Let `f:R->R` be a function satisfying `f((x y)/2)=(f(x)*f(y))/2,AAx , y in R and f(1)=f'(1)=!=0.` Then, `f(x)+f(1-x)` is (for all non-zero real values of `x`) a.) constant b.) can't be discussed c.) `x d.) 1/x`

Options

A. constant
B. can't be discussed
C. x
D. `(1)/(x)`

Answer

To solve the problem, we need to analyze the given functional equation and find the expression for \( f(x) + f(1 - x) \).

### Step 1: Analyze the functional equation
The functional equation given is:

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

for all \( x, y \in \mathbb{R} \).

### Step 2: Substitute specific values
Let's substitute \( x = 1 \) and \( y = 1 \):

\[ f\left(\frac{1 \cdot 1}{2}\right) = \frac{f(1)f(1)}{2} \]

This simplifies to:

\[ f\left(\frac{1}{2}\right) = \frac{f(1)^2}{2} \]

### Step 3: Assume a form for \( f(x) \)
To find a suitable form for \( f(x) \), we can assume \( f(x) = kx \) for some constant \( k \).

### Step 4: Substitute into the functional equation
Substituting \( f(x) = kx \) into the functional equation:

\[ f\left(\frac{xy}{2}\right) = k\left(\frac{xy}{2}\right) = \frac{kxy}{2} \]

Now substituting \( f(x) \) and \( f(y) \):

\[ \frac{f(x)f(y)}{2} = \frac{(kx)(ky)}{2} = \frac{k^2xy}{2} \]

### Step 5: Equate both sides
Setting both expressions equal gives:

\[ \frac{kxy}{2} = \frac{k^2xy}{2} \]

Assuming \( xy \neq 0 \), we can cancel \( \frac{xy}{2} \) from both sides:

\[ k = k^2 \]

This implies \( k(k - 1) = 0 \), so \( k = 0 \) or \( k = 1 \). Since \( f(1) \neq 0 \), we have \( k = 1 \).

### Step 6: Determine \( f(x) \)
Thus, we conclude:

\[ f(x) = x \]

### Step 7: Calculate \( f(x) + f(1 - x) \)
Now we compute:

\[ f(x) + f(1 - x) = x + (1 - x) = 1 \]

### Conclusion
The expression \( f(x) + f(1 - x) \) is constant and equals \( 1 \).

### Final Answer
The correct option is:
**a.) constant**