The problem is divided into two parts, A and B. Part A deals with the function $g(x) = -(x-1)^2 + 1 - \ln(x-1)$. We need to study its variations and determine the sign of $g(x)$ on specific intervals. Part B introduces the function $f(x) = -x + 3 + \frac{\ln(x-1)}{x-1}$. We need to find its domain, its derivative $f'(x)$, its variations, and discuss its asymptote. Finally, we restrict the function $f$ to an interval and analyze its bijectivity and inverse.

1. Problem Description

The problem is divided into two parts, A and B. Part A deals with the function $g(x) = -(x-1)^2 + 1 - \ln(x-1)$. We need to study its variations and determine the sign of $g(x)$ on specific intervals. Part B introduces the function $f(x) = -x + 3 + \frac{\ln(x-1)}{x-1}$. We need to find its domain, its derivative $f'(x)$, its variations, and discuss its asymptote. Finally, we restrict the function $f$ to an interval and analyze its bijectivity and inverse.

2. Solution Steps

Part A:

1) Study the variations of $g(x) = -(x-1)^2 + 1 - \ln(x-1)$.

First, determine the domain of $g$. Since $\ln(x-1)$ is defined only for $x-1 > 0$, the domain of $g$ is $D_g = (1, +\infty)$.

Next, calculate the derivative of $g(x)$:

$g'(x) = -2(x-1) - \frac{1}{x-1} = \frac{-2(x-1)^2 - 1}{x-1} = \frac{-2(x^2 - 2x + 1) - 1}{x-1} = \frac{-2x^2 + 4x - 3}{x-1}$.

Since $x > 1$, the denominator $x-1$ is positive. We analyze the numerator: $-2x^2 + 4x - 3$. The discriminant is $\Delta = 4^2 - 4(-2)(-3) = 16 - 24 = -8 < 0$. Since the leading coefficient is negative, the quadratic expression $-2x^2 + 4x - 3$ is always negative. Thus, $g'(x) < 0$ for all $x$ in $(1, +\infty)$. Therefore, $g(x)$ is strictly decreasing on $(1, +\infty)$.

2) Calculate $g(2)$ and show that $g(x) \geq 0$ on $(1, 2]$ and $g(x) \leq 0$ on $[2, +\infty)$.

$g(2) = -(2-1)^2 + 1 - \ln(2-1) = -1 + 1 - \ln(1) = 0$.

Since $g$ is decreasing on $(1, +\infty)$ and $g(2) = 0$, we have:

For $1 < x \leq 2$, $g(x) \geq g(2) = 0$. So, $g(x) \geq 0$ on $(1, 2]$.

For $x \geq 2$, $g(x) \leq g(2) = 0$. So, $g(x) \leq 0$ on $[2, +\infty)$.

Part B:

1) Determine $D_f$ and the limits at the boundaries of $D_f$.

$f(x) = -x + 3 + \frac{\ln(x-1)}{x-1}$.

The domain of $f$ is the same as the domain of $\ln(x-1)$, so $x-1 > 0$, which means $x > 1$. Thus, $D_f = (1, +\infty)$.

Now, let's calculate the limits:

$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (-x + 3 + \frac{\ln(x-1)}{x-1})$.

As $x \to 1^+$, $x-1 \to 0^+$, so $\ln(x-1) \to -\infty$ and $\frac{\ln(x-1)}{x-1} \to -\infty$.

Therefore, $\lim_{x \to 1^+} f(x) = -1 + 3 - \infty = -\infty$.

$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (-x + 3 + \frac{\ln(x-1)}{x-1})$.

As $x \to +\infty$, $\frac{\ln(x-1)}{x-1} \to 0$.

Therefore, $\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (-x + 3) = -\infty$.

2) Determine $f'(x)$ and verify that $f'(x) = \frac{g(x)}{(x-1)^2}$.

$f(x) = -x + 3 + \frac{\ln(x-1)}{x-1}$.

$f'(x) = -1 + \frac{\frac{1}{x-1}(x-1) - \ln(x-1)}{(x-1)^2} = -1 + \frac{1 - \ln(x-1)}{(x-1)^2} = \frac{-(x-1)^2 + 1 - \ln(x-1)}{(x-1)^2} = \frac{-(x-1)^2 + 1 - \ln(x-1)}{(x-1)^2} = \frac{g(x)}{(x-1)^2}$.

3) Draw the variation table of $f$.

Since $f'(x) = \frac{g(x)}{(x-1)^2}$, and $(x-1)^2 > 0$ for $x > 1$, the sign of $f'(x)$ is the same as the sign of $g(x)$.

