Sets, Relations & Functions
Periodic functions
Grade 11

Question:

<p>The period of \(f(x) = 2\cos\frac{x-\pi}{3}\) is</p>
<p>(a) \(2\pi/3\)</p>
<p>(b) \(4\pi/3\)</p>
<p>(c) \(2\pi\)</p>
<p>(d) \(6\pi\)</p>

Step-by-Step Solution

Key Concept: The period of f(x) = A·cos(Bx + C) is 2π/|B|. Here B = 1/3, so period = 2π/(1/3) = 6π. The horizontal shift (x - π)/3 does NOT affect the period.
<p><strong>Step 1:</strong> Rewrite f(x) in standard form f(x) = A·cos(Bx + C)</p><p>f(x) = 2cos((x - π)/3) = 2cos((1/3)x - π/3)</p><p>Here A = 2, B = 1/3, C = -π/3</p><p><strong>Step 2:</strong> Apply the period formula for cosine: Period = 2π/|B|</p><p>Period = 2π/|1/3| = 2π × 3 = 6π</p><p><strong>Step 3:</strong> Verify: f(x + 6π) = 2cos((x + 6π - π)/3) = 2cos((x - π)/3 + 2π) = 2cos((x - π)/3) = f(x) ✓</p><p>∴ Answer: <strong>6π</strong> (Option D)</p>
Correct Answer: D

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