If \[f: \mathbb{R} \rightarrow \mathbb{R}\] be a continuous function satisfying \[\int_\limits{0}^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x+\alpha \int_\limits{0}^{\frac{\pi}{4}} f(\cos 2 x) \cos x d x=0\], then the value of \[\alpha\] is :
Solutions · 1
Official solution
The integral equation is given by :
\[\int\limits_0^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0\]
Break the first integral into two parts : 1 : Break the first integral into two parts :
\[I = \int\limits_0^{\frac{\pi}{4}} f(\sin 2x) \sin x \, dx + \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx\]
3. Apply the King's property, a∫bf(x)dx=a∫bf(a+b−x)dx, to the first integral and substitute x−4π=t in the second integral.
This gives :
\[
\int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \sin(\frac{\pi}{4}-x) \, dx + \int\limits_0^{\frac{\pi}{4}} f(\cos 2t) \sin(\frac{\pi}{4}+t) \, dt + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0
\]
\[
\int\limits_0^{\frac{\pi}{4}} f(\cos 2x) [2 \sin \frac{\pi}{4} \cos x + \alpha \cos x] \, dx = 0
\]
Then, noticing that 2sin4π=2, you can factor out the term cosx:
\[
= (\alpha + \sqrt{2}) \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0
\]
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 α=−2, which corresponds to Option C.