A stationary point of a function $f$ is a real number $r$ such that $f ^ { \prime } ( r ) = 0$. A polynomial need not have a stationary point (e.g. $x ^ { 3 } + x$ has none). Consider a polynomial $p ( x )$.
(a) If $p ( x )$ is of degree 2022, then $p ( x )$ must have at least one stationary point.
(b) If the number of distinct real roots of $p ( x )$ is 2021, then $p ( x )$ must have at least 2020 stationary points.
(c) If the number of distinct real roots of $p ( x )$ is 2021, then $p ( x )$ can have at most 2020 stationary points.
(d) If $r$ is a stationary point of $p ( x )$ AND $p ^ { \prime \prime } ( r ) = 0$, then the point $( r , p ( r ) )$ is neither a local maximum nor a local minimum point on the graph of $p ( x )$.

A function $f ( x )$ defined for $x > a$ and a quartic function $g ( x )$ with leading coefficient $-1$ satisfy the following conditions. (Here, $a$ is a constant.)
(a) For all real numbers $x > a$, $$( x - a ) f ( x ) = g ( x ).$$ (b) For two distinct real numbers $\alpha , \beta$, the function $f ( x )$ has the same local maximum value $M$ at $x = \alpha$ and $x = \beta$. (Here, $M > 0$)
(c) The number of values of $x$ where the function $f ( x )$ has a local maximum or minimum is greater than the number of values of $x$ where the function $g ( x )$ has a local maximum or minimum. When $\beta - \alpha = 6 \sqrt { 3 }$, find the minimum value of $M$. [4 points]

We assume that $f$ is the zero application on $C(0,1)$ and that $u$ is an element of $\mathcal{D}_f$. For all $n \in \mathbb{N}$, we define the application $$u_n : \begin{array}{rll} \bar{D}(0,1) & \rightarrow & \mathbb{R} \\ (x,y) & \mapsto & u(x,y) + \dfrac{1}{n}(x^2 + y^2) \end{array}$$
Suppose that $u_n$ admits a local maximum at $(\tilde{x}, \tilde{y}) \in D(0,1)$.
a) By examining the behavior of the function $x \mapsto u_n(x, \tilde{y})$ show that, in this case, $\partial_{11} u_n(\tilde{x}, \tilde{y}) \leqslant 0$. Similarly, one can show that $\partial_{22} u_n(\tilde{x}, \tilde{y}) \leqslant 0$. Thus $\Delta u_n(\tilde{x}, \tilde{y}) \leqslant 0$. This result is admitted for the rest.
b) Deduce that $u_n$ does not admit a local maximum on $D(0,1)$.

Let $U$ be a non-empty bounded open set of $\mathbb{R}^n$ ($n \geqslant 2$) and $f: U \to \mathbb{R}$ of class $\mathcal{C}^2$. Let $f$ be a function continuous on $\bar{U}$. Show that $f$ attains a maximum at some point $x_0 \in \bar{U}$.

Let $U$ be a non-empty bounded open set of $\mathbb{R}^n$ ($n \geqslant 2$) and $f : U \rightarrow \mathbb{R}$ of class $\mathcal{C}^2$. Let $f$ be a function continuous on $\bar{U}$. Show that $f$ attains a maximum at some point $x_0 \in \bar{U}$.

Let $a < b$ be two real numbers and $f : [a,b] \rightarrow \mathbb{R}$ be an infinitely differentiable function. Let us call (H) the following hypothesis: there exists a unique point $x_0 \in [a,b]$ where $f$ attains its maximum, we have $a < x_0 < b$, and $f''(x_0) \neq 0$.
Show that under hypothesis (H), we have $f''(x_0) < 0$.

Let $a < b$ be two real numbers and $f : [ a , b ] \rightarrow \mathbb { R }$ be an infinitely differentiable function. Let us call (H) the following hypothesis: there exists a unique point $x _ { 0 } \in [ a , b ]$ where $f$ attains its maximum, we have $a < x _ { 0 } < b$, and $f ^ { \prime \prime } \left( x _ { 0 } \right) \neq 0$.
Show that under hypothesis $( H )$, we have $f ^ { \prime \prime } \left( x _ { 0 } \right) < 0$.

