Question

The range of the function $f(x) = \sqrt{9 - x^2} $ is

• A. $(0, 3)$
• B. $[0,3]$
• C. $(0, 3]$
• D. $[0, 3)$

Answer 1

Answer: (B)

We have, $f(x)=\sqrt{9-x^{2}}$
Let $ y=\sqrt{9-x^{2}} $
$ \Rightarrow y^{2}=9-x^{2} $
$ \Rightarrow x^{2}=9-y^{2} $
$ \Rightarrow x=\sqrt{9-y^{2}} $
Since, $f(x) \geq 0 $
$ \therefore 9-y^{2} \geq 0 $
$\Rightarrow (3-y)(3+y) \geq 0$

$\therefore y \in [-3, 3]$
But, $f(x) = \sqrt{9 - x^2}$
$\therefore f(x) \in [0, 3]$