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 $-\sqrt{2}\le \cos x-\sin x\le \sqrt{2}$
$\Rightarrow {\log }_{\sqrt{5}}\left[3+\sqrt{2}(-\sqrt{2})\le f(x)\le {\log }_{\sqrt{5}}[3+\sqrt{2}(\sqrt{2})]\right]$
$\Rightarrow {\log }_{\sqrt{5}}(1)\le f(x)\le {\log }_{\sqrt{5}}(5)$
So Range of $f(x)$ is $[0,2]$