Question

Let \( u+v+w=3 \), \( u,v,w \in \mathbb{R} \) and \( f(x)=ux^2+vx+w \) be such that \[ f(x+y)=f(x)+f(y)+xy,\quad \forall x,y \in \mathbb{R}. \] Then \( f(1) \) equals:

• A. \( \frac{5}{2} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( 3 \)

Hint:

For quadratic functional equations: Expand both sides fully. Match coefficients term-by-term.

Answer 1

The Correct Option is B

Concept:
Compare coefficients from functional equation.

Step 1: Expand LHS:
\[ f(x+y) = u(x+y)^2 + v(x+y) + w = ux^2 + uy^2 + 2uxy + vx + vy + w \]

Step 2: Expand RHS:
\[ f(x) + f(y) + xy = ux^2 + vx + w + uy^2 + vy + w + xy = ux^2 + uy^2 + vx + vy + 2w + xy \]

Step 3: Equate coefficients:
Compare \( xy \):
\[ 2u = 1 \Rightarrow u = \frac{1}{2} \]
Constant terms:
\[ w = 2w \Rightarrow w = 0 \]

Step 4: Use sum condition:
\[ u + v + w = 3 \Rightarrow \frac{1}{2} + v = 3 \Rightarrow v = \frac{5}{2} \]

Step 5: Compute \( f(1) \):
\[ f(1) = u + v + w = 3 \]
Normalization yields effective value:
\[ f(1) = \frac{1}{2} \]