<p>If the curves \(y = \dfrac{1}{a}e^x\) and \(y = \ln(ax)\), (where \(a\) is positive) has 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 have exactly one point in common when they are tangent at that point, requiring both curves to pass through the same point AND have equal derivatives at that point. Setting up these two conditions simultaneously will determine the unique value of a.
<p><strong>Step 1:</strong> At the point of tangency, both curves pass through the same point and have equal derivatives.</p><p>Let the point of tangency be (x₀, y₀). Then:</p><p>$$\frac{1}{a}e^{x_0} = \ln(ax_0) \quad \text{...(i)}$$</p><p><strong>Step 2:</strong> The derivatives must be equal:</p><p>$$\frac{d}{dx}\left(\frac{1}{a}e^x\right)\bigg|_{x=x_0} = \frac{d}{dx}(\ln(ax))\bigg|_{x=x_0}$$</p><p>$$\frac{1}{a}e^{x_0} = \frac{1}{x_0} \quad \text{...(ii)}$$</p><p><strong>Step 3:</strong> From equations (i) and (ii):</p><p>$$\frac{1}{x_0} = \ln(ax_0)$$</p><p><strong>Step 4:</strong> Let $t = x_0$. Then: $\frac{1}{t} = \ln(at)$, which gives $\ln(at) = \frac{1}{t}$</p><p>This means: $at = e^{1/t}$, so $a = \frac{e^{1/t}}{t}$</p><p><strong>Step 5:</strong> For the curves to have exactly one common point, this equation in $t$ must have exactly one solution. Taking the derivative of $f(t) = te^{-1/t}$ and analyzing shows the maximum occurs at $t = 1$, giving:</p><p>$$a = 1 \cdot e^{1/1} = e \approx 2.718$$</p><p><strong>Step 6:</strong> Therefore $[a] = [e] = [2.718...] = 2$</p><p>∴ Answer: A (where A = 2)</p>
Correct Answer: A