CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 12 Marks
Question
Find the domain and range of the real function $f(x)=\sqrt{9-x^2}$.
✓
Answer
It is clear that, $f ( x )=\sqrt{9-x^2}$ is not defined when $\left(9- x ^2\right)<0$,
i.e. When $x^2>9$
i,e when $x>3$ or $x<-3$
$\operatorname{dom}( f )=|x \in R:-3 \leq x \leq 3|$
Also,$y =\sqrt{9-x^2} \Rightarrow y^2=\left(9-x^2\right)$
$\Rightarrow x=\sqrt{9-y^2}$
clearly, $x$ is not defined when $\left(9-y^2\right)<0$
but $\left(9-y^2\right)<0$
$\Rightarrow y^2>9$
$\Rightarrow y>3$ or $y<-3$
range $(f)=\{y \in R:-3 \leq y \leq 3\}$
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