<p>If \(x = 1 + \dfrac{1}{3 + \dfrac{1}{2 + \dfrac{1}{3 + \dfrac{1}{2\ldots\infty}}}}\), then value of \(x\) is</p>
<p>\(\sqrt{\dfrac{5}{2}}\)</p>
<p>\(\sqrt{\dfrac{3}{2}}\)</p>
<p>\(\sqrt{\dfrac{7}{3}}\)</p>
<p>\(\sqrt{\dfrac{5}{3}}\)</p>
Step-by-Step Solution
Key Concept: The continued fraction has a repeating pattern (3, 2, 3, 2, ...). Let y represent the infinite tail starting from any position in the cycle, then use the periodicity to set up a quadratic equation for y.
<p><strong>Step 1:</strong> Identify the repeating pattern. After x = 1 + 1/(...), the pattern 3, 2, 3, 2, ... repeats infinitely.</p><p><strong>Step 2:</strong> Let y = 3 + 1/(2 + 1/(3 + 1/(2 + ...))) represent the infinite repeating part. By periodicity, y also equals the tail after the first 2, so: y = 3 + 1/(2 + 1/y)</p><p><strong>Step 3:</strong> Solve for y: y = 3 + 1/(2 + 1/y) → y = 3 + y/(2y + 1) → y(2y + 1) = 3(2y + 1) + y → 2y² + y = 6y + 3 + y → 2y² - 6y - 3 = 0</p><p><strong>Step 4:</strong> Using quadratic formula: y = (6 ± √(36 + 24))/4 = (6 ± √60)/4 = (6 ± 2√15)/4 = (3 ± √15)/2. Since y > 0, take y = (3 + √15)/2</p><p><strong>Step 5:</strong> Now find x = 1 + 1/y = 1 + 2/(3 + √15) = 1 + 2(3 - √15)/((3 + √15)(3 - √15)) = 1 + 2(3 - √15)/(9 - 15) = 1 + 2(3 - √15)/(-6) = 1 + (√15 - 3)/3 = (3 + √15 - 3)/3 = √15/3</p><p>∴ Answer: D</p>
Correct Answer: D