Applications of Derivatives
Monotonicity and Inequalities
Grade 12
Question:
<p>Function \(f(x)\) is such that \(f(x) = a\ln x + \dfrac{x^2}{2}\) where \(a > 0\) is a parameter. If \(\dfrac{f(x_1) - f(x_2)}{x_1 - x_2} \geq 2 \; \forall\, x_1, x_2 \in (0, \infty)\) and \(x_1 \neq x_2\), then possible value of \('a'\) can be:</p>
<p>(a) 0</p>
<p>(b) 1/2</p>
<p>(c) 3/2</p>
<p>(d) 1</p>
Step-by-Step Solution
Key Concept: The condition on the difference quotient means the slope between ANY two points on f(x) is at least 2, which occurs when the minimum value of f'(x) equals 2. Use f'(x) ≥ 2 for all x > 0 to find the constraint on parameter a.
<p><strong>Step 1:</strong> Recognize that the difference quotient condition implies f'(x) ≥ 2 for all x ∈ (0, ∞).</p><p>By the Mean Value Theorem, for any x₁, x₂, there exists c between them where f(c) = [f(x₁) - f(x₂)]/(x₁ - x₂). The given condition requires this derivative at every point to be ≥ 2.</p><p><strong>Step 2:</strong> Compute f'(x):</p><p>f(x) = a ln x + x²/2</p><p>f'(x) = a/x + x</p><p><strong>Step 3:</strong> Require f'(x) ≥ 2 for all x > 0:</p><p>a/x + x ≥ 2 for all x > 0</p><p><strong>Step 4:</strong> Find the minimum of f'(x). Taking d/dx[a/x + x] = -a/x² + 1 = 0:</p><p>x² = a, so x = √a (since x > 0)</p><p>Minimum value: f'(√a) = a/√a + √a = √a + √a = 2√a</p><p><strong>Step 5:</strong> For the condition f'(x) ≥ 2 to hold for all x > 0:</p><p>2√a ≥ 2</p><p>√a ≥ 1</p><p>a ≥ 1</p><p>∴ The possible values of a are those satisfying a ≥ 1</p>
Correct Answer: C,D