Let `A=[-1,1]` and `f:AtoA` be defined as `f(x)=x|x|` for all `x inA`, then f(x) is

Options

A. many-one and into function
B. one-one and into function
C. many-one and into function
D. one-one and onto function

Answer

To solve the problem, we need to analyze the function \( f(x) = x |x| \) defined on the set \( A = [-1, 1] \). We will determine whether the function is one-to-one (injective) and onto (surjective).

### Step 1: Understand the function
The function is defined as:
\[ f(x) = x |x| \]
This can be rewritten as:
- \( f(x) = x^2 \) when \( x \geq 0 \)
- \( f(x) = -x^2 \) when \( x < 0 \)

### Step 2: Analyze the function for \( x \in A \)
1. **For \( x \in [-1, 0] \)**:
- Here, \( f(x) = -x^2 \).
- As \( x \) moves from -1 to 0, \( f(x) \) will move from:
\[ f(-1) = -(-1)^2 = -1 \quad \text{to} \quad f(0) = -0^2 = 0 \]
- Therefore, the output values for this interval are from -1 to 0.

2. **For \( x \in [0, 1] \)**:
- Here, \( f(x) = x^2 \).
- As \( x \) moves from 0 to 1, \( f(x) \) will move from:
\[ f(0) = 0^2 = 0 \quad \text{to} \quad f(1) = 1^2 = 1 \]
- Therefore, the output values for this interval are from 0 to 1.

### Step 3: Determine the range of \( f(x) \)
Combining both intervals:
- From \( x \in [-1, 0] \), \( f(x) \) gives values in the range \([-1, 0]\).
- From \( x \in [0, 1] \), \( f(x) \) gives values in the range \([0, 1]\).

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

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

1. **For \( x < 0 \)**:
- \( f(x) = -x^2 \) is a decreasing function in this interval.

2. **For \( x \geq 0 \)**:
- \( f(x) = x^2 \) is an increasing function in this interval.

Since both parts of the function are either strictly increasing or strictly decreasing, \( f(x) \) is one-to-one.

### Step 5: Check if the function is onto
A function is onto if every element in the codomain is mapped by some element in the domain.

- The codomain is \( A = [-1, 1] \).
- The range of \( f(x) \) is also \([-1, 1]\).

Since the range and codomain are equal, \( f(x) \) is onto.

### Conclusion
The function \( f(x) = x |x| \) defined on \( A = [-1, 1] \) is both one-to-one and onto.

### Final Answer
Thus, \( f(x) \) is a bijective function.

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