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फलन $f(x)=\log _{\sqrt{5}}(3+\cos \left(\frac{3 \pi}{4}+x\right)+\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)$
$-\cos \left(\frac{3 \pi}{4}-x\right))$ का परिसर है
$(0, \sqrt{5})$
$[-2,2]$
$\left[\frac{1}{\sqrt{5}}, \sqrt{5}\right]$
$[0,2]$
Solution
$f(x)=\log _{\sqrt{5}}$
$(3+\cos \left(\frac{3 \pi}{4}+x\right)+\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)$
$-\cos \left(\frac{3 \pi}{4}-x\right))$
$f(x)=\log _{\sqrt{5}}\left[3+2 \cos \left(\frac{\pi}{4}\right) \cos (x)-2 \sin \left(\frac{3 \pi}{4}\right) \sin (x)\right]$
$f(x)=\log _{\sqrt{5}}[3+\sqrt{2}(\cos x-\sin x)]$
$\text { Since }-\sqrt{2} \leq \cos x-\sin x \leq \sqrt{2}$
$\Rightarrow \log _{\sqrt{5}}\left[3+\sqrt{2}(-\sqrt{2}) \leq f(x) \leq \log _{\sqrt{5}}[3+\sqrt{2}(\sqrt{2})]\right]$
$\Rightarrow \log _{\sqrt{5}}(1) \leq f(x) \leq \log _{\sqrt{5}}(5)$
So Range of $f(x)$ is $[0,2]$
