Basic Mathematics & Logarithm
Functional Equations
Grade 11
Question:
<p>Let <i>f</i> : ℕ → ℕ be such that <i>f</i>(<i>n</i> + 1) ≥ <i>f</i>(<i>f</i>(<i>n</i>)) for all <i>n</i> ∈ ℕ, then</p>
<p>(a) <i>f</i>(<i>x</i>) = <i>x</i></p>
<p>(b) <i>f</i>(<i>x</i>) = <i>x</i> − 1</p>
<p>(c) <i>f</i>(<i>x</i>) = <i>x</i>² + 1</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: We need to find which function f: ℕ → ℕ satisfies f(n+1) ≥ f(f(n)) for all n ∈ ℕ. By testing each option with the constraint and analyzing the behavior of the function iteratively, we can eliminate invalid functions.
<p><strong>Step 1: Check Option (b): f(x) = x - 1</strong></p><p>For f: ℕ → ℕ, we need f(n) ∈ ℕ for all n ∈ ℕ. If f(1) = 1 - 1 = 0 ∉ ℕ, this violates the codomain requirement. ✗</p><p><strong>Step 2: Check Option (c): f(x) = x² + 1</strong></p><p>Test n = 1: f(2) ≥ f(f(1))</p><p>f(2) = 4 + 1 = 5</p><p>f(1) = 1 + 1 = 2, so f(f(1)) = f(2) = 5</p><p>We need 5 ≥ 5 ✓</p><p>Test n = 2: f(3) ≥ f(f(2))</p><p>f(3) = 9 + 1 = 10</p><p>f(2) = 5, so f(f(2)) = f(5) = 25 + 1 = 26</p><p>We need 10 ≥ 26, which is FALSE ✗</p><p><strong>Step 3: Check Option (a): f(x) = x</strong></p><p>The constraint becomes: n + 1 ≥ f(n)</p><p>If f(n) = n, then: n + 1 ≥ n, which is TRUE for all n ∈ ℕ ✓</p><p>Also, f: ℕ → ℕ is clearly satisfied since f(n) = n ∈ ℕ for all n ∈ ℕ ✓</p><p><strong>Step 4: Verify Uniqueness</strong></p><p>From f(n+1) ≥ f(f(n)), applying to n=1: f(2) ≥ f(f(1))</p><p>The identity function f(x) = x is the only standard function that satisfies this for all n without growing too rapidly or violating the natural number constraint.</p><p><strong>∴ Answer: a</strong></p>
Correct Answer: a