Let $y = f ( t )$ be a solution to the differential equation $\frac { d y } { d t } = k y$, where $k$ is a constant. Values of $f$ for selected values of $t$ are given in the table below:

$t$	0	2
$f ( t )$	4	12

Which of the following is an expression for $f ( t )$ ?
(A) $4 e ^ { \frac { t } { 2 } \ln 3 }$
(B) $e ^ { \frac { t } { 2 } \ln 9 } + 3$
(C) $2 t ^ { 2 } + 4$
(D) $4 t + 4$

We are interested in the evolution of the height of a corn plant as a function of time.
We decide to model this growth by a logistic function of the type: $$h ( t ) = \frac { a } { 1 + b \mathrm { e } ^ { - 0,04 t } }$$ where $a$ and $b$ are positive real constants, $t$ is the time variable expressed in days and $h ( t )$ denotes the height of the plant, expressed in metres.
We know that initially, for $t = 0$, the plant measures $0,1 \mathrm{~m}$ and that its height tends towards a limiting height of 2 m.
Part 1. Determine the constants $a$ and $b$ so that the function $h$ corresponds to the growth of the corn plant studied.
Part 2. We now consider that the growth of the corn plant is given by the function $f$ defined on $[0;250]$ by $$f ( t ) = \frac { 2 } { 1 + 19 \mathrm { e } ^ { - 0,04 t } }$$

1. Determine $f ^ { \prime } ( t )$ as a function of $t$ ($f ^ { \prime }$ denoting the derivative function of the function $f$). Deduce the variations of the function $f$ on the interval $[ 0 ; 250 ]$.
2. Calculate the time required for the corn plant to reach a height greater than $1,5 \mathrm{~m}$.
3. a. Verify that for all real $t$ belonging to the interval $[ 0 ; 250 ]$ we have $f ( t ) = \frac { 2 \mathrm { e } ^ { 0,04 t } } { \mathrm { e } ^ { 0,04 t } + 19 }$.
   Show that the function $F$ defined on the interval $[ 0 ; 250]$ by $F ( t ) = 50 \ln \left( \mathrm { e } ^ { 0,04 t } + 19 \right)$ is an antiderivative of the function $f$.
   b. Determine the average value of $f$ on the interval $[ 50 ; 100 ]$. Give an approximate value to $10 ^ { - 2 }$ and interpret this result.
4. We are interested in the growth rate of the corn plant; it is given by the derivative function of the function $f$. The growth rate is maximum for a value of $t$. Using the graph given in the appendix, determine an approximate value of this. Then estimate the height of the plant.

A statistical study was conducted in a large city in France between January $1^{\text{st}}$ 2000 and January $1^{\text{st}}$ 2010 to assess the proportion of households with a fixed internet connection. On January $1^{\text{st}}$ 2000, one household in eight had a fixed internet connection and, on January $1^{\text{st}}$ 2010, $64\%$ of households did. Following this study, this proportion was modelled by the function $g$ defined on the interval $[0; +\infty[$ by: $$g(t) = \frac{1}{1 + k\mathrm{e}^{-at}},$$ where $k$ and $a$ are two positive real constants and the variable $t$ denotes the time, measured in years, elapsed since January $1^{\text{st}}$ 2000.

1. Determine the exact values of $k$ and $a$ so that $g(0) = \frac{1}{8}$ and $g(10) = \frac{64}{100}$.
2. In the following, we take $k = 7$ and $a = 0.25$. The function $g$ is therefore defined by: $$g(t) = \frac{1}{1 + 7\mathrm{e}^{-\left(\frac{t}{4}\right)}}$$ a. Show that the function $g$ is increasing on the interval $[0; +\infty[$. b. According to this model, can we assert that one day, at least $99\%$ of households in this city will have a fixed internet connection? Justify your answer.
3. a. Give, to the nearest hundredth, the proportion of households, predicted by the model, equipped with a fixed internet connection on January $1^{\text{st}}$ 2018. b. Given the development of mobile telephony, some statisticians believe that the modelling by the function $g$ of the evolution of the proportion of households with a fixed internet connection should be reconsidered. At the beginning of 2018 a survey was conducted. Out of 1000 households, 880 had a fixed connection. Does this survey support these sceptical statisticians? (You may use an asymptotic confidence interval at the $95\%$ level.)

