Let `f(x)` be a function satisfying `f(x) + f(x+2) = 10 AA x in R`, then

Options

A. `f(x)` is a periodic function
B. `f(x)` is aperiodic function
C. `underset(1)overset(7)intf(x) dx = 20`
D. `underset(1)overset(7)intf(x) dx = 20`

Answer

To solve the problem, we need to analyze the function \( f(x) \) that satisfies the equation:

\[ f(x) + f(x+2) = 10 \quad \text{for all } x \in \mathbb{R} \]

### Step 1: Substitute \( x \) with \( x + 2 \)

Let's start by substituting \( x \) with \( x + 2 \) in the original equation:

\[ f(x + 2) + f((x + 2) + 2) = 10 \]

This simplifies to:

\[ f(x + 2) + f(x + 4) = 10 \]

### Step 2: Express \( f(x + 2) \)

From the original equation \( f(x) + f(x + 2) = 10 \), we can express \( f(x + 2) \) in terms of \( f(x) \):

\[ f(x + 2) = 10 - f(x) \]

### Step 3: Substitute \( f(x + 2) \) into the new equation

Now, we substitute \( f(x + 2) \) into the equation we obtained in Step 1:

\[ (10 - f(x)) + f(x + 4) = 10 \]

### Step 4: Simplify the equation

This simplifies to:

\[ 10 - f(x) + f(x + 4) = 10 \]

Subtracting 10 from both sides gives:

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

### Step 5: Rearranging the equation

Rearranging the equation gives us:

\[ f(x + 4) = f(x) \]

### Conclusion: Periodicity of the function

The equation \( f(x + 4) = f(x) \) indicates that the function \( f(x) \) is periodic with a period of 4. Thus, the function repeats its values every 4 units.

### Final Answer

The function \( f(x) \) is periodic with period 4.

---