We denote $\|\cdot\|_2$ the canonical Euclidean norm on $\mathbb{R}^2$ and we denote $$\mathcal{C} := \left\{ x \in \mathbb{R}^2 \mid \|x\|_2 = 1 \right\}$$ We fix an arbitrary norm $\|\cdot\|$ on $\mathbb{R}^2$ and we denote $$\mathcal{K} = \left\{ A \in M_2(\mathbb{R}) \mid \forall x \in \mathbb{R}^2,\ \|x\|_2 \geq \|Ax\| \right\}.$$ a) Show that $\mathcal{K}$ is a compact and convex subset of $M_2(\mathbb{R})$. b) Show that there exists $A \in \mathcal{K}$ such that $\det A = \sup_{B \in \mathcal{K}} \det B$.

We fix an element $A$ of $\mathcal{K}$ such that $\det A = \sup_{B \in \mathcal{K}} \det B$. Show that $\det A > 0$ and that there exists $x \in \mathcal{C}$ such that $\|Ax\| = 1$.

We define $Q_{>0} = (\mathbb{R}_+^*)^{I \times J}$ and $\mathscr{L} : Q_{>0} \times (\mathbb{R}^I \times \mathbb{R}^J) \rightarrow \mathbb{R}$ defined by $$\mathscr{L}(\boldsymbol{q}, (f, g)) = J_\epsilon(\boldsymbol{q}) + \sum_{i \in I} f_i \left(\alpha_i - \sum_{j \in J} q_{ij}\right) + \sum_{j \in J} g_j \left(\beta_j - \sum_{i \in I} q_{ij}\right).$$ (a) Show that for all $(f, g) \in \mathbb{R}^I \times \mathbb{R}^J$, the minimum of $\boldsymbol{q} \mapsto \mathscr{L}(\boldsymbol{q}, (f, g))$ on $Q_{>0}$ is attained at $q(f,g)_{ij} = e^{(f_i + g_j - C_{ij})/\epsilon} p_{ij}$.
(b) Calculate the value of $G(f, g) = \mathscr{L}(q(f,g), (f,g))$.
(c) Verify that $G$ is concave on $\mathbb{R}^I \times \mathbb{R}^J$.

Verify that if $f_* : \mathbb{R}^J \rightarrow \mathbb{R}^I$ and $g_* : \mathbb{R}^I \rightarrow \mathbb{R}^J$ are defined by $$f_*(g)_i = -\epsilon \log\left(\sum_{j \in J} e^{(g_j - C_{ij})/\epsilon} \beta_j\right) \text{ and } g_*(f)_j = -\epsilon \log\left(\sum_{i \in I} e^{(f_i - C_{ij})/\epsilon} \alpha_i\right)$$ then for all $(f, g) \in \mathbb{R}^I \times \mathbb{R}^J$, we have $\frac{\partial G}{\partial f_i}(f_*(g), g) = \frac{\partial G}{\partial g_j}(f, g_*(f)) = 0$ for all $(i,j) \in I \times J$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
We admit that $f$ and $g _ { s }$ are of class $\mathcal { C } ^ { 1 }$ on $U _ { n }$. Give the expression of their gradient at a point $x = \left( x _ { 1 } , \ldots , x _ { n } \right)$ of $U _ { n }$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
Prove that the restriction of $f$ to $X _ { s }$ admits a maximum on $X _ { s }$ and that this maximum is in fact attained on $X _ { s } \cap U _ { n }$.
You may verify that $f$ is strictly positive at certain points of $X _ { s } \cap U _ { n }$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
We denote by $a = \left( a _ { 1 } , \ldots , a _ { n } \right)$ an element of $X _ { s } \cap U _ { n }$ at which the restriction of $f$ to $X _ { s }$ attains its maximum.
Prove that there exists a real number $\lambda > 0$ such that, for all $k$ in $\llbracket 1 , n \rrbracket , a _ { k } = \frac { f ( a ) } { \lambda }$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $h _ { n }$ the map from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad h _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \cdots + x _ { n } - 1 .$$
We denote by $H _ { n }$ the set $H _ { n } = \left\{ \left( x _ { 1 } , \ldots , x _ { n } \right) \in \mathbb { R } ^ { n } \mid x _ { 1 } + \cdots + x _ { n } = 1 \right\}$. We denote by $M _ { n }$ the maximum of $F _ { n }$ on $\overline { U _ { n } } \cap H _ { n }$ and we denote by $( a _ { 1 } , \ldots , a _ { n } )$ a point of $U _ { n } \cap H _ { n }$ at which it is attained. For $k$ between 1 and $n$, we denote $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$.
Prove that there exists a real number $\lambda > 0$ such that
$$\left\{ \begin{aligned} \gamma _ { 1 } + \frac { \gamma _ { 2 } } { 2 } + \cdots + \frac { \gamma _ { n } } { n } & = \lambda a _ { 1 } \\ \frac { \gamma _ { 2 } } { 2 } + \cdots + \frac { \gamma _ { n } } { n } & = \lambda a _ { 2 } \\ & \vdots \\ \frac { \gamma _ { n } } { n } & = \lambda a _ { n } \\ a _ { 1 } + a _ { 2 } + \cdots + a _ { n } & = 1 \end{aligned} \right.$$

