Let `f: R to R ` be given by
`f(x+(5)/(6))+f(x)=f(x+(1)/(2))+f(x+(1)/(3))` for all ` x in R ` . Then ,

Options

A. f(x) is periodic
B. f(x) is even
C. `f(x+2)-f(x+1)=f(x+1)-f(x)`
D. none of these

Explanation

To solve the functional equation given by

\[ f\left(x + \frac{5}{6}\right) + f(x) = f\left(x + \frac{1}{2}\right) + f\left(x + \frac{1}{3}\right) \]

for all \(x \in \mathbb{R}\), we can follow these steps:

### Step 1: Analyze the functional equation
We start with the equation:

\[ f\left(x + \frac{5}{6}\right) + f(x) = f\left(x + \frac{1}{2}\right) + f\left(x + \frac{1}{3}\right) \]

This suggests a relationship between the values of the function at different points.

### Step 2: Substitute specific values for \(x\)
Let's substitute \(x = 0\):

\[ f\left(0 + \frac{5}{6}\right) + f(0) = f\left(0 + \frac{1}{2}\right) + f\left(0 + \frac{1}{3}\right) \]

This simplifies to:

\[ f\left(\frac{5}{6}\right) + f(0) = f\left(\frac{1}{2}\right) + f\left(\frac{1}{3}\right) \]

### Step 3: Explore the implications of the equation
Now, we can analyze the implications of this equation. If we assume \(f\) is a linear function of the form \(f(x) = ax + b\), we can substitute this into our original functional equation.

### Step 4: Substitute \(f(x) = ax + b\)
Substituting \(f(x) = ax + b\) into the functional equation gives us:

\[ a\left(x + \frac{5}{6}\right) + b + ax + b = a\left(x + \frac{1}{2}\right) + b + a\left(x + \frac{1}{3}\right) + b \]

This simplifies to:

\[ ax + a\frac{5}{6} + 2b = ax + a\frac{1}{2} + b + ax + a\frac{1}{3} + b \]

### Step 5: Simplify and equate coefficients
Combining like terms, we have:

\[ a\frac{5}{6} + 2b = 2ax + b + a\left(\frac{1}{2} + \frac{1}{3}\right) \]

We can simplify the right side:

\[ a\frac{5}{6} + 2b = 2ax + b + a\left(\frac{5}{6}\right) \]

This leads to:

\[ 2b = b \implies b = 0 \]

### Step 6: Conclusion about the function
Thus, we find that \(f(x) = ax\) for some constant \(a\).

### Step 7: Check for properties of the function
1. **Evenness**: \(f(-x) = -ax = -f(x)\), hence \(f\) is an odd function.
2. **Periodicity**: Since \(f(x) = ax\) is a linear function, it is not periodic unless \(a = 0\).
3. **The condition \(f(x + 2) - f(x + 1) = f(x + 1) - f(x)\)** holds true.

### Final Answer
Thus, the correct option is:

**Option C**: \(f(x + 2) - f(x + 1) = f(x + 1) - f(x)\).

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