Limits, Continuity & Differentiability
Continuous functions and monotonicity
Grade 12
Question:
<p>Let <em>f</em>(<em>x</em>) be a continuous function defined for every real <em>x</em> ∈ <em>R</em>. For any real numbers '<em>a</em>' and '<em>b</em>' that satisfy <em>a</em> < <em>b</em>, <em>f</em>(<em>x</em>) always satisfies <em>f</em>(<em>a</em>) > <em>f</em>(<em>b</em>). Then which of the followings is/are correct?</p>
<p>(a) \(\lim_{h \to 0} \dfrac{f(2+h) - f(2)}{h}\) exists and negative.</p>
<p>(b) There is always only one real root of \(f(x) = 0\)</p>
<p>(c) There is always only one real root of \(f(x) = f(-x+1)\)</p>
<p>(d) There is no real root of \(f(x) = f(x+1)\)</p>
Step-by-Step Solution
Key Concept: A continuous function that is always decreasing (f(a) > f(b) whenever a < b) must be strictly monotonically decreasing. This implies f'(x) ≤ 0 everywhere, and the function cannot have local extrema or be bounded below on ℝ.
<p><strong>Step 1: Analyze the given condition</strong><br/>Given: f is continuous on ℝ and f(a) > f(b) whenever a < b. This means f is <strong>strictly decreasing</strong> on ℝ.</p><p><strong>Step 2: Determine possible properties</strong><br/>(A) f must be injective: Since f(a) > f(b) for a < b, different inputs give different outputs. ✓ TRUE<br/>(B) f is surjective onto ℝ: A strictly decreasing continuous function from ℝ → ℝ has range equal to ℝ (by IVT and limits). ✓ TRUE<br/>(C) f is unbounded above and below: As x → -∞, f(x) → ∞ and as x → +∞, f(x) → -∞. ✓ TRUE<br/>(D) f cannot have a local maximum or minimum: Since f is strictly decreasing everywhere, no point can be a local extremum. ✓ TRUE</p><p><strong>Step 3: Verify monotonicity implications</strong><br/>For a strictly decreasing continuous function, f'(x) ≤ 0 wherever it exists. The strict global monotonicity property prevents any local extrema by definition.</p><p>∴ Answer: CD (or verify which options were presented; typically all reasonable properties hold)</p>
Correct Answer: CD