Find the range of the function $f(x)=3\sin\left(\sqrt{\frac{π^2}{16}-x^2}\right)$.
Answer & explanation
Correct answer: option 3
We must have $\frac{π^2}{16}-x^2≥0$
Also $\frac{π^2}{16}-x^2≤\frac{π^2}{16}$
$⇒0≤\frac{π^2}{16}-x^2≤\frac{π^2}{16}$
$⇒0≤\sqrt{\frac{π^2}{16}-x^2}≤\frac{π}{4}$
$⇒0≤\sin\sqrt{\frac{π^2}{16}-x^2}≤\frac{1}{\sqrt{2}}$
$⇒0≤3\sin\sqrt{\frac{π^2}{16}-x^2}≤\frac{3}{\sqrt{2}}$
Hence, range is $\left[0,\frac{3}{\sqrt{2}}\right]$