Suppose, `f,g: R to R` be defined by `f(x) = ax + b, g(x) =cx + d`, where a,b,c,d `in R` and `ac ne 0`, if `f(g(x)) = g(f(x)) AA x in R`, then

Options

A. f(a)=g( c)
B. f(d) = g(b)
C. f( c) = g(d)
D. f(a) = g(d)

Answer

To solve the problem, we need to analyze the functions \( f \) and \( g \) defined as follows:

- \( f(x) = ax + b \)
- \( g(x) = cx + d \)

where \( a, b, c, d \in \mathbb{R} \) and \( ac \neq 0 \). We are given the condition \( f(g(x)) = g(f(x)) \) for all \( x \in \mathbb{R} \).

### Step-by-step Solution:

1. **Calculate \( f(g(x)) \)**:
\[ g(x) = cx + d \]
Now, substituting \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f(cx + d) = a(cx + d) + b = acx + ad + b \]

2. **Calculate \( g(f(x)) \)**:
\[ f(x) = ax + b \]
Now, substituting \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g(ax + b) = c(ax + b) + d = acx + bc + d \]

3. **Set the two expressions equal**:
Since \( f(g(x)) = g(f(x)) \), we have:
\[ acx + ad + b = acx + bc + d \]

4. **Eliminate \( acx \)**:
Since \( acx \) appears on both sides, we can subtract it from both sides:
\[ ad + b = bc + d \]

5. **Rearrange the equation**:
Rearranging gives us:
\[ ad - d = bc - b \]
Factoring out common terms:
\[ d(a - 1) = b(c - 1) \]

6. **Analyze the condition**:
This means that the ratio \( \frac{d}{b} = \frac{c - 1}{a - 1} \) if \( b \neq 0 \) and \( d \neq 0 \).

### Conclusion:
From the derived equation \( d(a - 1) = b(c - 1) \), we can conclude that this relationship must hold true for the functions \( f \) and \( g \) to satisfy the condition \( f(g(x)) = g(f(x)) \).