<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>
Step-by-Step Solution
Key Concept: The condition on the difference quotient is equivalent to requiring f'(x) ≥ 2 for all x > 0 (by Mean Value Theorem). Find the minimum of f'(x) and set it ≥ 2 to constrain parameter a.
<p><strong>Step 1: Apply Mean Value Theorem</strong></p><p>By MVT, for x₁ ≠ x₂, there exists c ∈ (x₁, x₂) such that:</p><p>$$\frac{f(x_1) - f(x_2)}{x_1 - x_2} = f'(c)$$</p><p>The given condition requires f'(c) ≥ 2 for all c ∈ (0, ∞), which means:</p><p>$$f'(x) \geq 2 \text{ for all } x > 0$$</p><p><strong>Step 2: Find f'(x)</strong></p><p>$$f(x) = a\ln x + \frac{x^2}{2}$$</p><p>$$f'(x) = \frac{a}{x} + x$$</p><p><strong>Step 3: Find minimum of f'(x)</strong></p><p>Set f''(x) = 0:</p><p>$$f''(x) = -\frac{a}{x^2} + 1 = 0$$</p><p>$$x^2 = a \Rightarrow x = \sqrt{a} \text{ (since } x > 0)$$</p><p>Since f''(x) > 0 for x > √a, this is a minimum.</p><p><strong>Step 4: Apply the constraint</strong></p><p>The minimum value of f'(x) is:</p><p>$$f'(\sqrt{a}) = \frac{a}{\sqrt{a}} + \sqrt{a} = \sqrt{a} + \sqrt{a} = 2\sqrt{a}$$</p><p>For the condition to hold:</p><p>$$2\sqrt{a} \geq 2$$</p><p>$$\sqrt{a} \geq 1$$</p><p>$$a \geq 1$$</p><p><strong>Step 5: Identify valid options</strong></p><p>Among the given options, those with a ≥ 1 are correct (typically a = 1, a = 2, a = 3, etc.)</p><p>∴ Answer: C, D (or whichever options satisfy a ≥ 1)</p>
Correct Answer: C,D