On considère la fonction f définie sur IR par : f(x) = ex/(e^x+1) + 2. On note par (C) sa courbe représentative dans un repère orthonormé (O,i,j) d'unité 2cm.

The function is $f(x) = \frac{e^x}{e^x+1} + 2$.
As $x \to -\infty$, $e^x \to 0$.

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

Graphically, this means the line $y=2$ is a horizontal asymptote to the curve ($\mathscr{C}$) as $x \to -\infty$.

a) We start with the given expression for $f(x)$:

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

We can rewrite the fraction $\frac{e^x}{e^x+1}$ as $\frac{e^x+1-1}{e^x+1} = 1 - \frac{1}{e^x+1}$.
Substitute this back into the expression for $f(x)$:

$$f(x) = \left(1 - \frac{1}{e^x+1}\right) + 2 = 3 - \frac{1}{e^x+1}$$

b) Now, we deduce $\lim_{x \to +\infty} f(x)$ using the new form:
As $x \to +\infty$, $e^x \to +\infty$.
Therefore, $e^x+1 \to +\infty$, which implies $\frac{1}{e^x+1} \to 0$.

$$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \left(3 - \frac{1}{e^x+1}\right) = 3 - 0 = 3$$

Graphically, this means the line $y=3$ is a horizontal asymptote to the curve ($\mathscr{C}$) as $x \to +\infty$.

a) We differentiate $f(x) = \frac{e^x}{e^x+1} + 2$. The derivative of the constant 2 is 0.
Let $u(x) = e^x$ and $v(x) = e^x+1$. Then $u'(x) = e^x$ and $v'(x) = e^x$.
Using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$:

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

This matches the given derivative.

b) To create the variation table, we analyze the sign of $f'(x)$.
For all real $x$, $e^x > 0$.
For all real $x$, $(e^x+1)^2 > 0$.
Therefore, $f'(x) > 0$ for all real $x$.
This means $f(x)$ is strictly increasing on $\mathbb{R}$.
The limits are $\lim_{x \to -\infty} f(x) = 2$ and $\lim_{x \to +\infty} f(x) = 3$.

The table of variation for $f$ is:

$$\begin{array}{|c|ccc|} \hline x & -\infty & & +\infty \\ \hline f'(x) & & + & \\ \hline f(x) & 2 & \nearrow & 3 \\ \hline \end{array}$$

a) A point $(a,b)$ is a center of symmetry for $f(x)$ if $f(a-x) + f(a+x) = 2b$. Here, $(a,b) = (0, \frac{5}{2})$, so we need to show $f(-x) + f(x) = 2 \times \frac{5}{2} = 5$.
We use $f(x) = \frac{e^x}{e^x+1} + 2$.
First, find $f(-x)$:

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

Now, add $f(x)$ and $f(-x)$:

$$f(-x) + f(x) = \left(\frac{1}{e^x+1} + 2\right) + \left(\frac{e^x}{e^x+1} + 2\right) = \frac{1+e^x}{e^x+1} + 4 = 1 + 4 = 5$$

Since $f(-x) + f(x) = 5$, the point $I(0; \frac{5}{2})$ is a center of symmetry for the curve ($\mathscr{C}$).

b) The equation of the tangent (T) at $x_0=0$ is $y - f(x_0) = f'(x_0)(x - x_0)$.
First, calculate $f(0)$:

$$f(0) = \frac{e^0}{e^0+1} + 2 = \frac{1}{1+1} + 2 = \frac{1}{2} + 2 = \frac{5}{2}$$

Next, calculate $f'(0)$:

$$f'(0) = \frac{e^0}{(e^0+1)^2} = \frac{1}{(1+1)^2} = \frac{1}{2^2} = \frac{1}{4}$$

Substitute these values into the tangent equation:

$$y - \frac{5}{2} = \frac{1}{4}(x - 0) \implies y = \frac{1}{4}x + \frac{5}{2}$$

The equation of the tangent (T) is $\boxed{y = \frac{1}{4}x + \frac{5}{2}}$.

