Q99 / 170 · Manipal (MET) 2019 · Mathematics
The domain of the function $f(x)=\frac{\sin ^{-1}(3-x)}{\log (|x|-2)}$ is :
- A$[2,4]$
- Correct answerB$(3,4]$
- C$[2, \infty)$
- D$(-\infty, 3) \cup[2, \infty)$
Solution
Given that, $f(x)=\frac{\sin ^{-1}(3-x)}{\log (|x|-2)}$
From numerator of the function, we have $\begin{aligned} & \sin ^{-1}(3-x), \text { here }-1 \leq 3-x \leq 1 \\ & \Rightarrow \quad-4 \leq-x \leq-2 \end{aligned}$ $\Rightarrow \quad 2 \leq x \leq 4$ ...(i) From denominator of the function, we have $|x|-2\gt1 \text { or }|x|\gt3$ $\Rightarrow \quad x \lt -3 \cup x\gt3$ ...(ii) From both cases, we get the domain of the given function. i.e. $\quad 3 \lt x \leq 4$ or $x \in(3,4]$.
From numerator of the function, we have $\begin{aligned} & \sin ^{-1}(3-x), \text { here }-1 \leq 3-x \leq 1 \\ & \Rightarrow \quad-4 \leq-x \leq-2 \end{aligned}$ $\Rightarrow \quad 2 \leq x \leq 4$ ...(i) From denominator of the function, we have $|x|-2\gt1 \text { or }|x|\gt3$ $\Rightarrow \quad x \lt -3 \cup x\gt3$ ...(ii) From both cases, we get the domain of the given function. i.e. $\quad 3 \lt x \leq 4$ or $x \in(3,4]$.
Hence, the correct answer is option: 'B'.