Sets, Relations & Functions
Types of Functions
Grade 11
Question:
<p>If \(f: \mathbb{R} \to \mathbb{R}\) satisfies \(f(x + y) = f(x) + f(y)\), for all \(x, y \in \mathbb{R}\) and \(f(1) = 7\), then \(\displaystyle\sum_{r=1}^{n} f(r)\) is</p>
<p>\(\dfrac{7n}{2}\)</p>
<p>\(\dfrac{7(n+1)}{2}\)</p>
<p>\(7n(n+1)\)</p>
<p>\(\dfrac{7n(n+1)}{2}\)</p>
Step-by-Step Solution
Key Concept: Cauchy's functional equation f(x+y) = f(x) + f(y) with f(1) = 7 implies f is linear (f(x) = 7x). Use this to convert the sum into an arithmetic series that telescopes or simplifies via standard formulas.
<p><strong>Step 1:</strong> Identify the nature of f from Cauchy's functional equation.</p><p>Given: f(x + y) = f(x) + f(y) for all x, y ∈ ℝ and f(1) = 7</p><p>Setting y = 1: f(x + 1) = f(x) + f(1) = f(x) + 7</p><p>This shows f is linear with constant difference.</p><p><strong>Step 2:</strong> Determine the explicit form of f(x).</p><p>For n ∈ ℕ: f(n) = f(1 + 1 + ... + 1) = nf(1) = 7n</p><p>Therefore, f(x) = 7x for all x ∈ ℝ</p><p><strong>Step 3:</strong> Calculate the required sum.</p><p>∑_{r=1}^{n} f(r) = ∑_{r=1}^{n} 7r = 7∑_{r=1}^{n} r = 7 · n(n+1)/2</p><p>∴ Answer: <strong>7n(n+1)/2</strong> or equivalent form (Option D)</p>
Correct Answer: D