| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | October |
| 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) change of sign method to verify the root to required accuracy. All techniques are routine C3/C4 content 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 | Mark | Guidance |
| \(x^5 + x^3 - 12x^2 - 8 = 0 \Rightarrow x^5 + x^3 = 12x^2 + 8\) | M1 | Attempts to write equation in form \(x^5 \pm x^3 = 12x^2 \pm 8\) or \(x^3(x^2 \pm 1) = 12x^2 \pm 8\) |
| \(x^3(x^2+1) = 12x^2+8 \Rightarrow x^3 = \frac{12x^2+8}{(x^2+1)}\) or e.g. \(x^3 = \frac{4(3x^2+2)}{(x^2+1)}\) | A1 | Intermediate line of \(x^3 = \frac{12x^2+8}{(x^2+1)}\) seen |
| \(\Rightarrow x = \sqrt[3]{\frac{4(3x^2+2)}{(x^2+1)}}\) or \(x = \sqrt[3]{\frac{4(2+3x^2)}{(x^2+1)}}\) | A1* | cso — factorisation of lhs seen explicitly and statement that \(f(x)=0\); beware other algebraic methods — if in doubt send to review |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_1 = \sqrt[3]{\frac{4(3\times 2^2+2)}{2^2+1}} = 2.237\) | M1A1 | Substitutes \(x_0\) into the iteration formula |
| \(x_2 = 2.246,\quad x_3 = 2.247\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Interval \([2.2465, 2.2475] \Rightarrow f(2.2465) = \ldots,\ f(2.2475) = \ldots\) | M1 | |
| \(f(2.2465) = -0.0057,\quad f(2.2475) = (+)0.083\) + Reason + Conclusion | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \sqrt[3]{\frac{4(3x^2+2)}{(x^2+1)}} \Rightarrow x^3(x^2+1) = 12x^2+8\) | M1 | Cubes the printed result and multiplies up |
| \(x^5 + x^3 - 12x^2 - 8 = 0\) | A1 | Obtains the required equation with no errors |
| Statement: Hence \(f(x) = 0\) | A1* | Makes a conclusion (may be minimal e.g. tick, QED) and \(x^3(x^2+1) = x^5+x^3\) seen explicitly |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^5 + x^3 - 12x^2 - 8 = 0 \Rightarrow x^5 + x^3 = 12x^2 + 8$ | M1 | Attempts to write equation in form $x^5 \pm x^3 = 12x^2 \pm 8$ or $x^3(x^2 \pm 1) = 12x^2 \pm 8$ |
| $x^3(x^2+1) = 12x^2+8 \Rightarrow x^3 = \frac{12x^2+8}{(x^2+1)}$ or e.g. $x^3 = \frac{4(3x^2+2)}{(x^2+1)}$ | A1 | Intermediate line of $x^3 = \frac{12x^2+8}{(x^2+1)}$ seen |
| $\Rightarrow x = \sqrt[3]{\frac{4(3x^2+2)}{(x^2+1)}}$ or $x = \sqrt[3]{\frac{4(2+3x^2)}{(x^2+1)}}$ | A1* | cso — factorisation of lhs seen explicitly and statement that $f(x)=0$; beware other algebraic methods — if in doubt send to review |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1 = \sqrt[3]{\frac{4(3\times 2^2+2)}{2^2+1}} = 2.237$ | M1A1 | Substitutes $x_0$ into the iteration formula |
| $x_2 = 2.246,\quad x_3 = 2.247$ | A1 | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Interval $[2.2465, 2.2475] \Rightarrow f(2.2465) = \ldots,\ f(2.2475) = \ldots$ | M1 | |
| $f(2.2465) = -0.0057,\quad f(2.2475) = (+)0.083$ + Reason + Conclusion | A1 | |
## Alt (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \sqrt[3]{\frac{4(3x^2+2)}{(x^2+1)}} \Rightarrow x^3(x^2+1) = 12x^2+8$ | M1 | Cubes the printed result and multiplies up |
| $x^5 + x^3 - 12x^2 - 8 = 0$ | A1 | Obtains the required equation with no errors |
| Statement: Hence $f(x) = 0$ | A1* | Makes a conclusion (may be minimal e.g. tick, QED) and $x^3(x^2+1) = x^5+x^3$ seen explicitly |1.
$$f ( x ) = x ^ { 5 } + x ^ { 3 } - 12 x ^ { 2 } - 8 , \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be written as
$$x = \sqrt [ 3 ] { \frac { 4 \left( 3 x ^ { 2 } + 2 \right) } { x ^ { 2 } + 1 } }$$
\item Use the iterative formula
$$x _ { n + 1 } = \sqrt [ 3 ] { \frac { 4 \left( 3 x _ { n } ^ { 2 } + 2 \right) } { x _ { n } ^ { 2 } + 1 } }$$
with $x _ { 0 } = 2$, to find $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ giving your answers to 3 decimal places.
The equation $\mathrm { f } ( x ) = 0$ has a single root, $\alpha$.
\item By choosing a suitable interval, prove that $\alpha = 2.247$ to 3 decimal places.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q1 [8]}}