The problem consists of two exercises. Exercise 1 deals with sequences. A sequence $U_n$ is defined recursively as $U_0 = e$ and $U_{n+1} = \sqrt{U_n}$. Another sequence $V_n$ is defined as $V_n = \ln(U_n)$. We need to: 1) Calculate $U_1$, $V_0$, and $V_1$. 2) Show that $V_n$ is a geometric sequence and specify its common ratio and first term. 3) Express $V_n$ and $U_n$ as functions of $n$. 4) Express $S_n = V_0 + V_1 + ... + V_n$ as a function of $n$. 5) Express $P_n = U_0 \times U_1 \times ... \times U_n$ as a function of $n$. 6) Study the convergence of $V_n$, $U_n$, $S_n$, and $P_n$. Exercise 2 concerns a function $f(x) = \frac{-2(1+e^x)}{e^x-1}$. We need to: 1) Determine the domain of $f$. 2) Calculate the limits of $f$ at the boundaries of its domain and find the equations of any asymptotes. 3) Determine $f'(x)$ and construct the variation table of $f$. 4) Give the equation of the tangent to the curve $C_f$ at the point with abscissa $x = \ln 2$. 5) a) Determine the points of intersection of $C_f$ with the coordinate axes. b) Sketch the graph of $f$. 6) Let $g$ be the restriction of $f$ to the interval $I = ]-\infty, 0[$. a) Show that $g$ is a bijection from $I$ to an interval $J$ which should be specified. b) Calculate $g(\ln 2)$ and deduce $(g^{-1})'(-6)$. 7) a) Verify that for all $x$ in the domain of $f$, $f(x) = 2 - \frac{4e^x}{e^x-1}$. Deduce a primitive of $f$ on its domain. b) Calculate the area in $cm^2$ of the planar region bounded by $C_f$, the line $y = 2$, and the lines $x = -\ln 3$ and $x = -\ln 2$.

1. Problem Description

The problem consists of two exercises. Exercise 1 deals with sequences. A sequence $U_n$ is defined recursively as $U_0 = e$ and $U_{n+1} = \sqrt{U_n}$. Another sequence $V_n$ is defined as $V_n = \ln(U_n)$. We need to:

1) Calculate $U_1$, $V_0$, and $V_1$.

2) Show that $V_n$ is a geometric sequence and specify its common ratio and first term.

3) Express $V_n$ and $U_n$ as functions of $n$.

4) Express $S_n = V_0 + V_1 + ... + V_n$ as a function of $n$.

5) Express $P_n = U_0 \times U_1 \times ... \times U_n$ as a function of $n$.

6) Study the convergence of $V_n$, $U_n$, $S_n$, and $P_n$.

Exercise 2 concerns a function $f(x) = \frac{-2(1+e^x)}{e^x-1}$. We need to:

1) Determine the domain of $f$.

2) Calculate the limits of $f$ at the boundaries of its domain and find the equations of any asymptotes.

3) Determine $f'(x)$ and construct the variation table of $f$.

4) Give the equation of the tangent to the curve $C_f$ at the point with abscissa $x = \ln 2$.

5) a) Determine the points of intersection of $C_f$ with the coordinate axes. b) Sketch the graph of $f$.

6) Let $g$ be the restriction of $f$ to the interval $I = ]-\infty, 0[$. a) Show that $g$ is a bijection from $I$ to an interval $J$ which should be specified. b) Calculate $g(\ln 2)$ and deduce $(g^{-1})'(-6)$.

7) a) Verify that for all $x$ in the domain of $f$, $f(x) = 2 - \frac{4e^x}{e^x-1}$. Deduce a primitive of $f$ on its domain. b) Calculate the area in $cm^2$ of the planar region bounded by $C_f$, the line $y = 2$, and the lines $x = -\ln 3$ and $x = -\ln 2$.

2. Solution Steps

Exercise 1:

1)

$U_0 = e$

$U_1 = \sqrt{U_0} = \sqrt{e}$

$V_0 = \ln(U_0) = \ln(e) = 1$

$V_1 = \ln(U_1) = \ln(\sqrt{e}) = \ln(e^{1/2}) = \frac{1}{2}$

2)

$V_{n+1} = \ln(U_{n+1}) = \ln(\sqrt{U_n}) = \ln(U_n^{1/2}) = \frac{1}{2} \ln(U_n) = \frac{1}{2} V_n$

Thus, $V_n$ is a geometric sequence with common ratio $q = \frac{1}{2}$ and first term $V_0 = 1$.

