| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show root in interval |
| Difficulty | Moderate -0.3 This is a standard A-level numerical methods question requiring routine application of sign change and iteration formula. Part (a) is straightforward substitution, (b) is simple algebraic rearrangement, (c) is calculator work following a given formula, and (d) requires comparing successive iterations. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 = x^3 + x - 3 \Rightarrow x^3 - x^2 + x - 3 = 0\), \(f(x) = x^3 - x^2 + x - 3\) | M1 (1.1a) | Rearranges to \(f(x) = 0\) and evaluates \(f(x)\) at least once in \([1.5, 1.6]\) |
| \(f(1.5) = -0.375 < 0\), \(f(1.6) = 0.136 > 0\), hence \(\alpha\) lies between \(1.5\) and \(1.6\) | R1 (2.1) | Correct evaluations either side of root with reference to change of sign |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^3 = x^2 - x + 3\), so \(x^2 = x - 1 + \frac{3}{x}\) | M1 (1.1a) | Isolates \(x^3\) or divides by \(x\); condone one slip in cancelling or one sign error; e.g. \(x = x^2 + 1 - \frac{3}{x}\) OE |
| Completes argument to show given result; three terms need not be in given order | R1 (2.1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_2 = 1.5811\), \(x_3 = 1.5743\), \(x_4 = 1.5748\) | M1 (1.1a) | Any one correct value to at least 3 d.p., ignoring labels |
| \(x_2, x_3, x_4\) all correct to 4 d.p. | A1 (1.1b) | If no labels, accept answers in correct order with no extras beyond \(x_4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.574 \leq \alpha \leq 1.575\) | R1 (2.2a) | Interval of correct width including \(1.5743\) and \(1.5748\); condone strict inequalities; condone correct inequality in words |
## Question 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 = x^3 + x - 3 \Rightarrow x^3 - x^2 + x - 3 = 0$, $f(x) = x^3 - x^2 + x - 3$ | M1 (1.1a) | Rearranges to $f(x) = 0$ and evaluates $f(x)$ at least once in $[1.5, 1.6]$ |
| $f(1.5) = -0.375 < 0$, $f(1.6) = 0.136 > 0$, hence $\alpha$ lies between $1.5$ and $1.6$ | R1 (2.1) | Correct evaluations either side of root with reference to change of sign |
---
## Question 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^3 = x^2 - x + 3$, so $x^2 = x - 1 + \frac{3}{x}$ | M1 (1.1a) | Isolates $x^3$ or divides by $x$; condone one slip in cancelling or one sign error; e.g. $x = x^2 + 1 - \frac{3}{x}$ OE |
| Completes argument to show given result; three terms need not be in given order | R1 (2.1) | |
---
## Question 7(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_2 = 1.5811$, $x_3 = 1.5743$, $x_4 = 1.5748$ | M1 (1.1a) | Any one correct value to at least 3 d.p., ignoring labels |
| $x_2, x_3, x_4$ all correct to 4 d.p. | A1 (1.1b) | If no labels, accept answers in correct order with no extras beyond $x_4$ |
---
## Question 7(d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.574 \leq \alpha \leq 1.575$ | R1 (2.2a) | Interval of correct width including $1.5743$ and $1.5748$; condone strict inequalities; condone correct inequality in words |
---7 The equation $x ^ { 2 } = x ^ { 3 } + x - 3$ has a single solution, $x = \alpha$\\
7
\begin{enumerate}[label=(\alph*)]
\item By considering a suitable change of sign, show that $\alpha$ lies between 1.5 and 1.6\\[0pt]
[2 marks]\\
7
\item Show that the equation $x ^ { 2 } = x ^ { 3 } + x - 3$ can be rearranged into the form
$$x ^ { 2 } = x - 1 + \frac { 3 } { x }$$
7
\item Use the iterative formula
$$x _ { n + 1 } = \sqrt { x _ { n } - 1 + \frac { 3 } { x _ { n } } }$$
with $x _ { 1 } = 1.5$, to find $x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving your answers to four decimal places.\\
7
\item Hence, deduce an interval of width 0.001 in which $\alpha$ lies.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2021 Q7 [7]}}