<p>If the curves \(y=\dfrac{1}{a}e^x\) and \(y=\ln(ax)\) (where \(a\) is positive) have only one point in common, then the value of \([a]\) is:<br>[Note: \([\cdot]\) denotes the greatest integer function.]</p>
Step-by-Step Solution
Key Concept: Two curves touch (have exactly one common point) when they intersect at a point AND have the same tangent line there. This requires both the function values and derivatives to be equal at that point, creating a system that determines the unique value of a.
<p><strong>Step 1:</strong> At the point of tangency, both curves pass through the same point and have the same slope:</p><p>Function equality: $\frac{1}{a}e^x = \ln(ax)$</p><p>Derivative equality: $\frac{1}{a}e^x = \frac{1}{x}$</p><p><strong>Step 2:</strong> From the derivative condition, $e^x = \frac{a}{x}$, so $x = \ln\left(\frac{a}{x}\right)$, which gives $xe^x = a$.</p><p><strong>Step 3:</strong> Substitute $e^x = \frac{a}{x}$ into the function equality:</p><p>$\frac{1}{a} \cdot \frac{a}{x} = \ln(ax)$</p><p>$\frac{1}{x} = \ln(ax)$</p><p><strong>Step 4:</strong> Let $u = ax$. Then $x = \frac{u}{a}$, so:</p><p>$\frac{a}{u} = \ln u$</p><p>This means $u = ae^{a/u}$, or equivalently $ue^{-u} = a e^{-a}$.</p><p><strong>Step 5:</strong> The function $f(t) = te^{-t}$ has maximum at $t=1$ with value $\frac{1}{e} \approx 0.368$. For the equation $ue^{-u} = ae^{-a}$ to have exactly one solution in $u$, we need $ae^{-a} = \frac{1}{e}$, giving $a = 1$.</p><p>Verification: At $a=1$, both conditions yield $x=1$, confirming tangency.</p><p>$\therefore [a] = [1] = \boxed{1}$</p>
Correct Answer: A