| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.3 This is a straightforward exponential modelling question requiring standard manipulations: converting log form to exponential form using index laws (part a), interpreting a constant (part b), and differentiating an exponential function using the chain rule (part c). All techniques are routine P3/C3 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.06c Logarithm definition: log_a(x) as inverse of a^x1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A correct equation for \(p\) or \(q\): e.g. \(\log_{10} p = 2.74\) or \(\log_{10} q = 0.079\) | M1 | Implied by correct value of either \(p\) or \(q\) |
| Either \(p = \text{awrt } 550\) or \(q = \text{awrt } 1.2\) | A1 | |
| Correct equation for both \(p\) and \(q\): e.g. \(p = 10^{2.74}\) and \(q = 10^{0.079}\) | dM1 | Implied by \(x = 550 \times 1.2^{-t}\) |
| Both \(p = \text{awrt } 550\) and \(q = \text{awrt } 1.2\) with proof | A1* | No incorrect working; "show that" question — proof can be shown via \(x = 10^{2.74} \times 10^{-0.079t}\) followed by correct equations and values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| "\(p\)" is the amount of antibiotic (in mg) in the patient's bloodstream at the start | B1 | Award for statement referring to amount of antibiotic when \(t=0\); condone "dose given to patient"; do not allow correct and incorrect answer together |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 400 \times 1.4^{-t} \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}t} = -400\ln 1.4 \times 1.4^{-t}\) | B1 | Alt: \(\ln x = \ln 400 - t\ln 1.4 \Rightarrow \frac{1}{x}\frac{\mathrm{d}x}{\mathrm{d}t} = \pm\ln 1.4\) |
| Substitutes \(t = 5\) into \(\frac{\mathrm{d}x}{\mathrm{d}t}\) | M1 | Must substitute into changed function of form \(k \times 1.4^{-t}\) where \(k \neq 400\) |
| \(\frac{\mathrm{d}x}{\mathrm{d}t} = \text{awrt } -25\) | A1 | Candidates losing negative sign get B1 M1 A0 |
# Question 8:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| A correct equation for $p$ or $q$: e.g. $\log_{10} p = 2.74$ or $\log_{10} q = 0.079$ | M1 | Implied by correct value of either $p$ or $q$ |
| Either $p = \text{awrt } 550$ or $q = \text{awrt } 1.2$ | A1 | |
| Correct equation for both $p$ and $q$: e.g. $p = 10^{2.74}$ and $q = 10^{0.079}$ | dM1 | Implied by $x = 550 \times 1.2^{-t}$ |
| Both $p = \text{awrt } 550$ and $q = \text{awrt } 1.2$ with proof | A1* | No incorrect working; "show that" question — proof can be shown via $x = 10^{2.74} \times 10^{-0.079t}$ followed by correct equations and values |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| "$p$" is the amount of antibiotic (in mg) in the patient's bloodstream at the start | B1 | Award for statement referring to amount of antibiotic when $t=0$; condone "dose given to patient"; do not allow correct and incorrect answer together |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 400 \times 1.4^{-t} \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}t} = -400\ln 1.4 \times 1.4^{-t}$ | B1 | Alt: $\ln x = \ln 400 - t\ln 1.4 \Rightarrow \frac{1}{x}\frac{\mathrm{d}x}{\mathrm{d}t} = \pm\ln 1.4$ |
| Substitutes $t = 5$ into $\frac{\mathrm{d}x}{\mathrm{d}t}$ | M1 | Must substitute into changed function of form $k \times 1.4^{-t}$ where $k \neq 400$ |
| $\frac{\mathrm{d}x}{\mathrm{d}t} = \text{awrt } -25$ | A1 | Candidates losing negative sign get B1 M1 A0 |
---8. A dose of antibiotics is given to a patient.
The amount of the antibiotic, $x$ milligrams, in the patient's bloodstream $t$ hours after the dose was given, is found to satisfy the equation
$$\log _ { 10 } x = 2.74 - 0.079 t$$
\begin{enumerate}[label=(\alph*)]
\item Show that this equation can be written in the form
$$x = p q ^ { - t }$$
where $p$ and $q$ are constants to be found. Give the value of $p$ to the nearest whole number and the value of $q$ to 2 significant figures.
\item With reference to the equation in part (a), interpret the value of the constant $p$.
When a different dose of the antibiotic is given to another patient, the values of $x$ and $t$ satisfy the equation
$$x = 400 \times 1.4 ^ { - t }$$
\item Use calculus to find, to 2 significant figures, the value of $\frac { \mathrm { d } x } { \mathrm {~d} t }$ when $t = 5$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q8 [8]}}