Question

The domain of the real-valued function$f\left(x\right)=\sqrt{1-2x}+2{\sin }^{-1}\left[\frac{3x-1}{2}\right]$ is

• A. $\left[-\frac{1}{3},1\right]$
• B. $\left[\frac{1}{2},1\right]$
• C. $\left[-\frac{1}{2},\frac{1}{3}\right]$
• D. $\left[-\frac{1}{3},\frac{1}{2}\right]$

Answer 1

The correct option is D

$\left[-\frac{1}{3},\frac{1}{2}\right]$

Explanation for correct answer Option(s)

Find the domain of the function $f\left(x\right)$

Consider the given function as,

$f\left(x\right)=\sqrt{1-2x}+2{\sin }^{-1}\left[\frac{3x-1}{2}\right]$

Defined the above function as,

Case: 1

$$\begin{gathered} \Rightarrow \sqrt{1-2x}\ge 0 \\ \Rightarrow 1-2x\ge 0 \\ \Rightarrow 2x\ge 1 \\ \Rightarrow x\ge \frac{1}{2} \\ \Rightarrow x\in \left[-\infty ,\frac{1}{2}\right] \end{gathered}$$

Case: 2

$$\begin{gathered} \Rightarrow -1\le {\sin }^{-1}x\le 1 \\ \Rightarrow -1\le {\sin }^{-1}\left[\frac{3x-1}{2}\right]\le 1 \\ \Rightarrow -1\le \left[\frac{3x-1}{2}\right]\le 1 \\ \Rightarrow -1\times 2\le 3x-1\le 1\times 2 \\ \Rightarrow -2\le 3x-1\le 2 \\ \Rightarrow -2+1\le 3x\le 2+1 \\ \Rightarrow -1\le 3x\le 3 \\ \Rightarrow -\frac{1}{3}\le 3x\le 1 \end{gathered}$$

Thus, $\Rightarrow x\in \left[-\frac{1}{3},\frac{1}{2}\right]$

Hence, the domain of the given function is $\left[-\frac{1}{3},\frac{1}{2}\right]$

Therefore, the correct answer is Option D.