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

The range of the real valued function $f(x)=\sqrt{9-x^2}$ is
  1. $[-3,3]$
  2. $[-3,0]$
  3. $[0,3]$
  4. $[-2,2]$

Solution

$ \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 $[0,3]$

Asked in: AP EAMCET 2023 (18 May Shift 2)

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