Key Concepts and Formulas

• Monotonicity of Functions: A function $h(x)$ is strictly increasing if $x_1 < x_2 \implies h(x_1) < h(x_2)$. A function $h(x)$ is strictly decreasing if $x_1 < x_2 \implies h(x_1) > h(x_2)$.
• Derivative Test for Monotonicity: For a differentiable function $h(x)$, $h'(x) > 0$ implies $h(x)$ is strictly increasing, and $h'(x) < 0$ implies $h(x)$ is strictly decreasing.
• Composition of Monotonic Functions:
  • If $f$ is strictly increasing and $g$ is strictly decreasing, then $f(g(x))$ is strictly decreasing.
  • If $f$ is strictly increasing and $g$ is strictly increasing, then $f(g(x))$ is strictly increasing.
  • If $f$ is strictly decreasing and $g$ is strictly decreasing, then $f(g(x))$ is strictly increasing.
  • If $f$ is strictly decreasing and $g$ is strictly increasing, then $f(g(x))$ is strictly decreasing.
• Solving Quadratic Inequalities: For a quadratic inequality of the form $ax^2 + bx + c < 0$ or $ax^2 + bx + c > 0$, we find the roots of the corresponding quadratic equation and use a sign chart or knowledge of the parabola's orientation to determine the intervals where the inequality holds.

Step-by-Step Solution

Step 1: Analyze the Monotonicity of $f(x)$ We are given $f(x) = \log_e(x^2 + 1) - e^{-x} + 1$. To determine its monotonicity, we compute its derivative $f'(x)$. Using the chain rule and standard derivative formulas: $\frac{d}{dx}(\log_e(x^2 + 1)) = \frac{1}{x^2 + 1} \cdot (2x) = \frac{2x}{x^2 + 1}$. $\frac{d}{dx}(-e^{-x}) = -e^{-x} \cdot (-1) = e^{-x}$. $\frac{d}{dx}(1) = 0$. Therefore, $f'(x) = \frac{2x}{x^2 + 1} + e^{-x}$.

We need to determine the sign of $f'(x)$ for all $x \in \mathbb{R}$. Consider the term $\frac{2x}{x^2 + 1}$. Its range is $[-1, 1]$. The term $e^{-x}$ is always positive.

If $x \ge 0$, then $\frac{2x}{x^2 + 1} \ge 0$. Since $e^{-x} > 0$, $f'(x) = \frac{2x}{x^2 + 1} + e^{-x} > 0$. If $x < 0$, let $x = -t$ where $t > 0$. Then $f'(x) = \frac{-2t}{t^2 + 1} + e^t$. We know that for $t > 0$, $e^t > 1$. Also, for $t > 0$, $0 < \frac{2t}{t^2 + 1} \le 1$. Thus, $e^t > 1 \ge \frac{2t}{t^2 + 1}$, which implies $e^t - \frac{2t}{t^2 + 1} > 0$. So, $f'(x) > 0$ for all $x \in \mathbb{R}$. This means $f(x)$ is a strictly increasing function.

Step 2: Analyze the Monotonicity of $g(x)$ We are given $g(x) = \frac{1 - 2e^{2x}}{e^x}$. We can rewrite $g(x)$ as: $g(x) = \frac{1}{e^x} - \frac{2e^{2x}}{e^x} = e^{-x} - 2e^x$. Now, we compute its derivative $g'(x)$: $\frac{d}{dx}(e^{-x}) = -e^{-x}$. $\frac{d}{dx}(-2e^x) = -2e^x$. Therefore, $g'(x) = -e^{-x} - 2e^x$.

For any $x \in \mathbb{R}$, $e^{-x} > 0$ and $e^x > 0$. Thus, $-e^{-x} < 0$ and $-2e^x < 0$. The sum of two negative numbers is always negative. So, $g'(x) < 0$ for all $x \in \mathbb{R}$. This means $g(x)$ is a strictly decreasing function.

Step 3: Simplify the Given Inequality using Monotonicity The given inequality is $f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha -{5 \over 3}} \right)} \right)$. Since $f(x)$ is strictly increasing, for $f(A) > f(B)$ to hold, it must be that $A > B$. Let $A = g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)$ and $B = g\left( {\alpha -{5 \over 3}} \right)$. So, the inequality simplifies to: $g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right) > g\left( {\alpha -{5 \over 3}} \right)$.

Now, since $g(x)$ is strictly decreasing, for $g(X) > g(Y)$ to hold, it must be that $X < Y$. Let $X = {{{{(\alpha - 1)}^2}} \over 3}$ and $Y = {\alpha -{5 \over 3}}$. The inequality further simplifies to: ${{{{(\alpha - 1)}^2}} \over 3} < \alpha - {5 \over 3}$.

Step 4: Solve the Resulting Inequality for $\alpha$ We need to solve the inequality: ${{{{(\alpha - 1)}^2}} \over 3} < \alpha - {5 \over 3}$. Multiply both sides by 3 to eliminate the denominators: $(\alpha - 1)^2 < 3(\alpha - {5 \over 3})$ $(\alpha - 1)^2 < 3\alpha - 5$. Expand the left side: ${\alpha ^2} - 2\alpha + 1 < 3\alpha - 5$. Rearrange the terms to form a quadratic inequality: ${\alpha ^2} - 2\alpha - 3\alpha + 1 + 5 < 0$ ${\alpha ^2} - 5\alpha + 6 < 0$. To solve this quadratic inequality, find the roots of the equation ${\alpha ^2} - 5\alpha + 6 = 0$. Factoring the quadratic, we get: $(\alpha - 2)(\alpha - 3) < 0$. The roots are $\alpha = 2$ and $\alpha = 3$. Since the coefficient of $\alpha^2$ is positive, the parabola opens upwards. The inequality $(\alpha - 2)(\alpha - 3) < 0$ holds for values of $\alpha$ between the roots. Therefore, the solution is $2 < \alpha < 3$.

Common Mistakes & Tips

• Incorrectly applying the composition rule: If $f$ were strictly decreasing, $f(A) > f(B)$ would imply $A < B$. Always be mindful of the monotonicity of both functions involved.
• Errors in differentiation: Double-check each step of the differentiation process to avoid sign errors or missed terms.
• Algebraic mistakes in solving inequalities: When rearranging terms or multiplying inequalities, ensure that the direction of the inequality sign is maintained correctly, especially when multiplying by negative numbers.

Summary The problem requires analyzing the monotonicity of the given functions $f(x)$ and $g(x)$. We found that $f(x)$ is strictly increasing and $g(x)$ is strictly decreasing. Using these properties, the original complex inequality involving function compositions was reduced to a simple quadratic inequality in terms of $\alpha$. Solving this quadratic inequality yielded the range of $\alpha$ for which the original inequality holds.

The final answer is $\boxed{(2, 3)}$.