Statement -1: Let f(x) be a function satisfying `f(x-1)+f(x+1)=sqrt(2)f(x)` for all ` x in R ` . Then f(x) is periodic with period 8. Statement-2: For every natural number n there exists a periodic functions with period n.

Options

A. Statement-1 is True, Statement-2 is True, statement-2 is a correct explanation for the statement-1 .
B. Statement-1 is True, Statement-2 is True, statement-2 is not a correct explanation for the statement-1 .
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False , Statement-2 is True.

Explanation

To solve the problem, we need to analyze the two statements provided.

### Step 1: Analyze Statement 1
We have the functional equation:
\[ f(x-1) + f(x+1) = \sqrt{2} f(x) \]
for all \( x \in \mathbb{R} \). We need to determine if \( f(x) \) is periodic with period 8.

### Step 2: Substitute Values
Let's substitute \( x \) with \( x + 1 \) in the original equation:
\[ f((x + 1) - 1) + f((x + 1) + 1) = \sqrt{2} f(x + 1) \]
This simplifies to:
\[ f(x) + f(x + 2) = \sqrt{2} f(x + 1) \] (Equation 2)

Now, let's substitute \( x \) with \( x - 1 \):
\[ f((x - 1) - 1) + f((x - 1) + 1) = \sqrt{2} f(x - 1) \]
This simplifies to:
\[ f(x - 2) + f(x) = \sqrt{2} f(x - 1) \] (Equation 3)

### Step 3: Combine Equations
Now we have three equations:
1. \( f(x-1) + f(x+1) = \sqrt{2} f(x) \) (Original)
2. \( f(x) + f(x+2) = \sqrt{2} f(x + 1) \) (Equation 2)
3. \( f(x - 2) + f(x) = \sqrt{2} f(x - 1) \) (Equation 3)

Let's add Equations 2 and 3:
\[ (f(x) + f(x + 2)) + (f(x - 2) + f(x)) = \sqrt{2} f(x + 1) + \sqrt{2} f(x - 1) \]
This simplifies to:
\[ 2f(x) + f(x + 2) + f(x - 2) = \sqrt{2} (f(x + 1) + f(x - 1)) \]

### Step 4: Rearranging
From the original equation, we can express \( f(x + 1) + f(x - 1) \) in terms of \( f(x) \):
\[ f(x + 1) + f(x - 1) = \sqrt{2} f(x) \]
Substituting this back into our combined equation gives us:
\[ 2f(x) + f(x + 2) + f(x - 2) = \sqrt{2} \cdot \sqrt{2} f(x) \]
\[ 2f(x) + f(x + 2) + f(x - 2) = 2f(x) \]
Thus, we have:
\[ f(x + 2) + f(x - 2) = 0 \]
This implies:
\[ f(x + 2) = -f(x - 2) \]

### Step 5: Finding Periodicity
Continuing this process, we can find that:
- \( f(x + 4) = -f(x) \)
- \( f(x + 8) = f(x) \)

This shows that \( f(x) \) is periodic with period 8.

### Conclusion for Statement 1
Thus, Statement 1 is true: \( f(x) \) is periodic with period 8.

### Step 6: Analyze Statement 2
Statement 2 states that for every natural number \( n \), there exists a periodic function with period \( n \). This is indeed true since we can construct functions like \( f(x) = \sin\left(\frac{2\pi}{n} x\right) \) which has period \( n \).

### Final Conclusion
Both statements are true, but Statement 2 does not explain Statement 1. Therefore, the correct conclusion is that both statements are true, but Statement 2 is not an explanation for Statement 1.