f (x) is differentiable function satisfying the relationship `f ^(2) (x) + f ^(2) (y)+2 (xy-1) = f^(2) (x+y) AA x, y in R`
Also `f (x) gt 0 AA x in R and f (sqrt2)=2.` Then which of the following statement (s) is/are correct about `f (x)` ?

Options

A. `[f (3)]=3([.]` denotes greatest integer function)
B. `f (sqrt7) =3`
C. ` f (x)` is even
D. `f '(0)=0`

Answer

To solve the problem, we start with the given relationship:

\[ f^2(x) + f^2(y) + 2(xy - 1) = f^2(x+y) \]

for all \( x, y \in \mathbb{R} \), along with the conditions \( f(x) > 0 \) for all \( x \in \mathbb{R} \) and \( f(\sqrt{2}) = 2 \).

### Step 1: Differentiate the given relationship

We differentiate both sides of the equation with respect to \( x \):

\[ \frac{d}{dx}[f^2(x) + f^2(y) + 2(xy - 1)] = \frac{d}{dx}[f^2(x+y)] \]

Using the chain rule and product rule, we get:

\[ 2f(x)f'(x) + 0 + 2y = 2f(x+y)f'(x+y) \]

### Step 2: Substitute \( x = 0 \)

Let’s substitute \( x = 0 \):

\[ 2f(0)f'(0) + 2y = 2f(y)f'(y) \]

Dividing through by 2 gives:

\[ f(0)f'(0) + y = f(y)f'(y) \]

### Step 3: Analyze the equation

Let \( f'(0) = k \). Then we have:

\[ kf(0) + y = f(y)f'(y) \]

This indicates that \( f(y)f'(y) \) is linear in \( y \).

### Step 4: Set \( y = 0 \)

Now, let’s set \( y = 0 \):

\[ kf(0) = f(0)f'(0) \implies kf(0) = kf(0) \]

This doesn't give new information, so we proceed to analyze the function further.

### Step 5: Substitute \( y = \sqrt{2} \)

Now, substitute \( y = \sqrt{2} \):

\[ kf(0) + \sqrt{2} = f(\sqrt{2})f'(\sqrt{2}) \]

Given \( f(\sqrt{2}) = 2 \):

\[ kf(0) + \sqrt{2} = 2f'(\sqrt{2}) \]

### Step 6: Solve for \( f(0) \)

We can also use the original equation by substituting \( x = y = 0 \):

\[ f^2(0) + f^2(0) + 2(0 \cdot 0 - 1) = f^2(0) \]

This simplifies to:

\[ 2f^2(0) - 2 = f^2(0) \implies f^2(0) = 2 \implies f(0) = \sqrt{2} \]

### Step 7: Find \( k \)

Now substituting \( f(0) = \sqrt{2} \) back into our earlier equation:

\[ k\sqrt{2} + \sqrt{2} = 2f'(\sqrt{2}) \implies (k + 1)\sqrt{2} = 2f'(\sqrt{2}) \]

### Step 8: Find \( f(y) \)

From the earlier steps, we hypothesized that:

\[ f(y) = \sqrt{y^2 + 2} \]

### Step 9: Verify \( f(y) \)

We can verify if this function satisfies the original equation:

1. Calculate \( f^2(x) + f^2(y) + 2(xy - 1) \).
2. Calculate \( f^2(x+y) \).

Both sides should equal.

### Step 10: Check the conditions

1. \( f(3) = \sqrt{3^2 + 2} = \sqrt{11} \) which is between 3 and 4.
2. \( f(\sqrt{7}) = \sqrt{7 + 2} = 3 \).
3. Check if \( f(x) \) is even: \( f(-x) = f(x) \).
4. Calculate \( f'(x) \) and check \( f'(0) = 0 \).

### Conclusion

The statements about \( f(x) \) that are correct are:

1. \( f(3) \) is between 3 and 4.
2. \( f(\sqrt{7}) = 3 \).
3. \( f(x) \) is an even function.
4. \( f'(0) = 0 \).