Let A = [-1, 1]. Which of the following functions on A is not a bijection?

Options

A. `f(x) = x[x]`
B. `f(x) = x^(3)`
C. `f(x) = sin(pi/2x)`
D. `f(x) = cos(pi/2.x)`

Answer

To determine which function on the set \( A = [-1, 1] \) is not a bijection, we need to analyze each function provided and check if they satisfy the conditions of being a bijection. A function is a bijection if it is both injective (one-to-one) and surjective (onto). This means that every element in the codomain must be mapped to by exactly one element in the domain.

Let's analyze each option step by step:

### Step 1: Analyze the first function \( f(x) = x \cdot \lfloor x \rfloor \)

1. **Domain**: The domain is \( A = [-1, 1] \).
2. **Behavior**:
- For \( x \in [-1, 0) \), \( \lfloor x \rfloor = -1 \), so \( f(x) = x \cdot (-1) = -x \).
- For \( x \in [0, 1) \), \( \lfloor x \rfloor = 0 \), so \( f(x) = x \cdot 0 = 0 \).
- At \( x = 1 \), \( \lfloor 1 \rfloor = 1 \), so \( f(1) = 1 \cdot 1 = 1 \).
3. **Range**: The range of \( f(x) \) is \( [-1, 0] \cup \{0\} \cup \{1\} = [-1, 1] \).
4. **Conclusion**: The function is not injective because multiple inputs give the same output (e.g., all values in \( [0, 1) \) map to 0). Thus, it is not a bijection.

### Step 2: Analyze the second function \( f(x) = x^3 \)

1. **Domain**: The domain is \( A = [-1, 1] \).
2. **Behavior**: The function \( f(x) = x^3 \) is continuous and strictly increasing on the interval.
3. **Range**: The range of \( f(x) \) for \( x \in [-1, 1] \) is also \( [-1, 1] \).
4. **Conclusion**: This function is both injective and surjective, hence it is a bijection.

### Step 3: Analyze the third function \( f(x) = \sin\left(\frac{\pi}{2} x\right) \)

1. **Domain**: The domain is \( A = [-1, 1] \).
2. **Behavior**: The sine function is continuous and maps the interval \( [-1, 1] \) to \( [-\sin\left(\frac{\pi}{2}\right), \sin\left(\frac{\pi}{2}\right)] = [-1, 1] \).
3. **Range**: The range of \( f(x) \) is \( [-1, 1] \).
4. **Conclusion**: This function is both injective and surjective, hence it is a bijection.

### Step 4: Analyze the fourth function \( f(x) = \cos\left(\frac{\pi}{2} x\right) \)

1. **Domain**: The domain is \( A = [-1, 1] \).
2. **Behavior**: The cosine function is even and continuous.
- At \( x = -1 \), \( f(-1) = \cos\left(-\frac{\pi}{2}\right) = 0 \).
- At \( x = 0 \), \( f(0) = \cos(0) = 1 \).
- At \( x = 1 \), \( f(1) = \cos\left(\frac{\pi}{2}\right) = 0 \).
3. **Range**: The range of \( f(x) \) is \( [0, 1] \).
4. **Conclusion**: The function is not injective because both \( x = -1 \) and \( x = 1 \) map to 0. Thus, it is not a bijection.

### Final Conclusion

The function that is not a bijection is the first function \( f(x) = x \cdot \lfloor x \rfloor \) and the fourth function \( f(x) = \cos\left(\frac{\pi}{2} x\right) \).