Let `f(x)=1/2[f(xy)+f(x/y)] " for " x,y in R^(+)` such that f(1)=0,f'(1)=2.`
f(e) is equal to

Options

A. 2
B. 1
C. 3
D. 4

Answer

To solve the problem, we start with the functional equation given:

\[ f(x) = \frac{1}{2} \left[ f(xy) + f\left(\frac{x}{y}\right) \right] \]

for all \( x, y \in \mathbb{R}^+ \), along with the conditions \( f(1) = 0 \) and \( f'(1) = 2 \).

### Step 1: Rewrite the functional equation

We can rewrite the functional equation as:

\[ 2f(x) = f(xy) + f\left(\frac{x}{y}\right) \]

### Step 2: Assume a form for \( f(x) \)

To solve this, we can assume that \( f(x) \) has the form \( f(x) = k \log x \), where \( k \) is a constant to be determined.

### Step 3: Substitute the assumed form into the equation

Substituting \( f(x) = k \log x \) into the functional equation:

\[ 2(k \log x) = k \log(xy) + k \log\left(\frac{x}{y}\right) \]

Using properties of logarithms, we can simplify the right-hand side:

\[ k \log(xy) = k(\log x + \log y) \quad \text{and} \quad k \log\left(\frac{x}{y}\right) = k(\log x - \log y) \]

Thus,

\[ k \log(xy) + k \log\left(\frac{x}{y}\right) = k(\log x + \log y + \log x - \log y) = k(2 \log x) \]

### Step 4: Equate both sides

Now we have:

\[ 2k \log x = k(2 \log x) \]

This holds true for all \( x \), confirming our assumption that \( f(x) = k \log x \) is valid.

### Step 5: Use the conditions to find \( k \)

Now we apply the conditions given in the problem:

1. From \( f(1) = 0 \):

\[ f(1) = k \log 1 = k \cdot 0 = 0 \quad \text{(which is satisfied for any } k\text{)} \]

2. From \( f'(1) = 2 \):

First, we differentiate \( f(x) = k \log x \):

\[ f'(x) = \frac{k}{x} \]

Now substituting \( x = 1 \):

\[ f'(1) = \frac{k}{1} = k \]

Setting this equal to 2 gives us:

\[ k = 2 \]

### Step 6: Write the final form of \( f(x) \)

Thus, we have:

\[ f(x) = 2 \log x \]

### Step 7: Find \( f(e) \)

Now we can find \( f(e) \):

\[ f(e) = 2 \log e \]

Since \( \log e = 1 \):

\[ f(e) = 2 \cdot 1 = 2 \]

### Final Answer

Thus, \( f(e) \) is equal to \( \boxed{2} \).

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