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(A) Given $f(x) = \log_{e}(x^{2}+1) - e^{-x} + 1$. Calculating the derivative,$f'(x) = \frac{2x}{x^{2}+1} + e^{-x}$. Since $x^{2}+1 \geq 2|x|$,we have

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$(2,3)$

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(A) Given $f(x) = \log_{e}(x^{2}+1) - e^{-x} + 1$. Calculating the derivative,$f'(x) = \frac{2x}{x^{2}+1} + e^{-x}$. Since $x^{2}+1 \geq 2|x|$,we have $\frac{2x}{x^{2}+1} \geq -1$. Also $e^{-x} > 0$. Actually,$f'(x) = \frac{2x}{x^{2}+1} + e^{-x}$. For $x \geq 0$,$f'(x) > 0$. For $x  0$,then $f'(-t) = \frac{-2t}{t^{2}+1} + e^{t}$. Since $e^{t} > 1$ and $\frac{2t}{t^{2}+1} \leq 1$,$f'(x) > 0$ for all $x \in R$. Thus,$f$ is strictly increasing. Given $g(x) = \frac{1-2e^{2x}}{e^{x}} = e^{-x} - 2e^{x}$. Calculating the derivative,$g'(x) = -e^{-x} - 2e^{x} = -(e^{-x} + 2e^{x}) Thus,$g$ is strictly decreasing. Given the inequality $f(g(\frac{(\alpha-1)^{2}}{3})) > f(g(\alpha-\frac{5}{3}))$. Since $f$ is strictly increasing,this implies $g(\frac{(\alpha-1)^{2}}{3}) > g(\alpha-\frac{5}{3})$. Since $g$ is strictly decreasing,this implies $\frac{(\alpha-1)^{2}}{3} Multiplying by $3$,we get $(\alpha-1)^{2} $\alpha^{2} - 2\alpha + 1 $\alpha^{2} - 5\alpha + 6 $(\alpha-2)(\alpha-3) Therefore,$\alpha \in (2, 3)$.

Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be two functions defined by $f(x)=\log _{e}(x^{2}+1)-e^{-x}+1$ and $g(x)=\frac{1-2e^{2x}}{e^{x}}$. Then,for which of the following range of $\alpha$,the inequality $f(g(\frac{(\alpha-1)^{2}}{3})) > f(g(\alpha-\frac{5}{3}))$ holds?

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