<p>If inequality \( \left(\dfrac{1}{x}\right)^{\lambda/x} \leq \dfrac{1}{9} \) has positive integer solution, then the minimum value of \( \lambda \) (using \( \ln 9 = 2.197 \)) is:</p>
Step-by-Step Solution
Key Concept: Rewrite the inequality as x^(-λ/x) ≤ 9^(-1), then take natural logarithm to convert the exponential inequality into an algebraic form. The critical insight is that for positive integers x, we need -λ/x · ln(x) ≤ -ln(9), which simplifies to λ ≥ (x·ln 9)/ln(x).
<p><strong>Step 1:</strong> Rewrite the inequality: $(1/x)^{λ/x} ≤ 1/9$ as $x^{-λ/x} ≤ 9^{-1}$</p><p><strong>Step 2:</strong> Take natural logarithm on both sides (inequality reverses): $-\frac{λ}{x}·\ln(x) ≤ -\ln(9)$</p><p><strong>Step 3:</strong> Multiply by -1 (flip inequality): $\frac{λ\ln(x)}{x} ≥ \ln(9)$</p><p><strong>Step 4:</strong> Solve for λ: $λ ≥ \frac{x·\ln(9)}{\ln(x)}$</p><p><strong>Step 5:</strong> For positive integer solutions to exist, find minimum λ by testing x = 2, 3, 4, ...</p><p>For <strong>x = 2:</strong> $λ ≥ \frac{2 × 2.197}{\ln(2)} = \frac{4.394}{0.693} ≈ 6.34$</p><p>For <strong>x = 3:</strong> $λ ≥ \frac{3 × 2.197}{\ln(3)} = \frac{6.591}{1.099} ≈ 5.99$</p><p>For <strong>x = 4:</strong> $λ ≥ \frac{4 × 2.197}{\ln(4)} = \frac{8.788}{1.386} ≈ 6.34$</p><p>The minimum value occurs at x = 3, giving <strong>λ ≈ 6</strong> (or exactly $\frac{3\ln(9)}{\ln(3)} = 6$)</p><p>∴ Answer: <strong>B</strong></p>
Correct Answer: B