The figure above shows the graph of $f$. If $f ( x ) = \int _ { 2 } ^ { x } g ( t ) d t$, which of the following could be the graph of $y = g ( x )$ ?
(A) [graph A]
(B) [graph B]
(C) [graph C]
(D) [graph D]
(E) [graph E]

The function $h$ is defined on $\mathbb{R}$ by $$h : \mathbb{R} \longrightarrow \mathbb{R}, \quad u \longmapsto u - [u] - 1/2$$ where $[u]$ denotes the integer part of $u$.
Carefully draw the graph of the application $h$ on the interval $[-1, 1]$.

We denote $\mathcal{D}$ the set of functions from $\mathbb{R}$ to $\mathbb{R}$ of class $\mathcal{C}^{\infty}$ and with compact support. We denote $\varphi$ the function defined by: $$\begin{cases} \varphi(x) = 0 & \text{if } |x| \geqslant 1 \\ \varphi(x) = \exp\left(-\frac{x^2}{1-x^2}\right) & \text{if } |x| < 1 \end{cases}$$
a) Study the variations of $\varphi$. b) Sketch the graph of $\varphi$. c) Show that $\varphi$ is $\mathcal{C}^{\infty}$. d) Show that $\mathcal{D}$ is a vector space over $\mathbb{R}$ not reduced to $\{0\}$.

We recall that the function $\phi$ is defined on $] - 1 , + \infty [$ by $\phi ( s ) = s - \ln ( 1 + s )$.
Sketch the graph of $\phi$. Show that $\phi$ defines by restriction to the intervals $] - 1,0 [$ and $] 0 , + \infty [$ respectively

• a bijection $\left. \phi _ { - } : \right] - 1,0 [ \rightarrow ] 0 , + \infty [$,
• a bijection $\left. \phi _ { + } : \right] 0 , + \infty [ \rightarrow ] 0 , + \infty [$.

We denote $\left. \phi _ { - } ^ { - 1 } : \right] 0 , + \infty [ \rightarrow ] - 1,0 \left[ \right.$ and $\left. \phi _ { + } ^ { - 1 } : \right] 0 , + \infty [ \rightarrow ] 0 , + \infty [$ the inverse bijections.

Study the variations of $g_{\sigma}$. Show that the second derivative of $g_{\sigma}$ vanishes and changes sign at exactly two points. Give the shape of the graph of $g_{\sigma}$ and mark the two points mentioned.

Give the shape of the representative curve of $\zeta(x) = \sum_{n=1}^{+\infty} \frac{1}{n^x}$.

For all $s \in [0,1]$, the function $k_s$ is defined by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ Let $s \in ]0,1[$. Sketch the graph of $k_s$ on $[0,1]$.

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $s \in [0,1]$, we define the function $k_s$ by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ Let $s \in ]0,1[$. Sketch the graph of $k_s$ on $[0,1]$.

Let $f(x) = xe^x$ and let $W$ denote the inverse of the bijection $f|_{[-1,+\infty[}$. Sketch, on the same diagram, the curves $\mathcal { C } _ { f }$ and $\mathcal { C } _ { W }$ representing the functions $f$ and $W$. Specify the tangent lines to the two curves at the point with abscissa 0 as well as the tangent line to $\mathcal { C } _ { W }$ at the point with abscissa $- \mathrm { e } ^ { - 1 }$.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Sketch the graph of $f$ by making best use of the previous results.

Draw the graph (on plain paper) of $f(x) = \min\{|x|-1, |x-1|-1, |x-2|-1\}$.

Let $h(x) = \frac{x^4}{(1-x)^4}$ and $g(x) = f(h(x)) = h(x) + 1/h(x)$. Sketch the graph of $g(x)$ and show that $g(x)$ has a root between $0$ and $1$.

Let $P$ be a variable point on the parabola $y = 4 x ^ { 2 } + 1$. Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y = x$ is:
(1) $( 3 x - y ) ^ { 2 } + ( x - 3 y ) + 2 = 0$
(2) $2 ( 3 x - y ) ^ { 2 } + ( x - 3 y ) + 2 = 0$
(3) $( 3 x - y ) ^ { 2 } + 2 ( x - 3 y ) + 2 = 0$
(4) $2 ( x - 3 y ) ^ { 2 } + ( 3 x - y ) + 2 = 0$

The combined equation of the two lines $ax + by + c = 0$ and $a'x + b'y + c' = 0$ can be written as $(ax + by + c)(a'x + b'y + c') = 0$. The equation of the angle bisectors of the lines represented by the equation $2x^2 + xy - 3y^2 = 0$ is
(1) $3x^2 + 5xy + 2y^2 = 0$
(2) $x^2 - y^2 + 10xy = 0$
(3) $3x^2 + xy - 2y^2 = 0$
(4) $x^2 - y^2 - 10xy = 0$