The domain of sin-1 2x is:

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Solution

Q.
The domain of $\sin^{- 1} 2 x$ is:

1 $[0, 1]$

2 $[- 1, 1]$

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

4 $[- 2, 2]$

Ans.
(3)

Ans.    $[- \frac{1}{2}, \frac{1}{2}]$

Explanation:

Let   $\sin^{- 1} 2 x = θ$

So,   $2 x = \sin θ$

Now, $- 1 \leq \sin θ \leq 1,$ i.e., $- 1 \leq 2 x \leq 1,$ which gives $- \frac{1}{2} \leq x \leq \frac{1}{2}$

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