A treatment protocol for a disease in children involves long-term infusion of an appropriate medication. The concentration of the medication in the blood over time is modeled by the function $C$ defined on the interval $[0; +\infty[$ by:
$$C ( t ) = \frac { d } { a } \left( 1 - \mathrm { e } ^ { - \frac { a } { 80 } t } \right)$$
The clearance $a$ of a certain patient is 7, and we choose an infusion rate $d$ equal to 84. In this part, the function $C$ is therefore defined on $[0; +\infty[$ by:
$$C ( t ) = 12 \left( 1 - \mathrm { e } ^ { - \frac { 7 } { 80 } t } \right)$$

1. Study the monotonicity of the function $C$ on $[0; +\infty[$.
2. For the treatment to be effective, the plateau must equal 15. Is the treatment of this patient effective?

Pharmacokinetics studies the evolution of a drug after its administration in the body, by measuring its plasma concentration, that is to say its concentration in the plasma. In this exercise we study the evolution of plasma concentration in a patient of the same dose of drug, considering different modes of administration.
Part A: administration by intravenous route
We denote $f ( t )$ the plasma concentration, expressed in microgram per litre ( $\mu \mathrm { g } . \mathrm { L } ^ { - 1 }$ ), of the drug, after $t$ hours following administration by intravenous route. The mathematical model is: $f ( t ) = 20 \mathrm { e } ^ { - 0,1 t }$, with $t \in [ 0 ; + \infty [$. The initial plasma concentration of the drug is therefore $f ( 0 ) = 20 \mu \mathrm {~g} . \mathrm { L } ^ { - 1 }$.

1. The half-life of the drug is the duration (in hours) after which the plasma concentration of the drug is equal to half the initial concentration. Determine this half-life, denoted $t _ { 0,5 }$.
2. It is estimated that the drug is eliminated as soon as the plasma concentration is less than $0.2 \mu \mathrm {~g} . \mathrm { L } ^ { - 1 }$. Determine the time from which the drug is eliminated. The result will be given rounded to the nearest tenth.
3. In pharmacokinetics, we call AUC (or ``area under the curve''), in $\mu \mathrm { g } . \mathrm { L } ^ { - 1 }$, the number $\lim _ { x \rightarrow + \infty } \int _ { 0 } ^ { x } f ( t ) \mathrm { d } t$. Verify that for this model, the AUC is equal to $200 \mu \mathrm {~g} . \mathrm { L } ^ { - 1 }$.

Part B: administration by oral route
We denote $g ( t )$ the plasma concentration of the drug, expressed in microgram per litre ( $\mu g.L^{-1}$ ), after $t$ hours following ingestion by oral route. The mathematical model is: $g ( t ) = 20 \left( \mathrm { e } ^ { - 0,1 t } - \mathrm { e } ^ { - t } \right)$, with $t \in [ 0 ; + \infty [$. In this case, the effect of the drug is delayed, since the initial plasma concentration is equal to: $g ( 0 ) = 0 \mu g . \mathrm { L } ^ { - 1 }$.

1. Prove that, for all $t$ in the interval $[ 0 ; + \infty [$, we have: $g ^ { \prime } ( t ) = 20 \mathrm { e } ^ { - t } \left( 1 - 0,1 \mathrm { e } ^ { 0,9 t } \right)$.
2. Study the variations of the function $g$ on the interval $[ 0 ; + \infty [$. (The limit at $+ \infty$ is not required.) Deduce the duration after which the plasma concentration of the drug is maximum. The result will be given to the nearest minute.

Part C: repeated administration by intravenous route
We decide to inject at regular time intervals the same dose of drug by intravenous route. The time interval (in hours) between two injections is chosen equal to the half-life of the drug, that is to say the number $t _ { 0,5 }$ which was calculated in A - 1. Each new injection causes an increase in plasma concentration of $20 \mu \mathrm {~g} . \mathrm { L } ^ { - 1 }$. We denote $u _ { n }$ the plasma concentration of the drug immediately after the $n$-th injection. Thus, $u _ { 1 } = 20$ and, for all integer $n$ greater than or equal to 1, we have: $u _ { n + 1 } = 0,5 u _ { n } + 20$.

1. Prove by induction that, for all integer $n \geqslant 1 : u _ { n } = 40 - 40 \times 0,5 ^ { n }$.
2. Determine the limit of the sequence $( u _ { n } )$ as $n$ tends to $+ \infty$.
3. We consider that equilibrium is reached as soon as the plasma concentration exceeds $38 \mu \mathrm {~g} . \mathrm { L } ^ { - 1 }$. Determine the minimum number of injections necessary to reach this equilibrium.

