Let `A={-1 le x le 1} and f:A to A` such that `f(x)=x|x|` then f is:

Options

A. a bijection
B. injective but not surjective
C. surjective but not injective
D. neither injective nor surjective

Answer

To determine the nature of the function \( f: A \to A \) defined by \( f(x) = x |x| \) where \( A = \{ x \mid -1 \leq x \leq 1 \} \), we will analyze the function step by step.

### Step 1: Define the function
The function is given as:
\[ f(x) = x |x| \]
We need to evaluate this function over the interval \( A = [-1, 1] \).

### Step 2: Analyze the function based on the sign of \( x \)
The absolute value function \( |x| \) behaves differently based on whether \( x \) is positive or negative. Therefore, we will consider two cases:

1. **Case 1: \( x \in [0, 1] \)**
- Here, \( |x| = x \).
- Thus, \( f(x) = x \cdot x = x^2 \).

2. **Case 2: \( x \in [-1, 0) \)**
- Here, \( |x| = -x \).
- Thus, \( f(x) = x \cdot (-x) = -x^2 \).

### Step 3: Write the piecewise function
From the analysis, we can express \( f(x) \) as a piecewise function:
\[ f(x) = \begin{cases} x^2 & \text{if } 0 \leq x \leq 1 \\ -x^2 & \text{if } -1 \leq x < 0 \end{cases} \]

### Step 4: Determine if the function is one-to-one (injective)
To check if \( f \) is one-to-one, we need to see if different inputs yield different outputs.

- For \( x \in [0, 1] \), \( f(x) = x^2 \) is a strictly increasing function (as the derivative \( f'(x) = 2x > 0 \) for \( x > 0 \)).
- For \( x \in [-1, 0) \), \( f(x) = -x^2 \) is a strictly decreasing function (as the derivative \( f'(x) = -2x < 0 \) for \( x < 0 \)).

Since both pieces are strictly monotonic in their respective intervals, \( f \) is one-to-one.

### Step 5: Determine if the function is onto (surjective)
Next, we need to check if every value in the codomain \( A = [-1, 1] \) is achieved by \( f(x) \).

- For \( x \in [0, 1] \), \( f(x) \) takes values from \( 0 \) to \( 1 \) (inclusive).
- For \( x \in [-1, 0) \), \( f(x) \) takes values from \( -1 \) to \( 0 \) (inclusive).

Thus, the range of \( f \) is \( [-1, 1] \), which matches the codomain.

### Step 6: Conclusion
Since \( f \) is both one-to-one and onto, we conclude that \( f \) is a bijective function.

### Final Answer
The function \( f \) is a **bijection**.

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