| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a standard C3/C4 fixed-point iteration question with routine steps: sign change verification, algebraic rearrangement (given structure to follow), calculator-based iteration, and interval checking. All techniques are textbook exercises requiring careful arithmetic but no problem-solving insight, making it slightly easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09b 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 | Mark | Guidance |
| \(f(1)=-3\), \(f(2)=2\); sign change, \(f\) continuous, so root \(\alpha\) in \([1,2]\) | M1 A1 | M1: attempts both \(f(1)\) and \(f(2)\), achieves at least one correct value; A1: both values correct, sign change stated, conclusion given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x)=0 \Rightarrow -x^3+4x^2-6=0 \Rightarrow x^2(4-x)=6\) | M1 | Must state \(f(x)=0\) or equivalent before writing \(\pm x^2(x-4)=\pm 6\) |
| \(\Rightarrow x^2=\left(\frac{6}{4-x}\right)\) and so \(x=\sqrt{\left(\frac{6}{4-x}\right)}\) | A1* | Completely correct with all signs correct; no requirement to show \(\frac{-6}{4-x}\to\frac{6}{x-4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_2=\sqrt{\left(\frac{6}{4-1.5}\right)}\) | M1 | |
| \(x_2 = \text{awrt } 1.5492\) | A1 | |
| \(x_3 = \text{awrt } 1.5647\) and \(x_4 = \text{awrt } 1.5696/1.5697\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(1.5715)=-0.00254665\ldots\), \(f(1.5725)=0.0026157969\); sign change, \(f\) continuous, so \(\alpha\in[1.5715,1.5725]\Rightarrow\alpha=1.572\) (3 dp) | M1A1 | Both values needed with sign change and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(x_0 = 1.5\) into iterative formula. Sight of \(\sqrt{\left(\frac{6}{4-1.5}\right)}\) or \(x_2 = \text{awrt } 1.55\) | M1 | Attempt to substitute \(x_0 = 1.5\) |
| \(x_2 = \text{awrt } 1.5492\) | A1 | |
| Both \(x_3 = \text{awrt } 1.5647\) and \(x_4 = \text{awrt } 1.5696\) *or* \(1.5697\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Choose suitable interval for \(x\), e.g. \([1.5715, 1.5725]\) and at least one attempt to evaluate \(f(x)\) not the iterative formula | M1 | Continued iteration is M0 |
| (i) Both evaluations correct to 1sf (rounded or truncated), e.g. \(f(1.5715) = -0.003\) rounded, \(f(1.5715) = -0.002\) truncated; (ii) sign change stated, \(f(a) \times f(b) < 0\); (iii) conclusion e.g. "so result shown" | A1 | All three conditions needed |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1)=-3$, $f(2)=2$; sign change, $f$ continuous, so root $\alpha$ in $[1,2]$ | M1 A1 | M1: attempts both $f(1)$ and $f(2)$, achieves at least one correct value; A1: both values correct, sign change stated, conclusion given |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x)=0 \Rightarrow -x^3+4x^2-6=0 \Rightarrow x^2(4-x)=6$ | M1 | Must state $f(x)=0$ or equivalent before writing $\pm x^2(x-4)=\pm 6$ |
| $\Rightarrow x^2=\left(\frac{6}{4-x}\right)$ and so $x=\sqrt{\left(\frac{6}{4-x}\right)}$ | A1* | Completely correct with all signs correct; no requirement to show $\frac{-6}{4-x}\to\frac{6}{x-4}$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2=\sqrt{\left(\frac{6}{4-1.5}\right)}$ | M1 | |
| $x_2 = \text{awrt } 1.5492$ | A1 | |
| $x_3 = \text{awrt } 1.5647$ **and** $x_4 = \text{awrt } 1.5696/1.5697$ | A1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(1.5715)=-0.00254665\ldots$, $f(1.5725)=0.0026157969$; sign change, $f$ continuous, so $\alpha\in[1.5715,1.5725]\Rightarrow\alpha=1.572$ (3 dp) | M1A1 | Both values needed with sign change and conclusion |
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $x_0 = 1.5$ into iterative formula. Sight of $\sqrt{\left(\frac{6}{4-1.5}\right)}$ or $x_2 = \text{awrt } 1.55$ | M1 | Attempt to substitute $x_0 = 1.5$ |
| $x_2 = \text{awrt } 1.5492$ | A1 | |
| **Both** $x_3 = \text{awrt } 1.5647$ **and** $x_4 = \text{awrt } 1.5696$ *or* $1.5697$ | A1 | |
## Question 5(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Choose suitable interval for $x$, e.g. $[1.5715, 1.5725]$ and at least one attempt to evaluate $f(x)$ not the iterative formula | M1 | Continued iteration is M0 |
| (i) Both evaluations correct to 1sf (rounded or truncated), e.g. $f(1.5715) = -0.003$ rounded, $f(1.5715) = -0.002$ truncated; (ii) sign change stated, $f(a) \times f(b) < 0$; (iii) conclusion e.g. "so result shown" | A1 | All three conditions needed |
---5.
$$f ( x ) = - x ^ { 3 } + 4 x ^ { 2 } - 6$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ has a root between $x = 1$ and $x = 2$
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be rewritten as
$$x = \sqrt { \left( \frac { 6 } { 4 - x } \right) }$$
\item Starting with $x _ { 1 } = 1.5$ use the iteration $x _ { n + 1 } = \sqrt { \left( \frac { 6 } { 4 - x _ { n } } \right) }$ to calculate the values of $x _ { 2 }$, $x _ { 3 }$ and $x _ { 4 }$ giving all your answers to 4 decimal places.
\item Using a suitable interval, show that 1.572 is a root of $\mathrm { f } ( x ) = 0$ correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2016 Q5 [9]}}