt 2 - 1} \right)$$
$$\left( {{\pi \over 4} - \sqrt 2 + 1} \right)$$
$$\left( {1 - {\pi \over 4} - \sqrt 2 } \right)$$
$$\left( {1 - {\pi \over 4} + \sqrt 2 } \right)$$
Explanation
Given that
$$\int\limits_{\pi /4}^\beta {f\left( x \right)} dx = \beta \sin \beta + {\pi \over 4}\cos \,\beta + \sqrt 2 \beta $$
Differentiating $$w.r.t$$ $$\beta $$
$$f\left( \beta \right) = \beta \cos \beta + \sin \beta - {\pi \over 4}\sin \beta + \sqrt 2 $$
$$f\left( {{\pi \over 2}} \right) = \left( {1 - {\pi \over 4}} \right)\sin {\pi \over 2} + \sqrt 2 $$
$$ = 1 - {\pi \over 4} + \sqrt 2 $$
$$\int\limits_{\pi /4}^\beta {f\left( x \right)} dx = \beta \sin \beta + {\pi \over 4}\cos \,\beta + \sqrt 2 \beta $$
Differentiating $$w.r.t$$ $$\beta $$
$$f\left( \beta \right) = \beta \cos \beta + \sin \beta - {\pi \over 4}\sin \beta + \sqrt 2 $$
$$f\left( {{\pi \over 2}} \right) = \left( {1 - {\pi \over 4}} \right)\sin {\pi \over 2} + \sqrt 2 $$
$$ = 1 - {\pi \over 4} + \sqrt 2 $$
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