Exercise 1 -- Part A

We consider the function $f$ defined on the set $\mathbb{R}$ of real numbers by: $$f(x) = \frac{7}{2} - \frac{1}{2}\left(\mathrm{e}^{x} + \mathrm{e}^{-x}\right)$$

1. a. Determine the limit of the function $f$ at $+\infty$. b. Show that the function $f$ is strictly decreasing on the interval $[0; +\infty[$. c. Show that the equation $f(x) = 0$ admits, on the interval $[0; +\infty[$, a unique solution, which we denote $\alpha$.
2. By noting that, for all real $x$, $f(-x) = f(x)$, justify that the equation $f(x) = 0$ admits exactly two solutions in $\mathbb{R}$ and that they are opposite.

Part B

The plane is given an orthonormal coordinate system with unit 1 meter. The function $f$ and the real number $\alpha$ are defined in Part A. In the rest of the exercise, we model a greenhouse arch by the curve $\mathscr{C}$ of the function $f$ on the interval $[-\alpha; +\alpha]$.

1. Calculate the height of an arch.
2. a. In this question, we propose to calculate the exact value of the length of the curve $\mathscr{C}$ on the interval $[0; \alpha]$. It is admitted that this length is given, in meters, by the integral: $$I = \int_{0}^{\alpha} \sqrt{1 + \left(f'(x)\right)^{2}} \, dx$$ Show that, for all real $x$, we have: $1 + \left(f'(x)\right)^{2} = \frac{1}{4}\left(\mathrm{e}^{x} + \mathrm{e}^{-x}\right)^{2}$ b. Deduce the value of the integral $I$ as a function of $\alpha$. Justify that the length of an arch, in meters, is equal to: $\mathrm{e}^{\alpha} - \mathrm{e}^{-\alpha}$.

Part C

We wish to build a garden greenhouse in the shape of a tunnel. We fix four metal arches to the ground, whose shape is that described in the previous part, spaced 1.5 meters apart. On the south facade, we plan an opening modeled by the rectangle $ABCD$ with width 1 meter and length 2 meters.

1. Show that the quantity of sheet necessary to cover the south and north facades is given, in $m^2$, by: $$\mathscr{A} = 4\int_{0}^{\alpha} f(x)\,dx - 2$$
2. We take 1.92 as an approximate value of $\alpha$. Determine, to the nearest $\mathrm{m}^2$, the total area of plastic sheet necessary to build this greenhouse.

Let $a$ and $b$ be two real numbers. If $a > 0$ and $a ^ { 2 } - b ^ { 2 } = 1$ show that there exists a unique $\theta \in \mathbb { R }$ such that $a = \operatorname { ch } \theta$ and $b = \operatorname { sh } \theta$.

Deduce the equality: $$O ^ { + } ( 1,1 ) = \left\{ \left( \begin{array} { c c } \operatorname { ch } \gamma & \operatorname { sh } \gamma \\ \operatorname { sh } \gamma & \operatorname { ch } \gamma \end{array} \right) , \gamma \in \mathbb { R } \right\} \cup \left\{ \left( \begin{array} { c c } - \operatorname { ch } \gamma & \operatorname { sh } \gamma \\ \operatorname { sh } \gamma & - \operatorname { ch } \gamma \end{array} \right) , \gamma \in \mathbb { R } \right\}$$

Let $k$ be a non-negative integer. We define the function $f _ { k }$ from the interval $[ - 1,1 ]$ to itself by $f _ { k } ( x ) = \cos ( k \arccos x )$, and $T_k$ is the polynomial of degree $k$ that identifies with $f_k$ on $[-1,1]$. We denote by arcosh the inverse function of the hyperbolic cosine, defined from $[ 1 , + \infty [$ to $[ 0 , + \infty [$.
Show that $$\forall x \in ] - \infty , - 1 ] \quad T _ { k } ( x ) = ( - 1 ) ^ { k } \cosh ( k \operatorname { arcosh } ( - x ) ) .$$

