Trigonometry & Inverse Trigonometry
Equation Solving and Intersection Points
Grade 11
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/>Number of point(s) where the graphs of the two functions, <i>y</i> = <i>f</i>(<i>x</i>) and <i>y</i> = <i>g</i>(<i>x</i>) intersects in [0, π], is:</p>
<p>(a) 2</p>
<p>(b) 3</p>
<p>(c) 4</p>
<p>(d) 5</p>
Step-by-Step Solution
Key Concept: Set f(x) = g(x), rearrange to get a trigonometric equation, use sin addition formulas, and find all solutions in [0, π].
<p><strong>Solution:</strong> We need to find where \(f(x) = g(x)\):</p><p>\(\sin 3x + \cos x = \cos 3x + \sin x\)</p><p>\(\sin 3x - \cos 3x = \sin x - \cos x\)</p><p>\(\sqrt{2}\sin(3x - \frac{\pi}{4}) = \sqrt{2}\sin(x - \frac{\pi}{4})\)</p><p>\(\sin(3x - \frac{\pi}{4}) = \sin(x - \frac{\pi}{4})\)</p><p>This gives two cases:</p><p><strong>Case 1:</strong> \(3x - \frac{\pi}{4} = x - \frac{\pi}{4} + 2\pi k\) → \(x = \pi k\)</p><p>In [0, π]: <i>x</i> = 0, π (2 points)</p><p><strong>Case 2:</strong> \(3x - \frac{\pi}{4} = \pi - (x - \frac{\pi}{4}) + 2\pi k\) → \(4x = \pi + 2\pi k\) → \(x = \frac{\pi}{4} + \frac{\pi k}{2}\)</p><p>In [0, π]: <i>x</i> = π/4, 3π/4 (2 points)</p><p>Total: 4 points</p><p>∴ Answer is (c)</p>
Correct Answer: C