Let $x_0$ be the real number such that $e^{x_0} + x_0 = 0$. For a given real number $\alpha$, define $g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)}$ for all real numbers $x$. Then which one of the following statements is TRUE?

Let x₀ be the real number such that e^(x₀) + x₀ = 0. For a given real number α, define g(x) = (3xeˣ + 3x - α eˣ - α x)/(3(eˣ + 1)) for all real numbers x. Then which one of the following statements is TRUE? $e^{x_0} + x_0 = 0 \Rightarrow e^{x_0} = -x_0$ $g(x) = \dfrac{3xe^{x} + 3x - \alpha e^{x} - \alpha x}{3(e^{x} + 1)}$ $\text{To find: } \lim_{x \to x_0} \left| \dfrac{g(x) + e^{x_0}}{x - x_0} \right|$ $\text{Plan: }1.\ \text{simplify } g(x)$ $2.\ g(x_0) + e^{x_0} = 0 \Rightarrow \text{limit} = |g'(x_0)|$ $3.\ \text{find } g'(x_0) \text{ and test } \alpha = 2, 3$ $3xe^{x} + 3x - \alpha e^{x} - \alpha x = 3x(e^{x}+1) - \alpha(e^{x}+x)$ $g(x) = x - \dfrac{\alpha}{3}\cdot\dfrac{e^{x}+x}{e^{x}+1}$ $g(x_0) = x_0 - \dfrac{\alpha}{3}\cdot\dfrac{e^{x_0}+x_0}{e^{x_0}+1} = x_0 - 0 = x_0$ $g(x_0) + e^{x_0} = x_0 + e^{x_0} = 0$ $\dfrac{g(x)+e^{x_0}}{x-x_0} = \dfrac{g(x)-g(x_0)}{x-x_0} \longrightarrow g'(x_0)$ $\text{Hence } L = |g'(x_0)|$ $h(x) = \dfrac{e^{x}+x}{e^{x}+1},\qquad g'(x) = 1 - \dfrac{\alpha}{3}\,h'(x)$ $h'(x) = \dfrac{(e^{x}+1)^{2} - e^{x}(e^{x}+x)}{(e^{x}+1)^{2}} = \dfrac{2e^{x}+1-xe^{x}}{(e^{x}+1)^{2}}$ $2e^{x_0}+1-x_0e^{x_0} = e^{x_0}(2-x_0)+1 = (x_0-1)^{2}$ $h'(x_0) = \dfrac{(x_0-1)^{2}}{(1-x_0)^{2}} = 1$ $g'(x_0) = 1 - \dfrac{\alpha}{3}$ $L = \left| 1 - \dfrac{\alpha}{3} \right|$ $\alpha = 2:\quad L = \left| 1 - \dfrac{2}{3} \right| = \dfrac{1}{3} \neq 0,\ 1$ $\Rightarrow (1), (2) \text{ are false}$ $\alpha = 3:\quad L = |1 - 1| = 0$ $\Rightarrow (3) \text{ is TRUE},\ (4) \text{ is false}$ $\boxed{\text{option (3)}}$