To solve the problem step by step, we start with the given functions and the equation:
1. **Define the Functions**:
- Let \( f(x) = ax + b \)
- Let \( g(x) = bx + a \)
2. **Calculate \( f(g(x)) \)**:
- Substitute \( g(x) \) into \( f(x) \):
\[
f(g(x)) = f(bx + a) = a(bx + a) + b = abx + a^2 + b
\]
3. **Calculate \( g(f(x)) \)**:
- Substitute \( f(x) \) into \( g(x) \):
\[
g(f(x)) = g(ax + b) = b(ax + b) + a = abx + b^2 + a
\]
4. **Evaluate \( f(g(20)) \) and \( g(f(20)) \)**:
- First, find \( g(20) \):
\[
g(20) = b(20) + a = 20b + a
\]
- Now, substitute \( g(20) \) into \( f \):
\[
f(g(20)) = f(20b + a) = a(20b + a) + b = 20ab + a^2 + b
\]
- Next, find \( f(20) \):
\[
f(20) = a(20) + b = 20a + b
\]
- Now, substitute \( f(20) \) into \( g \):
\[
g(f(20)) = g(20a + b) = b(20a + b) + a = 20ab + b^2 + a
\]
5. **Set up the equation**:
- We know from the problem statement:
\[
f(g(20)) - g(f(20)) = 28
\]
- Substitute the expressions we derived:
\[
(20ab + a^2 + b) - (20ab + b^2 + a) = 28
\]
- Simplifying this gives:
\[
a^2 + b - b^2 - a = 28
\]
- Rearranging terms:
\[
a^2 - b^2 + b - a = 28
\]
6. **Factor the equation**:
- The equation can be factored as:
\[
(a - b)(a + b) + (b - a) = 28
\]
- This simplifies to:
\[
(a - b)(a + b - 1) = 28
\]
7. **Find integer solutions**:
- The factors of 28 are \( (1, 28), (2, 14), (4, 7) \).
- We can set \( a - b = k \) and \( a + b - 1 = \frac{28}{k} \) for each factor \( k \).
- **Case 1**: \( k = 1 \)
- \( a - b = 1 \)
- \( a + b - 1 = 28 \)
- Solving gives \( a = 15, b = 14 \).
- **Case 2**: \( k = 2 \)
- \( a - b = 2 \)
- \( a + b - 1 = 14 \)
- Solving gives \( a = 8, b = 6 \).
- **Case 3**: \( k = 4 \)
- \( a - b = 4 \)
- \( a + b - 1 = 7 \)
- Solving gives \( a = 6, b = 2 \).
8. **Evaluate the options**:
- We have the following pairs:
- \( (a, b) = (15, 14) \)
- \( (a, b) = (8, 6) \)
- \( (a, b) = (6, 2) \)
- Now, check the options:
- a. \( a = 15 \) (True)
- b. \( a = 6 \) (True)
- c. \( b = 14 \) (True)
- d. \( b = 3 \) (Not true, since \( b = 14 \) or \( b = 6 \) or \( b = 2 \))
Thus, the answer is that option **d** is not true.
