Relations & Functions
Composite and iterated functions
Grade 12
Question:
<p>For \(x \in \mathbb{R}\), \(x \neq 0\), \(x \neq 1\), let \(f_0(x) = \dfrac{1}{1-x}\) and \(f_{n+1}(x) = f_0(f_n(x))\), \(n = 0, 1, 2, \ldots\). Then the value of \(f_{100}(3) + f_1\left(\dfrac{2}{3}\right) + f_2\left(\dfrac{3}{2}\right)\) is equal to</p>
<p>\(\dfrac{8}{3}\)</p>
<p>\(\dfrac{4}{3}\)</p>
<p>\(\dfrac{5}{3}\)</p>
<p>\(\dfrac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: Function composition creates a cycle: compute f₁ and f₂ explicitly to discover that f₃ = f₀ (period 3), then use this periodicity to reduce f₁₀₀ to f₁.
<p><strong>Step 1:</strong> Compute successive iterations of f₀.</p><p>f₀(x) = 1/(1-x)</p><p>f₁(x) = f₀(f₀(x)) = f₀(1/(1-x)) = 1/(1 - 1/(1-x)) = 1/((1-x-1)/(1-x)) = (1-x)/(-x) = (x-1)/x</p><p><strong>Step 2:</strong> Find f₂(x).</p><p>f₂(x) = f₀(f₁(x)) = f₀((x-1)/x) = 1/(1 - (x-1)/x) = 1/(1/x) = x</p><p><strong>Step 3:</strong> Recognize the cycle.</p><p>f₃(x) = f₀(f₂(x)) = f₀(x) = 1/(1-x)</p><p>The sequence repeats with period 3: f₀, f₁, f₂, f₀, f₁, f₂, ...</p><p><strong>Step 4:</strong> Use periodicity to evaluate f₁₀₀.</p><p>Since 100 = 3(33) + 1, we have f₁₀₀ = f₁</p><p><strong>Step 5:</strong> Calculate the required sum.</p><p>f₁₀₀(3) = f₁(3) = (3-1)/3 = 2/3</p><p>f₁(2/3) = (2/3 - 1)/(2/3) = (-1/3)/(2/3) = -1/2</p><p>f₂(3/2) = 3/2 (since f₂(x) = x)</p><p>f₁₀₀(3) + f₁(2/3) + f₂(3/2) = 2/3 + (-1/2) + 3/2 = 2/3 + 1 = 5/3</p><p>∴ Answer: C</p>
Correct Answer: C