If `f(x)=(x-1)/(x+1)` , then `f(f(a x))` in terms of `f(x)` is equal to `(a)(f(x)-1)/(a(f(x)-1))` (b) `(f(x)+1)/(a(f(x)-1))` `(f(x)-1)/(a(f(x)+1))` (d) `(f(x)+1)/(a(f(x)+1))`

Options

A. `(f(x)+a)/(1+af(x))`
B. `((a-1)f(x)+a+1)/((a+1)f(x)+a-1)`
C. `((a+1)f(x)+a-1)/((a-1)f(x)+a+1)`
D. None of these

Answer

To solve the problem, we need to find \( f(f(ax)) \) in terms of \( f(x) \), where \( f(x) = \frac{x-1}{x+1} \).

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

Given:
\[ f(x) = \frac{x-1}{x+1} \]
We first find \( f(ax) \):
\[ f(ax) = \frac{ax - 1}{ax + 1} \]

### Step 2: Substitute \( f(ax) \) into \( f \)

Next, we need to find \( f(f(ax)) \):
\[ f(f(ax)) = f\left(\frac{ax - 1}{ax + 1}\right) \]

### Step 3: Use the definition of \( f \)

Using the definition of \( f \):
\[ f\left(\frac{ax - 1}{ax + 1}\right) = \frac{\frac{ax - 1}{ax + 1} - 1}{\frac{ax - 1}{ax + 1} + 1} \]

### Step 4: Simplify the numerator

The numerator becomes:
\[ \frac{ax - 1}{ax + 1} - 1 = \frac{ax - 1 - (ax + 1)}{ax + 1} = \frac{ax - 1 - ax - 1}{ax + 1} = \frac{-2}{ax + 1} \]

### Step 5: Simplify the denominator

The denominator becomes:
\[ \frac{ax - 1}{ax + 1} + 1 = \frac{ax - 1 + (ax + 1)}{ax + 1} = \frac{ax - 1 + ax + 1}{ax + 1} = \frac{2ax}{ax + 1} \]

### Step 6: Combine the results

Now we can combine the numerator and denominator:
\[ f(f(ax)) = \frac{\frac{-2}{ax + 1}}{\frac{2ax}{ax + 1}} = \frac{-2}{2ax} = \frac{-1}{ax} \]

### Step 7: Substitute \( x \) in terms of \( f(x) \)

From the earlier steps, we found that:
\[ x = \frac{1 + f(x)}{1 - f(x)} \]
Substituting this into our expression for \( f(f(ax)) \):
\[ f(f(ax)) = \frac{-1}{a \cdot \frac{1 + f(x)}{1 - f(x)}} = \frac{-(1 - f(x))}{a(1 + f(x))} \]

### Final Result

Thus, we can express \( f(f(ax)) \) in terms of \( f(x) \):
\[ f(f(ax)) = \frac{f(x) - 1}{a(f(x) + 1)} \]

### Conclusion

The correct answer is:
\[ \boxed{\frac{f(x) - 1}{a(f(x) + 1)}} \]