If `f : {x : -1 le x le 1} rarr {x : -1 le x le 1}`, then which is/are bijective ?

Options

A. f(x) = [x]
B. `f(x) = sin.(pi x)/(2)`
C. `f(x) = |x|`
D. `f(x) = x|x|`

Answer

To determine which of the given functions is bijective, we need to analyze each function based on the definitions of one-to-one (injective) and onto (surjective) functions.

### Step 1: Understand the Definitions
A function \( f: A \to B \) is bijective if:
1. **Injective (One-to-One)**: For every \( a_1, a_2 \in A \), if \( f(a_1) = f(a_2) \), then \( a_1 = a_2 \).
2. **Surjective (Onto)**: For every \( b \in B \), there exists at least one \( a \in A \) such that \( f(a) = b \).

### Step 2: Analyze Each Function

**Option 1: \( f(x) = \lfloor x \rfloor \) (Greatest Integer Function)**
- **Injective Check**: For \( x = 0.5 \) and \( x = 0.6 \), both yield \( f(x) = 0 \). Hence, it is not injective (many-to-one).
- **Conclusion**: Not bijective.

**Option 2: \( f(x) = \sin\left(\frac{\pi}{2} x\right) \)**
- **Injective Check**: The sine function is increasing in the interval \([-1, 1]\). Thus, it is one-to-one.
- **Surjective Check**: The range of \( f(x) \) is \([-1, 1]\), which matches the codomain.
- **Conclusion**: This function is bijective.

**Option 3: \( f(x) = |x| \)**
- **Injective Check**: For \( x = 0.5 \) and \( x = -0.5 \), both yield \( f(x) = 0.5 \). Hence, it is not injective (many-to-one).
- **Conclusion**: Not bijective.

**Option 4: \( f(x) = x \cdot |x| \)**
- **Injective Check**: Break into cases:
- For \( x \in [-1, 0] \): \( f(x) = -x^2 \) (decreasing).
- For \( x \in [0, 1] \): \( f(x) = x^2 \) (increasing).
- Both parts are one-to-one.
- **Surjective Check**: The range is \([-1, 1]\), which matches the codomain.
- **Conclusion**: This function is bijective.

### Final Conclusion
The functions that are bijective are:
- **Option 2: \( f(x) = \sin\left(\frac{\pi}{2} x\right) \)**
- **Option 4: \( f(x) = x \cdot |x| \)**