The functions `f:[-1/2, 1/2] to [-pi/2, pi/2]` defined by `f(x)=sin^(-1)(3x-4x^(3))` is

Options

A. bijection
B. injection but not a surjection
C. surjection but not and injection
D. neither an injection nor a surjection

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^{-1}(3x - 4x^3) \) defined on the interval \( x \in [-\frac{1}{2}, \frac{1}{2}] \).

### Step-by-Step Solution:

1. **Identify the function and its domain**:
We have \( f(x) = \sin^{-1}(3x - 4x^3) \) and the domain is \( x \in [-\frac{1}{2}, \frac{1}{2}] \).

2. **Substitute \( x \) with \( \sin \theta \)**:
Let \( x = \sin \theta \). Then, we need to determine how \( \theta \) varies when \( x \) is in the given interval.
- When \( x = -\frac{1}{2} \), \( \theta = \sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6} \).
- When \( x = \frac{1}{2} \), \( \theta = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \).
Thus, \( \theta \) varies from \( -\frac{\pi}{6} \) to \( \frac{\pi}{6} \).

3. **Rewrite the function in terms of \( \theta \)**:
Substitute \( x = \sin \theta \) into the function:
\[ f(x) = \sin^{-1}(3\sin \theta - 4\sin^3 \theta) \]
We can use the identity for \( \sin 3\theta \):
\[ \sin 3\theta = 3\sin \theta - 4\sin^3 \theta \]
Therefore, we have:
\[ f(x) = \sin^{-1}(\sin 3\theta) \]

4. **Determine the range of \( 3\theta \)**:
Since \( \theta \) varies from \( -\frac{\pi}{6} \) to \( \frac{\pi}{6} \), we can find the range of \( 3\theta \):
\[ 3\left(-\frac{\pi}{6}\right) < 3\theta < 3\left(\frac{\pi}{6}\right) \]
This simplifies to:
\[ -\frac{\pi}{2} < 3\theta < \frac{\pi}{2} \]

5. **Apply the property of the inverse sine function**:
Since \( \sin^{-1}(\sin x) = x \) for \( x \) in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), we have:
\[ f(x) = 3\theta \]
Substituting back for \( \theta \):
\[ f(x) = 3\sin^{-1}(x) \]

6. **Conclusion about the function**:
The function \( f(x) = 3\sin^{-1}(x) \) is defined for \( x \in [-\frac{1}{2}, \frac{1}{2}] \) and maps to the range \( [-\frac{3\pi}{6}, \frac{3\pi}{6}] = [-\frac{\pi}{2}, \frac{\pi}{2}] \).
Thus, \( f \) is a bijective function (one-to-one and onto).