If `1/2 le x le 1 then sin^(-1)3x-4x^(3)` equals

Options

A. `3 sin^(-1)x`
B. `pi-3 sin^(-1)x`
C. `-pi - 3 sin^(-1)x`
D. none of these

Answer

To solve the problem, we need to find the value of \( \sin^{-1}(3x - 4x^3) \) given that \( \frac{1}{2} \leq x \leq 1 \).

### Step-by-Step Solution:

1. **Identify the Range for \( x \)**:
We are given that \( x \) lies between \( \frac{1}{2} \) and \( 1 \):
\[ \frac{1}{2} \leq x \leq 1 \]

2. **Use the Formula for \( \sin^{-1}(3x - 4x^3) \)**:
There is a known identity for \( \sin^{-1}(3x - 4x^3) \) when \( x \) is in the range \( \left[\frac{1}{2}, 1\right] \):
\[ 3 \sin^{-1}(x) = \pi - \sin^{-1}(3x - 4x^3) \]
Therefore, we can rearrange this to express \( \sin^{-1}(3x - 4x^3) \):
\[ \sin^{-1}(3x - 4x^3) = \pi - 3 \sin^{-1}(x) \]

3. **Calculate \( \sin^{-1}(x) \)**:
Since \( x \) is in the range \( \left[\frac{1}{2}, 1\right] \), we can find \( \sin^{-1}(x) \) for specific values of \( x \):
- If \( x = \frac{1}{2} \), then \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \).
- If \( x = 1 \), then \( \sin^{-1}(1) = \frac{\pi}{2} \).

4. **Substituting Values**:
Now, substituting \( \sin^{-1}(x) \) back into our equation:
- For \( x = \frac{1}{2} \):
\[ \sin^{-1}(3 \cdot \frac{1}{2} - 4 \cdot \left(\frac{1}{2}\right)^3) = \pi - 3 \cdot \frac{\pi}{6} = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]
- For \( x = 1 \):
\[ \sin^{-1}(3 \cdot 1 - 4 \cdot 1^3) = \pi - 3 \cdot \frac{\pi}{2} = \pi - \frac{3\pi}{2} = -\frac{\pi}{2} \]

5. **Conclusion**:
Therefore, for \( x \) in the interval \( \left[\frac{1}{2}, 1\right] \), we conclude that:
\[ \sin^{-1}(3x - 4x^3) = \pi - 3 \sin^{-1}(x) \]

### Final Answer:
The value of \( \sin^{-1}(3x - 4x^3) \) is \( \pi - 3 \sin^{-1}(x) \).