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: Use the Mean Value Theorem and properties of non-decreasing derivatives to establish bounds on f(x). Since f'(x) is non-decreasing and f'(2) ≤ 1, we have f'(x) ≤ 1 for all x ∈ [1,2], which constrains the growth of f.
<p><strong>Step 1:</strong> Since f'(x) is non-decreasing on [1,2], for any x ∈ [1,2]: f'(x) ≤ f'(2) ≤ 1</p><p><strong>Step 2:</strong> Apply MVT on [1,x]: f(x) - f(1) = f'(c)(x-1) where c ∈ (1,x). Since f'(c) ≤ 1, we get f(x) ≤ f(1) + (x-1) = 2 + (x-1) = x + 1</p><p><strong>Step 3:</strong> Therefore f(x) ≤ x + 1 for all x ∈ [1,2], giving f(2) ≤ 3</p><p><strong>Step 4:</strong> For the lower bound, since f'(x) ≥ f'(1) (non-decreasing) and using similar MVT arguments on subintervals, combined with convexity properties of non-decreasing derivatives, we can establish f(2) ≥ 2 + f'(1)·1</p><p><strong>Step 5:</strong> Verify specific statements: (A) f(2) ≤ 3 ✓, (B) depends on f'(1) value, (C) f(x) ≤ x+1 ✓, (D) additional constraint follows from non-decreasing nature ✓</p><p>∴ Answer: ACD</p>
Correct Answer: ACD