Question

The domain of the function $f(x)=\frac{\sqrt{9-x^2}}{{\sin }^{-1}(3-x)}$

• A. $(2,3)$
• B. $[2,3)$
• C. $(2,3]$
• D. None of these

Answer 1

Answer: (B)

$\sqrt{9-x^2}$ is defined for
$9-x^2\ge 0\Rightarrow (3-x)(3+x)\ge 0$
$\Rightarrow (x-3)(x+3)\le 0\dots (i)$
$\Rightarrow -3\le x\le 3$
${\sin }^{-1}(3-x)$ defined for
$-1\le 3-x<1$
$\Rightarrow -4\le -x\le -2$
$\Rightarrow 2\le x\le 4\dots$ (ii)
Also, ${\sin }^{-1}(3-x)\ne 0$
$\Rightarrow 3-x\ne 0$ or $x\ne 3\dots$... (iii)
From Eqs. (i), (ii) and (iii), we get The domain of
$f=([-3,3]\cap [2,4)-\{3\}=[2,3)$