Question

The range of $f\left(x\right)=\sqrt{16-x^2}$ is

• A. $[-4,4]$
• B. $[0,4]$
• C. $[0,\infty )$
• D. $(-\infty ,\infty )$

Answer 1

The correct option is B $[0,4]$

Since square root can only take positive value so
$-4\le x\le 4\Rightarrow \sqrt{16-x^2}\in [0,4]$
Hence, option 'B' is correct.