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

Options

A. injective but not surjective
B. surjective but not injective
C. bijective
D. None of these

Explanation

To determine the properties of the function \( f: A \to A \) defined by \( f(x) = x |x| \) where \( A = \{ x \mid -1 \leq x \leq 1 \} \), we need to check if the function is injective (one-to-one), surjective (onto), or bijective (both).

### Step 1: Understand the function

The function \( f(x) = x |x| \) can be expressed piecewise:
- For \( x \geq 0 \): \( f(x) = x^2 \)
- For \( x < 0 \): \( f(x) = -x^2 \)

### Step 2: Determine the range of \( f \)

Since \( A = [-1, 1] \):
- For \( x \in [0, 1] \): \( f(x) = x^2 \) ranges from \( 0 \) to \( 1 \).
- For \( x \in [-1, 0) \): \( f(x) = -x^2 \) ranges from \( -1 \) to \( 0 \).

Thus, the overall range of \( f(x) \) is \( [-1, 1] \).

### Step 3: Check if \( f \) is injective

To check if \( f \) is injective, we need to see if \( f(a) = f(b) \) implies \( a = b \).

1. **For \( a, b \geq 0 \)**:
- If \( f(a) = f(b) \), then \( a^2 = b^2 \).
- This implies \( a = b \) or \( a = -b \). Since \( a, b \geq 0 \), we conclude \( a = b \).

2. **For \( a, b < 0 \)**:
- If \( f(a) = f(b) \), then \( -a^2 = -b^2 \).
- This implies \( a^2 = b^2 \), leading to \( a = b \) or \( a = -b \). Since \( a, b < 0 \), we conclude \( a = b \).

3. **For \( a \geq 0 \) and \( b < 0 \)**:
- Here, \( f(a) = a^2 \) and \( f(b) = -b^2 \).
- Since \( a^2 \geq 0 \) and \( -b^2 \leq 0 \), \( f(a) \) cannot equal \( f(b) \).

Thus, \( f \) is injective.

### Step 4: Check if \( f \) is surjective

To check if \( f \) is surjective, we need to see if for every \( y \in A \) (i.e., \( y \in [-1, 1] \)), there exists an \( x \in A \) such that \( f(x) = y \).

1. For \( y \in [0, 1] \):
- We can find \( x = \sqrt{y} \) which is in \( [0, 1] \) such that \( f(x) = y \).

2. For \( y \in [-1, 0) \):
- We can find \( x = -\sqrt{-y} \) which is in \( [-1, 0) \) such that \( f(x) = y \).

Since we can find \( x \) for every \( y \in A \), \( f \) is surjective.

### Conclusion

Since \( f \) is both injective and surjective, we conclude that \( f \) is bijective.

### Final Answer

The function \( f \) is bijective.

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