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 need to analyze the given functional equation and derive the function \( f(x) \).

### Step 1: Analyze the given functional equation
The functional equation given is:
\[ f^2(x) + f^2(y) + 2(xy - 1) = f^2(x+y) \]
for all \( x, y \in \mathbb{R} \).

### Step 2: Differentiate with respect to \( x \)
We differentiate both sides of the equation with respect to \( x \):
\[ \frac{d}{dx}[f^2(x)] + 0 + 2y = \frac{d}{dx}[f^2(x+y)] \]
Using the chain rule, we have:
\[ 2f(x)f'(x) + 2y = 2f(x+y)f'(x+y) \]
Dividing through by 2 gives:
\[ f(x)f'(x) + y = f(x+y)f'(x+y) \]

### Step 3: Substitute \( x = 0 \)
Now, let’s substitute \( x = 0 \):
\[ f(0)f'(0) + y = f(y)f'(y) \]
Let \( c = f(0)f'(0) \). Thus, we can rewrite the equation as:
\[ f(y)f'(y) = y + c \]

### Step 4: Integrate the equation
We can rearrange this to:
\[ f(y)f'(y) = y + c \]
Integrating both sides with respect to \( y \):
\[ \int f(y)f'(y) \, dy = \int (y + c) \, dy \]
The left side becomes:
\[ \frac{f^2(y)}{2} = \frac{y^2}{2} + cy + d \]
where \( d \) is the constant of integration.

### Step 5: Rearranging the equation
Multiplying through by 2 gives:
\[ f^2(y) = y^2 + 2cy + 2d \]

### Step 6: Use the condition \( f(\sqrt{2}) = 2 \)
We know \( f(\sqrt{2}) = 2 \):
\[ f^2(\sqrt{2}) = 4 = (\sqrt{2})^2 + 2c\sqrt{2} + 2d \]
This simplifies to:
\[ 4 = 2 + 2c\sqrt{2} + 2d \implies 2 = 2c\sqrt{2} + 2d \]
Dividing by 2:
\[ 1 = c\sqrt{2} + d \tag{1} \]

### Step 7: Use the condition \( f(0) \)
Substituting \( x = 0 \) and \( y = 0 \) in the original equation gives:
\[ f^2(0) = 0 + 0 + 2d \implies f^2(0) = 2d \]
Let \( f(0) = k \), then:
\[ k^2 = 2d \tag{2} \]

### Step 8: Solve equations (1) and (2)
From (2), we have \( d = \frac{k^2}{2} \). Substituting into (1):
\[ 1 = c\sqrt{2} + \frac{k^2}{2} \]
We also know \( f(0) = k \) and \( f(0)f'(0) = c \). Since \( f(0) > 0 \), we can conclude \( c = 0 \) (as \( f'(0) \) must be 0 for the product to be zero).

### Step 9: Substitute \( c = 0 \) into the equations
Substituting \( c = 0 \) into (1):
\[ 1 = 0 + d \implies d = 1 \]
Now substituting \( d = 1 \) into (2):
\[ k^2 = 2 \implies k = \sqrt{2} \]

### Step 10: Final function form
Thus, we have:
\[ f^2(y) = y^2 + 2 \implies f(y) = \sqrt{y^2 + 2} \]

### Step 11: Check statements
Now we can check the statements:
1. \( f(3) = \sqrt{3^2 + 2} = \sqrt{11} \) (integral part is 3).
2. \( f(\sqrt{7}) = \sqrt{7 + 2} = 3 \).
3. \( f(x) \) is even since \( f(-x) = f(x) \).
4. \( f'(0) = 0 \) since \( f'(y) = \frac{y}{\sqrt{y^2 + 2}} \) evaluated at \( y = 0 \) gives 0.

All statements are correct.