Question:

Step 1: Determine $f(0)$. Given the functional equation $f(x+y) = f(x)+f(y)+2xy-1$. Set $x=0$ and $y=0$: $$f(0+0) = f(0)+f(0)+2(0)(0)-1$$ $$f(0) = 2f(0)-1$$ $$f(0) = 1$$ Step 2: Determine $f'(x)$. Differentiate the functional equation with respect to $x$, treating $y$ as a constant: $$\frac{d}{dx}f(x+y) = \frac{d}{dx}f(x) + \frac{d}{dx}f(y) + \frac{d}{dx}(2xy) - \frac{d}{dx}(1)$$ $$f'(x+y) = f'(x) + 0 + 2y - 0$$ $$f'(x+y) = f'(x) + 2y$$ Set $x=0$: $$f'(y) = f'(0) + 2y$$ Given $f'(0) = \cos\alpha$: $$f'(y) = \cos\alpha + 2y$$ Replacing $y$ with $x$: $$f'(x) = 2x + \cos\alpha$$ Step 3: Determine $f(x)$. Integrate $f'(x)$ with respect to $x$: $$f(x) = \int (2x + \cos\alpha) dx$$ $$f(x) = x^2 + x\cos\alpha + C$$ Use the value of $f(0)$ from Step 1: $$f(0) = 0^2 + (0)\cos\alpha + C$$ $$1 = C$$ Therefore, the function is: $$f(x) = x^2 + x\cos\alpha + 1$$ Based on the derived function and its derivative, the following statement is true: $$f'(x) = 2x + \cos\alpha$$