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 sinh⁻¹(x) = ln(x + √(x²+1)), so f(x) = 1 + x·sinh⁻¹(x). Then use the derivative relationship: d/dx[sinh⁻¹(x)] = 1/√(1+x²) = g(x) to establish the connection between f and g.
<p><strong>Step 1:</strong> Recognize the special form. Note that sinh⁻¹(x) = ln(x + √(x²+1)), so f(x) = 1 + x·sinh⁻¹(x).</p><p><strong>Step 2:</strong> Recall the derivative of inverse hyperbolic sine: d/dx[sinh⁻¹(x)] = 1/√(1+x²) = g(x).</p><p><strong>Step 3:</strong> Differentiate f(x): f'(x) = d/dx[x·sinh⁻¹(x)] = sinh⁻¹(x) + x·g(x).</p><p><strong>Step 4:</strong> This reveals that f'(x) involves g(x), and f(x) is continuous and differentiable for all x ∈ ℝ since sinh⁻¹(x) and g(x) are defined everywhere.</p><p><strong>Step 5:</strong> The relationship f'(x) = sinh⁻¹(x) + x·g(x) connects f and g through differentiation, establishing that f and g satisfy a differential relationship where g(x) = √(1+x²) is the derivative of sinh⁻¹(x).</p><p>∴ Answer: A</p>
Correct Answer: A