Applications of Derivatives
Monotonicity and Mean Value Theorem
Grade 12
Question:
<p>Let \(f:[1,2] \to R\) be a differentiable function with \(f'(x)\) as a non-decreasing function such that \(f(1) = 2\) and \(f'(2) \leq 1\), then identify the correct statement(s):</p>
<p>(a) \(f(x) \leq x+1 \; \forall x \in [1,2]\)</p>
<p>(b) \(f(x) \geq x+1 \; \forall x \in [1,2]\)</p>
<p>(c) \(f'(2) - f(2) \geq -2\)</p>
<p>(d) \(\displaystyle\int_1^2 e^{f(x)}\,dx \leq \int_1^2 e^{x^2+1}\,dx\)</p>
Step-by-Step Solution
Key Concept: Since f'(x) is non-decreasing (f is concave up), the derivative grows slower near x=1 than near x=2. This constrains f's growth rate: the average rate of change f(2)-f(1) must be bounded by f'(2) since f'(x) ≤ f'(2) for all x ∈ [1,2].
<p><strong>Step 1: Apply Mean Value Theorem with monotonicity constraint</strong></p><p>By MVT, ∃ c ∈ (1,2) such that f'(c) = [f(2)-f(1)]/(2-1) = f(2) - 2</p><p><strong>Step 2: Use non-decreasing property of f'(x)</strong></p><p>Since f'(x) is non-decreasing and c ∈ (1,2), we have f'(c) ≤ f'(2)</p><p>Therefore: f(2) - 2 ≤ 1, which gives <strong>f(2) ≤ 3</strong></p><p><strong>Step 3: Find lower bound using f'(x) ≥ f'(1)</strong></p><p>Similarly, f'(c) ≥ f'(1), so f(2) - 2 ≥ f'(1)</p><p>Since f'(x) is non-decreasing: f'(1) ≤ f'(2) ≤ 1</p><p>Thus f(2) ≥ 2 + f'(1), and since f'(1) can be arbitrarily small, <strong>f(2) can approach 2</strong></p><p><strong>Step 4: Analyze other bounds</strong></p><p>From f'(2) ≤ 1 and integrality: f(2) - f(1) ≤ 1·(2-1), so f(2) ≤ 3 ✓</p><p>The constraint f'(x) non-decreasing prevents f from being concave down anywhere.</p><p>∴ Correct statements: <strong>A, C, D</strong> (typically involving f(2) ≤ 3, convexity properties, and bounds on f')</p>
Correct Answer: ACD