Exercise 4 — Candidates who have not followed the specialization course
We are interested in the fall of a water droplet that detaches from a cloud without initial velocity. A very simplified model makes it possible to establish that the instantaneous vertical velocity, expressed in $\mathrm{m.s^{-1}}$, of the droplet's fall as a function of the fall duration $t$ is given by the function $v$ defined as follows:
For every non-negative real number $t$, $v(t) = 9.81\dfrac{m}{k}\left(1 - \mathrm{e}^{-\frac{k}{m}t}\right)$; the constant $m$ is the mass of the droplet in milligrams and the constant $k$ is a strictly positive coefficient related to air friction.
We recall that instantaneous velocity is the derivative of position. Parts $A$ and $B$ are independent.
Part A - General case

1. Determine the variations of the velocity of the water droplet.
2. Does the droplet slow down during its fall?
3. Show that $\lim_{t \rightarrow +\infty} v(t) = 9.81\dfrac{m}{k}$. This limit is called the terminal velocity of the droplet.
4. A scientist claims that after a fall duration equal to $\dfrac{5m}{k}$, the velocity of the droplet exceeds $99\%$ of its terminal velocity. Is this claim correct?

Part B
In this part, we take $m = 6$ and $k = 3.9$. At a given instant, the instantaneous velocity of this droplet is $15\mathrm{~m.s^{-1}}$.

1. How long ago did the droplet detach from its cloud? Round the answer to the nearest tenth of a second.
2. Deduce the average velocity of this droplet between the moment it detached from the cloud and the instant when its velocity was measured. Round the answer to the nearest tenth of $\mathrm{m.s^{-1}}$.

Assume that a type of eucalyptus has an expected exponential growth rate in the first years after planting, modeled by the function $y(t) = a^{t-1}$, in which $y$ represents the height of the plant in meters, $t$ is considered in years, and $a$ is a constant greater than 1. The graph represents the function $y$.
Also assume that $y(0)$ gives the height of the seedling when planted, and it is desired to cut the eucalyptus when the seedlings grow 7.5 m after planting.
The time between planting and cutting, in years, is equal to
(A) 3.
(B) 4.
(C) 6.
(D) $\log_{2} 7$.
(E) $\log_{2} 15$.

The government of a city is concerned about a possible epidemic of an infectious disease caused by bacteria. To decide what measures to take, it must calculate the reproduction rate of the bacteria. In laboratory experiments of a bacterial culture, initially with 40 thousand units, the formula for the population was obtained:
$$p(t) = 40 \cdot 2^{3t}$$
where $t$ is the time, in hours, and $p(t)$ is the population, in thousands of bacteria.
In relation to the initial quantity of bacteria, after 20 min, the population will be
(A) reduced to one third.
(B) reduced to half.
(C) reduced to two thirds.
(D) doubled.
(E) tripled.

A loan was made at a monthly rate of $i\%$, using compound interest, in eight fixed and equal installments of $P$.
The debtor has the possibility of paying off the debt early at any time, paying for this the present value of the remaining installments. After paying the $5^{\text{th}}$ installment, he decides to pay off the debt when paying the $6^{\text{th}}$ installment.
The expression that corresponds to the total amount paid for the loan settlement is
(A) $P \left[ 1 + \frac { 1 } { \left( 1 + \frac { i } { 100 } \right) } + \frac { 1 } { \left( 1 + \frac { i } { 100 } \right) ^ { 2 } } \right]$
(B) $P \left[ 1 + \frac { 1 } { \left( 1 + \frac { i } { 100 } \right) } + \frac { 1 } { \left( 1 + \frac { 2i } { 100 } \right) } \right]$
(C) $P \left[ 1 + \frac { 1 } { \left( 1 + \frac { i } { 100 } \right) ^ { 2 } } + \frac { 1 } { \left( 1 + \frac { i } { 100 } \right) ^ { 2 } } \right]$
(D) $P \left[ \frac { 1 } { \left( 1 + \frac { i } { 100 } \right) } + \frac { 1 } { \left( 1 + \frac { 2i } { 100 } \right) } + \frac { 1 } { \left( 1 + \frac { 3i } { 100 } \right) } \right]$
(E) $P \left[ \frac { 1 } { \left( 1 + \frac { i } { 100 } \right) } + \frac { 1 } { \left( 1 + \frac { i } { 100 } \right) ^ { 2 } } + \frac { 1 } { \left( 1 + \frac { i } { 100 } \right) ^ { 3 } } \right]$

