If `f:RrarrR` is a function satisfying the property `f(x+1)+f(x+3)=2" for all" x in R` than `f` is

Options

A. periodic with period 3
B. periodic with period 4
C. non periodic
D. periodic with period 5

Explanation

To solve the problem, we need to analyze the function \( f: \mathbb{R} \to \mathbb{R} \) that satisfies the equation:

\[ f(x+1) + f(x+3) = 2 \quad \text{for all } x \in \mathbb{R} \]

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

We start by substituting \( x \) with \( x + 2 \) in the original equation:

\[ f((x + 2) + 1) + f((x + 2) + 3) = 2 \]

This simplifies to:

\[ f(x + 3) + f(x + 5) = 2 \]

Let’s label this as Equation (2).

### Step 2: Write down both equations

Now we have two equations:

1. \( f(x + 1) + f(x + 3) = 2 \) (Equation 1)
2. \( f(x + 3) + f(x + 5) = 2 \) (Equation 2)

### Step 3: Subtract Equation 1 from Equation 2

Next, we subtract Equation 1 from Equation 2:

\[ (f(x + 3) + f(x + 5)) - (f(x + 1) + f(x + 3)) = 2 - 2 \]

This simplifies to:

\[ f(x + 5) - f(x + 1) = 0 \]

Thus, we can conclude:

\[ f(x + 5) = f(x + 1) \quad \text{(Equation 3)} \]

### Step 4: Substitute \( x \) with \( x - 1 \)

Now, we will replace \( x \) with \( x - 1 \) in Equation 3:

\[ f((x - 1) + 5) = f((x - 1) + 1) \]

This simplifies to:

\[ f(x + 4) = f(x) \quad \text{(Equation 4)} \]

### Step 5: Conclusion about periodicity

From Equation 4, we see that the function \( f \) satisfies the property:

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

This indicates that the function \( f \) is periodic with a period of 4.

### Final Answer

Thus, the function \( f \) is periodic with period 4.

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