Limits, Continuity & Differentiability
Differentiation with Differential Equations
Grade 12
Question:
<p>Let <i>f</i> be a twice differentiable function such that \(f''(x) = -f(x)\) and \(f'(x) = g(x)\). If \(h(x) = \{f(x)\}^2 + \{g(x)\}^2\), where \(h(5) = 11\), find \(h(10)\).</p>
<p>(a) 0</p>
<p>(b) 9</p>
<p>(c) 11</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Recognize that differentiating h(x) yields 0 by substituting the given differential equation conditions, making h(x) constant.
<p><strong>Step 1:</strong> Given $h(x) = \{f(x)\}^2 + \{g(x)\}^2$.</p><p><strong>Step 2:</strong> Differentiate both sides with respect to $x$:</p><p>$h'(x) = 2f(x)·f'(x) + 2g(x)·g'(x)$</p><p><strong>Step 3:</strong> Since $f'(x) = g(x)$, we have $f''(x) = g'(x)$.</p><p><strong>Step 4:</strong> From the given condition $f''(x) = -f(x)$, we get $g'(x) = -f(x)$.</p><p><strong>Step 5:</strong> Substitute into the derivative:</p><p>$h'(x) = 2f(x)·g(x) + 2g(x)·(-f(x)) = 2f(x)g(x) - 2f(x)g(x) = 0$</p><p><strong>Step 6:</strong> Since $h'(x) = 0$, $h(x)$ is constant.</p><p><strong>Step 7:</strong> Given $h(5) = 11$, therefore $h(10) = 11$.</p><p>∴ Answer is (c) 11.</p>
Correct Answer: C