Let `f(x)=(ax+b)/(cx+d)`. Then the fof (x) =x provided that

Options

A. d=-a
B. d=a
C. a=b=c=d=1
D. None of these

Answer

To solve the problem, we need to find the conditions under which \( f(f(x)) = x \) for the function \( f(x) = \frac{ax + b}{cx + d} \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \frac{ax + b}{cx + d} \]

2. **Find \( f(f(x)) \)**:
We need to substitute \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{ax + b}{cx + d}\right) \]
Substituting into the function:
\[ f(f(x)) = \frac{a\left(\frac{ax + b}{cx + d}\right) + b}{c\left(\frac{ax + b}{cx + d}\right) + d} \]

3. **Simplify the numerator**:
\[ = \frac{\frac{a(ax + b)}{cx + d} + b}{\frac{c(ax + b)}{cx + d} + d} \]
This simplifies to:
\[ = \frac{\frac{a^2x + ab + b(cx + d)}{cx + d}}{\frac{c(ax + b) + d(cx + d)}{cx + d}} \]
Which further simplifies to:
\[ = \frac{(a^2 + bc)x + (ab + bd)}{(ca + dc)x + (cb + d^2)} \]

4. **Set \( f(f(x)) = x \)**:
For \( f(f(x)) \) to equal \( x \), we need:
\[ \frac{(a^2 + bc)x + (ab + bd)}{(ca + dc)x + (cb + d^2)} = x \]

5. **Cross-multiply**:
\[ (a^2 + bc)x + (ab + bd) = x((ca + dc)x + (cb + d^2)) \]
This expands to:
\[ (a^2 + bc)x + (ab + bd) = (ca + dc)x^2 + (cb + d^2)x \]

6. **Rearrange the equation**:
\[ 0 = (ca + dc)x^2 + (cb + d^2 - (a^2 + bc))x - (ab + bd) \]

7. **Compare coefficients**:
For the equation to hold for all \( x \), the coefficients of \( x^2 \), \( x \), and the constant term must all be zero:
- Coefficient of \( x^2 \): \( ca + dc = 0 \)
- Coefficient of \( x \): \( cb + d^2 - (a^2 + bc) = 0 \)
- Constant term: \( ab + bd = 0 \)

8. **Solve the system of equations**:
From \( ca + dc = 0 \):
\[ c = -\frac{d}{a} \quad \text{(if } a \neq 0\text{)} \]
From \( ab + bd = 0 \):
\[ b(a + d) = 0 \]
This gives us two cases:
- \( b = 0 \)
- \( a + d = 0 \) (which implies \( d = -a \))

9. **Final conditions**:
The conditions derived are:
\[ ac + cd = 0, \quad a^2 = d^2, \quad ab + bd = 0 \]
Thus, we conclude that:
\[ A = -D \quad \text{or} \quad D = -A \]