Mathematics - Functions Question with Solution | TestHub

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

We know that $1\le 16{\sin }^{2}x+1\le 17$.

Therefore, $0\le {\log }_2\left(16{\sin }^{2}x+1\right)\le {\log }_{2}17$.

This implies $2-{\log }_{2}17\le 2-{\log }_2\left(16{\sin }^{2}x+1\right)\le 2$.

Now consider:

$$0<2-{\log }_2\left(16{\sin }^{2}x+1\right)\le 2$$

$$-\infty <{\log }_{\sqrt{2}}\left[2-{\log }_2\left(16{\sin }^{2}x+1\right)\right]\le {\log }_{\sqrt{2}}2=2$$

The range is $(-\infty ,2]$.