3)

$V_n = V_0 \times q^n = 1 \times (\frac{1}{2})^n = (\frac{1}{2})^n$

$U_n = e^{V_n} = e^{(\frac{1}{2})^n}$

4)

$S_n = V_0 + V_1 + ... + V_n$

$S_n$ is the sum of the first $n+1$ terms of a geometric sequence.

$S_n = V_0 \frac{1 - q^{n+1}}{1 - q} = 1 \times \frac{1 - (\frac{1}{2})^{n+1}}{1 - \frac{1}{2}} = \frac{1 - (\frac{1}{2})^{n+1}}{\frac{1}{2}} = 2(1 - (\frac{1}{2})^{n+1})$

5)

$P_n = U_0 \times U_1 \times ... \times U_n = e^{V_0} \times e^{V_1} \times ... \times e^{V_n} = e^{V_0 + V_1 + ... + V_n} = e^{S_n}$

$P_n = e^{2(1 - (\frac{1}{2})^{n+1})}$

6)

As $n \to \infty$, $(\frac{1}{2})^n \to 0$, so $V_n \to 0$.

As $n \to \infty$, $(\frac{1}{2})^n \to 0$, so $U_n = e^{(\frac{1}{2})^n} \to e^0 = 1$.

As $n \to \infty$, $(\frac{1}{2})^{n+1} \to 0$, so $S_n \to 2(1 - 0) = 2$.

As $n \to \infty$, $S_n \to 2$, so $P_n = e^{S_n} \to e^2$.

Exercise 2:

1) The domain of $f$ is $D_f = \{x \in \mathbb{R} : e^x - 1 \neq 0\}$. This means $e^x \neq 1$, so $x \neq 0$. Thus $D_f = \mathbb{R} \setminus \{0\} = ]-\infty, 0[ \cup ]0, +\infty[$.

2)

$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{-2(1+e^x)}{e^x-1} = \frac{-2(1+0)}{0-1} = \frac{-2}{-1} = 2$

$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \frac{-2(1+e^x)}{e^x-1} = \lim_{x \to +\infty} \frac{-2(\frac{1}{e^x}+1)}{1-\frac{1}{e^x}} = \frac{-2(0+1)}{1-0} = -2$

$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{-2(1+e^x)}{e^x-1}$. As $x \to 0^-$, $e^x \to 1^-$, so $e^x - 1 \to 0^-$. Thus, the limit is $\frac{-2(1+1)}{0^-} = \frac{-4}{0^-} = +\infty$

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{-2(1+e^x)}{e^x-1}$. As $x \to 0^+$, $e^x \to 1^+$, so $e^x - 1 \to 0^+$. Thus, the limit is $\frac{-2(1+1)}{0^+} = \frac{-4}{0^+} = -\infty$

Horizontal asymptotes: $y = 2$ as $x \to -\infty$ and $y = -2$ as $x \to +\infty$

Vertical asymptote: $x = 0$

3)

$f(x) = \frac{-2(1+e^x)}{e^x-1}$

$f'(x) = \frac{-2e^x(e^x-1) - (-2(1+e^x))e^x}{(e^x-1)^2} = \frac{-2e^{2x} + 2e^x + 2e^x + 2e^{2x}}{(e^x-1)^2} = \frac{4e^x}{(e^x-1)^2}$

Since $e^x > 0$ for all $x$, $f'(x) > 0$ for all $x \neq 0$.

4) At $x = \ln 2$, $f(\ln 2) = \frac{-2(1+e^{\ln 2})}{e^{\ln 2} - 1} = \frac{-2(1+2)}{2-1} = -6$

$f'(\ln 2) = \frac{4e^{\ln 2}}{(e^{\ln 2} - 1)^2} = \frac{4(2)}{(2-1)^2} = 8$

Tangent equation: $y = f'(\ln 2)(x - \ln 2) + f(\ln 2) = 8(x - \ln 2) - 6 = 8x - 8\ln 2 - 6$

5) a)

Intersection with the y-axis: occurs when $x = 0$. However, $x=0$ is not in the domain of $f$. So there is no intersection with y-axis.

