Question

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

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

Hint:

Always compare the coefficients carefully when matching expansions. Special attention to cross terms like \(xy\) is key!

Answer 1

The Correct Option is D

Step 1: Expand \( f(x+y) \).
Given \( f(x) = ux^2 + vx + w \), so \[ f(x+y) = u(x+y)^2 + v(x+y) + w = u(x^2 + 2xy + y^2) + v(x+y) + w, \] \[ = ux^2 + 2uxy + uy^2 + vx + vy + w. \]
Step 2: Expand \( f(x) + f(y) + xy \).
Also, \[ f(x) = ux^2 + vx + w, \quad f(y) = uy^2 + vy + w, \] thus \[ f(x) + f(y) + xy = ux^2 + uy^2 + vx + vy + 2w + xy. \]
Step 3: Compare coefficients from both sides.
From \[ ux^2 + 2uxy + uy^2 + vx + vy + w = ux^2 + uy^2 + vx + vy + 2w + xy, \] equating the coefficients: Coefficient of \(x^2\): \( u = u \) (ok).
Coefficient of \(y^2\): \( u = u \) (ok).
Coefficient of \(xy\): \( 2u = 1 \quad \Rightarrow \quad u = \frac{1}{2} \).
Coefficient of \(x\): \( v = v \) (ok).
Coefficient of \(y\): \( v = v \) (ok).
Constant term: \( w = 2w \quad \Rightarrow \quad w = 0 \).

Step 4: Find \(v\) using the given condition.
Given: \[ u + v + w = 3. \] Substituting \( u = \frac{1}{2} \) and \( w = 0 \), \[ \frac{1}{2} + v + 0 = 3 \quad \Rightarrow \quad v = \frac{5}{2}. \]
Step 5: Find \( f(1) \).
Now, \[ f(1) = u(1)^2 + v(1) + w = \frac{1}{2} + \frac{5}{2} + 0 = 3. \]