The range of the function,

$$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$$ is :

$$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$$<br><br>$$f(x) = {\log _{\sqrt 5 }}\left[ {3 + 2\cos \left( {{\pi \over 4}} \right)\cos (x) - 2\sin \left( {{{3\pi } \over 4}} \right)\sin (x)} \right]$$<br><br>$f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$<br><br>Since $- \sqrt 2 \le \cos x - \sin x \le \sqrt 2$<br><br>$$ \Rightarrow {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 \left( { - \sqrt 2 } \right) \le f(x) \le {{\log }_{\sqrt 5 }}\left[ {3 + \sqrt 2 \left( {\sqrt 2 } \right)} \right]} \right]$$<br><br>$\Rightarrow {\log _{\sqrt 5 }}(1) \le f(x) \le {\log _{\sqrt 5 }}(5)$<br><br>So, Range of f(x) is [0, 2]<br><br>Option (d)