Functions and Graphs
Inverse Functions / Tangency
GRB_1000_SCQ
Grade Class 11

Question:

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: [Note: \([\cdot]\) denotes the greatest integer function.]
1
2
3
4

Step-by-Step Solution

Key Concept: Inverse functions intersect on \(y=x\); tangency condition for exactly one common point.
Step 1: Set up the intersection condition for the two curves. For the curves $y=\dfrac{1}{a}e^x$ and $y=\ln(ax)$ to have a common point, we need: $$\frac{1}{a}e^x = \ln(ax)$$ Step 2: Recognize that the two curves are inverse functions of each other. Consider the curve $y=\dfrac{1}{a}e^x$. If we swap $x$ and $y$ to find the inverse function: $$x = \frac{1}{a}e^y$$ $$ax = e^y$$ $$y = \ln(ax)$$ This is exactly the second curve. Therefore, the two curves are inverse functions of each other. Step 3: Use the property that inverse functions intersect on the line $y=x$. For two inverse functions to intersect, the intersection points must lie on the line $y=x$. Therefore, at any intersection point: $$\frac{1}{a}e^x = x$$ Multiplying both sides by $a$: $$e^x = ax$$ Step 4: Determine the condition for exactly one intersection point. For the curves to have exactly one point in common, the equation $e^x = ax$ must have exactly one solution. Geometrically, this means the line $y=ax$ (passing through the origin) must be tangent to the curve $y=e^x$. At the point of tangency, two conditions must hold: - The curves intersect: $e^x = ax$ - The slopes are equal: $\dfrac{d}{dx}(e^x) = a$, which gives $e^x = a$ Step 5: Solve for the value of $a$. From $e^x = a$ and $e^x = ax$, we have: $$a = ax$$ $$x = 1$$ Substituting $x=1$ into $e^x = a$: $$e^1 = a$$ $$a = e$$ Step 6: Find the greatest integer function value of $a$. Since $a = e \approx 2.718$, we have: $$[a] = [e] = 2$$ The value of $[a]$ is **2**, which corresponds to **Option 2**.
Correct Answer: 2

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