Statement 2: The equation $x - \cos x = 0$ has a unique solution in the interval $\left[ 0 ; \frac{\pi}{2} \right]$.
Indicate whether this statement is true or false, justifying your answer.

The rate (as a percentage) of $\mathrm{CO}_2$ contained in a room after $t$ minutes of hood operation is modelled by the function $f$ defined for all real $t$ in the interval $[0;20]$ by: $$f(t) = (0{,}8t + 0{,}2)\mathrm{e}^{-0{,}5t} + 0{,}03.$$ It is desired that the rate of $\mathrm{CO}_2$ in the room returns to a value $V$ less than or equal to $3.5\%$. a. Justify that there exists a unique instant $T$ satisfying this condition. b. Consider the following algorithm: \begin{verbatim} $t \leftarrow 1,75$ $p \leftarrow 0,1$ $V \leftarrow 0,7$ While $V > 0,035$ $t \leftarrow t + p$ $V \leftarrow ( 0,8 t + 0,2 ) \mathrm { e } ^ { - 0,5 t } + 0,03$ End While \end{verbatim} What is the value of the variable $t$ at the end of the algorithm? What does this value represent in the context of the exercise?

We consider the function $f$ defined on the set $] 0 ; + \infty [$ by
$$f ( x ) = 1 + x ^ { 2 } - 2 x ^ { 2 } \ln ( x )$$
We admit that $f$ is differentiable on the interval and we denote $f ^ { \prime }$ its derivative function.

1. Justify that $\lim _ { x \rightarrow 0 } f ( x ) = 1$ and, by noting that $f ( x ) = 1 + x ^ { 2 } [ 1 - 2 \ln ( x ) ]$, justify that $\lim _ { x \rightarrow + \infty } f ( x ) = - \infty$.
2. Show that for all real $x$ in the interval $] 0 ; + \infty \left[ , f ^ { \prime } ( x ) = - 4 x \ln ( x ) \right.$.
3. Study the sign of $f ^ { \prime } ( x )$ on the interval $] 0$; $+ \infty [$, then draw up the table of variations of the function on the interval $] 0 ; + \infty [$.
4. Prove that the equation $f ( x ) = 0$ admits a unique solution $\alpha$ in the interval $[ 1 ; + \infty [$ and that $\alpha \in [ 1 ; \mathrm { e } ]$.

We admit in the rest of the exercise that the equation $f ( x ) = 0$ has no solution on the interval $] 0 ; 1]$.
5. We are given the function below written in Python. The instruction from lycee import* allows access to the function $\ln$.
\begin{verbatim} from lycee import * def f(x) : return 1 + x**2 - 2*x**2*ln(x) def dichotomie(p) a=1 b=2.7 while b - a > 10**(-p) : if f(a)*f((a+b)/2) < 0 : | b = (a+b)/2 else : |a =(a+b)/2 return (a,b) \end{verbatim}
It writes in the execution console:
\begin{verbatim} >>> dichotomie(1) \end{verbatim}
Among the four propositions below, copy the one displayed by the previous instruction. Justify your answer (you may proceed by elimination).
Proposition A: $\quad ( 1.75,1.9031250000000002 )$ Proposition B : ( $1.85,1.9031250000000002 )$ Proposition C : $\quad ( 2.75,2.9031250000000002 )$ Proposition D : (2.85, 2.9031250000000002)

Let two real numbers $a$ and $b$ with $a < b$. Consider a function $f$ defined, continuous, strictly increasing on the interval $[a; b]$ and which vanishes at a real number $\alpha$. Among the following propositions, the function in Python language that allows giving an approximate value of $\alpha$ to 0.001 is:
a. \begin{verbatim} def racine(a, b): while abs(b - a) >= 0.001: m = (a + b) / 2 if f(m) < 0: b = m else: a = m return m \end{verbatim}
c. \begin{verbatim} def racine(a, b): m = (a + b) / 2 while abs(b - a) <= 0.001: if f(m) < 0: a = m else: b = m return m \end{verbatim}
b. \begin{verbatim} def racine(a, b): m = (a + b) / 2 while abs(b - a) >= 0.001: if f(m) < 0: a = m else: b = m return m \end{verbatim}
d. \begin{verbatim} def racine(a, b): while abs(b - a) >= 0.001: m = (a + b) / 2 if f(m) < 0: a = m else: b = m return m \end{verbatim}

