વિધેય mathrm{f}(mathrm{x})=log _{sqrt{5}}(3+c | vedclass

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Question

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

Vedclass · Accepted Answer

$[0,2]$

Vedclass · Answer

$f(x)=\log _{\sqrt{5}}$

$(3+\cos \left(\frac{3 \pi}{4}+xight)+\cos \left(\frac{\pi}{4}+xight)+\cos \left(\frac{\pi}{4}-xight)$

$-\cos \left(\frac{3 \pi}{4}-xight))$

$f(x)=\log _{\sqrt{5}}\left[3+2 \cos \left(\frac{\pi}{4}ight) \cos (x)-2 \sin \left(\frac{3 \pi}{4}ight) \sin (x)ight]$

$f(x)=\log _{\sqrt{5}}[3+\sqrt{2}(\cos x-\sin x)]$

$	ext { Since }-\sqrt{2} \leq \cos x-\sin x \leq \sqrt{2}$

$\Rightarrow \log _{\sqrt{5}}\left[3+\sqrt{2}(-\sqrt{2}) \leq f(x) \leq \log _{\sqrt{5}}[3+\sqrt{2}(\sqrt{2})]ight]$

$\Rightarrow \log _{\sqrt{5}}(1) \leq f(x) \leq \log _{\sqrt{5}}(5)$

So Range of $f(x)$ is $[0,2]$

વિધેય $\mathrm{f}(\mathrm{x})=\log _{\sqrt{5}}(3+\cos \left(\frac{3 \pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}-\mathrm{x}\right)$

$-\cos \left(\frac{3 \pi}{4}-\mathrm{x}\right))$ નો વિસ્તાર મેળવો.

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વિધેય $\mathrm{f}(\mathrm{x})=\log _{\sqrt{5}}(3+\cos \left(\frac{3 \pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}-\mathrm{x}\right)$ $-\cos \left(\frac{3 \pi}{4}-\mathrm{x}\right))$ નો વિસ્તાર મેળવો.