If `f (x) =(ax +b)/( cx +d)`and (fof) x= x, `then

Options

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

Answer

To solve the problem, we need to find the conditions under which the function \( f(f(x)) = x \) holds true for the function defined as:

\[ f(x) = \frac{ax + b}{cx + d} \]

### Step 1: Find \( f(f(x)) \)

First, we need to compute \( f(f(x)) \). We start by substituting \( f(x) \) into itself:

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

Using the definition of \( f(x) \):

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

### Step 2: Simplify the expression

Now we simplify the numerator and denominator separately.

**Numerator:**

\[ = \frac{a(ax + b) + b(cx + d)}{cx + d} = \frac{a^2x + ab + bcx + bd}{cx + d} = \frac{(a^2 + bc)x + (ab + bd)}{cx + d} \]

**Denominator:**

\[ = c\left(\frac{ax + b}{cx + d}\right) + d = \frac{c(ax + b) + d(cx + d)}{cx + d} = \frac{acx + bc + dcx + d^2}{cx + d} = \frac{(ac + dc)x + (bc + d^2)}{cx + d} \]

### Step 3: Combine and simplify \( f(f(x)) \)

Thus, we have:

\[ f(f(x)) = \frac{(a^2 + bc)x + (ab + bd)}{(ac + dc)x + (bc + d^2)} \]

### Step 4: Set \( f(f(x)) = x \)

Now, we set this equal to \( x \):

\[ \frac{(a^2 + bc)x + (ab + bd)}{(ac + dc)x + (bc + d^2)} = x \]

### Step 5: Cross multiply

Cross multiplying gives us:

\[ (a^2 + bc)x + (ab + bd) = x((ac + dc)x + (bc + d^2)) \]

Expanding the right side:

\[ (a^2 + bc)x + (ab + bd) = (ac + dc)x^2 + (bc + d^2)x \]

### Step 6: Rearranging the equation

Rearranging the equation, we get:

\[ 0 = (ac + dc)x^2 + (bc + d^2 - a^2 - bc)x - (ab + bd) \]

### Step 7: Coefficients must equal zero

For this quadratic equation to hold for all \( x \), the coefficients of \( x^2 \), \( x \), and the constant term must all equal zero:

1. \( ac + dc = 0 \)
2. \( bc + d^2 - a^2 - bc = 0 \)
3. \( - (ab + bd) = 0 \)

### Step 8: Solve the equations

From the first equation:

\[ c(a + d) = 0 \]

This implies either \( c = 0 \) or \( a + d = 0 \).

From the second equation:

\[ d^2 - a^2 = 0 \implies d^2 = a^2 \implies d = a \text{ or } d = -a \]

From the third equation:

\[ ab + bd = 0 \implies b(a + d) = 0 \]

This implies either \( b = 0 \) or \( a + d = 0 \).

### Conclusion

Thus, the conditions derived are:

1. If \( c \neq 0 \), then \( a + d = 0 \) (i.e., \( d = -a \)).
2. If \( b = 0 \), the conditions still hold.