Question

The range of the function, $f(x)=\log _{\sqrt{5}}\left(3+\cos \left(\frac{3 \pi}{4}+x\right)+\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)-\cos \left(\frac{3 \pi}{4}-x\right)\right)$is :

• A. $(0, \sqrt{5})$
• B. $[-2,2]$
• C. $\left[\frac{1}{\sqrt{5}}, \sqrt{5}\right]$
• D. $[0,2]$

Answer 1

Answer: (D)

$f(x)=\log _{\sqrt{5}}$
$\left(3+\cos \left(\frac{3 \pi}{4}+x\right)+\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)-\cos \left(\frac{3 \pi}{4}-x\right)\right)$
$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 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 $[0,2]$