| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a standard P3 fixed point iteration question requiring routine application of learned techniques: sign change to show root existence, algebraic rearrangement (given structure to follow), and calculator iteration. All steps are procedural with no novel insight 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 | Mark | Guidance |
| \(f(2) = 2^4 - 5(2)^2 + 4(2) - 7 = -3 < 0\) and \(f(3) = 3^4 - 5(3)^2 + 4(3) - 7 = 41 > 0\) | M1 | Attempts both \(f(2)\) and \(f(3)\), or narrower interval containing root; considers signs |
| Change of sign, \(f(x)\) continuous, therefore root in interval | A1* | Requires: both values correct, reference to sign change, reference to continuity, minimal conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x^3 = \dfrac{5x^2 - 4x + 7}{x} \Rightarrow x = \sqrt[3]{\dfrac{5x^2 - 4x + 7}{x}}\) | B1* | Proceeds to given answer with no errors; \(x^3 = \ldots\) must be seen at some stage |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x_2 = \sqrt[3]{\dfrac{5(2)^2 - 4(2) + 7}{2}}\) | M1 | Attempts \(x_2\) using iteration formula; must see correct cube root index |
| \(= \text{awrt } 2.1179\) | A1 | May be implied by awrt 2.118 or awrt 2.147 \((= x_3)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\alpha = \text{awrt } 2.1565\) | A1 | Provided M1 scored in (c)(i) |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(2) = 2^4 - 5(2)^2 + 4(2) - 7 = -3 < 0$ and $f(3) = 3^4 - 5(3)^2 + 4(3) - 7 = 41 > 0$ | M1 | Attempts both $f(2)$ and $f(3)$, or narrower interval containing root; considers signs |
| Change of sign, $f(x)$ continuous, therefore root in interval | A1* | Requires: both values correct, reference to sign change, reference to continuity, minimal conclusion |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x^3 = \dfrac{5x^2 - 4x + 7}{x} \Rightarrow x = \sqrt[3]{\dfrac{5x^2 - 4x + 7}{x}}$ | B1* | Proceeds to given answer with no errors; $x^3 = \ldots$ must be seen at some stage |
## Part (c)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x_2 = \sqrt[3]{\dfrac{5(2)^2 - 4(2) + 7}{2}}$ | M1 | Attempts $x_2$ using iteration formula; must see correct cube root index |
| $= \text{awrt } 2.1179$ | A1 | May be implied by awrt 2.118 or awrt 2.147 $(= x_3)$ |
## Part (c)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\alpha = \text{awrt } 2.1565$ | A1 | Provided M1 scored in (c)(i) |
---\begin{enumerate}
\item A curve has equation $y = \mathrm { f } ( x )$ where
\end{enumerate}
$$\mathrm { f } ( x ) = x ^ { 4 } - 5 x ^ { 2 } + 4 x - 7 \quad x \in \mathbb { R }$$
(a) Show that the equation $\mathrm { f } ( x ) = 0$ has a root, $\alpha$, in the interval [2,3]\\
(b) Show that the equation $\mathrm { f } ( x ) = 0$ can be written as
$$x = \sqrt [ 3 ] { \frac { 5 x ^ { 2 } - 4 x + 7 } { x } }$$
The iterative formula
$$x _ { n + 1 } = \sqrt [ 3 ] { \frac { 5 x _ { n } ^ { 2 } - 4 x _ { n } + 7 } { x _ { n } } }$$
is used to find $\alpha$\\
(c) Starting with $x _ { 1 } = 2$ and using the iterative formula,\\
(i) find, to 4 decimal places, the value of $x _ { 2 }$\\
(ii) find, to 4 decimal places, the value of $\alpha$
\hfill \mbox{\textit{Edexcel P3 2024 Q2 [6]}}