Limits, Continuity & Differentiability
Properties of continuous functions
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 satisfying f(a) > f(b) whenever a < b is strictly decreasing everywhere. Use this monotonicity property along with continuity to evaluate logical statements about f's behavior.
<p><strong>Key Observation:</strong> Given that f(a) > f(b) whenever a < b, the function f is <strong>strictly decreasing</strong> on ℝ and is continuous.</p><p><strong>Properties that follow:</strong></p><p><strong>Property 1:</strong> f is <strong>injective (one-one)</strong> — If f(x₁) = f(x₂), then neither x₁ < x₂ nor x₁ > x₂ (by strict monotonicity), so x₁ = x₂. ✓</p><p><strong>Property 2:</strong> f is <strong>surjective (onto)</strong> — By the Intermediate Value Theorem, since f is continuous and strictly decreasing, for any y ∈ ℝ, there exists x ∈ ℝ such that f(x) = y. ✓</p><p><strong>Property 3:</strong> f is <strong>bijective</strong> — Follows from Properties 1 and 2, so f has an inverse f⁻¹: ℝ → ℝ. ✓</p><p><strong>Property 4:</strong> The inverse f⁻¹ is also <strong>strictly decreasing and continuous</strong> — Since f is strictly decreasing, if y₁ < y₂, then f⁻¹(y₁) > f⁻¹(y₂). Continuity of f⁻¹ follows from continuity of f and monotonicity. ✓</p><p><strong>Property 5:</strong> f is NOT necessarily differentiable — Continuity doesn't guarantee differentiability (e.g., f(x) = -|x| is continuous and strictly decreasing but not differentiable at x = 0). ✗</p><p><strong>∴ Answer: CD</strong> (Statements C and D are correct — typically these refer to: C = f is bijective, D = f⁻¹ exists and is continuous/strictly decreasing)</p>
Correct Answer: CD