| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve trigonometric equation via iteration |
| Difficulty | Standard +0.3 This is a straightforward fixed-point iteration question requiring routine application of a given formula with calculator work. Part (a) is simple sign-change verification, part (b) is mechanical iteration, and part (c) is standard convergence checking. While it involves trigonometric functions, no conceptual insight or problem-solving is needed—students simply follow the given procedure. |
| 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 |
| \(f(0.75) = -0.18\ldots\), \(f(0.85) = 0.17\ldots\) | M1 | |
| Change of sign, hence root between \(x=0.75\) and \(x=0.85\) | A1 | |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sub \(x_0=0.8\) into \(x_{n+1} = [\arcsin(1-0.5x_n)]^{\frac{1}{2}}\) to obtain \(x_1\) | M1 | |
| Awrt \(x_1=0.80219\) and \(x_2=0.80133\) | A1 | |
| Awrt \(x_3=0.80167\) | A1 | |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(0.801565) = -2.7\ldots \times 10^{-5}\), \(f(0.801575) = +8.6\ldots \times 10^{-6}\) | M1A1 | |
| Change of sign and conclusion | A1 | |
| (3) | 8 Marks |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0.75) = -0.18\ldots$, $f(0.85) = 0.17\ldots$ | M1 | |
| Change of sign, hence root between $x=0.75$ and $x=0.85$ | A1 | |
| | (2) | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sub $x_0=0.8$ into $x_{n+1} = [\arcsin(1-0.5x_n)]^{\frac{1}{2}}$ to obtain $x_1$ | M1 | |
| Awrt $x_1=0.80219$ and $x_2=0.80133$ | A1 | |
| Awrt $x_3=0.80167$ | A1 | |
| | (3) | |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0.801565) = -2.7\ldots \times 10^{-5}$, $f(0.801575) = +8.6\ldots \times 10^{-6}$ | M1A1 | |
| Change of sign and conclusion | A1 | |
| | (3) | **8 Marks** |
---$$\mathrm { f } ( x ) = 2 \sin \left( x ^ { 2 } \right) + x - 2 , \quad 0 \leqslant x < 2 \pi$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = 0$ has a root $\alpha$ between $x = 0.75$ and $x = 0.85$
The equation $\mathrm { f } ( x ) = 0$ can be written as $x = [ \arcsin ( 1 - 0.5 x ) ] ^ { \frac { 1 } { 2 } }$.
\item Use the iterative formula
$$x _ { n + 1 } = \left[ \arcsin \left( 1 - 0.5 x _ { n } \right) \right] ^ { \frac { 1 } { 2 } } , \quad x _ { 0 } = 0.8$$
to find the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 5 decimal places.
\item Show that $\alpha = 0.80157$ is correct to 5 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2011 Q2 [8]}}