| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | October |
| 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 standard P3 fixed point iteration question requiring routine application of the intermediate value theorem and calculator-based iteration. Part (a) involves straightforward substitution to show sign change, while part (b) is purely mechanical computation with no conceptual challenges or 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 |
| \(f(1) = 1 - 5 + e = -1.281... < 0\) | M1 | Attempts \(f(1)\) and \(f(2)\) with substitution seen or at least one correct to 1 d.p. rounded or truncated and considers their signs. Note showing \(f(1)f(2) < 0\) is sufficient for sign change reasoning |
| \(f(2) = 2^2 - 5\times2 + e^2 = 1.389... > 0\) | ||
| As there is a change of sign and \(f(x)\) is continuous over \([1,2]\), then there is a root | A1* | Must have: both \(f(1)\) and \(f(2)\) correct; reason mentioning continuity and sign change; conclusion "hence root" or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_2 = \sqrt{5\times1 - e^1} = \text{awrt } 1.5105\) | M1A1 | M1 for attempting to find \(x_2\) using iteration formula, implied by sight of 1 embedded or awrt 1.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha = \text{awrt } 1.7340\) | A1 | Provided M1 has been scored. Allow 1.734 with trailing zero omitted. 1.7340 from graphical calculator alone scores M0A0A0 |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1) = 1 - 5 + e = -1.281... < 0$ | M1 | Attempts $f(1)$ and $f(2)$ with substitution seen or at least one correct to 1 d.p. rounded or truncated **and considers their signs**. Note showing $f(1)f(2) < 0$ is sufficient for sign change reasoning |
| $f(2) = 2^2 - 5\times2 + e^2 = 1.389... > 0$ | | |
| As there is a change of sign and $f(x)$ is continuous over $[1,2]$, then there is a root | A1* | Must have: both $f(1)$ and $f(2)$ correct; reason mentioning continuity and sign change; conclusion "hence root" or equivalent |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2 = \sqrt{5\times1 - e^1} = \text{awrt } 1.5105$ | M1A1 | M1 for attempting to find $x_2$ using iteration formula, implied by sight of 1 embedded or awrt 1.5 |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha = \text{awrt } 1.7340$ | A1 | Provided M1 has been scored. Allow 1.734 with trailing zero omitted. 1.7340 from graphical calculator alone scores M0A0A0 |
---\begin{enumerate}
\item A curve has equation $y = \mathrm { f } ( x )$ where
\end{enumerate}
$$\mathrm { f } ( x ) = x ^ { 2 } - 5 x + \mathrm { e } ^ { x } \quad x \in \mathbb { R }$$
(a) Show that the equation $\mathrm { f } ( x ) = 0$ has a root, $\alpha$, in the interval [1,2]
The iterative formula
$$x _ { n + 1 } = \sqrt { 5 x _ { n } - \mathrm { e } ^ { x _ { n } } }$$
with $x _ { 1 } = 1$ is used to find an approximate value for the root $\alpha$.\\
(b) (i) Find the value of $x _ { 2 }$ to 4 decimal places.\\
(ii) Find, by repeated iteration, the value of $\alpha$, giving your answer to 4 decimal places.
\hfill \mbox{\textit{Edexcel P3 2023 Q1 [5]}}