| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| 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 quotient rule differentiation, setting the derivative to zero, sign-change verification, and applying a given iterative formula. All techniques are routine for P2/C3 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Attempt use of quotient rule or equivalent | M1 | |
| Obtain \(\frac{2(x + 2)\cos 2x - \sin 2x}{(x + 2)^2}\) or equivalent | A1 | |
| Equate numerator to zero and attempt rearrangement | M1 | |
| Confirm given result \(\tan 2x = 2x + 4\) | A1 | [4] |
| (ii) Consider sign of \(\tan 2x - 2x - 4\) for \(0.6\) and \(0.7\) or equivalent | M1 | |
| Obtain \(-2.63\) and \(0.40\) or equivalents and justify conclusion | A1 | [2] |
| (iii) Use iteration process correctly at least once | M1 | |
| Obtain final answer \(0.694\) | A1 | |
| Show sufficient iterations to 5 decimal places to justify answer or show a sign change in the interval \((0.6935, 0.6945)\) | A1 | [3] |
**(i)** Attempt use of quotient rule or equivalent | M1 |
Obtain $\frac{2(x + 2)\cos 2x - \sin 2x}{(x + 2)^2}$ or equivalent | A1 |
Equate numerator to zero and attempt rearrangement | M1 |
Confirm given result $\tan 2x = 2x + 4$ | A1 | [4]
**(ii)** Consider sign of $\tan 2x - 2x - 4$ for $0.6$ and $0.7$ or equivalent | M1 |
Obtain $-2.63$ and $0.40$ or equivalents and justify conclusion | A1 | [2]
**(iii)** Use iteration process correctly at least once | M1 |
Obtain final answer $0.694$ | A1 |
Show sufficient iterations to 5 decimal places to justify answer or show a sign change in the interval $(0.6935, 0.6945)$ | A1 | [3]
Example iterations shown:
- $[0.6 \to 0.69040 \to 0.69352 \to 0.69363$
- $0.65 \to 0.69215 \to 0.69358 \to 0.69363$
- $0.7 \to 0.69384 \to 0.69364 \to 0.69363]$
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{48ab71ff-c37b-4e0b-b031-d99b0cf517a8-3_421_976_251_580}
The diagram shows the curve $y = \frac { \sin 2 x } { x + 2 }$ for $0 \leqslant x \leqslant \frac { 1 } { 2 } \pi$. The $x$-coordinate of the maximum point $M$ is denoted by $\alpha$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and show that $\alpha$ satisfies the equation $\tan 2 x = 2 x + 4$.\\
(ii) Show by calculation that $\alpha$ lies between 0.6 and 0.7 .\\
(iii) Use the iterative formula $x _ { n + 1 } = \frac { 1 } { 2 } \tan ^ { - 1 } \left( 2 x _ { n } + 4 \right)$ to find the value of $\alpha$ correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
\hfill \mbox{\textit{CAIE P2 2012 Q6 [9]}}