We know $g(x) \geq 0$ on $(1, 2]$ and $g(x) \leq 0$ on $[2, +\infty)$.

Therefore, $f'(x) \geq 0$ on $(1, 2]$ and $f'(x) \leq 0$ on $[2, +\infty)$.

Thus, $f$ is increasing on $(1, 2]$ and decreasing on $[2, +\infty)$. Also, $f(2) = -2 + 3 + \frac{\ln(2-1)}{2-1} = 1 + \frac{\ln(1)}{1} = 1$.

4) a- Show that the line (D): $y = -x + 3$ is an oblique asymptote to $C_f$.

We need to show that $\lim_{x \to +\infty} [f(x) - (-x + 3)] = 0$.

$\lim_{x \to +\infty} [f(x) - (-x + 3)] = \lim_{x \to +\infty} [\frac{\ln(x-1)}{x-1}] = 0$.

Therefore, the line $y = -x + 3$ is an oblique asymptote to $C_f$.

b- Determine the coordinates of the intersection point A of $C_f$ with (D) and study their relative positions.

For the intersection, $f(x) = -x + 3$.

$-x + 3 + \frac{\ln(x-1)}{x-1} = -x + 3$.

$\frac{\ln(x-1)}{x-1} = 0$.

$\ln(x-1) = 0$.

$x-1 = 1$.

$x = 2$.

$y = -2 + 3 = 1$.

So the intersection point is $A(2, 1)$.

To study the relative positions, we analyze the sign of $f(x) - (-x + 3) = \frac{\ln(x-1)}{x-1}$.

If $x > 2$, then $x-1 > 1$, so $\ln(x-1) > 0$, and $\frac{\ln(x-1)}{x-1} > 0$. So $C_f$ is above (D).

If $1 < x < 2$, then $0 < x-1 < 1$, so $\ln(x-1) < 0$, and $\frac{\ln(x-1)}{x-1} < 0$. So $C_f$ is below (D).

5) Let $h$ be the restriction of $f$ to $I = [2, +\infty)$.

a- Show that $h$ is bijective from $I$ to $J$. Specify $J$.

Since $f$ is continuous and strictly decreasing on $[2, +\infty)$, its restriction $h$ to $I = [2, +\infty)$ is also continuous and strictly decreasing.

Thus, $h$ is a bijection from $I = [2, +\infty)$ to $J = [\lim_{x \to +\infty} f(x), f(2)] = (-\infty, 1]$.

b- Calculate $h(3)$ and deduce $(h^{-1})'(\frac{\ln 2}{2})$ then $(h^{-1})'(\frac{\ln 2}{2})$.

$h(3) = f(3) = -3 + 3 + \frac{\ln(3-1)}{3-1} = \frac{\ln 2}{2}$.

Since $h(3) = \frac{\ln 2}{2}$, then $h^{-1}(\frac{\ln 2}{2}) = 3$.

We have the formula $(h^{-1})'(y) = \frac{1}{h'(h^{-1}(y))}$.

Thus, $(h^{-1})'(\frac{\ln 2}{2}) = \frac{1}{h'(3)} = \frac{1}{f'(3)} = \frac{1}{\frac{g(3)}{(3-1)^2}} = \frac{4}{g(3)}$.

$g(3) = -(3-1)^2 + 1 - \ln(3-1) = -4 + 1 - \ln 2 = -3 - \ln 2$.

Therefore, $(h^{-1})'(\frac{\ln 2}{2}) = \frac{4}{-3 - \ln 2} = \frac{-4}{3 + \ln 2}$.

3. Final Answer

Part A:

1) $g(x)$ is strictly decreasing on $(1, +\infty)$.

2) $g(2) = 0$, $g(x) \geq 0$ on $(1, 2]$, $g(x) \leq 0$ on $[2, +\infty)$.

Part B:

1) $D_f = (1, +\infty)$, $\lim_{x \to 1^+} f(x) = -\infty$, $\lim_{x \to +\infty} f(x) = -\infty$.

2) $f'(x) = \frac{g(x)}{(x-1)^2}$.

3) $f$ is increasing on $(1, 2]$ and decreasing on $[2, +\infty)$. $f(2) = 1$.

4) a- $y = -x + 3$ is an oblique asymptote to $C_f$.

b- $A(2, 1)$. $C_f$ is below (D) for $1 < x < 2$ and above (D) for $x > 2$.

5) a- $h$ is a bijection from $I = [2, +\infty)$ to $J = (-\infty, 1]$.

b- $h(3) = \frac{\ln 2}{2}$, $(h^{-1})'(\frac{\ln 2}{2}) = \frac{-4}{3 + \ln 2}$.