If f is a function of real variable x satisfying `f(x+4)-f(x+2)+f(x)=0` , then f is periodic function with period:

Options

A. 6
B. 8
C. 10
D. 12

Explanation

To solve the problem, we need to determine the period of the function \( f(x) \) given the functional equation:

\[ f(x+4) - f(x+2) + f(x) = 0 \]

### Step-by-Step Solution:

1. **Rewrite the original equation**:
We start with the given equation:
\[ f(x+4) - f(x+2) + f(x) = 0 \]
Rearranging gives:
\[ f(x+4) = f(x+2) - f(x) \]

2. **Substitute \( x \) with \( x + 2 \)**:
Now, we replace \( x \) with \( x + 2 \) in the original equation:
\[ f((x+2)+4) - f((x+2)+2) + f(x+2) = 0 \]
This simplifies to:
\[ f(x+6) - f(x+4) + f(x+2) = 0 \]
Rearranging gives:
\[ f(x+6) = f(x+4) - f(x+2) \]

3. **Substitute \( x \) with \( x + 4 \)**:
Next, we replace \( x \) with \( x + 4 \) in the original equation:
\[ f((x+4)+4) - f((x+4)+2) + f(x+4) = 0 \]
This simplifies to:
\[ f(x+8) - f(x+6) + f(x+4) = 0 \]
Rearranging gives:
\[ f(x+8) = f(x+6) - f(x+4) \]

4. **Establish a pattern**:
We have the following equations:
- From step 1: \( f(x+4) = f(x+2) - f(x) \)
- From step 2: \( f(x+6) = f(x+4) - f(x+2) \)
- From step 3: \( f(x+8) = f(x+6) - f(x+4) \)

Continuing this process, we can see that we can express \( f(x+n) \) in terms of previous values of \( f \).

5. **Finding periodicity**:
If we continue this substitution process, we will eventually find that:
\[ f(x+12) = f(x) \]
This indicates that the function \( f(x) \) is periodic with a period of 12.

### Conclusion:
Thus, the function \( f(x) \) is periodic with period:

\[ \text{Period} = 12 \]