Question

The range of the function
$f\left(x\right)={\log }_{\sqrt{2}}\left(2-{\log }_2\left(16{\sin }^{2}x+1\right)\right)$ is

• A. $\left(-\infty ,1\right)$
• B. $\left(-\infty ,2\right)$
• C. $(-\infty ,1]$
• D. $(-\infty ,2]$

Answer 1

The correct option is C $(-\infty ,1]$

Given:$f\left(x\right)={\log }_2(2-{\log }_{\sqrt{2}}(16{\sin }^{2}x+1))$

f is defined for

$\left(i\right)2-{\log }_{\sqrt{2}}(16{\sin }^{2}x+1)>0$

$\left(ii\right)16{\sin }^{2}x+1>0$

which is true for all $x$ in $R$

Now, from $\left(i\right)$, we get,

$16{\sin }^{2}x+1<2$

$0\le 16{\sin }^{2}x<1$

$1\le 16{\sin }^{2}x+1<2$

$\Rightarrow {\log }_{\sqrt{2}}1\le {\log }_{\sqrt{2}}16{\sin }^{2}x+1<{\log }_{\sqrt{2}}2$

$\Rightarrow 0\le {\log }_{\sqrt{2}}16{\sin }^{2}x+1<2$

$\Rightarrow -2\le -{\log }_{\sqrt{2}}16{\sin }^{2}x+1\le 0$

$\Rightarrow 2-2\le 2-{\log }_{\sqrt{2}}16{\sin }^{2}x+1\le 2$

$\Rightarrow 0\le 2-{\log }_{\sqrt{2}}16{\sin }^{2}x+1\le 2$

$\Rightarrow {\log }_{2}0\le {\log }_2(2-{\log }_{\sqrt{2}}16{\sin }^{2}x+1)\le {\log }_{2}2$

$\Rightarrow -\infty \le {\log }_2(2-{\log }_{\sqrt{2}}16{\sin }^{2}x+1)\le 1$

Thus,Range $=(-\infty ,1]$