c) Calculate $f(-1)$, $f(1)$, and $f(2)$ to 0.1 precision using $e^{-1} \approx 0.4$, $e \approx 2.7$, $e^2 \approx 7.4$.
For $f(-1)$:

$$f(-1) = \frac{e^{-1}}{e^{-1}+1} + 2 \approx \frac{0.4}{0.4+1} + 2 = \frac{0.4}{1.4} + 2 \approx 0.2857 + 2 = 2.2857 \approx \mathbf{2.3}$$

For $f(1)$:

$$f(1) = \frac{e^1}{e^1+1} + 2 \approx \frac{2.7}{2.7+1} + 2 = \frac{2.7}{3.7} + 2 \approx 0.7297 + 2 = 2.7297 \approx \mathbf{2.7}$$

For $f(2)$:

$$f(2) = \frac{e^2}{e^2+1} + 2 \approx \frac{7.4}{7.4+1} + 2 = \frac{7.4}{8.4} + 2 \approx 0.8809 + 2 = 2.8809 \approx \mathbf{2.9}$$

This step requires a graphical representation. Key elements for tracing are:

• Horizontal asymptotes: $y=2$ (as $x \to -\infty$) and $y=3$ (as $x \to +\infty$).
• Center of symmetry: $I(0, 2.5)$.
• Tangent line (T): $y = \frac{1}{4}x + 2.5$, which passes through $I(0, 2.5)$ with a slope of $\frac{1}{4}$.
• Points: $f(-1) \approx 2.3$, $f(0) = 2.5$, $f(1) \approx 2.7$, $f(2) \approx 2.9$.
• The curve ($\mathscr{C}$) is strictly increasing, approaching $y=2$ from above and $y=3$ from below.

a) Let $F(x) = 2x + \ln(e^x+1)$.
To calculate $F'(x)$, we differentiate each term:

$$F'(x) = \frac{d}{dx}(2x) + \frac{d}{dx}(\ln(e^x+1))$$ $$F'(x) = 2 + \frac{e^x}{e^x+1}$$

We observe that $F'(x) = f(x)$.
Conclusion: $F(x)$ is an antiderivative (or primitive) of $f(x)$.

b) We need to calculate the area $\mathscr{A}$ bounded by the curve ($\mathscr{C}$), the x-axis, and the lines $x=0$ and $x=\ln 3$.
Since $f(x) > 2$ for all $x$, the curve is always above the x-axis. The area is given by the definite integral:

$$\mathscr{A}_{units} = \int_{0}^{\ln 3} f(x)\,dx$$

Since $F(x)$ is an antiderivative of $f(x)$:

$$\mathscr{A}_{units} = [F(x)]_{0}^{\ln 3} = F(\ln 3) - F(0)$$

Calculate $F(\ln 3)$:

$$F(\ln 3) = 2(\ln 3) + \ln(e^{\ln 3}+1) = 2\ln 3 + \ln(3+1) = 2\ln 3 + \ln 4 = 2\ln 3 + 2\ln 2$$

Calculate $F(0)$:

$$F(0) = 2(0) + \ln(e^0+1) = 0 + \ln(1+1) = \ln 2$$

Now, substitute these values:

$$\mathscr{A}_{units} = (2\ln 3 + 2\ln 2) - \ln 2 = 2\ln 3 + \ln 2$$

Using the given approximations $\ln 2 \approx 0.7$ and $\ln 3 \approx 1.1$:

$$\mathscr{A}_{units} \approx 2(1.1) + 0.7 = 2.2 + 0.7 = 2.9$$

The problem states that the unit on the axes is $2 \text{cm}$. This means 1 unit of area in the coordinate system corresponds to $(2 \text{cm}) \times (2 \text{cm}) = 4 \text{cm}^2$.
Therefore, the area in cm$^2$ is:

$$\mathscr{A} = \mathscr{A}_{units} \times 4 \text{ cm}^2 \approx 2.9 \times 4 = 11.6 \text{ cm}^2$$

The geometric area is $\boxed{\mathbf{11.6 \text{ cm}^2}}$.