| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a standard fixed-point iteration question with routine algebraic rearrangement (part a), straightforward calculator work applying the iteration formula (part b), and a standard change-of-sign verification (part c). All techniques are textbook exercises requiring no novel insight, making it slightly easier than average for A-level. |
| Spec | 1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = x^3 - 5x + 16 = 0\) so \(x^3 = 5x - 16\) | M1 | Must state \(f(x)=0\) or imply by writing \(x^3-5x+16=0\) and reach \(x^3 = \pm5x \pm 16\) |
| \(\Rightarrow x = \sqrt[3]{5x-16}\) | A1 (2) | Completely correct including \(f(x)=0\) stated or implied |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_2 = \sqrt[3]{5\times-3-16}\) | M1 | Attempt to substitute \(x_1=-3\) into iterative formula |
| \(x_2 = -3.141\) awrt | A1 | |
| \(x_3 = -3.165\) awrt and \(x_4 = -3.169\) awrt | A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(-3.175) = -0.130984...\) , \(f(-3.165) = 0.120482...\) Sign change (and \(f(x)\) continuous) therefore root \(\alpha\) lies in \([-3.175, -3.165] \Rightarrow \alpha = -3.17\) (2 dp) | M1 A1 (2) | M1: Choose interval e.g. \([-3.175,-3.165]\) and at least one evaluation. A1: Both evaluations correct to 1 sf, sign change stated, conclusion given |
# Question 2:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = x^3 - 5x + 16 = 0$ so $x^3 = 5x - 16$ | M1 | Must state $f(x)=0$ or imply by writing $x^3-5x+16=0$ and reach $x^3 = \pm5x \pm 16$ |
| $\Rightarrow x = \sqrt[3]{5x-16}$ | A1 (2) | Completely correct including $f(x)=0$ stated or implied |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_2 = \sqrt[3]{5\times-3-16}$ | M1 | Attempt to substitute $x_1=-3$ into iterative formula |
| $x_2 = -3.141$ awrt | A1 | |
| $x_3 = -3.165$ awrt **and** $x_4 = -3.169$ awrt | A1 (3) | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(-3.175) = -0.130984...$ , $f(-3.165) = 0.120482...$ Sign change (and $f(x)$ continuous) therefore root $\alpha$ lies in $[-3.175, -3.165] \Rightarrow \alpha = -3.17$ (2 dp) | M1 A1 (2) | M1: Choose interval e.g. $[-3.175,-3.165]$ and at least one evaluation. A1: Both evaluations correct to 1 sf, sign change stated, conclusion given |
---2.
$$f ( x ) = x ^ { 3 } - 5 x + 16$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be rewritten as
$$x = ( a x + b ) ^ { \frac { 1 } { 3 } }$$
giving the values of the constants $a$ and $b$.
The equation $\mathrm { f } ( x ) = 0$ has exactly one real root $\alpha$, where $\alpha = - 3$ to one significant figure.
\item Starting with $x _ { 1 } = - 3$, use the iteration
$$x _ { n + 1 } = \left( a x _ { n } + b \right) ^ { \frac { 1 } { 3 } }$$
with the values of $a$ and $b$ found in part (a), to calculate the values of $x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving all your answers to 3 decimal places.
\item Using a suitable interval, show that $\alpha = - 3.17$ correct to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q2 [7]}}