If f(x) and g(x) are two real functions such that `f(x)+g(x)=e^(x) and f(x)-g(x)=e^(-x)` , then

Options

A. f(x) is an odd function
B. g(x) is an even function
C. f(x) and g(x) are periodic functions.
D. none of these

Answer

To solve the problem, we have two equations involving functions \( f(x) \) and \( g(x) \):

1. \( f(x) + g(x) = e^x \) (Equation 1)
2. \( f(x) - g(x) = e^{-x} \) (Equation 2)

### Step 1: Add the two equations

We start by adding Equation 1 and Equation 2:

\[ (f(x) + g(x)) + (f(x) - g(x)) = e^x + e^{-x} \]

This simplifies to:

\[ 2f(x) = e^x + e^{-x} \]

### Step 2: Solve for \( f(x) \)

Now, divide both sides by 2 to isolate \( f(x) \):

\[ f(x) = \frac{e^x + e^{-x}}{2} \]

### Step 3: Find \( f(-x) \)

Next, we need to find \( f(-x) \):

\[ f(-x) = \frac{e^{-x} + e^{x}}{2} \]

Notice that \( f(-x) = f(x) \). This indicates that \( f(x) \) is an even function.

### Step 4: Subtract the two equations

Now, we subtract Equation 2 from Equation 1:

\[ (f(x) + g(x)) - (f(x) - g(x)) = e^x - e^{-x} \]

This simplifies to:

\[ 2g(x) = e^x - e^{-x} \]

### Step 5: Solve for \( g(x) \)

Now, divide both sides by 2 to isolate \( g(x) \):

\[ g(x) = \frac{e^x - e^{-x}}{2} \]

### Step 6: Find \( g(-x) \)

Next, we need to find \( g(-x) \):

\[ g(-x) = \frac{e^{-x} - e^{x}}{2} = -\frac{e^{x} - e^{-x}}{2} = -g(x) \]

This indicates that \( g(x) \) is an odd function.

### Conclusion

1. **\( f(x) \)** is an even function.
2. **\( g(x) \)** is an odd function.

Given the options:
- The first option states \( f(x) \) is an odd function (incorrect).
- The second option states \( g(x) \) is an even function (incorrect).
- The third option states both functions are periodic (incorrect).
- The fourth option states none of the above (correct).

Thus, the correct answer is **None of the above**.