Mathematics

 If $$ f \left ( x \right )= \dfrac{\sin x}{x}\:\forall \:x\:\in \left [ 0,\pi  \right ]$$, then $$ \displaystyle \frac{\pi }{2}\int_{0}^{\frac{\pi}{2}}f \left ( x \right )f \left ( \frac{\pi }{2}-x \right )\:dx$$ is equal to


ANSWER

$$\displaystyle \int_{0}^{\pi }f \left ( x \right )\:dx $$


SOLUTION
Let, $$I=\displaystyle \frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}f(x)f\left ( \frac{\pi}{2}-x \right )dx$$
$$=\displaystyle \frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{x}\frac{\cos x}{\dfrac{\pi}{2}-x}dx=\displaystyle\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin2x}{x\left ( \pi -2x \right )}dx$$
$$=\displaystyle\frac{\pi}{2}\int_{0}^{\pi}\frac{\sin z}{z(\pi-z)}dz$$ $$($$Let $$2x=z\Rightarrow 2dx = dz)$$
$$=\displaystyle\frac{1}{2}\int_{0}^{\pi}\sin z\left ( \frac{1}{z}+\frac{1}{\pi-z} \right )dz$$
$$=\displaystyle\frac{1}{2}\left(\int_{0}^{\pi}\frac{\sin z}{z}dz+\int_{0}^{\pi}\frac{\sin z}{\pi-z}dz\right)=\displaystyle\int_{0}^{\pi}\frac{\sin z}{z}dz$$

$$\because\displaystyle\int_{0}^{a}f(x)=\int_{0}^{a}f(a-x)dx$$

$$\therefore\dfrac{\pi}2\displaystyle\int_0^{\frac{\pi}2}f(x)f\left(\dfrac{\pi}2 -x\right)\>dx =\displaystyle\int_{0}^{\pi}f(x)dx$$

Hence, option C.
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Single Correct Hard Published on 17th 09, 2020
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