Question

If $f\left(x\right)=3\sin \sqrt{\frac{{\pi }^2}{16}-x^2},$ then its range is

• A. $[-\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}]$
• B. $[0,\frac{3}{\sqrt{2}}]$
• C. $[-\frac{3}{\sqrt{2}},0]$
• D. $[-1,0]$

Answer 1

The correct option is B $[0,\frac{3}{\sqrt{2}}]$

$f\left(x\right)=3\sin \sqrt{\frac{{\pi }^2}{16}-x^2}$
$\Rightarrow$Domain is $x\in [-\frac{\pi }{4},\frac{\pi }{4}]$
In the given domain, $x^2\in [0,\frac{{\pi }^2}{16}]$
$\Rightarrow \frac{{\pi }^2}{16}-x^2\in [0,\frac{{\pi }^2}{16}]$
$\Rightarrow \sqrt{\frac{{\pi }^2}{16}-x^2}\in [0,\frac{\pi }{4}]$
$\Rightarrow \sin \sqrt{\frac{{\pi }^2}{16}-x^2}\in [\sin 0,\sin \frac{\pi }{4}](\because \sin x$ is increasing function in $[0,\frac{\pi }{4}])$
$\Rightarrow 3\sin \sqrt{\frac{{\pi }^2}{16}-x^2}\in [0,\frac{3}{\sqrt{2}}]$

Hence Range is $[0,\frac{3}{\sqrt{2}}]$