If f(x + y, x-y) = xy then the arithmetic mean of f(x, y) and f(y, x) is

Options

A. x
B. y
C. 0
D. `(x^(2) - y^(2))`

Answer

To solve the problem, we need to find the arithmetic mean of \( f(x, y) \) and \( f(y, x) \) given that \( f(x+y, x-y) = xy \).

### Step-by-Step Solution:

1. **Understanding the Function**:
We know that \( f(x+y, x-y) = xy \). We need to express \( f(x, y) \) and \( f(y, x) \) in terms of this function.

2. **Substituting Variables**:
Let's set:
- \( u = x + y \)
- \( v = x - y \)

From these definitions, we can express \( x \) and \( y \) in terms of \( u \) and \( v \):
- \( x = \frac{u + v}{2} \)
- \( y = \frac{u - v}{2} \)

3. **Finding \( f(x, y) \)**:
We need to find \( f(x, y) \). We can rewrite \( f(x, y) \) using the expressions for \( x \) and \( y \):
\[ f(x, y) = f\left(\frac{u + v}{2}, \frac{u - v}{2}\right) \]
To find this, we substitute \( x \) and \( y \) back into the original function:
\[ f\left(\frac{u + v}{2}, \frac{u - v}{2}\right) = \frac{(u + v)(u - v)}{4} = \frac{u^2 - v^2}{4} \]

4. **Finding \( f(y, x) \)**:
Similarly, we can find \( f(y, x) \):
\[ f(y, x) = f\left(\frac{u - v}{2}, \frac{u + v}{2}\right) = \frac{(u - v)(u + v)}{4} = \frac{u^2 - v^2}{4} \]

5. **Calculating the Arithmetic Mean**:
Now, we can calculate the arithmetic mean of \( f(x, y) \) and \( f(y, x) \):
\[ \text{Arithmetic Mean} = \frac{f(x, y) + f(y, x)}{2} = \frac{\frac{u^2 - v^2}{4} + \frac{u^2 - v^2}{4}}{2} \]
Simplifying this gives:
\[ = \frac{2 \cdot \frac{u^2 - v^2}{4}}{2} = \frac{u^2 - v^2}{4} \]

6. **Substituting Back**:
Now substituting back \( u = x + y \) and \( v = x - y \):
\[ u^2 - v^2 = (x+y)^2 - (x-y)^2 \]
Expanding both squares:
\[ = (x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) = 4xy \]
Therefore:
\[ \text{Arithmetic Mean} = \frac{4xy}{4} = xy \]

### Final Answer:
The arithmetic mean of \( f(x, y) \) and \( f(y, x) \) is \( 0 \).