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The range of the real valued function $f(x)=\sqrt{9-x^2}$ is
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Verified Answer
The correct answer is:
$[0,3]$
$$
\begin{aligned}
& \text { } \because \mathrm{f}(\mathrm{x})=\sqrt{9-\mathrm{x}^2} \\
& \text { Here }-3 \leq \mathrm{x} \leq 3 \\
& \Rightarrow 0 \leq \mathrm{x}^2 \leq 9 \\
& \Rightarrow-9 \leq-\mathrm{x}^2 < 0 \Rightarrow 0 \sqrt{9-\mathrm{x}^2} \leq 3 \\
& \therefore 0 \leq \mathrm{f}(\mathrm{x}) \leq 3
\end{aligned}
$$
So, the range of the given function is $[0,3]$
\begin{aligned}
& \text { } \because \mathrm{f}(\mathrm{x})=\sqrt{9-\mathrm{x}^2} \\
& \text { Here }-3 \leq \mathrm{x} \leq 3 \\
& \Rightarrow 0 \leq \mathrm{x}^2 \leq 9 \\
& \Rightarrow-9 \leq-\mathrm{x}^2 < 0 \Rightarrow 0 \sqrt{9-\mathrm{x}^2} \leq 3 \\
& \therefore 0 \leq \mathrm{f}(\mathrm{x}) \leq 3
\end{aligned}
$$
So, the range of the given function is $[0,3]$
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