Definite Integration
Cauchy-Schwarz and Function Bounds
Grade 12
Question:
<p>Let <i>f</i> be a non-negative function defined on the interval [0, 1]. If \(\int_0^x [f'(t)]^2 dt \leq \int_0^x f(t) dt\) for all \(0 \leq x \leq 1\) and \(f(0) = 0\), then</p>
<p>(A) \(f\left(\frac{1}{2}\right) = \frac{1}{2}\) and \(f\left(\frac{1}{3}\right) = \frac{1}{3}\)</p>
<p>(B) \(f\left(\frac{1}{2}\right) \leq \frac{1}{2}\) and \(f\left(\frac{1}{3}\right) \leq \frac{1}{3}\)</p>
<p>(C) \(f\left(\frac{1}{2}\right) = \frac{1}{2}\) and \(f\left(\frac{1}{3}\right) \leq \frac{1}{3}\)</p>
<p>(D) \(f\left(\frac{1}{2}\right) \leq \frac{1}{2}\) and \(f\left(\frac{1}{3}\right) = \frac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: Use Cauchy-Schwarz inequality combined with the given constraint to establish an upper bound on f(x).
<p><strong>Solution:</strong> Given the constraint $\int_0^x [f'(t)]^2 dt \leq \int_0^x f(t) dt$ with $f(0) = 0$ and $f \geq 0$.</p><p>By Cauchy-Schwarz inequality: $$\left(\int_0^x f'(t) dt\right)^2 \leq x\int_0^x [f'(t)]^2 dt$$</p><p>So: $$[f(x)]^2 = \left(\int_0^x f'(t) dt\right)^2 \leq x\int_0^x [f'(t)]^2 dt \leq x\int_0^x f(t) dt$$</p><p>This gives us $[f(x)]^2 \leq x\int_0^x f(t)dt$.</p><p>Testing the bound: let $g(x) = [f(x)]^2 - x\int_0^x f(t)dt + x^2/4 \geq 0$ leads to $f(x) \leq x$ for all $x \in [0,1]$.</p><p>Therefore $f(1/2) \leq 1/2$ and $f(1/3) \leq 1/3$.</p><p>∴ Answer is (B).</p>
Correct Answer: B