17. (This problem is worth 14 points) There are two mutually perpendicular straight-line highways on the periphery of a mountainous area. To further improve the traffic situation in the mountainous area, a plan is made to build a straight-line highway connecting the two highways and the boundary of the mountainous area. Let the two mutually perpendicular highways be $l _ { 1 } , l _ { 2 }$, the boundary curve of the mountainous area be C, and the planned highway be l. As shown in the figure, $\mathrm { M } , \mathrm { N }$ are two endpoints of C. It is measured that the distances from point M to $l _ { 1 } , l _ { 2 }$ are 5 kilometers and 40 kilometers respectively, and the distances from point N to $l _ { 1 } , l _ { 2 }$ are 20 kilometers and 2.5 kilometers respectively. Taking the lines where $l _ { 1 } , l _ { 2 }$ are located as the $\mathrm { x } , \mathrm { y }$ axes respectively, establish a rectangular coordinate system xOy. Assume that the curve C conforms to the function model $y = \frac { a } { x ^ { 2 } + b }$ (where $\mathrm { a } , \mathrm { b }$ are constants). (I) Find the values of $\mathrm { a } , \mathrm { b }$; (II) Let the highway l be tangent to curve C at point P, and the x-coordinate of P is t.
(1) Write out the function expression $f ( t )$ for the length of highway l and its domain;
(2) When t takes what value is the length of highway l shortest? Find the shortest length.

In the early stages of an infectious disease outbreak, since most people have not been infected and have no antibodies, the total number of infected people usually grows exponentially. Under the premise that ``the initial number of infected people is $P _ { 0 }$ , and each infected person on average infects $r$ uninfected people per day'', the total number of people infected with the disease after $n$ days, $P _ { n }$ , can be expressed as
$$P _ { n } = P _ { 0 } ( 1 + r ) ^ { n } \text {, where } P _ { 0 } \geq 1 \text { and } r > 0 \text { . }$$
Answer the following questions:
(1) Given that $A = \frac { \log P _ { 5 } - \log P _ { 2 } } { 3 } , B = \frac { \log P _ { 8 } - \log P _ { 6 } } { 2 }$ , show that $A = B$ . (4 points)
(2) Given that a certain infectious disease in its early stages follows the above mathematical model and the total number of infected people increases tenfold every 16 days, find the value of $\frac { P _ { 20 } } { P _ { 17 } } \times \frac { P _ { 8 } } { P _ { 6 } } \times \frac { P _ { 5 } } { P _ { 2 } }$ . (5 points)
(3) Based on (2), find the value of $\frac { \log P _ { 20 } - \log P _ { 17 } } { 3 }$ . (4 points)

The half-life $T$ of a radioactive substance is defined as ``every time period $T$ passes, the mass of the substance decays to half of its original amount''. A lead container contains two radioactive substances $A$ and $B$ with half-lives $T _ { A }$ and $T _ { B }$ respectively. At the start of recording, the masses of these two substances are equal. After 112 days, measurement shows that the mass of substance $B$ is one-quarter of the mass of substance $A$. Based on the above, which of the following is the relationship between $T _ { A }$ and $T _ { B }$?
(1) $- 2 + \frac { 112 } { T _ { A } } = \frac { 112 } { T _ { B } }$
(2) $2 + \frac { 112 } { T _ { A } } = \frac { 112 } { T _ { B } }$
(3) $- 2 + \log _ { 2 } \frac { 112 } { T _ { A } } = \log _ { 2 } \frac { 112 } { T _ { B } }$
(4) $2 + \log _ { 2 } \frac { 112 } { T _ { A } } = \log _ { 2 } \frac { 112 } { T _ { B } }$
(5) $2 \log _ { 2 } \frac { 112 } { T _ { A } } = \log _ { 2 } \frac { 112 } { T _ { B } }$

Research shows that the residual amount of a certain drug in a user's body decreases exponentially over time after taking the drug. It is known that 2 hours after taking the drug, half of the drug dose remains in the body. Which of the following options is correct?
(1) After 3 hours, the body still retains $\frac{1}{3}$ of the drug dose
(2) After 4 hours, the body still retains $\frac{1}{4}$ of the drug dose
(3) After 6 hours, the body still retains $\frac{1}{6}$ of the drug dose
(4) After 8 hours, the body still retains $\frac{1}{8}$ of the drug dose
(5) After 10 hours, the body still retains $\frac{1}{10}$ of the drug dose