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WBJEE · Maths · Functions

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

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

Answer & Solution

Correct Answer

(D) \(3\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & u+v+w=3 \\ & f(x)=u x^2+v x+w \\ & f(x+y)=f(x)+f(y)+x y \\ & f(1)=3\end{aligned}\)