Relations & Functions
Functional Equations
Grade 12
Question:
<p>If a function satisfies \((x-y)f(x+y) - (x+y)f(x-y) = 2(x^2y - y^3)\ \forall\ x,\ y \in \mathbb{R}\) and \(f(1) = 2\), then</p>
<p>(a) \(f(x)\) must be polynomial function</p>
<p>(b) \(f(3) = 12\)</p>
<p>(c) \(f(0) = 0\)</p>
<p>(d) \(f(x)\) may not be differentiable</p>
Step-by-Step Solution
Key Concept: Substitute specific values of x and y to extract information about f, then use the functional equation structure to determine that f must be a polynomial of specific form (typically odd or even components).
<p><strong>Step 1:</strong> Substitute y = 0 in the given equation:<br/>(x - 0)f(x + 0) - (x + 0)f(x - 0) = 2(0 - 0)<br/>xf(x) - xf(x) = 0 ✓ (identity, no information)</p><p><strong>Step 2:</strong> Substitute x = 0:<br/>(0 - y)f(y) - (0 + y)f(-y) = 2(0 - y³)<br/>-yf(y) - yf(-y) = -2y³<br/>f(y) + f(-y) = 2y² ... (i)</p><p><strong>Step 3:</strong> Substitute y = x:<br/>(x - x)f(2x) - (x + x)f(0) = 2(x³ - x³)<br/>0 = 0 (no information)</p><p><strong>Step 4:</strong> Substitute x = y:<br/>(x - x)f(2x) - (2x)f(0) = 2(x³ - x³)<br/>-2xf(0) = 0 ⟹ f(0) = 0</p><p><strong>Step 5:</strong> From equation (i) with f(0) = 0:<br/>f(y) + f(-y) = 2y² means f is even part + odd part structure</p><p><strong>Step 6:</strong> Assume f(x) = x² + cx (combining even and odd parts). Verify:<br/>f(-x) = x² - cx, so f(x) + f(-x) = 2x² ✓</p><p><strong>Step 7:</strong> Use f(1) = 2:<br/>1 + c = 2 ⟹ c = 1<br/>Therefore f(x) = x² + x</p><p><strong>Step 8:</strong> Verify in original equation with f(x) = x² + x:<br/>(x-y)(x²+2xy+y²+x+y) - (x+y)(x²-2xy+y²+x-y) = 2(x²y - y³)<br/>Expanding confirms this works.</p><p>∴ Answer: f(x) = x² + x (and related true statements about f)
Correct Answer: A,B,C