Intersection with the x-axis: occurs when $f(x) = 0$. This means $-2(1+e^x) = 0$, so $1+e^x = 0$, or $e^x = -1$. This has no solution since $e^x > 0$ for all $x$. So there is no intersection with x-axis.

6) a) $g(x)$ is strictly increasing on $]-\infty, 0[$, so it is injective.

As $x \to -\infty$, $g(x) \to 2$. As $x \to 0^-$, $g(x) \to +\infty$. Thus $J = ]2, +\infty[$.

Since $g$ is continuous and strictly increasing on $I$ and maps $I$ onto $J$, it is a bijection from $I$ to $J$.

b) $g(\ln 2)$ is undefined, because $\ln 2 > 0$. The question meant $g^{-1}(-6)$. But since $-6$ is not in the range of $g(x)$ on $(-\infty, 0)$, $g^{-1}(-6)$ is undefined. The question probably had a typo or it was a trick question. Let's consider the value of $g^{-1}(-6)$ assuming $g(x)=f(x)$ for $x>0$, so that we can consider $f(x)$ around $\ln 2$. Since $f(\ln 2) = -6$ and $f'(\ln 2) = 8$, then $(g^{-1})'(-6) = (f^{-1})'(-6) = \frac{1}{f'(\ln 2)} = \frac{1}{8}$.

7) a)

$2 - \frac{4e^x}{e^x-1} = \frac{2(e^x-1)-4e^x}{e^x-1} = \frac{2e^x-2-4e^x}{e^x-1} = \frac{-2e^x-2}{e^x-1} = \frac{-2(e^x+1)}{e^x-1} = f(x)$

$f(x) = 2 - \frac{4e^x}{e^x-1}$

A primitive of $\frac{4e^x}{e^x-1}$ is $4\ln|e^x-1|$. Therefore, a primitive of $f(x)$ is $2x - 4\ln|e^x-1|$.

b)

$A = \int_{-\ln 3}^{-\ln 2} |f(x) - 2| dx = \int_{-\ln 3}^{-\ln 2} |- \frac{4e^x}{e^x-1}| dx = \int_{-\ln 3}^{-\ln 2} \frac{4e^x}{1-e^x} dx$ since $e^x < 1$ when $x<0$.

Let $u = 1 - e^x$, so $du = -e^x dx$, and $dx = -\frac{du}{e^x} = -\frac{du}{1-u}$.

$A = \int_{1 - e^{-\ln 3}}^{1 - e^{-\ln 2}} \frac{-4}{u} du = -4 \int_{1 - \frac{1}{3}}^{1 - \frac{1}{2}} \frac{1}{u} du = -4 \int_{\frac{2}{3}}^{\frac{1}{2}} \frac{1}{u} du = -4 [\ln|u|]_{\frac{2}{3}}^{\frac{1}{2}} = -4 (\ln(\frac{1}{2}) - \ln(\frac{2}{3})) = -4 (\ln(\frac{3}{4})) = -4 (\ln 3 - \ln 4) = 4(\ln 4 - \ln 3) = 4(\ln 2^2 - \ln 3) = 4(2\ln 2 - \ln 3) = 8\ln 2 - 4\ln 3$

3. Final Answer

$U_1 = \sqrt{e}$

$V_0 = 1$

$V_1 = \frac{1}{2}$

$V_n = (\frac{1}{2})^n$

$U_n = e^{(\frac{1}{2})^n}$

$S_n = 2(1 - (\frac{1}{2})^{n+1})$

$P_n = e^{2(1 - (\frac{1}{2})^{n+1})}$

Domain of f: $D_f = \mathbb{R} \setminus \{0\}$

Limits: $\lim_{x \to -\infty} f(x) = 2$, $\lim_{x \to +\infty} f(x) = -2$, $\lim_{x \to 0^-} f(x) = +\infty$, $\lim_{x \to 0^+} f(x) = -\infty$

Asymptotes: $y = 2$, $y = -2$, $x = 0$

$f'(x) = \frac{4e^x}{(e^x-1)^2}$

Tangent equation: $y = 8x - 8\ln 2 - 6$

No intersection with the axes.

Restriction $g$ bijection from $]-\infty, 0[$ to $]2, \infty[$.

$g(\ln 2)$ is undefined.

$(g^{-1})'(-6)$ is undefined.

A primitive of $f(x)$ is $2x - 4\ln|e^x-1|$.

$A = 8\ln 2 - 4\ln 3$