| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Apply iteration to find root (pure fixed point) |
| Difficulty | Moderate -0.3 This is a straightforward application of fixed point iteration requiring only substitution and calculator work. Part (a) is routine sign-change verification, while part (b) involves mechanical iteration with no conceptual challenges—slightly easier than average due to minimal problem-solving required. |
| 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 | Mark | Guidance |
| \(g(3) = -265\), \(g(4) = 3104\) | M1 | Attempts value of g at 3 and 4 with one correct (accept any value for the other as an attempt). Narrower ranges possible but must contain root and lie in \([3,4]\) |
| States change of sign, continuous and hence root in \([3,4]\) | A1 | Both values correct with reason (sign change stated or indicated, continuous function) and minimal conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_2 = \sqrt[6]{1000 - 2\times3} = 3.1591\) | M1 A1 | M1: Attempts to substitute \(x_1 = 3\) into formula. Implied by sight of expression, awrt 3.159 |
| \((\alpha =)\ 3.1589\) | A1 | cao - must be to 4 d.p. Sight of this value implies M1 even if \(x_2\) not seen |
## Question 1:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g(3) = -265$, $g(4) = 3104$ | M1 | Attempts value of g at 3 and 4 with one correct (accept any value for the other as an attempt). Narrower ranges possible but must contain root and lie in $[3,4]$ |
| States change of sign, continuous and hence root in $[3,4]$ | A1 | Both values correct with reason (sign change stated or indicated, continuous function) and minimal conclusion |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2 = \sqrt[6]{1000 - 2\times3} = 3.1591$ | M1 A1 | M1: Attempts to substitute $x_1 = 3$ into formula. Implied by sight of expression, awrt 3.159 |
| $(\alpha =)\ 3.1589$ | A1 | cao - must be to 4 d.p. Sight of this value implies M1 even if $x_2$ not seen |
---1.
$$g ( x ) = x ^ { 6 } + 2 x - 1000$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { g } ( x ) = 0$ has a root $\alpha$ in the interval [3,4]
Using the iteration formula
$$x _ { n + 1 } = \sqrt [ 6 ] { 1000 - 2 x _ { n } } \quad \text { with } x _ { 1 } = 3$$
\item \begin{enumerate}[label=(\roman*)]
\item find, to 4 decimal places, the value of $x _ { 2 }$
\item find, by repeated iteration, the value of $\alpha$.
Give your answer to 4 decimal places.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2023 Q1 [5]}}