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $M _ { n }$ the maximum of $F _ { n }$ on $\overline { U _ { n } } \cap H _ { n }$ and we denote by $( a _ { 1 } , \ldots , a _ { n } )$ a point of $U _ { n } \cap H _ { n }$ at which it is attained. For $k$ between 1 and $n$, we denote $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$. There exists a real number $\lambda > 0$ satisfying the system in Q17.
Deduce that:
a) $\lambda = \gamma _ { 1 } + \gamma _ { 2 } + \cdots + \gamma _ { n } = M _ { n }$;
b) for all $k$ in $\llbracket 1 , n \rrbracket , \gamma _ { k } = \lambda \omega _ { k } a _ { k }$, where
$$\left\{ \begin{array} { l } \omega _ { k } = k \left( 1 - \frac { a _ { k + 1 } } { a _ { k } } \right) \text { if } k \in \llbracket 1 , n - 1 \rrbracket \\ \omega _ { n } = n \end{array} \right.$$

Let $a \in E$ and let $U$ be an open set of $E$ containing $a$. Let $f : U \rightarrow E$ be an application of class $\mathscr{C}^1$ on $U$ such that $df(a) = \operatorname{Id}_E$. We fix a real number $r > 0$ such that $\overline{B(a,r)} \subset U$ and $$\forall x_1, x_2 \in \overline{B(a,r)}, \quad \left\|f(x_1) - f(x_2)\right\| \geqslant \frac{1}{2} \left\|x_1 - x_2\right\|.$$ Let $y_0 \in E$ such that $\left\|y_0 - f(a)\right\| \leqslant \frac{r}{4}$.
Show that the application $$\begin{array}{ccc} \overline{B(a,r)} & \longrightarrow & \mathbb{R} \\ x & \longmapsto & \left\|y_0 - f(x)\right\|^2 \end{array}$$ admits a minimum attained at a point $x_0$ of $B(a,r)$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. Establish an upper bound for the sequence with general term $a_n := f(x_n) - f(x_*)$. Conclude that $f(x_n) \rightarrow f(x_*)$ when $n \rightarrow \infty$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. Show that $\frac{\tau}{2}(2 - \tau L)\sum_{0 \leq i < n}\left|f'(x_i)\right|^2 \leq \left(f(x_0) - f(x_n)\right)$ for all $n \in \mathbb{N}^*$. Deduce that $f'(x_n) \rightarrow 0$ when $n \rightarrow \infty$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. a) Show that the sequence $\left(x_n\right)_{n \in \mathbb{N}}$ admits a convergent subsequence. We denote by $\varphi : \mathbb{N} \rightarrow \mathbb{N}$ the extractor and $x_{**}$ the corresponding limit, so that $x_{\varphi(n)} \rightarrow x_{**}$ when $n \rightarrow \infty$. Hint. We may use without proof the Bolzano-Weierstrass theorem: from any bounded sequence in $\mathbb{R}$, we can extract a convergent subsequence. b) Show that $f'(x_{**}) = 0$. c) Deduce that $x_{**}$ is a minimizer of $f$, then that $\left|x_n - x_{**}\right| \rightarrow 0$ when $n \rightarrow \infty$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$. Let also $\tau > 0$. Show that the function $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$ admits a unique minimizer on $\mathbb{R}$, which we will denote $p_f(x_0)$. Hint: We may consider minimizers $x_1$ and $x_2$ of $F_{x_0}$, and note that $$\left|\frac{1}{2}(x_1 + x_2) - x_0\right|^2 < \frac{1}{2}|x_1 - x_0|^2 + \frac{1}{2}|x_2 - x_0|^2 \quad \text{if } x_1 \neq x_2$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Show that $x_0 \in \mathbb{R}$ is a minimizer of $f$ if and only if $p_f(x_0) = x_0$. Hint: consider the quantity $F_{x_0}((1-t)x_0 + tx_*)$ when $t \rightarrow 0^+$.

