Question

Given the function $f\left(x\right)$ such that $2f\left(x\right)+xf\left(\frac{1}{x}\right)-2f(\mid \sqrt{2}\sin \pi (x+\frac{1}{4})\mid )=4{\cos }^2\frac{\pi x}{2}+x\cos \left(\frac{\pi }{x}\right)$, then which one of the following is correct ?

• A. $f\left(2\right)+f\left(\frac{1}{2}\right)=1$
• B. $f\left(1\right)=-1$, but the values of $f\left(2\right),f\left(\frac{1}{2}\right)$ cannot be determined
• C. $f\left(2\right)+f\left(1\right)=f\left(\frac{1}{2}\right)$
• D. $f\left(2\right)+f\left(1\right)=0$

Answer 1

Answer: (A), (C), (D)

$f\left(2\right)+f\left(1\right)=f\left(\frac{1}{2}\right)$
$f\left(2\right)+f\left(\frac{1}{2}\right)=1$
$f\left(2\right)+f\left(1\right)=0$

Given $2f\left(x\right)+xf\left(\frac{1}{x}\right)-2f(\mid \sqrt{2}\sin \pi (x+\frac{1}{4})\mid )=4{\cos }^2\frac{\pi x}{2}+x\cos \left(\frac{\pi }{x}\right)$ .....(1)

Put $x=1$ in (1)

$2f\left(1\right)+1f\left(1\right)-2f(\mid \sqrt{2}sin\frac{5\pi }{4}\mid )=-1$

$\Rightarrow 3f\left(1\right)-2f\left(1\right)=-1$

$\Rightarrow f\left(1\right)=-1$

Now put $x=2$ in (1)

$2f\left(2\right)+2f\left(\frac{1}{2}\right)-2f\left(1\right)=4{\cos }^2\pi +2\cos \frac{\pi }{2}$

$\Rightarrow 2f\left(2\right)+2f\left(\frac{1}{2}\right)-2f\left(1\right)=4$

$\Rightarrow f\left(2\right)+f\left(\frac{1}{2}\right)=1$ .......... (2)

Now put $x=\frac{1}{2}$ in (1), we get

$2f\left(\frac{1}{2}\right)+\frac{1}{2}f\left(2\right)-2f(\mid \sqrt{2}\sin \left(\frac{3\pi }{4}\right)\mid )=4{\cos }^2\frac{\pi }{4}+\frac{1}{2}\cos 2\pi$

$4f\left(\frac{1}{2}\right)+f\left(2\right)=1$ .......... (3)

From (2) and (3)

$f\left(\frac{1}{2}\right)=0$ & $f\left(2\right)=1$

Hence, $f\left(2\right)+f\left(1\right)=0=f\left(\frac{1}{2}\right)$