Question

If a function satisfies $f(x+1)+f(x-1)=\sqrt{2}f\left(x\right)$ and $f\left(x\right)\ne 0$, then period of $f\left(x\right)$ can be

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

$f\left(x\right)=\frac{f(x+1)+f(x-1)}{\sqrt{2}}$

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

$\Rightarrow 2f\left(x\right)=f\left(x+2\right)+2f\left(x\right)+f\left(x-2\right)$

$\Rightarrow 0=f\left(x+2\right)+f\left(x-2\right)$
Hence,

$f\left(x\right)=-f(x+4)$

And

$-f(x+4)=f(x+8)$

$\therefore f\left(x\right)=f(x+8)$
The period of $f\left(x\right)$ is $8$