<p>Let \(S\) be infinite sum of the series \(2 + 3\cos x + 4\cos^2 x + 5\cos^3 x + \ldots\infty\), where \(x\) satisfies the equation \(|5\cos x + 4| + |5\cos x - 2| = 6\). If the least value of \(S\) is equal to \(\left(\dfrac{a}{b}\right)\) where \(a\) and \(b\) are co-prime numbers, then find the value of \((a + b)\).</p>
Step-by-Step Solution
Key Concept: First, solve the absolute value equation to find the range of cos x, then recognize the series as a derivative of a geometric series, and finally optimize S over the valid range.
<p><strong>Step 1: Solve the absolute value equation.</strong></p><p>We have |5cos x + 4| + |5cos x - 2| = 6.</p><p>Let u = 5cos x. Since -1 ≤ cos x ≤ 1, we have -5 ≤ u ≤ 5.</p><p>The equation becomes |u + 4| + |u - 2| = 6.</p><p><strong>Step 2: Analyze the absolute value equation by cases.</strong></p><p>• If u < -4: |u + 4| + |u - 2| = -(u + 4) - (u - 2) = -2u - 2 = 6 ⟹ u = -4 (boundary)</p><p>• If -4 ≤ u < 2: |u + 4| + |u - 2| = (u + 4) - (u - 2) = 6 ✓ (always true)</p><p>• If u ≥ 2: |u + 4| + |u - 2| = (u + 4) + (u - 2) = 2u + 2 = 6 ⟹ u = 2 (boundary)</p><p>Combined with -5 ≤ u ≤ 5, we get -4 ≤ 5cos x ≤ 2, so -4/5 ≤ cos x ≤ 2/5.</p><p><strong>Step 3: Express the series as a function of cos x.</strong></p><p>S = 2 + 3cos x + 4cos²x + 5cos³x + ... = Σ(n+1)cosⁿx (n = 1 to ∞)</p><p>Recognize that Σ(n+1)tⁿ = d/dt[Σtⁿ⁺¹] = d/dt[t/(1-t)] = 1/(1-t)² for |t| < 1.</p><p>Therefore: S = 1/(1 - cos x)² (valid since |cos x| ≤ 2/5 < 1)</p><p><strong>Step 4: Find the minimum of S over the valid range.</strong></p><p>S(c) = 1/(1 - c)² where c ∈ [-4/5, 2/5].</p><p>dS/dc = 2/(1 - c)³ > 0 for all c in the valid range (since 1 - c > 0).</p><p>Therefore S is strictly increasing on [-4/5, 2/5].</p><p>The minimum occurs at c = -4/5:</p><p>S_min = 1/(1 - (-4/5))² = 1/(1 + 4/5)² = 1/(9/5)² = 1/(81/25) = 25/81</p><p><strong>Step 5: Verify the answer is in lowest terms.</strong></p><p>gcd(25, 81) = gcd(5², 3⁴) = 1 ✓</p><p>Therefore a = 25, b = 81, and a + b = 106.</p><p><strong>∴ Answer: 106</strong></p>
Correct Answer: 106