A person intends to install a natural gas vehicle (NGV) kit in his car. At the store he chose to make the purchase and installation of this kit, there were five models of cylinders for gas storage, whose capacities, in cubic meters, were, respectively: $10, 14, 17, 21$, and 25. The price of the cylinder is proportional to its capacity. This car will travel 30 km daily, 7 days a week, and the NGV consumption is $1\,\mathrm{m}^3$ for every 13 km traveled. The person will choose the cylinder model with the lowest price and that guarantees only one refueling per week.
Under these conditions, what will be the capacity, in cubic meters, of the cylinder chosen by this person?
(A) 10
(B) 14
(C) 17
(D) 21
(E) 25

Consider the polynomial $$p(x) = x^6 + 10x^5 + 11x^4 + 12x^3 + 13x^2 - 12x - 11.$$ Find an integer $n$ with the least possible absolute value such that $p(x)$ has a real root between $n$ and $n+1$. Write this number along with your reason as per the given instruction. [2 points]
Instruction for (6): Write two numbers separated by a comma: value of $n$, number of the theorem below that justifies this answer. E.g., if you think that $n=5$ because of the factor theorem, then type $\mathbf{5,1}$ as your answer with no space, full stop or any other punctuation.

1. Factor theorem
2. Mean value theorem
3. Intermediate value theorem
4. Fundamental theorem of algebra
5. Fundamental theorem of calculus

Find the value of $\lim _ { x \rightarrow 0 } \sqrt { 2 x + 9 }$. [3 points]

What is the value of $\lim _ { n \rightarrow \infty } \frac { 4 n ^ { 2 } + 6 } { n ^ { 2 } + 3 n }$? [2 points]
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5

Let $[ a , b ]$ be a closed bounded interval of $\mathbb { R }$. If $\phi : [ a , b ] \rightarrow [ a , b ]$ is continuous, show that $\phi$ has at least one fixed point.

The largest value of the nonnegative integer $a$ for which $$\lim_{x \rightarrow 1} \left\{\frac{-ax + \sin(x-1) + a}{x + \sin(x-1) - 1}\right\}^{\frac{1-x}{1-\sqrt{x}}} = \frac{1}{4}$$ is

The real number $k$ for which the equation, $2x^3 + 3x + k = 0$ has two distinct real roots in $[0,1]$ belongs to
(1) lies between - 1 and 0 .
(2) does not exist.
(3) lies between 1 and 2 .
(4) lies between 2 and 3 .

$\lim _ { x \rightarrow 0 } \frac { ( 1 - \cos 2 x ) ( 3 + \cos x ) } { x \tan 4 x } =$
(1) $\frac { 1 } { 2 }$
(2) 4
(3) 3
(4) 2

$\lim_{x \rightarrow 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}$ equals
(1) $4\sqrt{2}$
(2) $2\sqrt{2}$
(3) $\sqrt{2}$
(4) 4

For each $t \in R$, let $[ t ]$ be the greatest integer less than or equal to $t$. Then, $\lim _ { x \rightarrow 1 ^ { + } } \frac { ( 1 - | x | + \sin | 1 - x | ) \sin \left( [ 1 - x ] \frac { \pi } { 2 } \right) } { | 1 - x | [ 1 - x ] }$
(1) equals 0
(2) equals - 1
(3) does not exist
(4) equal 1

For each $x \in R$, let $[x]$ be the greatest integer less than or equal to $x$. Then $\lim_{x \rightarrow 0^-} \frac{x([x] + |x|)\sin[x]}{|x|}$ is equal to
(1) 1
(2) 0
(3) $-\sin 1$
(4) $\sin 1$

