| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Standard +0.3 This is a straightforward exponential decay model with standard techniques: finding k from given information, solving for t using logarithms, and differentiation for rate of change. The shifted asymptote (400) adds minimal complexity. All steps are routine C3 procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context1.07j Differentiate exponentials: e^(kx) and a^(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = 0\), \(m = 480\) | B1 | |
| \(\therefore t = 10\), \(m = 0.998 \times 480 = 479.04\) | M1 | |
| \(\therefore 479.04 = 400 + 80e^{-10k}\) | ||
| \(e^{-10k} = \frac{79.04}{80}\) | A1 | |
| \(k = -\frac{1}{10}\ln\frac{79.04}{80} = 0.00121\) (3sf) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(475 = 400 + 80e^{-kt}\), \(\quad e^{-kt} = \frac{75}{80}\) | M1 | |
| \(t = -\frac{1}{k}\ln\frac{75}{80} = 53.5\) (3sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dm}{dt} = -80ke^{-kt}\) | M1 A1 | |
| \(t = 100\), \(\frac{dm}{dt} = -80ke^{-100k} = -0.0856\) | M1 | |
| \(\therefore\) decreasing at rate of \(0.0856\) g yr\(^{-1}\) (3sf) | A1 | (11) |
# Question 8:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 0$, $m = 480$ | B1 | |
| $\therefore t = 10$, $m = 0.998 \times 480 = 479.04$ | M1 | |
| $\therefore 479.04 = 400 + 80e^{-10k}$ | | |
| $e^{-10k} = \frac{79.04}{80}$ | A1 | |
| $k = -\frac{1}{10}\ln\frac{79.04}{80} = 0.00121$ (3sf) | M1 A1 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $475 = 400 + 80e^{-kt}$, $\quad e^{-kt} = \frac{75}{80}$ | M1 | |
| $t = -\frac{1}{k}\ln\frac{75}{80} = 53.5$ (3sf) | A1 | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dm}{dt} = -80ke^{-kt}$ | M1 A1 | |
| $t = 100$, $\frac{dm}{dt} = -80ke^{-100k} = -0.0856$ | M1 | |
| $\therefore$ decreasing at rate of $0.0856$ g yr$^{-1}$ (3sf) | A1 | **(11)** |
---8. A rock contains a radioactive substance which is decaying. The mass of the rock, $m$ grams, at time $t$ years after initial observation is given by
$$m = 400 + 80 \mathrm { e } ^ { - k t }$$
where $k$ is a positive constant.\\
Given that the mass of the rock decreases by $0.2 \%$ in the first 10 years, find\\
(i) the value of $k$,\\
(ii) the value of $t$ when $m = 475$,\\
(iii) the rate at which the mass of the rock is decreasing when $t = 100$.\\
\hfill \mbox{\textit{OCR C3 Q8 [11]}}