Question

The range of $f(x)=8\sqrt{2}\sin \sqrt{\frac{{\pi }^2}{16}-x^2}$ is

• A. [-1 ,1]
• B. [0 ,1]
• C. [0 ,8]
• D. [0 ,4]

Answer 1

Answer: (C)

$-\frac{\pi }{4}\le x\le \frac{\pi }{4}\Rightarrow 0\le \sqrt{\frac{{\pi }^2}{16}-x^2}\le \frac{\pi }{4}$
$\Rightarrow 8\sqrt{2}\sin \sqrt{\frac{{\pi }^2}{16}-x^2}\in \left[0,8\right]$