| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a straightforward application of fixed point iteration with clear scaffolding. Part (i) is algebraic rearrangement (shown), part (ii) is calculator work applying a given formula three times, part (iii) tests understanding of initial values, and part (b) is equation solving. All steps are routine for A-level Further Maths students with no novel problem-solving required. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\therefore 10 + 100\left(\frac{T}{30}\right)^3 - 50\left(\frac{T}{30}\right)^4 = 0\) | M1 (AO3.4) | Uses model to form equation with \(V=0\) |
| \(\Rightarrow 50\left(\frac{T}{30}\right)^4 = 10 + 100\left(\frac{T}{30}\right)^3\) | M1 (AO1.1a) | Rearranges to isolate \(T^4\) term |
| \(\Rightarrow \frac{T^4}{16200} = 10 + \frac{T^3}{270}\) | ||
| \(\Rightarrow \frac{T^3}{16200} = \frac{10}{T} + \frac{T^2}{270}\) | ||
| \(\Rightarrow T = \sqrt[3]{\frac{162000}{T} + 60T^2}\) | R1 (AO2.1) | Completes rigorous and convincing argument. Must show evidence of division by \(T\) to isolate \(T^3\) term. Must be an equation throughout. AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_1 = 44.963\) | M1 (AO1.1a) | Calculates \(T_1\). (44.96345...) |
| \(T_2 = 49.987\) | ||
| \(T_3 = 53.504\) | A1 (AO1.1b) | Calculates \(T_2\) and \(T_3\). Condone greater than 3dp. (49.98742...) (53.50407...) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 38 represents current year 2018 | B1 (AO3.2a) | Explains 38 in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Translates 2029 into \(t=49\) | B1 (AO3.3) | |
| \(10 + 100\left(\frac{t}{30}\right)^3 - 50\left(\frac{t}{30}\right)^4 = 4.5 \times 1.063^t\) | M1 (AO3.4) | Uses models to set up equation or evaluate both models at one value of \(t\) |
| \(\Rightarrow t = 49.009\) | ||
| \(1980 + 49 = 2029\) | A1 (AO1.1b) | Obtains correct values for both models for two appropriate values of \(t\), \(t \in [49,50]\). e.g. \(t=49\) and \(t=50\). \(t=49\) gives: 89.89 and 89.81; \(t=50\) gives: 87.16 and 95.47 |
| Therefore use of oil and production of oil will be equal in the year 2029 | E1 (AO2.4) | Explains that the use of oil and the production of oil are equal when \(t = 49.009\), or uses a change of sign argument to explain that the value of each model for two appropriate values of \(t\) shows that the production of oil and the use of oil are the same for \(t \in (49,50)\) |
## Question 11:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\therefore 10 + 100\left(\frac{T}{30}\right)^3 - 50\left(\frac{T}{30}\right)^4 = 0$ | M1 (AO3.4) | Uses model to form equation with $V=0$ |
| $\Rightarrow 50\left(\frac{T}{30}\right)^4 = 10 + 100\left(\frac{T}{30}\right)^3$ | M1 (AO1.1a) | Rearranges to isolate $T^4$ term |
| $\Rightarrow \frac{T^4}{16200} = 10 + \frac{T^3}{270}$ | | |
| $\Rightarrow \frac{T^3}{16200} = \frac{10}{T} + \frac{T^2}{270}$ | | |
| $\Rightarrow T = \sqrt[3]{\frac{162000}{T} + 60T^2}$ | R1 (AO2.1) | Completes rigorous and convincing argument. Must show evidence of division by $T$ to isolate $T^3$ term. Must be an equation throughout. AG |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_1 = 44.963$ | M1 (AO1.1a) | Calculates $T_1$. (44.96345...) |
| $T_2 = 49.987$ | | |
| $T_3 = 53.504$ | A1 (AO1.1b) | Calculates $T_2$ and $T_3$. Condone greater than 3dp. (49.98742...) (53.50407...) |
### Part (a)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 38 represents current year 2018 | B1 (AO3.2a) | Explains 38 in context |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Translates 2029 into $t=49$ | B1 (AO3.3) | |
| $10 + 100\left(\frac{t}{30}\right)^3 - 50\left(\frac{t}{30}\right)^4 = 4.5 \times 1.063^t$ | M1 (AO3.4) | Uses models to set up equation or evaluate both models at one value of $t$ |
| $\Rightarrow t = 49.009$ | | |
| $1980 + 49 = 2029$ | A1 (AO1.1b) | Obtains correct values for both models for two appropriate values of $t$, $t \in [49,50]$. e.g. $t=49$ and $t=50$. $t=49$ gives: 89.89 and 89.81; $t=50$ gives: 87.16 and 95.47 |
| Therefore use of oil and production of oil will be equal in the year 2029 | E1 (AO2.4) | Explains that the use of oil and the production of oil are equal when $t = 49.009$, or uses a change of sign argument to explain that the value of each model for two appropriate values of $t$ shows that the production of oil and the use of oil are the same for $t \in (49,50)$ |
---11 The daily world production of oil can be modelled using
$$V = 10 + 100 \left( \frac { t } { 30 } \right) ^ { 3 } - 50 \left( \frac { t } { 30 } \right) ^ { 4 }$$
where $V$ is volume of oil in millions of barrels, and $t$ is time in years since 1 January 1980.
11
\begin{enumerate}[label=(\alph*)]
\item (i) The model is used to predict the time, $T$, when oil production will fall to zero.\\
Show that $T$ satisfies the equation
$$T = \sqrt [ 3 ] { 60 T ^ { 2 } + \frac { 162000 } { T } }$$
11 (a) (ii) Use the iterative formula $T _ { n + 1 } = \sqrt [ 3 ] { 60 T _ { n } { } ^ { 2 } + \frac { 162000 } { T _ { n } } }$, with $T _ { 0 } = 38$, to find the values of $T _ { 1 } , T _ { 2 }$, and $T _ { 3 }$, giving your answers to three decimal places.\\
11 (a) (iii) Explain the relevance of using $T _ { 0 } = 38$
11
\item From 1 January 1980 the daily use of oil by one technologically developing country can be modelled as
$$V = 4.5 \times 1.063 ^ { t }$$
Use the models to show that the country's use of oil and the world production of oil will be equal during the year 2029.\\[0pt]
[4 marks]\\
$12 \quad \mathrm { p } ( x ) = 30 x ^ { 3 } - 7 x ^ { 2 } - 7 x + 2$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2018 Q11 [10]}}