Applications of Derivatives
Convex Functions and Inequalities
Grade 12
Question:
<p><strong>Ex. 10:</strong> Let f(x) be a twice differentiable function such that f''(x) ≥ 0 in [0, 2]. Then which statement is correct?</p><p>(a) \(f(0) + f(2) ≥ 2f(c), 0 ≤ c ≤ 2\)</p><p>(b) \(f(0) + f(2) ≥ 2f(1)\)</p><p>(c) \(f(0) + f(2) ≤ 2f(1)\)</p><p>(d) \(f(0) + f(2) ≤ 2f(1)\)</p>
<p>(a) \(f(0) + f(2) ≥ 2f(c), 0 ≤ c ≤ 2\)</p>
<p>(b) \(f(0) + f(2) ≥ 2f(1)\)</p>
<p>(c) \(f(0) + f(2) ≤ 2f(1)\)</p>
<p>(d) \(f(0) + f(2) ≤ 2f(1)\)</p>
Step-by-Step Solution
Key Concept: When f''(x) ≥ 0, the function is convex, which implies that the chord connecting any two points lies above the curve. Apply Lagrange's MVT and properties of second derivatives.
<p><strong>Solution (a, d):</strong> By the Intermediate Mean Value Theorem:</p><p>$$\frac{f(0) + f(2)}{2} = f(c), 0 ≤ c ≤ 2 \quad \text{...(i)}$$</p><p>By Lagrange's Mean Value Theorem:</p><p>$$f(1) - f(0) = f'(c_1), 0 ≤ c_1 ≤ 1 \quad \text{...(ii)}$$</p><p>$$f(2) - f(1) = f'(c_2), 1 ≤ c_2 ≤ 2 \quad \text{...(iii)}$$</p><p>Subtracting Eq. (ii) from Eq. (iii):</p><p>$$f(2) - f(0) - 2f(1) = f'(c_2) - f'(c_1) \quad \text{...(iv)}$$</p><p>By Lagrange's Mean Value Theorem again:</p><p>$$f''(c_3) = \frac{f'(c_2) - f'(c_1)}{c_2 - c_1}, \text{ for some } c_3 \in (c_1, c_2)$$</p><p>$$f'(c_2) - f'(c_1) = (c_2 - c_1)f''(c_3) ≥ 0 \quad \text{[since } f''(x) ≥ 0\text{]} \quad \text{...(v)}$$</p><p>From Eqs. (iv) and (v):</p><p>$$f(2) - f(0) - 2f(1) ≥ 0 \implies f(0) + f(2) ≥ 2f(1)$$</p><p>∴ Statements (a) and (d) are correct.</p>
Correct Answer: a, d