| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Rate of change in exponential model |
| Difficulty | Moderate -0.8 Part (a) requires straightforward substitution of t=0.5 into the given exponential function. Part (b) involves routine differentiation of an exponential function and simple algebraic manipulation to show the derivative is proportional to m, with k=-0.05. Both parts are standard textbook exercises requiring only direct application of basic exponential calculus with no problem-solving insight needed. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(t=0.5\) into \(m = 25e^{-0.05t} \Rightarrow m = 25e^{-0.05 \times 0.5}\) | M1 | Correct substitution |
| \(\Rightarrow m = 24.4\text{g}\) | A1 | Answer of \(24.4\text{g}\) with no working scores both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses \(\frac{d}{dt}(e^{-0.05t}) = \pm C e^{-0.05t}\) | M1 | Apply rule \(\frac{d}{dt}(e^{kx}) = ke^{kx}\) |
| \(\frac{dm}{dt} = -0.05 \times 25e^{-0.05t} = -0.05m \Rightarrow k = -0.05\) | A1 | Correct derivative and value of \(k\) |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $t=0.5$ into $m = 25e^{-0.05t} \Rightarrow m = 25e^{-0.05 \times 0.5}$ | M1 | Correct substitution |
| $\Rightarrow m = 24.4\text{g}$ | A1 | Answer of $24.4\text{g}$ with no working scores both marks |
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $\frac{d}{dt}(e^{-0.05t}) = \pm C e^{-0.05t}$ | M1 | Apply rule $\frac{d}{dt}(e^{kx}) = ke^{kx}$ |
| $\frac{dm}{dt} = -0.05 \times 25e^{-0.05t} = -0.05m \Rightarrow k = -0.05$ | A1 | Correct derivative and value of $k$ |
---5. The mass, $m$ grams, of a radioactive substance, $t$ years after first being observed, is modelled by the equation
$$m = 25 \mathrm { e } ^ { - 0.05 t }$$
According to the model,
\begin{enumerate}[label=(\alph*)]
\item find the mass of the radioactive substance six months after it was first observed,
\item show that $\frac { \mathrm { d } m } { \mathrm {~d} t } = k m$, where $k$ is a constant to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 Q5 [4]}}