If `0 le x le 1 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. none of these

Answer

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

### Step-by-Step Solution:

1. **Identify the Formula**: We know from trigonometric identities that:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
This means that if we let \( \theta = \cos^{-1}(x) \), then:
\[ \cos(2\theta) = 2x^2 - 1 \]
Therefore, we can express \( \cos^{-1}(2x^2 - 1) \) in terms of \( \theta \).

2. **Set Up the Equation**: From the identity, we have:
\[ \cos^{-1}(2x^2 - 1) = 2\theta \]
where \( \theta = \cos^{-1}(x) \).

3. **Substitute Back**: Since \( \theta = \cos^{-1}(x) \), we can write:
\[ \cos^{-1}(2x^2 - 1) = 2\cos^{-1}(x) \]

4. **Conclusion**: Therefore, the final result is:
\[ \cos^{-1}(2x^2 - 1) = 2\cos^{-1}(x) \]

### Final Answer:
\[ \cos^{-1}(2x^2 - 1) = 2\cos^{-1}(x) \]