Question

The range of ${\log }_{\sqrt{5}}\left[\sqrt{2}\right(\sin x-\cos x)+3]$ is

• A. $[0,2]$
• B. $[1,2]$
• C. $[0,3]$
• D. None

Answer 1

Answer: (A)

$[0,2]$

Given,

${\log }_{\sqrt{5}}\left[\sqrt{2}\right(\sin x-\cos x)+3]$

$\begin{array}{cc}& \because \sqrt{2}\frac{1}{\sqrt{2}}(\sin x-\cos x)\\ =& \sqrt{2}[\sin x\cos \frac{\pi }{4}-\cos x\sin \frac{\pi }{4}]\\ =& \sqrt{2}\left[\sin (x-\frac{\pi }{4})\right]\\ & \because \sin x\in [-1,1]\\ & \sqrt{2}\sin (x-\frac{\pi }{2})\in [-\sqrt{2},\sqrt{2}]\end{array}$

$\text{Now}{\log }_{\sqrt{5}}(\sqrt{2}(\sqrt{2}\sin (x-\frac{\pi }{4})+3$

$\Rightarrow {\log }_{\sqrt{5}}(-2,2]+3$

$\therefore {\log }_{\sqrt{5}}^{-2+3}\le {\log }_{\sqrt{5}}\left[\sqrt{2}\right(\sin \alpha -\cos x)+3]⩽{\log }_{\sqrt{5}}5$

$\Rightarrow 0⩽{\log }_{\sqrt{5}}\sqrt{2}(\sin x-\cos x)+3]⩽2$

Therefore Option A