Question

The domain and range of the function $y=f\left(x\right)=\sin \sqrt{\frac{{\pi }^2}{9}-x^2}$ is

• A. $D:\left[-\frac{\pi }{3}\frac{\pi }{3}\right];R:\left[0,\frac{\sqrt{3}}{2}\right]$
• B. $D:\left(-\frac{\pi }{3}\frac{\pi }{3}\right);R:\left[0,\frac{\sqrt{3}}{2}\right]$
• C. $D:\left[-\frac{\pi }{3}\frac{\pi }{3}\right];R:\left(0,\frac{\sqrt{3}}{2}\right)$
• D. None of these

Answer 1

The correct option is A $D:\left[-\frac{\pi }{3}\frac{\pi }{3}\right];R:\left[0,\frac{\sqrt{3}}{2}\right]$

$y=\sin \sqrt{\frac{{\pi }^2}{9}-x^2}$
For $f\left(x\right)$ to be defined $\frac{{\pi }^2}{9}-x^2\ge 0$
$\Rightarrow (x-\frac{\pi }{3})(x+\frac{\pi }{3})\le 0\therefore$ Domain $D:[-\frac{\pi }{3},\frac{\pi }{3}]$
And range $R:[\sin 0,\sin \frac{\pi }{3}]=[0,\frac{\sqrt{3}}{2}]$