MCQ
Domain of $\sqrt{\text{a}^2-\text{x}^2}(\text{a}>0)$ is.
- A$(-a, a)$
- ✓$[-a, a]$
- C$[0, a]$
- D$(-a, 0]$
✓
Answer
Correct option: B.
$[-a, a]$
We have $\text{f(x)}\sqrt{\text{a}^2-\text{x}^2}$
Clearly $f(x)$ is defined, if ${\text{a}^2-\text{x}^2}\geq0$
$\Rightarrow\text{x}^2\leq\text{a}^2$
$\Rightarrow-\text{a}\leq\text{x}\leq\text{a} \ [\therefore\text{a}>0]$
$\therefore$ Domain of $f$ is $[-a, a]$
Clearly $f(x)$ is defined, if ${\text{a}^2-\text{x}^2}\geq0$
$\Rightarrow\text{x}^2\leq\text{a}^2$
$\Rightarrow-\text{a}\leq\text{x}\leq\text{a} \ [\therefore\text{a}>0]$
$\therefore$ Domain of $f$ is $[-a, a]$
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