Let $f , g : [ - 1,2 ] \rightarrow \mathbb { R }$ be continuous functions which are twice differentiable on the interval $( - 1,2 )$. Let the values of $f$ and $g$ at the points $- 1,0$ and 2 be as given in the following table:

	$x = - 1$	$x = 0$	$x = 2$
$f ( x )$	3	6	0
$g ( x )$	0	1	- 1

In each of the intervals $( - 1,0 )$ and $( 0,2 )$ the function $( f - 3 g ) ^ { \prime \prime }$ never vanishes. Then the correct statement(s) is(are)
(A) $\quad f ^ { \prime } ( x ) - 3 g ^ { \prime } ( x ) = 0$ has exactly three solutions in $( - 1,0 ) \cup ( 0,2 )$
(B) $f ^ { \prime } ( x ) - 3 g ^ { \prime } ( x ) = 0$ has exactly one solution in $( - 1,0 )$
(C) $f ^ { \prime } ( x ) - 3 g ^ { \prime } ( x ) = 0$ has exactly one solution in $( 0,2 )$
(D) $f ^ { \prime } ( x ) - 3 g ^ { \prime } ( x ) = 0$ has exactly two solutions in ( $- 1,0$ ) and exactly two solutions in ( 0,2 )

For every twice differentiable function $f : \mathbb { R } \rightarrow [ - 2,2 ]$ with $( f ( 0 ) ) ^ { 2 } + \left( f ^ { \prime } ( 0 ) \right) ^ { 2 } = 85$, which of the following statement(s) is (are) TRUE?
(A) There exist $r , s \in \mathbb { R }$, where $r < s$, such that $f$ is one-one on the open interval ( $r , s$ )
(B) There exists $x _ { 0 } \in ( - 4,0 )$ such that $\left| f ^ { \prime } \left( x _ { 0 } \right) \right| \leq 1$
(C) $\lim _ { x \rightarrow \infty } f ( x ) = 1$
(D) There exists $\alpha \in ( - 4,4 )$ such that $f ( \alpha ) + f ^ { \prime \prime } ( \alpha ) = 0$ and $f ^ { \prime } ( \alpha ) \neq 0$

Let $S$ be the set of all twice differentiable functions $f$ from $\mathbb { R }$ to $\mathbb { R }$ such that $\frac { d ^ { 2 } f } { d x ^ { 2 } } ( x ) > 0$ for all $x \in ( - 1,1 )$. For $f \in S$, let $X _ { f }$ be the number of points $x \in ( - 1,1 )$ for which $f ( x ) = x$. Then which of the following statements is(are) true?
(A) There exists a function $f \in S$ such that $X _ { f } = 0$
(B) For every function $f \in S$, we have $X _ { f } \leq 2$
(C) There exists a function $f \in S$ such that $X _ { f } = 2$
(D) There does NOT exist any function $f$ in $S$ such that $X _ { f } = 1$