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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 $x = π / 4 and x = β > π / 4,$ is $(βsin β + \frac{π}{4} \cos β + \sqrt{2} β)$ then, $f (\frac{π}{2})$ is equal to

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$(\frac{π}{4} + \sqrt{2} - 1)$

$(\frac{π}{4} - \sqrt{2} + 1)$

$(1 - \frac{π}{4} - \sqrt{2})$

$(1 - \frac{π}{4} + \sqrt{2})$

answer is D.

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Detailed Solution

We have,

$\int_{π / 4}^{β} f (x) dx = βsin β + \frac{π}{4} \cos β + \sqrt{2} β$

On differentiating both sides w.r.t. $β$ (by Leibnitz rule), we get

$\begin{array}{l} f (β) = βcos β + \sin β - \frac{π}{4} \sin β + \sqrt{2} \\ f (\frac{π}{2}) = 1 - \frac{π}{4} + \sqrt{2} \end{array}$

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