AP EAMCET · Maths · Functions
The range of the real valued function \(f(x)=\sqrt{9-x^2}\) is
- A \([-3,3]\)
- B \([-3,0]\)
- C \([0,3]\)
- D \([-2,2]\)
Answer & Solution
Correct Answer
(C) \([0,3]\)
Step-by-step Solution
Detailed explanation
\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…
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