Question

If a differentiable function satisfies \[ (x-y)f(x+y) - (x+y)f(x-y) = 2(x^2y-y^3), \qquad \forall x,y\in\mathbb{R} \] and \(f(1)=2\), then:

• A. \(f(x)\) must be a polynomial function
• B. \(f(3)=13\)
• C. \(f(3)=12\)
• D. \(f(0)=0\)

Hint:

For functional equations containing both \(f(x+y)\) and \(f(x-y)\), the substitution \[ u=x+y, \qquad v=x-y \] usually converts the equation into a separable algebraic form.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a functional equation problem. We transform the independent variables \( x, y \) into new variables \( u, v \) to simplify the structure of the equation and derive the general form of the function \( f(x) \).
Step 2: Key Formula or Approach:
1. Let \( u = x + y \) and \( v = x - y \).
2. Express \( x \) and \( y \) in terms of \( u \) and \( v \): \( x = \frac{u+v}{2}, y = \frac{u-v}{2} \).
Step 3: Detailed Explanation:
Substitute \( u \) and \( v \) into the equation:
\[ v f(u) - u f(v) = 2 \left( \left( \frac{u+v}{2} \right)^2 \left( \frac{u-v}{2} \right) - \left( \frac{u-v}{2} \right)^3 \right) \]
Factor out \( \frac{u-v}{8} \) from the right side:
\[ v f(u) - u f(v) = 2 \cdot \frac{u-v}{8} \left[ (u+v)^2 - (u-v)^2 \right] \]
\[ v f(u) - u f(v) = \frac{u-v}{4} [4uv] = uv(u - v) = u^2v - uv^2 \]
Divide the entire equation by \( uv \):
\[ \frac{f(u)}{u} - \frac{f(v)}{v} = u - v \implies \frac{f(u)}{u} - u = \frac{f(v)}{v} - v \]
Since this holds for all \( u, v \), both sides must be equal to a constant \( C \).
\[ \frac{f(x)}{x} - x = C \implies f(x) = x^2 + Cx \]
Given \( f(1) = 2 \):
\[ 1^2 + C(1) = 2 \implies C = 1 \]
The function is \( f(x) = x^2 + x \).
- (A) \( f(x) = x^2 + x \) is a polynomial. Correct.
- (D) \( f(0) = 0^2 + 0 = 0 \). Correct.
- (C) \( f(3) = 3^2 + 3 = 12 \). Correct.
Step 4: Final Answer:
The function is \( f(x) = x^2 + x \). Options (A), (C), and (D) are correct.