Relations & Functions
Functional equations
Grade 12
Question:
<p>A function \(f: R \to R\) satisfies the equation \(f(x)f(y) - f(xy) = x + y,\ \forall\ x, y \in R\) and \(f(1) > 0\), then:</p>
<p>(a) \(f(x)f^{-1}(x) = x^2 - 4\)</p>
<p>(b) \(f(x)f^{-1}(x) = x^2 - 6\)</p>
<p>(c) \(f(x)f^{-1}(x) = x^2 - 1\)</p>
<p>(d) \(f(x)f^{-1}(x) = x^2\)</p>
Step-by-Step Solution
Key Concept: Substitute specific values (x=0, y=1, x=y=1) into the functional equation to find f(0) and f(1), then use f(1)>0 to determine the function uniquely.
<p><strong>Step 1:</strong> Substitute x=0, y=0:</p><p>f(0)·f(0) - f(0) = 0 + 0</p><p>f(0)² - f(0) = 0 → f(0)[f(0)-1] = 0</p><p>So f(0) = 0 or f(0) = 1</p><p><strong>Step 2:</strong> Substitute x=1, y=1:</p><p>f(1)·f(1) - f(1) = 1 + 1</p><p>f(1)² - f(1) - 2 = 0 → [f(1)-2][f(1)+1] = 0</p><p>So f(1) = 2 or f(1) = -1</p><p>Since f(1) > 0, we have <strong>f(1) = 2</strong></p><p><strong>Step 3:</strong> Substitute y=1:</p><p>f(x)·f(1) - f(x) = x + 1</p><p>2f(x) - f(x) = x + 1</p><p>f(x) = x + 1</p><p><strong>Step 4:</strong> Verify: f(x)f(y) - f(xy) = (x+1)(y+1) - (xy+1) = xy + x + y + 1 - xy - 1 = x + y ✓</p><p>Check f(0)=1 ✓ and f(1)=2>0 ✓</p><p>∴ Answer: f(x) = x + 1 (Answer C)</p>
Correct Answer: C