Applications of Derivatives
Monotonicity Intervals
Grade 12
Question:
<p>Consider <i>f</i>, <i>g</i> and <i>h</i> be three real valued functions defined on ℝ.<br/>Let <i>f</i>(<i>x</i>) = sin 3<i>x</i> + cos <i>x</i>, <i>g</i>(<i>x</i>) = cos 3<i>x</i> + sin <i>x</i> and <i>h</i>(<i>x</i>) = <i>f</i>²(<i>x</i>) + <i>g</i>²(<i>x</i>)<br/><br/>The length of a longest interval in which the function <i>y</i> = <i>h</i>(<i>x</i>) is increasing, is:</p>
<p>(a) \(\frac{\pi}{8}\)</p>
<p>(b) \(\frac{\pi}{4}\)</p>
<p>(c) \(\frac{\pi}{6}\)</p>
<p>(d) \(\frac{\pi}{2}\)</p>
Step-by-Step Solution
Key Concept: Simplify h(x) by expanding the squares and using trigonometric identities, then find where h'(x) ≥ 0.
<p><strong>Solution:</strong> First simplify \(h(x) = f^2(x) + g^2(x) = (\sin 3x + \cos x)^2 + (\cos 3x + \sin x)^2\)</p><p>Expanding: \(h(x) = \sin^2 3x + 2\sin 3x \cos x + \cos^2 x + \cos^2 3x + 2\cos 3x \sin x + \sin^2 x\)</p><p>\(= (\sin^2 3x + \cos^2 3x) + (\cos^2 x + \sin^2 x) + 2(\sin 3x \cos x + \cos 3x \sin x)\)</p><p>\(= 1 + 1 + 2\sin(3x + x) = 2 + 2\sin 4x\)</p><p>For <i>h</i>(<i>x</i>) to be increasing: \(h'(x) = 8\cos 4x \geq 0\)</p><p>This gives \(-\frac{\pi}{8} + \frac{k\pi}{2} \leq x \leq \frac{\pi}{8} + \frac{k\pi}{2}\)</p><p>The longest interval length is \(\frac{\pi}{8} - (-\frac{\pi}{8}) = \frac{\pi}{4}\)</p><p>∴ Answer is (b)</p>
Correct Answer: B