Given that $\frac{d}{d x} \int_0^{\phi(x)} f(t) d t=f(\phi(x)) \phi^{\prime}(x)$. For all $x \in\left(0, \frac{\pi}{2}\right)$, if $\int_1^{\cos x} t^2 f(t) d t=\cos 2 x$, then $f\left(\frac{1}{\sqrt{2}}\right)=$

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext {} \int_1^{\cos x} t^2 f(t) d t-\cos 2 x \ & \Rightarrow(\cos x)^2 f(\cos x)\left(\frac{d}{d x} \cos xight)=-2 \sin 2 x \ & \Rightarrow \cos ^2 x f(\cos x)(-\sin x)=-2 \sin 2 x \ & \Rightarrow f(\cos x)=\frac{-2 \sin 2 x}{-\cos ^2 x \sin x} \ & 	ext { At } x=\frac{\pi}{4} ; f\left(\cos \frac{\pi}{4}ight)=\frac{-2 \sin \frac{\pi}{2}}{-\cos ^2 \frac{\pi}{4} \sin \frac{\pi}{4}} \ & \Rightarrow f\left(\frac{1}{\sqrt{2}}ight)=\frac{2 	imes 1}{\frac{1}{2} \cdot \frac{1}{\sqrt{2}}} \Rightarrow f\left(\frac{1}{\sqrt{2}}ight)=4 \sqrt{2}\end{aligned}$

Given that $\frac{d}{d x} \int_0^{\phi(x)} f(t) d t=f(\phi(x)) \phi^{\prime}(x)$. For all $x \in\left(0,…

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