Sequences & Series
Infinite geometric series
Grade 11
Question:
<p>Let \(S \subset (0, \pi)\) denote the set of values of <em>x</em> satisfying the equation \(8^{1+|\cos x| + \cos^2 x + |\cos^3 x| + \cdots \text{ to } \infty} = 4^3\). Then \(S =\)</p>
<p>(1) \(\{\pi/3\}\)</p>
<p>(2) \(\{\pi/6, 5\pi/6\}\)</p>
<p>(3) \(\{\pi/3, 5\pi/6\}\)</p>
<p>(4) \(\{\pi/3, 2\pi/3\}\)</p>
Step-by-Step Solution
Key Concept: The infinite series in the exponent is a geometric series with first term 1 and ratio |cos x|. For convergence, |cos x| < 1, and the sum equals 1/(1 - |cos x|). Setting 8^(1/(1-|cos x|)) = 64 and solving for |cos x| gives the x values in (0, π).
<p><strong>Step 1:</strong> Recognize the exponent as a geometric series: 1 + |cos x| + cos²x + |cos³x| + ... = 1 + |cos x|(1 + |cos x| + |cos x|² + ...) = 1/(1 - |cos x|), valid when |cos x| < 1.</p><p><strong>Step 2:</strong> Rewrite the equation: 8^(1/(1-|cos x|)) = 4³ = 64. Since 8 = 2³ and 64 = 2⁶, we have 2^(3/(1-|cos x|)) = 2⁶.</p><p><strong>Step 3:</strong> Equate exponents: 3/(1 - |cos x|) = 6, so 1 - |cos x| = 1/2, giving |cos x| = 1/2.</p><p><strong>Step 4:</strong> In (0, π), |cos x| = 1/2 means cos x = ±1/2. This gives x = π/3 or x = 2π/3.</p><p><strong>Step 5:</strong> Verify convergence: both values satisfy |cos x| = 1/2 < 1 ✓</p><p>∴ Answer: S = {π/3, 2π/3}</p>
Correct Answer: D