| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a standard A-level calculus question requiring differentiation using the product rule, algebraic manipulation to rearrange into the given form, and routine application of fixed-point iteration. The steps are methodical and follow well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differentiate using the product rule to obtain \(ax^2\cos 2x - bx^3\sin 2x\) | M1 | |
| Obtain \(3x^2\cos 2x - 2x^3\sin 2x\) | A1 | |
| Equate first derivative to zero and confirm \(x = \sqrt[3]{1.5x^2\cot 2x}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Consider sign of \(x - \sqrt[3]{1.5x^2\cot 2x}\) or equivalent for 0.59 and 0.60 | M1 | |
| Obtain \(-0.009...\) and \(0.005...\) or equivalents and justify conclusion | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use iteration correctly at least once | M1 | |
| Obtain final answer 0.596 | A1 | |
| Show sufficient iterations to 5 sf to justify answer or show sign change in interval \([0.5955, 0.5965]\) | A1 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate using the product rule to obtain $ax^2\cos 2x - bx^3\sin 2x$ | M1 | |
| Obtain $3x^2\cos 2x - 2x^3\sin 2x$ | A1 | |
| Equate first derivative to zero and confirm $x = \sqrt[3]{1.5x^2\cot 2x}$ | A1 | AG |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Consider sign of $x - \sqrt[3]{1.5x^2\cot 2x}$ or equivalent for 0.59 and 0.60 | M1 | |
| Obtain $-0.009...$ and $0.005...$ or equivalents and justify conclusion | A1 | |
---
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use iteration correctly at least once | M1 | |
| Obtain final answer 0.596 | A1 | |
| Show sufficient iterations to 5 sf to justify answer or show sign change in interval $[0.5955, 0.5965]$ | A1 | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{8bdd1285-9e39-465a-8c09-bbe410504f9d-06_442_698_260_721}
The diagram shows part of the curve with equation $y = x ^ { 3 } \cos 2 x$. The curve has a maximum at the point $M$.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of $M$ satisfies the equation $x = \sqrt [ 3 ] { 1.5 x ^ { 2 } \cot 2 x }$.
\item Use the equation in part (a) to show by calculation that the $x$-coordinate of $M$ lies between 0.59 and 0.60.
\item Use an iterative formula, based on the equation in part (a), to find the $x$-coordinate of $M$ correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2020 Q5 [8]}}