If `f(x)=(2^(x)+2^(-x))/(2)`, then `f(x+y).f(x-y)` is equal to

Options

A. `(1)/(2) [f(x+y) +f(x-y)]`
B. `(1)/(2)[f(2x) +f(2y)]`
C. `(1)/(2) [f(x+y) .f(x-y)]`
D. None of these

Answer

To solve the problem, we need to find the value of \( f(x+y) \cdot f(x-y) \) given that \( f(x) = \frac{2^x + 2^{-x}}{2} \).

### Step-by-Step Solution:

1. **Calculate \( f(x+y) \)**:
\[ f(x+y) = \frac{2^{x+y} + 2^{-(x+y)}}{2} \]

2. **Calculate \( f(x-y) \)**:
\[ f(x-y) = \frac{2^{x-y} + 2^{-(x-y)}}{2} \]

3. **Multiply \( f(x+y) \) and \( f(x-y) \)**:
\[ f(x+y) \cdot f(x-y) = \left(\frac{2^{x+y} + 2^{-(x+y)}}{2}\right) \cdot \left(\frac{2^{x-y} + 2^{-(x-y)}}{2}\right) \]
\[ = \frac{(2^{x+y} + 2^{-(x+y)})(2^{x-y} + 2^{-(x-y)})}{4} \]

4. **Expand the product**:
\[ = \frac{2^{(x+y) + (x-y)} + 2^{(x+y) - (x-y)} + 2^{-(x+y) + (x-y)} + 2^{-(x+y) - (x-y)}}{4} \]
\[ = \frac{2^{2x} + 2^{2y} + 2^{-2y} + 2^{-2x}}{4} \]

5. **Rearranging the expression**:
\[ = \frac{2^{2x} + 2^{-2x} + 2^{2y} + 2^{-2y}}{4} \]

6. **Recognize the function form**:
Notice that:
\[ f(2x) = \frac{2^{2x} + 2^{-2x}}{2} \]
and
\[ f(2y) = \frac{2^{2y} + 2^{-2y}}{2} \]

7. **Final expression**:
Therefore, we can express:
\[ f(x+y) \cdot f(x-y) = \frac{1}{4} \left( f(2x) + f(2y) \right) \]

### Conclusion:
Thus, we have:
\[ f(x+y) \cdot f(x-y) = \frac{1}{4} \left( f(2x) + f(2y) \right) \]