Common Tangent Mania 2

$f(x) = \lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n (\dfrac{1}{x})^{n - j} =$ $\lim_{n \rightarrow \infty} \sum_{j = 0}^{n - 1} (1 - \dfrac{j}{n})^n (\dfrac{1}{x})^{j} =$ $\sum_{j = 0}^{\infty} (\dfrac{1}{ex})^j = \dfrac{ex}{ex - 1}$ on $|x| > \dfrac{1}{e}$ .

Let $g(x) = ax^2 + bx + c$ .

$f(1) = g(1) \implies \boxed{a + b + c = \dfrac{e}{e - 1}}$

$f(e) = g(e) \implies \boxed{e^2a + eb + c = \dfrac{e^2}{e^2 - 1}}$

$\dfrac{d}{dx}(f(x)) = -\dfrac{e}{(ex - 1)^2}$ and $\dfrac{d}{dx}(g(x)) = 2ax + b$

$\dfrac{d}{dx}(f(x))|_{x = e} = \dfrac{d}{dx}(g(x))|_{x = e} \implies$ $\boxed{2ea + b = -\dfrac{-e}{(e^2 - 1)^2}}$

Solving the system we obtain:

$(e + 1)a + b = -\dfrac{e}{(e - 1)^2 (e + 1)}$

$2ea + b = -\dfrac{e}{(e - 1)^2 (e + 1)^2}$

$\implies a = \dfrac{e^2}{(e - 1)^3 (e + 1)^2} \implies b = \dfrac{-2e^2 + e}{(e - 1)^3 (e + 1)} \implies c = \dfrac{e^5}{(e - 1)^3 (e + 1)^2}$

$\implies \lfloor{a + b + c}\rfloor = \lfloor{\dfrac{e(e^2 - 1)^2}{(e - 1)^3 (e + 1)^2}}\rfloor =$
$\lfloor{\dfrac{e}{e - 1}}\rfloor = \lfloor{1 + \dfrac{1}{e - 1}}\rfloor = \boxed{1}$ .

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Note: The common tangent line to both curves is: $(e^2 - 1)^2 y + ex - e^4 = 0$ .