Question

The domain of the function $f\left(x\right)=\frac{\sqrt{4-x^2}}{si{n}^{-1}(2-x)}$ is

• A. $[0,2]$
• B. $[0,2)$
• C. $[1,2)$
• D. $[1,2]$

Answer 1

The correct option is C $[1,2)$

Given $f\left(x\right)=\frac{\sqrt{4-x^2}}{si{n}^{-1}(2-x)}$

$\sqrt{4-x^2}$ is defined for $4-x^2\ge 0$.
$\Rightarrow x^2-4\le 0$

$\Rightarrow (x+2)(x-2)\le 0$
$\Rightarrow -2\le x\le 2$
and $si{n}^{-1}(2-x)$ is defined for $-1\le 2-x\le 1$ .... [$\because -1\le \sin \theta \le 1$]
$\Rightarrow -3\le -x\le -1$

$\Rightarrow 3\ge x\ge 1$
$1\le x\le 3$
Also,for $x=2$ $\Rightarrow$ $si{n}^{-1}(2-x)=0\Rightarrow f\left(x\right)$ is not defined so $x\ne 2$

$\therefore$ Domain of $f\left(x\right)=[-2,2]\cap [1,3]-\left\{2\right\}=[1,2)$
Hence, option c is correct.