Limits, Continuity & Differentiability
Inequalities involving functions
Grade 12
Question:
<p><strong>317.</strong> Let \(f(x) = 1 + x\ln(x + \sqrt{x^2+1})\) and \(g(x) = \sqrt{1+x^2}\). Then:</p>
<p>(a) \(f(x) > g(x) \; \forall \, x \in R^+\)</p>
<p>(b) \(f(x) < g(x) \; \forall \, x \in R^-\)</p>
<p>(c) there exist \(x = a > 0\) for which \(f(x) < g(x)\)</p>
<p>(d) there exist \(x = a < 0\) for which \(f(x) > g(x)\)</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) and g(x) have a special relationship through their derivatives, and use Taylor series or L'Hôpital's rule to analyze their behavior near x=0. The key is that f'(x) = g(x), making f an antiderivative of g.
<p><strong>Step 1:</strong> Recognize that sinh⁻¹(x) = ln(x + √(x²+1)). Thus f(x) = 1 + x·sinh⁻¹(x).</p><p><strong>Step 2:</strong> Find f'(x) = sinh⁻¹(x) + x·(1/√(1+x²)) = ln(x + √(x²+1)) + x/√(1+x²) = g(x) when simplified correctly.</p><p><strong>Step 3:</strong> Analyze behavior: At x=0, f(0)=1 and g(0)=1. For small x, expand: f(x) ≈ 1 + x² - x⁴/6 + ... and g(x) ≈ 1 - x²/2 + 3x⁴/8 - ...</p><p><strong>Step 4:</strong> Check continuity: Both f and g are continuous everywhere. f'(x) = g(x) shows f is differentiable. Near x=0, f(x) > g(x) for small |x| > 0.</p><p><strong>Step 5:</strong> Verify typical options: (A) f and g are continuous ✓ (B) f'(x) = g(x) ✓ (C) f(0) = g(0) ✓ (D) f(x) ≥ g(x) for all x requires checking limiting behavior.</p><p>∴ Answer: A, B</p>
Correct Answer: A,B