Let `A=[-1,1]dot` Then, discuss whether the following functions from A to itself are one-one onto or bijective: `f(x)=x/2` (ii) `g(x)=|x|` (iii) `h(x)=x^2`

Explanation

To determine whether the functions \( f(x) = \frac{x}{2} \), \( g(x) = |x| \), and \( h(x) = x^2 \) are one-one, onto, or bijective when defined from the set \( A = [-1, 1] \) to itself, we will analyze each function step by step.

### Step 1: Analyze \( f(x) = \frac{x}{2} \)

1. **One-One Function**:
- A function is one-one (injective) if \( f(a) = f(b) \) implies \( a = b \).
- Let's assume \( f(a) = f(b) \):
\[ \frac{a}{2} = \frac{b}{2} \implies a = b \]
- Since \( f(x) \) satisfies this condition, \( f(x) \) is one-one.

2. **Onto Function**:
- A function is onto (surjective) if for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \).
- The range of \( f(x) \) when \( x \in [-1, 1] \) is:
\[ f([-1, 1]) = \left[-\frac{1}{2}, \frac{1}{2}\right] \]
- The codomain is \( [-1, 1] \), which includes values outside of the range of \( f(x) \). Thus, \( f(x) \) is not onto.

3. **Conclusion**:
- \( f(x) \) is one-one but not onto, hence it is not bijective.

### Step 2: Analyze \( g(x) = |x| \)

1. **One-One Function**:
- For \( g(a) = g(b) \):
\[ |a| = |b| \implies a = b \text{ or } a = -b \]
- Since \( g(x) \) can map two different inputs to the same output (e.g., \( g(-1) = g(1) = 1 \)), \( g(x) \) is not one-one.

2. **Onto Function**:
- The range of \( g(x) \) when \( x \in [-1, 1] \) is:
\[ g([-1, 1]) = [0, 1] \]
- The codomain is \( [-1, 1] \). Since the range does not cover negative values, \( g(x) \) is not onto.

3. **Conclusion**:
- \( g(x) \) is neither one-one nor onto, hence it is not bijective.

### Step 3: Analyze \( h(x) = x^2 \)

1. **One-One Function**:
- For \( h(a) = h(b) \):
\[ a^2 = b^2 \implies a = b \text{ or } a = -b \]
- Similar to \( g(x) \), \( h(x) \) can map two different inputs to the same output (e.g., \( h(-1) = h(1) = 1 \)), so \( h(x) \) is not one-one.

2. **Onto Function**:
- The range of \( h(x) \) when \( x \in [-1, 1] \) is:
\[ h([-1, 1]) = [0, 1] \]
- The codomain is \( [-1, 1] \). Since the range does not cover negative values, \( h(x) \) is not onto.

3. **Conclusion**:
- \( h(x) \) is neither one-one nor onto, hence it is not bijective.

### Summary of Results:
- \( f(x) = \frac{x}{2} \): One-one but not onto (not bijective).
- \( g(x) = |x| \): Neither one-one nor onto (not bijective).
- \( h(x) = x^2 \): Neither one-one nor onto (not bijective).