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(C) Given $e^{x_0}+x_0=0$. $g(x) = \frac{3x(e^x+1) - \alpha(e^x+x)}{3(e^x+1)} = x - \frac{\alpha(e^x+x)}{3(e^x+1)}$. Since $e^{x_0}+x_0=0$,we have $g(

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For $\alpha=3, \lim _{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=0$

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(C) Given $e^{x_0}+x_0=0$. $g(x) = \frac{3x(e^x+1) - \alpha(e^x+x)}{3(e^x+1)} = x - \frac{\alpha(e^x+x)}{3(e^x+1)}$. Since $e^{x_0}+x_0=0$,we have $g(x_0) = x_0 - 0 = x_0$. Thus,$g(x_0) + e^{x_0} = x_0 + e^{x_0} = 0$. Using $L$'Hopital's rule,$\lim _{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right| = |g'(x_0)|$. $g'(x) = 1 - \frac{\alpha}{3} \left( \frac{(e^x+1)(e^x+1) - (e^x+x)e^x}{(e^x+1)^2} \right)$. At $x=x_0$,$e^{x_0}+x_0=0$,so $g'(x_0) = 1 - \frac{\alpha}{3} \left( \frac{(e^{x_0}+1)^2 - 0}{(e^{x_0}+1)^2} \right) = 1 - \frac{\alpha}{3}$. For $\alpha=3$,$|g'(x_0)| = |1 - \frac{3}{3}| = 0$. Therefore,the statement for $\alpha=3$ is true.

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

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