If `f(2x+y/8,2x-y/8)=x y` , then `f(m , n)=0` only when `m=n` only when `m!=n` `on l yw h e nm=-n` (d) `fora l lma n dn`

Options

A. only when `m = n`
B. only when `m ne n`
C. only when `m = -n`
D. for all m and n

Explanation

To solve the problem, we start with the given function:

\[ f(2x + \frac{y}{8}, 2x - \frac{y}{8}) = xy \]

We need to find the conditions under which \( f(m, n) = 0 \).

### Step 1: Define Variables
Let:
- \( \alpha = 2x + \frac{y}{8} \)
- \( \beta = 2x - \frac{y}{8} \)

From the given equation, we know:

\[ f(\alpha, \beta) = xy \]

### Step 2: Solve for \( x \) and \( y \)
We can derive \( x \) and \( y \) from \( \alpha \) and \( \beta \).

1. **Adding the equations**:
\[ \alpha + \beta = (2x + \frac{y}{8}) + (2x - \frac{y}{8}) = 4x \]
Thus, we have:
\[ x = \frac{\alpha + \beta}{4} \]

2. **Subtracting the equations**:
\[ \alpha - \beta = (2x + \frac{y}{8}) - (2x - \frac{y}{8}) = \frac{y}{4} \]
Thus, we have:
\[ y = 4(\alpha - \beta) \]

### Step 3: Substitute Back into the Function
Now substituting \( x \) and \( y \) back into the function:

\[ f(\alpha, \beta) = \left(\frac{\alpha + \beta}{4}\right) \cdot (4(\alpha - \beta)) \]

This simplifies to:

\[ f(\alpha, \beta) = (\alpha + \beta)(\alpha - \beta) = \alpha^2 - \beta^2 \]

### Step 4: Express \( f(m, n) \)
Now, if we let \( m = \alpha \) and \( n = \beta \), we have:

\[ f(m, n) = m^2 - n^2 \]

### Step 5: Analyze \( f(m, n) = 0 \)
Setting \( f(m, n) = 0 \):

\[ m^2 - n^2 = 0 \]

This implies:

\[ m^2 = n^2 \]

### Step 6: Solve for Conditions
The equation \( m^2 = n^2 \) can be factored as:

\[ (m - n)(m + n) = 0 \]

This gives us two cases:

1. \( m - n = 0 \) → \( m = n \)
2. \( m + n = 0 \) → \( m = -n \)

### Conclusion
Thus, \( f(m, n) = 0 \) under the conditions:
- \( m = n \)
- \( m = -n \)

However, since the problem asks for when \( f(m, n) = 0 \), we conclude that it holds true for all values of \( m \) and \( n \) because both conditions can occur simultaneously.

### Final Answer
The correct answer is:
(d) for all \( m \) and \( n \).