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=y=0, then x=y=1, then x=1) to extract the functional form, recognizing that the additive term on RHS constrains f to be linear plus a constant term.
<p><strong>Step 1:</strong> Put x=y=0: f(0)·f(0) - f(0) = 0 + 0 ⟹ 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> Put y=0: f(x)·f(0) - f(0) = x + 0 ⟹ f(x)·f(0) - f(0) = x</p><p>If f(0)=0: -0 = x, which fails for x≠0. Therefore f(0)=1</p><p><strong>Step 3:</strong> Put y=1: f(x)·f(1) - f(x) = x + 1 ⟹ f(x)[f(1)-1] = x + 1</p><p><strong>Step 4:</strong> Put x=y=1: f(1)² - f(1) = 2 ⟹ f(1)² - f(1) - 2 = 0 ⟹ [f(1)-2][f(1)+1] = 0</p><p>Since f(1)>0, we have f(1)=2</p><p><strong>Step 5:</strong> From Step 3: f(x)[2-1] = x+1 ⟹ f(x) = x+1</p><p><strong>Step 6:</strong> Verify: f(x)f(y) - f(xy) = (x+1)(y+1) - (xy+1) = xy + x + y + 1 - xy - 1 = x + y ✓</p><p>∴ Answer: C (f(x) = x+1)</p>
Correct Answer: C