| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Apply iteration to find root (pure fixed point) |
| Difficulty | Moderate -0.3 This is a standard fixed-point iteration question requiring routine application of given formulas: sign change verification, iterative calculation with a calculator, and interval verification for accuracy. All techniques are textbook exercises with no problem-solving or derivation 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 | Marks | Guidance |
| \(f(1.5) = -1.75\), \(f(2) = 8\) | M1 | Attempts to evaluate both \(f(1.5)\) and \(f(2)\); finds at least one of \(f(1.5) =\) awrt \(-1.8\) or truncated \(-1.7\), or \(f(2) = 8\) |
| Sign change (and \(f(x)\) is continuous) therefore there is a root \(\alpha\) lies in the interval \([1.5, 2]\) | A1 | Both \(f(1.5) =\) awrt \(-1.8\) or truncated \(-1.7\) and \(f(2) = 8\), states sign change and conclusion. Must use given interval or sub-interval e.g. \([1.55, 1.95]\), not one outside e.g. \([1.6, 2.1]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_1 = \left(5 - \frac{1}{2}(1.5)\right)^{\frac{1}{3}}\) | M1 | Attempt to substitute \(x_0 = 1.5\) into iterative formula; can be implied by \(x_1 =\) awrt \(1.6\) |
| \(x_1 = 1.6198\) | A1cao | This exact answer to 4 decimal places required |
| \(x_2 = 1.612159576...\), \(x_3 = 1.612649754...\) | A1 | \(x_2 =\) awrt \(1.6122\) and \(x_3 =\) awrt \(1.6126\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(1.61255) = -0.001166022687...\), \(f(1.61265) = 0.0004942645692...\) Sign change and as \(f(x)\) is continuous therefore a root \(\alpha\) lies in the interval \([1.61255, 1.61265] \Rightarrow \alpha = 1.6126\) (4 dp) | M1A1 | M1: Choose suitable interval e.g. \([1.61255, 1.61265]\) and at least one attempt to evaluate \(f(x)\). A1: needs (i) both evaluations correct to 1 sf, (ii) sign change stated, (iii) conclusion \(\Rightarrow \alpha = 1.6126\) |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1.5) = -1.75$, $f(2) = 8$ | M1 | Attempts to evaluate both $f(1.5)$ and $f(2)$; finds at least one of $f(1.5) =$ awrt $-1.8$ or truncated $-1.7$, **or** $f(2) = 8$ |
| Sign change (and $f(x)$ is continuous) therefore there is a root $\alpha$ lies in the interval $[1.5, 2]$ | A1 | **Both** $f(1.5) =$ awrt $-1.8$ or truncated $-1.7$ **and** $f(2) = 8$, states sign change and conclusion. Must use given interval or sub-interval e.g. $[1.55, 1.95]$, not one outside e.g. $[1.6, 2.1]$ |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_1 = \left(5 - \frac{1}{2}(1.5)\right)^{\frac{1}{3}}$ | M1 | Attempt to substitute $x_0 = 1.5$ into iterative formula; can be implied by $x_1 =$ awrt $1.6$ |
| $x_1 = 1.6198$ | A1cao | This exact answer to 4 decimal places required |
| $x_2 = 1.612159576...$, $x_3 = 1.612649754...$ | A1 | $x_2 =$ awrt $1.6122$ **and** $x_3 =$ awrt $1.6126$ |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1.61255) = -0.001166022687...$, $f(1.61265) = 0.0004942645692...$ Sign change and as $f(x)$ is continuous therefore a root $\alpha$ lies in the interval $[1.61255, 1.61265] \Rightarrow \alpha = 1.6126$ (4 dp) | M1A1 | M1: Choose suitable interval e.g. $[1.61255, 1.61265]$ and at least one attempt to evaluate $f(x)$. A1: needs (i) both evaluations correct to 1 sf, (ii) sign change stated, (iii) conclusion $\Rightarrow \alpha = 1.6126$ |
---1.
$$f ( x ) = 2 x ^ { 3 } + x - 10$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ has a root $\alpha$ in the interval $[ 1.5,2 ]$
The only real root of $\mathrm { f } ( x ) = 0$ is $\alpha$
The iterative formula
$$x _ { n + 1 } = \left( 5 - \frac { 1 } { 2 } x _ { n } \right) ^ { \frac { 1 } { 3 } } , \quad x _ { 0 } = 1.5$$
can be used to find an approximate value for $\alpha$
\item Calculate $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 4 decimal places.
\item By choosing a suitable interval, show that $\alpha = 1.6126$ correct to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q1 [7]}}