JEE Main 2023 — Definite Integration Question with Solution

The integral equation is given by :

$$\underset{0}{\overset{\frac{\pi }{2}}{\int }}f(\sin 2x)\sin xdx+\alpha \underset{0}{\overset{\frac{\pi }{4}}{\int }}f(\cos 2x)\cos xdx=0$$

Step 1 : Break the first integral into two parts :

$$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}f(\sin 2x)\sin xdx+\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{2}}{\int }}f(\sin 2x)\sin xdx+\alpha \underset{0}{\overset{\frac{\pi }{4}}{\int }}f(\cos 2x)\cos xdx$$

3. Apply the King's property, $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$, to the first integral and substitute $x-\frac{\pi }{4}=t$ in the second integral.

This gives :

$$\underset{0}{\overset{\frac{\pi }{4}}{\int }}f(\cos 2x)\sin (\frac{\pi }{4}-x)dx+\underset{0}{\overset{\frac{\pi }{4}}{\int }}f(\cos 2t)\sin (\frac{\pi }{4}+t)dt+\alpha \underset{0}{\overset{\frac{\pi }{4}}{\int }}f(\cos 2x)\cos xdx=0$$

$$\underset{0}{\overset{\frac{\pi }{4}}{\int }}f(\cos 2x)[2\sin \frac{\pi }{4}\cos x+\alpha \cos x]dx=0$$

Then, noticing that $2\sin \frac{\pi }{4}=\sqrt{2}$, you can factor out the term $\cos x$:

$$=(\alpha +\sqrt{2})\underset{0}{\overset{\frac{\pi }{4}}{\int }}f(\cos 2x)\cos xdx=0$$

In order for this equation to hold true, either the integral of the function is zero, or the term outside the integral is zero. Since we have no reason to assume that the integral of the function is zero, we set the term outside the integral to zero, yielding the solution:

$$\alpha +\sqrt{2}=0\Rightarrow \alpha =-\sqrt{2}$$

So, the correct answer to the original problem is $\alpha =-\sqrt{2}$, which corresponds to Option C.