Let `A={x in R :-1lt=xlt=1}=B` . Then, the mapping `f: A->B` given by `f(x)=x|x|` is (a) injective but not surjective (b) surjective but not injective (c) bijective (d) none of these

Explanation

To solve the problem, we need to analyze the mapping \( f: A \to B \) defined by \( f(x) = x |x| \) for the set \( A = B = \{ x \in \mathbb{R} : -1 < x < 1 \} \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function is defined as \( f(x) = x |x| \). This can be rewritten based on the value of \( x \):
- For \( x \geq 0 \): \( f(x) = x \cdot x = x^2 \)
- For \( x < 0 \): \( f(x) = x \cdot (-x) = -x^2 \)
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