The range of the function,

$\mathrm{f}(\mathrm{x})=\log _{\sqrt{5}}$

$\left(3+\cos \left(\frac{3 \pi}{4}+x\right)+\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)-\cos \left(\frac{3 \pi}{4}-x\right)\right)$

$f(x)=\log _{\sqrt{5}}\left[3+2 \cos \left(\frac{\pi}{4}\right) \cos (x)-2 \sin \left(\frac{3 \pi}{4}\right) \sin (x)\right]$

$f(x)=\log _{\sqrt{5}}[3+\sqrt{2}(\cos x-\sin x)]$

Since $-\sqrt{2} \leq \cos x-\sin x \leq \sqrt{2}$

$\Rightarrow \log _{\sqrt{5}}\left[3+\sqrt{2}(-\sqrt{2}) \leq \mathrm{f}(\mathrm{x}) \leq \log _{\sqrt{5}}[3+\sqrt{2}(\sqrt{2})]\right]$

$\Rightarrow \log _{\sqrt{5}}(1) \leq \mathrm{f}(\mathrm{x}) \leq \log _{\sqrt{5}}(5)$

So Range of $f(x)$ is $[0,2]$

Option (4)