| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a straightforward multi-part exponential question requiring basic substitution (parts i-ii), algebraic manipulation to derive an iterative formula (part iii), and mechanical application of iteration (part iv). All techniques are standard C3 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(P = 30 + 50e^{0.002\times 30} = 83.1\) | M1 | |
| \(\therefore\) population \(= 83\,100\) (3sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(30 + 50e^{0.002t} > 84\), \(e^{0.002t} > \frac{54}{50}\) | M1 | |
| \(t > \frac{1}{0.002}\ln\frac{54}{50}\), \(t > 38.5\) \(\therefore 2018\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(30 + 50e^{0.002t} = 26 + 50e^{0.003t}\), \(50e^{0.003t} - 50e^{0.002t} = 4\) | M1 | |
| \(e^{0.003t} - e^{0.002t} = 0.08\), \(e^{0.002t}(e^{0.001t} - 1) = 0.08\) | M1 | |
| \(e^{0.001t} - 1 = 0.08e^{-0.002t}\) | ||
| \(0.001t = \ln(1 + 0.08e^{-0.002t})\) | A1 | |
| \(t = 1000\ln(1 + 0.08e^{-0.002t})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(t_1 = 69.887\), \(t_2 = 67.251\), \(t_3 = 67.595\) | M1 A1 | |
| \(\therefore 2047\) | A1 | (11) |
# Question 8:
## Part (i):
| Answer/Working | Mark | Notes |
|---|---|---|
| $P = 30 + 50e^{0.002\times 30} = 83.1$ | M1 | |
| $\therefore$ population $= 83\,100$ (3sf) | A1 | |
## Part (ii):
| Answer/Working | Mark | Notes |
|---|---|---|
| $30 + 50e^{0.002t} > 84$, $e^{0.002t} > \frac{54}{50}$ | M1 | |
| $t > \frac{1}{0.002}\ln\frac{54}{50}$, $t > 38.5$ $\therefore 2018$ | M1 A1 | |
## Part (iii):
| Answer/Working | Mark | Notes |
|---|---|---|
| $30 + 50e^{0.002t} = 26 + 50e^{0.003t}$, $50e^{0.003t} - 50e^{0.002t} = 4$ | M1 | |
| $e^{0.003t} - e^{0.002t} = 0.08$, $e^{0.002t}(e^{0.001t} - 1) = 0.08$ | M1 | |
| $e^{0.001t} - 1 = 0.08e^{-0.002t}$ | | |
| $0.001t = \ln(1 + 0.08e^{-0.002t})$ | A1 | |
| $t = 1000\ln(1 + 0.08e^{-0.002t})$ | | |
## Part (iv):
| Answer/Working | Mark | Notes |
|---|---|---|
| $t_1 = 69.887$, $t_2 = 67.251$, $t_3 = 67.595$ | M1 A1 | |
| $\therefore 2047$ | A1 | **(11)** |
---8. The population in thousands, $P$, of a town at time $t$ years after $1 ^ { \text {st } }$ January 1980 is modelled by the formula
$$P = 30 + 50 \mathrm { e } ^ { 0.002 t }$$
Use this model to estimate\\
(i) the population of the town on $1 ^ { \text {st } }$ January 2010,\\
(ii) the year in which the population first exceeds 84000 .
The population in thousands, $Q$, of another town is modelled by the formula
$$Q = 26 + 50 \mathrm { e } ^ { 0.003 t }$$
(iii) Show that the value of $t$ when $P = Q$ is a solution of the equation
$$t = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t } \right)$$
(iv) Use the iterative formula
$$t _ { n + 1 } = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t _ { n } } \right)$$
with $t _ { 0 } = 50$ to find $t _ { 1 } , t _ { 2 }$ and $t _ { 3 }$ and hence, the year in which the populations of these two towns will be equal according to these models.\\
\hfill \mbox{\textit{OCR C3 Q8 [11]}}