Question

The domain of ${\sin }^{-1}\left(\frac{2x+1}{3}\right)$ is:

• A. $(2,-1)$
• B. $[-2,1]$
• C. R
• D. $(-1,1)$
• E. $(-2,0)$

Answer 1

Answer: (B)

Let $y={\sin }^{-1}\left(\frac{2x+1}{3}\right)$ $\therefore$ $-1\le \frac{2x+1}{3}\le 1$ $(\because -1\le {\sin }^{-1}x\le 1)$ $\Rightarrow$ $-3\le 2x+1\le 3$ $\Rightarrow$ $-4\le 2x\le 2$ $\Rightarrow$ $-2\le x\le 1$ $\therefore$ $x\in [-2,1]$