| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 8 |
| 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 standard parts: (a) algebraic rearrangement to show equivalence, (b) calculator work applying the iteration formula three times, and (c) sign-change verification for a given root. All techniques are routine for P3 level 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 |
| VIIIV SIHI NI JIIIM ION OC | VIIV SIHI NI JAHAM ION OO | VJ4V SIHIL NI JIIIM IONOO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^3 + 3x^2 + 4x - 12 = 0 \Rightarrow x^3 + 3x^2 = 12 - 4x\) | ||
| \(\Rightarrow x^2(x+3) = 12 - 4x\) | M1 | Separating terms and factorising; no need to factorise RHS |
| \(\Rightarrow x^2 = \frac{12-4x}{(x+3)}\) | dM1 | Divides by \((x+3)\), then takes square root; no need for RHS to be factorised |
| \(\Rightarrow x = \sqrt{\frac{4(3-x)}{(x+3)}}\) | A1* | CSO; \(12-4x\) must be factorised; do not allow root on numerator only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_1 = \sqrt{\frac{4(3-1)}{(3+1)}} = 1.41\) | M1 A1 | Attempt to substitute \(x_0 = 1\); \(\sqrt{2}\) is A0 |
| awrt \(x_2 = 1.20\), \(x_3 = 1.31\) | A1 | Condone 1.2 for \(x_2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(1.2725) = (+)0.00827...\), \(f(1.2715) = -0.00821...\) | M1 | Calculates both values; accept 1sf rounded or truncated |
| Change of sign with \(f(x)\) continuous, \(\Rightarrow \alpha = 1.272\) | A1 | Both values correct; valid reason (change of sign or \(f(1.2715) \times f(1.2725) < 0\)); minimal conclusion |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^3 + 3x^2 + 4x - 12 = 0 \Rightarrow x^3 + 3x^2 = 12 - 4x$ | | |
| $\Rightarrow x^2(x+3) = 12 - 4x$ | M1 | Separating terms and factorising; no need to factorise RHS |
| $\Rightarrow x^2 = \frac{12-4x}{(x+3)}$ | dM1 | Divides by $(x+3)$, then takes square root; no need for RHS to be factorised |
| $\Rightarrow x = \sqrt{\frac{4(3-x)}{(x+3)}}$ | A1* | CSO; $12-4x$ must be factorised; do not allow root on numerator only |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1 = \sqrt{\frac{4(3-1)}{(3+1)}} = 1.41$ | M1 A1 | Attempt to substitute $x_0 = 1$; $\sqrt{2}$ is A0 |
| awrt $x_2 = 1.20$, $x_3 = 1.31$ | A1 | Condone 1.2 for $x_2$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1.2725) = (+)0.00827...$, $f(1.2715) = -0.00821...$ | M1 | Calculates both values; accept 1sf rounded or truncated |
| Change of sign with $f(x)$ continuous, $\Rightarrow \alpha = 1.272$ | A1 | Both values correct; valid reason (change of sign or $f(1.2715) \times f(1.2725) < 0$); minimal conclusion |
---2.
$$f ( x ) = x ^ { 3 } + 3 x ^ { 2 } + 4 x - 12$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be written as
$$x = \sqrt { \left( \frac { 4 ( 3 - x ) } { ( 3 + x ) } \right) } \quad x \neq - 3$$
The equation $x ^ { 3 } + 3 x ^ { 2 } + 4 x - 12 = 0$ has a single root which is between 1 and 2
\item Use the iteration formula
$$x _ { n + 1 } = \sqrt { \left( \frac { 4 \left( 3 - x _ { n } \right) } { \left( 3 + x _ { n } \right) } \right) } \quad n \geqslant 0$$
with $x _ { 0 } = 1$ to find, to 2 decimal places, the value of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$
The root of $\mathrm { f } ( x ) = 0$ is $\alpha$.
\item By choosing a suitable interval, prove that $\alpha = 1.272$ to 3 decimal places.\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIIIV SIHI NI JIIIM ION OC & VIIV SIHI NI JAHAM ION OO & VJ4V SIHIL NI JIIIM IONOO \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2018 Q2 [8]}}