If `-1 le x le 0` then `cos^(-1)(2x^(2)-1)` equals

Options

A. `2 cos^(-1)x`
B. `pi-2 cos^(-1)x`
C. `2pi-2 cos^(-1)x`
D. `-2 cos^(-1)x`

Answer

To solve the problem, we need to find the value of \( \cos^{-1}(2x^2 - 1) \) for \( -1 \leq x \leq 0 \).

### Step-by-Step Solution:

1. **Identify the expression**: We start with the expression \( \cos^{-1}(2x^2 - 1) \).

2. **Use the double angle formula**: We recall the trigonometric identity:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
From this, we can see that if we let \( \theta = \cos^{-1}(x) \), then:
\[ 2\cos^2(\theta) - 1 = \cos(2\theta) \]
This implies:
\[ \cos^{-1}(2x^2 - 1) = 2\cos^{-1}(x) \quad \text{if } x \in [0, 1] \]

3. **Determine the correct case**: Since \( x \) is in the range \( -1 \leq x \leq 0 \), we need to use the alternative case of the formula:
\[ \cos^{-1}(2x^2 - 1) = 2\pi - 2\cos^{-1}(x) \]

4. **Rearranging the equation**: We can express this as:
\[ \cos^{-1}(2x^2 - 1) = 2\pi - 2\cos^{-1}(x) \]

5. **Final expression**: Thus, the value of \( \cos^{-1}(2x^2 - 1) \) for \( -1 \leq x \leq 0 \) is:
\[ \cos^{-1}(2x^2 - 1) = 2\pi - 2\cos^{-1}(x) \]

### Conclusion:
The final answer is:
\[ \cos^{-1}(2x^2 - 1) = 2\pi - 2\cos^{-1}(x) \]