Find domain of `sin^(-1) ( 2 x^(2) - 1)`

Options

A. `rArr x in [-1,1]`
B. `rArr x in [0,1]`
C. `rArr x in [-1,0]`
D. `rArr x in [-2,0]`

Answer

To find the domain of the function \( \sin^{-1}(2x^2 - 1) \), we need to determine the values of \( x \) for which the expression \( 2x^2 - 1 \) lies within the range of the inverse sine function. The range of the inverse sine function is from -1 to 1, inclusive. Therefore, we need to solve the inequality:

\[ -1 \leq 2x^2 - 1 \leq 1 \]

### Step 1: Solve the left side of the inequality

Starting with the left side of the inequality:

\[ -1 \leq 2x^2 - 1 \]

Add 1 to both sides:

\[ 0 \leq 2x^2 \]

This inequality is always true for all real \( x \) since \( 2x^2 \) is always non-negative.

### Step 2: Solve the right side of the inequality

Now, we solve the right side of the inequality:

\[ 2x^2 - 1 \leq 1 \]

Add 1 to both sides:

\[ 2x^2 \leq 2 \]

Divide both sides by 2:

\[ x^2 \leq 1 \]

### Step 3: Solve for \( x \)

Taking the square root of both sides gives us:

\[ -1 \leq x \leq 1 \]

### Conclusion

Thus, the domain of \( \sin^{-1}(2x^2 - 1) \) is:

\[ [-1, 1] \]