Let `f: R rarr R and g:R rarr R` be two functions defined by `f(x) = log_(e) (x^2+1)-e^(-x) +1` and `g(x)=(1-2e^(2x))/(e^x)` Then, for which of the following range of a, the inequality `f(g((alpha-1)^2)/3)gtf(g(alpha=5/3))` holds ?

Options

A. (2, 3)
B. (-2, -1)
C. (1, 2)
D. (-1, 1)

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) and determine the range of \( \alpha \) for which the inequality \( f\left(g\left(\frac{(\alpha - 1)^2}{3}\right)\right) > f\left(g\left(\frac{5}{3}\right)\right) \) holds.

### Step 1: Analyze the function \( f(x) \)

The function \( f(x) \) is defined as:
\[ f(x) = \log_e(x^2 + 1) - e^{-x} + 1 \]

To analyze its behavior, we compute the derivative \( f'(x) \):
\[ f'(x) = \frac{2x}{x^2 + 1} + e^{-x} \]

### Step 2: Determine the sign of \( f'(x) \)

Since \( \frac{2x}{x^2 + 1} \) is positive for \( x > 0 \) and \( e^{-x} \) is always positive, we conclude that:
- \( f'(x) > 0 \) for all \( x \in \mathbb{R} \)

This means that \( f(x) \) is a strictly increasing function.

### Step 3: Analyze the function \( g(x) \)

The function \( g(x) \) is defined as:
\[ g(x) = \frac{1 - 2e^{2x}}{e^x} \]

To analyze its behavior, we compute the derivative \( g'(x) \):
\[ g'(x) = \frac{-2e^{2x} \cdot e^x - (1 - 2e^{2x}) \cdot e^x}{(e^x)^2} = \frac{-2e^{3x} - e^x + 2e^{2x}}{e^{2x}} \]
\[ g'(x) = -2e^x - 1 + 2e^{-x} \]

### Step 4: Determine the sign of \( g'(x) \)

The expression \( g'(x) \) is negative for all \( x \) since both terms \( -2e^x \) and \( -1 \) are negative, and \( 2e^{-x} \) is always less than \( 2 \). Thus, \( g(x) \) is a strictly decreasing function.

### Step 5: Analyze the inequality

Given that \( f \) is increasing and \( g \) is decreasing, we can rewrite the inequality:
\[ f\left(g\left(\frac{(\alpha - 1)^2}{3}\right)\right) > f\left(g\left(\frac{5}{3}\right)\right) \]
This implies:
\[ g\left(\frac{(\alpha - 1)^2}{3}\right) < g\left(\frac{5}{3}\right) \]

### Step 6: Solve for \( \alpha \)

Since \( g \) is decreasing, we have:
\[ \frac{(\alpha - 1)^2}{3} > \frac{5}{3} \]
Multiplying both sides by 3:
\[ (\alpha - 1)^2 > 5 \]
Taking square roots:
\[ |\alpha - 1| > \sqrt{5} \]

This gives us two inequalities:
1. \( \alpha - 1 > \sqrt{5} \) → \( \alpha > 1 + \sqrt{5} \)
2. \( \alpha - 1 < -\sqrt{5} \) → \( \alpha < 1 - \sqrt{5} \)

### Step 7: Final Range of \( \alpha \)

Thus, the range of \( \alpha \) for which the inequality holds is:
\[ \alpha < 1 - \sqrt{5} \quad \text{or} \quad \alpha > 1 + \sqrt{5} \]