Let $k$ be a non-negative integer. We define the function $f _ { k }$ from the interval $[ - 1,1 ]$ to itself by $f _ { k } ( x ) = \cos ( k \arccos x )$, and $T_k$ is the polynomial of degree $k$ that identifies with $f_k$ on $[-1,1]$. We recall that $A \in \mathcal{S}_N^+(\mathbb{R})$ is not proportional to the identity, $\lambda_1$ is the smallest eigenvalue of $A$, $\lambda_N$ is the largest eigenvalue of $A$, and $$\Lambda _ { k } = \{ Q \in \mathbb { R } [ X ] \mid \operatorname { deg } ( Q ) \leq k , Q ( 0 ) = 1 \}$$
We set $\omega _ { k } = \frac { 1 } { T _ { k } \left( - \frac { \lambda _ { N } + \lambda _ { 1 } } { \lambda _ { N } - \lambda _ { 1 } } \right) }$. Show that $\omega _ { k }$ is well defined, that the polynomial $$Q _ { k } ( X ) = \omega _ { k } T _ { k } \left( \frac { 2 X - \lambda _ { 1 } - \lambda _ { N } } { \lambda _ { N } - \lambda _ { 1 } } \right)$$ is an element of $\Lambda _ { k }$, and that the maximum of $| Q _ { k } ( t ) |$ on $\left[ \lambda _ { 1 } , \lambda _ { N } \right]$ is $\left| \omega _ { k } \right|$.

We set $\theta = \operatorname { arcosh } \left( \frac { \lambda _ { N } + \lambda _ { 1 } } { \lambda _ { N } - \lambda _ { 1 } } \right) > 0$ and $\alpha = e ^ { - \theta }$. Show that $\alpha$ is a root of the polynomial $$X ^ { 2 } - 2 \frac { \lambda _ { N } + \lambda _ { 1 } } { \lambda _ { N } - \lambda _ { 1 } } X + 1$$ and deduce the expression of $\alpha$ in terms of the quantity $\beta = \frac { \lambda _ { N } + \lambda _ { 1 } } { \lambda _ { N } - \lambda _ { 1 } }$.

For all $(t,\theta)\in\mathbb{R}_+\times[0,2\pi]$, define $$F(t,\theta) = \begin{pmatrix} \frac{1}{\sqrt{3}}\operatorname{sh}(t)\cos(\theta) \\ \frac{1}{\sqrt{3}}\operatorname{sh}(t)\sin(\theta) \\ \operatorname{ch}(t) \end{pmatrix}.$$ Show that $F$ takes values in $\mathcal{H}$ and that $F:\mathbb{R}_+\times[0,2\pi]\rightarrow\mathcal{H}$ is surjective.

For all $(t,\theta)\in\mathbb{R}_+\times[0,2\pi]$, define $$F(t,\theta) = \begin{pmatrix} \frac{1}{\sqrt{3}}\operatorname{sh}(t)\cos(\theta) \\ \frac{1}{\sqrt{3}}\operatorname{sh}(t)\sin(\theta) \\ \operatorname{ch}(t) \end{pmatrix}.$$ Calculate, for all $\theta\in[0,2\pi]$, the hyperbolic length of the path $$\begin{array}{rcl} \gamma:[0,b] & \rightarrow & \mathcal{H} \\ t & \mapsto & F(t,\theta). \end{array}$$

We denote by $I$ a subset of $\mathbb { N }$ having at least two elements and by $u = \left( u _ { i } \right) _ { i \in I }$ a sequence of unit vectors of $\mathcal { M } _ { n , 1 } ( \mathbb { R } )$. $C(u)$ denotes the coherence parameter $C ( u ) = \sup \left\{ \left| \left\langle u _ { i } \mid u _ { j } \right\rangle \right| , ( i , j ) \in I ^ { 2 } , i \neq j \right\}$.
Show that if $C ( u ) = 0$, then the set $\left\{ u _ { i } , i \in I \right\}$ is finite and give an upper bound for its cardinality.

Prove that, for every real number $t$, $\operatorname { ch } ( t ) \leqslant \exp \left( \frac { t ^ { 2 } } { 2 } \right)$.