In the following functions defined from [–1, 1] to [–1, 1] the functions which are not bijective are

Options

A. `sin (sin^(-1) x)`
B. `(2)/(pi) sin^(-1) (sin x)`
C. `(sgn x ) ln e^(x)`
D. `x^(3) sgn x`

Answer

To determine which of the given functions defined from \([-1, 1]\) to \([-1, 1]\) are not bijective, we need to check if they are both one-to-one (injective) and onto (surjective). A function is bijective if it is both injective and surjective.

### Step-by-Step Solution:

1. **Understanding Bijective Functions**:
- A function \( f: A \to B \) is **one-to-one** (injective) if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
- A function is **onto** (surjective) if for every \( b \in B \), there exists at least one \( a \in A \) such that \( f(a) = b \).
- A function is **bijective** if it is both injective and surjective.

2. **Analyzing Each Function**:
- **Function 1: \( f(x) = \sin(\sin^{-1}(x)) \)**:
- This simplifies to \( f(x) = x \).
- This function is both injective and surjective on the interval \([-1, 1]\).
- **Conclusion**: This function is bijective.

- **Function 2: \( f(x) = \frac{2}{\pi} \sin^{-1}(\sin(x)) \)**:
- This simplifies to \( f(x) = \frac{2x}{\pi} \).
- This function is injective because different inputs yield different outputs.
- However, it is not onto because the range of \( \frac{2x}{\pi} \) for \( x \in [-1, 1] \) is \([-2/\pi, 2/\pi]\), which does not cover the entire interval \([-1, 1]\).
- **Conclusion**: This function is not bijective.

- **Function 3: \( f(x) = \text{signum}(x) \)**:
- The signum function is defined as \( \text{signum}(x) = \frac{|x|}{x} \) for \( x \neq 0 \), and it is \( 0 \) for \( x = 0 \).
- This function maps \( -1 \) to \( -1 \), \( 0 \) to \( 0 \), and \( 1 \) to \( 1 \).
- It is not injective because both \( 1 \) and \( -1 \) map to \( 1 \).
- **Conclusion**: This function is not bijective.

- **Function 4: \( f(x) = x^3 \cdot \text{signum}(x) \)**:
- This simplifies to \( f(x) = x^2 \) for \( x \neq 0 \).
- The function \( x^2 \) is not injective because \( f(1) = 1 \) and \( f(-1) = 1 \).
- **Conclusion**: This function is not bijective.

3. **Final Conclusion**:
- The functions that are **not bijective** are:
- Function 2: \( \frac{2}{\pi} \sin^{-1}(\sin(x)) \)
- Function 3: \( \text{signum}(x) \)
- Function 4: \( x^3 \cdot \text{signum}(x) \)

### Summary of Non-Bijective Functions:
- **Functions that are not bijective**:
- \( \frac{2}{\pi} \sin^{-1}(\sin(x)) \)
- \( \text{signum}(x) \)
- \( x^3 \cdot \text{signum}(x) \)