Applications of Derivatives
Local extrema
Grade 12
Question:
<p>Let <span class="math">f(x) = \int_0^x \cos\left(\frac{t^2 + 2t + 1}{5}\right) dt</span>, <span class="math">0 \leq x \leq 2</span>. Then <span class="math">f(x)</span></p>
<p>(a) increases monotonically</p>
<p>(b) decreasing monotonically</p>
<p>(c) has one point of local maximum</p>
<p>(d) has one point of local minima</p>
Step-by-Step Solution
Key Concept: Use the Fundamental Theorem of Calculus to find f'(x), then analyze where f'(x) = 0 and its sign changes.
<p><strong>Step 1:</strong> By Fundamental Theorem of Calculus:</p><p><span class="math">f'(x) = \cos\left(\frac{x^2 + 2x + 1}{5}\right) = \cos\left(\frac{(x+1)^2}{5}\right)</span></p><p><strong>Step 2:</strong> For <span class="math">0 \leq x \leq 2</span>, we have <span class="math">1 \leq (x+1)^2 \leq 9</span>, so <span class="math">\frac{1}{5} \leq \frac{(x+1)^2}{5} \leq \frac{9}{5}</span></p><p><strong>Step 3:</strong> Since <span class="math">\cos\left(\frac{(x+1)^2}{5}\right) = 0</span> when <span class="math">\frac{(x+1)^2}{5} = \frac{\pi}{2}</span>, and this value lies in the range <span class="math">\left[\frac{1}{5}, \frac{9}{5}\right]</span>, there is exactly one critical point.</p><p><strong>Step 4:</strong> The sign of <span class="math">f'(x)</span> changes from positive to negative, indicating a local maximum.</p><p>∴ Answer is (c).</p>
Correct Answer: C