Let u+v+w=3, where u, v, w in mathbb{R} and f | vedclass

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(D) Given $f(x) = u x^2 + v x + w$. Since $f(x+y) = f(x) + f(y) + x y$, we substitute the expression for $f$: $u(x+y)^2 + v(x+y) + w = (u x^2 + v x +

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$3$

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(D) Given $f(x) = u x^2 + v x + w$. Since $f(x+y) = f(x) + f(y) + x y$, we substitute the expression for $f$: $u(x+y)^2 + v(x+y) + w = (u x^2 + v x + w) + (u y^2 + v y + w) + x y$. Expanding the left side: $u(x^2 + 2 x y + y^2) + v x + v y + w = u x^2 + u y^2 + v x + v y + 2 w + x y$. $u x^2 + 2 u x y + u y^2 + v x + v y + w = u x^2 + u y^2 + v x + v y + 2 w + x y$. Comparing the coefficients of $x y$ on both sides, we get $2 u = 1$, which implies $u = \frac{1}{2}$. Comparing the constant terms, we get $w = 2 w$, which implies $w = 0$. Given $u + v + w = 3$, we substitute $u = \frac{1}{2}$ and $w = 0$: $\frac{1}{2} + v + 0 = 3 \implies v = 3 - \frac{1}{2} = \frac{5}{2}$. Thus, $f(x) = \frac{1}{2} x^2 + \frac{5}{2} x$. Calculating $f(1)$: $f(1) = \frac{1}{2}(1)^2 + \frac{5}{2}(1) = \frac{1}{2} + \frac{5}{2} = \frac{6}{2} = 3$.

Let $u+v+w=3$, where $u, v, w \in \mathbb{R}$ and $f(x)=u x^2+v x+w$ be such that $f(x+y)=f(x)+f(y)+x y$ for all $x, y \in \mathbb{R}$. Then $f(1)$ is equal to:

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