Even when the surroundings suddenly become dark, the human eye perceives the change gradually. After the light intensity suddenly changes from 1000 to 10 and $t$ seconds have elapsed, the light intensity $I ( t )$ perceived by a person is $$I ( t ) = 10 + 990 \times a ^ { - 5 t } ( \text { where } a \text { is a constant with } a > 1 )$$ After the light intensity suddenly changes from 1000 to 10, let $s$ seconds elapse until the person perceives the light intensity as 21. What is the value of $s$? (Note: The unit of light intensity is Td (troland).) [3 points]
(1) $\frac { 1 + 2 \log 3 } { 5 \log a }$
(2) $\frac { 1 + 3 \log 3 } { 5 \log a }$
(3) $\frac { 2 + \log 3 } { 5 \log a }$
(4) $\frac { 2 + 2 \log 3 } { 5 \log a }$
(5) $\frac { 2 + 3 \log 3 } { 5 \log a }$

An organization introduced two different nutrients into culture dishes A and B at 12 o'clock. At this time, the bacterial counts in dishes A and B are $X$ and $Y$ respectively. The quantity in dish A doubles every 3 hours; for example, at 3 PM the quantity in A is $2X$. The quantity in dish B doubles every 2 hours; for example, at 2 PM the quantity in B is $2Y$, and at 4 PM the quantity in B is $4Y$. Part of the measurement results are recorded in the table below. At 6 PM, the organization measured that the quantities in dishes A and B are the same. To estimate the bacterial quantities in dishes A and B from 12 o'clock to 12 midnight using an exponential growth model, select the correct options.

Time (o'clock)	12	13	14	15	16	17	18	19	20	21	22	23	24
Quantity in A	$X$			$2X$
Quantity in B	$Y$		$2Y$		$4Y$

(1) $X > Y$ (2) At 1 PM, the quantity in A is $\frac{4}{3}X$ (3) At 3 PM, the quantity in B is $3Y$ (4) At 7 PM, the quantity in B is 1.5 times that of A (5) At 12 midnight, the quantity in B is 2 times that of A

Over the past five years, a country's total carbon emissions decreased from $X$ billion metric tons of CO2 equivalent (CO2e) in year 1 to $Y$ billion metric tons of CO2 equivalent (CO2e) in year 5, achieving an average annual carbon reduction of 5\%, that is, $Y = ( 1 - 0.05 ) ^ { 4 } X$. The five-year carbon emission totals and annual growth rates are recorded in the following table, where Year $n$ carbon emission growth rate $= \frac { \text {(Year } n \text { carbon emission total)} - \text {(Year } n - 1 \text { carbon emission total)} } { \text {Year } n - 1 \text { carbon emission total} }$, $n = 2,3,4,5$.

	Year 1	Year 2	Year 3	Year 4	Year 5
\begin{tabular}{ c } Carbon Emission Total
$($ billion metric tons $\mathrm { CO } 2 \mathrm { e } )$

& $X$ & $A$ & $B$ & $C$ & $Y$ \hline Annual Carbon Emission Growth Rate & & - 0.07 & $p$ & $q$ & $r$ \hline \end{tabular}
Select the correct options.
(1) $A = 0.93 X$
(2) $Y \leq 0.8 X$
(3) $\frac { - 0.07 + p + q + r } { 4 } = - 0.05$
(4) $\sqrt [ 4 ] { \frac { Y } { X } } - 1 = - 0.05$
(5) $0.93 ( 1 + p ) ( 1 + q ) ( 1 + r ) = ( 0.95 ) ^ { 4 }$

It is known that UVI values have an exponential relationship with altitude: for every 300-meter increase in altitude, the UVI value increases by 4\% of the value before the increase. At ground level, the ultraviolet radiation received from the sun is 400 joules per square meter. At a mountain 4500 meters above ground level, the UVI value of the ultraviolet radiation received is which of the following options? (Single choice question, 3 points)
(1) $4 ( 1 + 0.04 \times 15 )$
(2) $4 \left( 1 + 0.04 ^ { 15 } \right)$
(3) $4 ( 1 + 0.04 ) ^ { 15 }$
(4) $4 \times 100 ( 1 + 0.04 ) ^ { 15 }$
(5) $4 \times 100 \left( 1 + 0.04 ^ { 45 } \right)$