Statement-1: Let f be a differentiable function satisfying `f(x+y)=f(x)+f(y)+2xy-1" for all "x,y in R` and `f'(0)=a "where"0 lt a lt 1" then ",f(x) gt 0` for all x.
Statement-2: f(x) is statement-1 is of the form `x^(2)+ax+1`

Options

A. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.
B. Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for 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 will analyze the given functional equation and derive the form of the function \( f(x) \).

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

### Step 2: Substitute \( x = 0 \) and \( y = 0 \)
Substituting \( x = 0 \) and \( y = 0 \) into the equation gives:
\[ f(0 + 0) = f(0) + f(0) + 2 \cdot 0 \cdot 0 - 1 \]
This simplifies to:
\[ f(0) = 2f(0) - 1 \]
Rearranging this, we find:
\[ f(0) = 2f(0) - 1 \implies f(0) - 2f(0) = -1 \implies -f(0) = -1 \implies f(0) = 1 \]

### Step 3: Differentiate the functional equation
Next, we differentiate the functional equation with respect to \( y \):
\[ \frac{d}{dy}[f(x+y)] = \frac{d}{dy}[f(x) + f(y) + 2xy - 1] \]
Using the chain rule on the left side, we have:
\[ f'(x+y) = f'(y) + 2x \]
Now, substituting \( y = 0 \):
\[ f'(x+0) = f'(0) + 2x \implies f'(x) = f'(0) + 2x \]
Let \( f'(0) = a \), where \( 0 < a < 1 \). Thus, we have:
\[ f'(x) = a + 2x \]

### Step 4: Integrate to find \( f(x) \)
Now, we integrate \( f'(x) \):
\[ f(x) = \int (a + 2x) \, dx = ax + x^2 + C \]
where \( C \) is the constant of integration.

### Step 5: Use the value of \( f(0) \)
We know \( f(0) = 1 \):
\[ f(0) = a \cdot 0 + 0^2 + C = C = 1 \]
Thus, we have:
\[ f(x) = ax + x^2 + 1 \]

### Step 6: Verify the form of \( f(x) \)
The function can be rewritten as:
\[ f(x) = x^2 + ax + 1 \]
This confirms that \( f(x) \) is of the form \( x^2 + ax + 1 \), which satisfies Statement 2.

### Step 7: Show that \( f(x) > 0 \)
To show that \( f(x) > 0 \) for all \( x \):
The quadratic function \( f(x) = x^2 + ax + 1 \) has a minimum value determined by its vertex. The vertex \( x_v \) is given by:
\[ x_v = -\frac{b}{2a} = -\frac{a}{2} \]
Substituting \( x_v \) into \( f(x) \):
\[ f\left(-\frac{a}{2}\right) = \left(-\frac{a}{2}\right)^2 + a\left(-\frac{a}{2}\right) + 1 = \frac{a^2}{4} - \frac{a^2}{2} + 1 = \frac{a^2}{4} - \frac{2a^2}{4} + 1 = 1 - \frac{a^2}{4} \]
Since \( 0 < a < 1 \), we have \( 1 - \frac{a^2}{4} > 0 \). Therefore, \( f(x) \) has a minimum value greater than 0, implying \( f(x) > 0 \) for all \( x \).

### Conclusion
Thus, both statements are true, and Statement 2 is indeed the correct explanation for Statement 1.