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Q.
The domain of the function $f(x)=\frac{\sqrt{9-x^{2}}}{\sin ^{-1}(3-x)}$
ManipalManipal 2010
Solution:
$\sqrt{9-x^{2}}$ is defined for
$9-x^{2} \geq 0 \Rightarrow (3-x)(3+x) \geq 0$
$\Rightarrow (x-3)(x+3) \leq 0 \ldots( i )$
$\Rightarrow -3 \leq x \leq 3$
$\sin ^{-1}(3-x)$ defined for
$-1 \leq 3-x<1$
$\Rightarrow -4 \leq-x \leq-2$
$\Rightarrow 2 \leq x \leq 4 \ldots$ (ii)
Also, $\sin ^{-1}(3-x) \neq 0$
$\Rightarrow 3-x \neq 0$ or $x \neq 3 \ldots$... (iii)
From Eqs. (i), (ii) and (iii), we get The domain of
$f=([-3,3] \cap[2,4)-\{3\}=[2,3)$