Let `f(x)=(ax + b )/(cx+d)`. Then the `fof (x)=x`, provided that : `(a!=0, b!= 0, c!=0,d!=0)`

Options

A. `d=-a`
B. `d=a`
C. `a=b=1`
D. `a=b=c=d=1,`

Explanation

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:
\[ f(f(x)) = f\left(\frac{ax + b}{cx + d}\right) = \frac{a\left(\frac{ax + b}{cx + d}\right) + b}{c\left(\frac{ax + b}{cx + d}\right) + d} \]

3. **Simplify the numerator**:
The numerator becomes:
\[ a\left(\frac{ax + b}{cx + d}\right) + b = \frac{a(ax + b) + b(cx + d)}{cx + d} = \frac{a^2x + ab + bcx + bd}{cx + d} \]
Thus, the numerator simplifies to:
\[ (a^2 + bc)x + (ab + bd) \]

4. **Simplify the denominator**:
The denominator becomes:
\[ c\left(\frac{ax + b}{cx + d}\right) + d = \frac{c(ax + b) + d(cx + d)}{cx + d} = \frac{cax + cb + dcx + d^2}{cx + d} \]
Thus, the denominator simplifies to:
\[ (ca + dc)x + (cb + d^2) \]

5. **Combine the results**:
Therefore, we have:
\[ f(f(x)) = \frac{(a^2 + bc)x + (ab + bd)}{(ca + dc)x + (cb + d^2)} \]

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

7. **Cross-multiply**:
This gives us:
\[ (a^2 + bc)x + (ab + bd) = x((ca + dc)x + (cb + d^2)) \]
Expanding the right side:
\[ (a^2 + bc)x + (ab + bd) = (ca + dc)x^2 + (cb + d^2)x \]

8. **Rearranging terms**:
Rearranging gives us:
\[ 0 = (ca + dc)x^2 + (cb + d^2 - (a^2 + bc))x - (ab + bd) \]

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

10. **Solving the equations**:
From \( ca + dc = 0 \), we can express \( d = -\frac{ca}{c} = -a \).

Substitute \( d = -a \) into the other equations:
- From \( ab + bd = 0 \):
\[ ab - ab = 0 \quad \text{(always true)} \]
- From \( cb + d^2 - (a^2 + bc) = 0 \):
\[ cb + a^2 - (a^2 + bc) = 0 \implies cb - bc = 0 \quad \text{(always true)} \]

Thus, the condition we derived is:
\[ d = -a \]

### Final Answer:
The condition for \( f(f(x)) = x \) is:
\[ d = -a \]