QJEEAdv24Ma2592: JEE Advanced 2024 Mathematics - Class 12 JEEAdv24Ma2592 Solution

Now $$I_1=\underset{0}{\overset{\frac{\pi }{2}}{\int }}f(x)\cdot g(x)dx=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\int }}g(x)dx$$

i.e. $\frac{1}{2} \int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} d x$

Using $\int \sqrt{a^2-x^2}=\frac{1}{2}\left(x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)\right)+C$

$\begin{aligned}& \Rightarrow \frac{1}{2}\left[\frac{\left(x-\frac{\pi}{4}\right)}{2} \sqrt{\frac{\pi x}{2}-x^2}+\frac{\frac{\pi^2}{16}}{2} \sin ^{-1}\left(\frac{x-\frac{\pi}{4}}{\frac{\pi}{4}}\right)\right]_0^{\pi / 2} \\& \Rightarrow \frac{1}{2}\left[\left(0+\frac{\pi^3}{64}\right)-\left(0+\left(\frac{-\pi^3}{64}\right)\right)\right] \\& \Rightarrow \frac{1}{2} \times \frac{\pi^3}{32}\end{aligned}$

Now $\frac{16}{\pi^3} \times \frac{\pi^3}{64}=\frac{1}{4}=0.25$