$\lim _ { n \rightarrow \infty } \left[ \frac { 1 } { n } + \frac { n } { ( n + 1 ) ^ { 2 } } + \frac { n } { ( n + 2 ) ^ { 2 } } + \ldots + \frac { n } { ( 2 n - 1 ) ^ { 2 } } \right]$ is equal to
(1) $\frac { 1 } { 2 }$
(2) $\frac { 1 } { 4 }$
(3) $\frac { 1 } { 3 }$
(4) 1

The number of distinct real roots of the equation $x ^ { 5 } \left( x ^ { 3 } - x ^ { 2 } - x + 1 \right) + x \left( 3 x ^ { 3 } - 4 x ^ { 2 } - 2 x + 4 \right) - 1 = 0$ is $\_\_\_\_$.

The number of elements in the set $\{n \in \mathbb{Z} : n^2 - 10n + 19 < 6\}$ is $\_\_\_\_$.

The number of real solutions of the equation $x\left(x^2 + 3|x| + 5|x-1| + 6|x-2|\right) = 0$ is $\underline{\hspace{1cm}}$.

The number of real solutions of the equation $x | x + 5 | + 2 | x + 7 | - 2 = 0$ is $\_\_\_\_$

Let $\mathrm { f } ( \mathrm { x } ) = 2 ^ { \mathrm { x } } - \mathrm { x } ^ { 2 } , \mathrm { x } \in \mathrm { R }$. If m and n are respectively the number of points at which the curves $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$ and $\mathrm { y } = \mathrm { f } ^ { \prime } ( \mathrm { x } )$ intersects the x-axis, then the value of $\mathrm { m } + \mathrm { n }$ is

$\lim _ { x \rightarrow \infty } \frac { \left( 2 x ^ { 2 } - 3 x + 5 \right) ( 3 x - 1 ) ^ { \frac { x } { 2 } } } { \left( 3 x ^ { 2 } + 5 x + 4 \right) \sqrt { ( 3 x + 2 ) ^ { x } } }$ is equal to :
(1) $\frac { 2 } { \sqrt { 3 \mathrm { e } } }$
(2) $\frac { 2 \mathrm { e } } { \sqrt { 3 } }$
(3) $\frac { 2 } { 3 \sqrt { e } }$
(4) $\frac { 2 e } { 3 }$

If the function $f ( x ) = \left\{ \begin{array} { l } \frac { 2 } { x } \left\{ \sin \left( k _ { 1 } + 1 \right) x + \sin \left( k _ { 2 } - 1 \right) x \right\} , \quad x < 0 \\ 4 , \quad x = 0 \end{array} \quad \right.$ is continuous at $\mathrm { x } = 0$, then $\mathrm { k } _ { 1 } ^ { 2 } + \mathrm { k } _ { 2 } ^ { 2 }$ is equal to
(1) 20
(2) 5
(3) 8
(4) 10

$\lim_{x \rightarrow 0} \operatorname{cosec} x \left(\sqrt{2\cos^2 x + 3\cos x} - \sqrt{\cos^2 x + \sin x + 4}\right)$ is:
(1) 0
(2) $\frac{1}{\sqrt{15}}$
(3) $\frac{1}{2\sqrt{5}}$
(4) $-\frac{1}{2\sqrt{5}}$

Let $[t]$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbf{N}$ for which $$\lim_{x \rightarrow 0^+}\left(x\left(\left[\frac{1}{x}\right] + \left[\frac{2}{x}\right] + \ldots + \left[\frac{\mathrm{p}}{x}\right]\right) - x^2\left(\left[\frac{1}{x^2}\right] + \left[\frac{2^2}{x^2}\right] + \ldots + \left[\frac{9^2}{x^2}\right]\right)\right) \geq 1$$ is equal to \_\_\_\_ .