I saw a question today. $$2f(x)+xf(\frac{1}{x})-2f(|\sqrt2 \sin(\pi(x+\frac{1}{4}))|) = 4 \cos^2 \frac{\pi x}{2}+x\cos\frac{\pi}{x}$$
It had options like this (one or more than one may be correct):
A. $f(2) + f(\frac{1}{2}) =1$
B. $f(1)=-1$, but the values of $f(2), f(1/2)$ cannot be determined
C. $f(2) + f (1) = f(\frac{1}{2})$
D. $f(2) + f (1) = 0$
How to approach these kinds of questions? I tried putting $\frac{1}{x}$ instead of $x$ but it didn't help.
Is there any way to get the answers other than plugging these values?
The correct options are
A, C, D
HINT.-$|\sqrt 2(\sin (\pi x+\frac {\pi}{4}))|=|\sin \pi x+\cos \pi x|$ it follows $$2f(x)+xf(\frac 1x)-2f(|\sin \pi x+\cos \pi x|)=4\cos (\frac{\pi x}{2})^2+x\cos(\frac{\pi}{x})$$ Hence $$2f(2)+2f(\frac 12)-2f(|0+1|)=4(-1)^2+2\cdot 0$$ $$f(2)=-f(\frac 12)+f(1)+2$$ Furthermore $$2f(1) +f(\frac 11)-2f(|0-1|)=4(\cos (\frac{\pi}{2}))^2+\cos \pi=0-1$$ $$3f(1)-2f(1)=f(1)=-1 $$
From this I deduced only option B is correct.