If $f:R\to [0, \infty )$ is a function such that $\text{f(x-1)+f(x+1) = }\sqrt{\text{3}}f\left(x\right)\to 1$, then the period f(x) is

$\text{f(x-1)+f(x+1) = }\sqrt{\text{3}}f\left(x\right)\to 1$

$\begin{array}{l}\text{putting x+2 in x}\\ \text{f(x-1)+f(x+3) = }\sqrt{\text{3}}f(x+2)\to 2\end{array}$

$\text{From 1 and 2 }$

$\text{f(x-13)+2f(x+1)+f(x+3)=}\sqrt{\text{3}}\left(\text{f(x)+f(x+2)}\right)$

$\text{=}\sqrt{3}\left(\sqrt{\text{3}}f(x+1)\right)\Rightarrow \text{= 3 f(x+1)}$

$\begin{array}{l}\text{f(x-1)+f(x+3)=f(x+1)}\to \text{3}\\ \text{Putting x+2 in eq - 3}\\ \text{f(x+1)+f(x+5)= f(x+3)}\to \text{4}\\ \text{Adding 3 and 4 we get f (x-1) = - f(x+5) now put x+1 for x, f(x+6)}\to \text{5}\end{array}$

$\begin{array}{l}\text{putting x+6 in plane of x in 5}\\ \text{f(x+6) = - f(x+12)}\\ \text{From eq 5 again , f(x) = - }\left[\text{-f(x+12)}\right]=\text{f}(x+12)\end{array}$

$\text{Hence the period of f(x) is 12}$