If `f(x)f[xy]=f[x]f[y]` then `f[t]` may be of the form:

Options

A. t+k
B. ct+k
C. `t^(k)+c`
D. `t^(k)`

Explanation

To solve the problem where \( f(x)f(xy) = f(x)f(y) \), we will analyze the functional equation step by step.

### Step 1: Understanding the Functional Equation
We start with the equation given:
\[ f(x)f(xy) = f(x)f(y) \]
This suggests a multiplicative property of the function \( f \).

### Step 2: Substituting Values
Let’s substitute \( y = 1 \) into the equation:
\[ f(x)f(x \cdot 1) = f(x)f(1) \]
This simplifies to:
\[ f(x)f(x) = f(x)f(1) \]
Assuming \( f(x) \neq 0 \), we can divide both sides by \( f(x) \):
\[ f(x) = f(1) \]
This implies that \( f(x) \) is constant for all \( x \).

### Step 3: General Form of the Function
Let’s denote \( f(1) = c \), where \( c \) is a constant. Thus, we can express:
\[ f(x) = c \quad \text{for all } x \]

### Step 4: Testing the Constant Function
Now, we need to check if this constant function satisfies the original equation:
\[ f(x)f(xy) = c \cdot c = c^2 \]
And on the right-hand side:
\[ f(x)f(y) = c \cdot c = c^2 \]
Both sides are equal, confirming that \( f(x) = c \) is indeed a solution.

### Step 5: Exploring Other Possible Forms
Next, we explore if there are other forms of \( f(t) \). We can also consider a function of the form:
\[ f(t) = \frac{t}{k} \]
for some constant \( k \).

### Step 6: Verifying the Form \( f(t) = \frac{t}{k} \)
Substituting \( f(t) = \frac{t}{k} \) into the original equation:
\[ f(x)f(xy) = \frac{x}{k} \cdot \frac{xy}{k} = \frac{xy^2}{k^2} \]
And for the right-hand side:
\[ f(x)f(y) = \frac{x}{k} \cdot \frac{y}{k} = \frac{xy}{k^2} \]
Both sides match, confirming that \( f(t) = \frac{t}{k} \) is also a valid solution.

### Conclusion
Thus, the function \( f(t) \) may be of the form:
\[ f(t) = \frac{t}{k} \quad \text{or} \quad f(t) = c \quad \text{where } c \text{ is a constant.} \]