On the graph below, we have drawn, in an orthonormal coordinate system $(\mathrm{O}, \vec{\imath}, \vec{\jmath})$, a curve $\mathscr{C}$ and the line $(\mathrm{AB})$ where A and B are the points with coordinates $(0;1)$ and $(-1;3)$ respectively.
We denote by $f$ the function differentiable on $\mathbb{R}$ whose representative curve is $\mathscr{C}$. We further assume that there exists a real number $a$ such that for all real $x$, $$f(x) = x + 1 + ax\mathrm{e}^{-x^{2}}$$
- a. Justify that the curve $\mathscr{C}$ passes through point A. b. Determine the slope of the line (AB). c. Prove that for all real $x$, $$f^{\prime}(x) = 1 - a\left(2x^{2} - 1\right)\mathrm{e}^{-x^{2}}$$ d. We assume that the line $(\mathrm{AB})$ is tangent to the curve $\mathscr{C}$ at point A. Determine the value of the real number $a$.
- According to the previous question, for all real $x$, $$f(x) = x + 1 - 3x\mathrm{e}^{-x^{2}} \text{ and } f^{\prime}(x) = 1 + 3\left(2x^{2} - 1\right)\mathrm{e}^{-x^{2}}.$$ a. Prove that for all real $x$ in the interval $]-1;0]$, $f(x) > 0$. b. Prove that for all real $x$ less than or equal to $-1$, $f^{\prime}(x) > 0$. c. Prove that there exists a unique real number $c$ in the interval $\left[-\frac{3}{2};-1\right]$ such that $f(c) = 0$. Justify that $c < -\frac{3}{2} + 2 \cdot 10^{-2}$.
- We denote by $\mathscr{A}$ the area, expressed in square units, of the region defined by: $$c \leqslant x \leqslant 0 \quad \text{and} \quad 0 \leqslant y \leqslant f(x)$$ a. Write $\mathscr{A}$ in the form of an integral. b. We admit that the integral $I = \int_{-\frac{3}{2}}^{0} f(x)\,\mathrm{d}x$ is an approximate value of $\mathscr{A}$ to within $10^{-3}$. Calculate the exact value of the integral $I$.