Limits, Continuity & Differentiability
Chain Rule and Product Rule
Grade 12
Question:
<p>If <span class="math">f'(x) = g(x)</span> and <span class="math">g'(x) = -f(x)</span> for all <span class="math">x</span> and <span class="math">f(2) = 4 = f'(2)</span>, then <span class="math">(f(24))^2 + (g(24))^2</span> is</p>
<p>(a) 32</p>
<p>(b) 24</p>
<p>(c) 64</p>
<p>(d) 48</p>
Step-by-Step Solution
Key Concept: If the derivative of a sum is zero, that sum is constant throughout the domain.
<p>We have, <span class="math">\frac{d}{dx}\{(f(x))^2 + (g(x))^2\} = 2f(x) \cdot f'(x) + 2g(x) \cdot g'(x)</span></p><p><span class="math">= 2f(x)g(x) + 2g(x)(-f(x)) = 2f(x)g(x) - 2f(x)g(x) = 0</span></p><p>Therefore, <span class="math">(f(x))^2 + (g(x))^2</span> is constant.</p><p>At <span class="math">x = 2</span>: <span class="math">(f(2))^2 + (g(2))^2 = (4)^2 + (4)^2 = 16 + 16 = 32</span></p><p>Therefore, <span class="math">(f(24))^2 + (g(24))^2 = 32</span></p><p>∴ Answer is (a).</p>
Correct Answer: A