Let `A = {x: -1lexle1}` = B for each of the following functions from A to B. Find whether it is surjective, injective or bijective
(i) f(x) = `(x)/(2)`
(ii) g(x) = |x|
(iii) h(x) = x|x|
(iv) `k(x) = x^(2)`
(v) l(x) = `sin pi x`

Explanation

To determine whether the given functions are surjective, injective, or bijective, we will analyze each function step by step.

### Given:
Let \( A = B = \{ x : -1 \leq x \leq 1 \} \).

### (i) Function: \( f(x) = \frac{x}{2} \)

1. **Injective (One-to-One)**:
- To check if \( f \) is injective, we assume \( f(x_1) = f(x_2) \).
- This implies \( \frac{x_1}{2} = \frac{x_2}{2} \).
- Cancelling \( \frac{1}{2} \) gives \( x_1 = x_2 \).
- Therefore, \( f \) is injective.

2. **Surjective (Onto)**:
- We need to check if for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \).
- Let \( y = \frac{x}{2} \). Then \( x = 2y \).
- For \( y \) in the range \([-1, 1]\), \( x = 2y \) will be in the range \([-2, 2]\).
- Since \( x \) must be in \([-1, 1]\), \( f \) is not surjective (e.g., \( y = 1 \) gives \( x = 2 \), which is outside \( A \)).

3. **Conclusion**:
- \( f(x) \) is injective but not surjective.

### (ii) Function: \( g(x) = |x| \)

1. **Injective**:
- Check if \( g(x_1) = g(x_2) \) implies \( x_1 = x_2 \).
- For \( g(1) = g(-1) = 1 \), \( g \) is not injective.

2. **Surjective**:
- For \( y = |x| \), \( x \) can be either positive or negative.
- However, \( g(x) \) can only take values from \( [0, 1] \), so it does not cover all of \( B \) (e.g., no \( x \) gives \( g(x) = -1 \)).
- Thus, \( g \) is not surjective.

3. **Conclusion**:
- \( g(x) \) is neither injective nor surjective.

### (iii) Function: \( h(x) = x|x| \)

1. **Injective**:
- For \( x \geq 0 \), \( h(x) = x^2 \) and for \( x < 0 \), \( h(x) = -x^2 \).
- Since \( h(1) = 1 \) and \( h(-1) = -1 \), different inputs yield different outputs.
- Therefore, \( h \) is injective.

2. **Surjective**:
- The range of \( h(x) \) is \([-1, 1]\) as \( h(x) \) takes all values from \( -1 \) to \( 1 \) (e.g., \( h(1) = 1 \) and \( h(-1) = -1 \)).
- Thus, \( h \) is surjective.

3. **Conclusion**:
- \( h(x) \) is both injective and surjective, hence bijective.

### (iv) Function: \( k(x) = x^2 \)

1. **Injective**:
- \( k(1) = 1 \) and \( k(-1) = 1 \), hence it is not injective.

2. **Surjective**:
- The range of \( k(x) \) is \([0, 1]\), which does not cover all of \( B \) (e.g., no \( x \) gives \( k(x) = -1 \)).
- Thus, \( k \) is not surjective.

3. **Conclusion**:
- \( k(x) \) is neither injective nor surjective.

### (v) Function: \( l(x) = \sin(\pi x) \)

1. **Injective**:
- \( l(1) = 0 \) and \( l(-1) = 0 \), hence it is not injective.

2. **Surjective**:
- The range of \( l(x) \) is \([-1, 1]\), which covers all of \( B \).
- Thus, \( l \) is surjective.

3. **Conclusion**:
- \( l(x) \) is surjective but not injective.

### Summary of Results:
- \( f(x) = \frac{x}{2} \): Injective, not surjective.
- \( g(x) = |x| \): Neither injective nor surjective.
- \( h(x) = x|x| \): Bijective.
- \( k(x) = x^2 \): Neither injective nor surjective.
- \( l(x) = \sin(\pi x) \): Surjective, not injective.