| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.8 This question requires finding a stationary point by differentiating a product involving trigonometric functions, then manipulating the resulting equation into a specific form for iteration. The differentiation uses product rule with cos(3x) and requires careful algebraic manipulation to isolate the tan^(-1) form. While the iteration itself is routine, deriving the equation requires multiple non-trivial steps and careful handling of trigonometric identities, making it moderately challenging but within reach of a well-prepared P3 student. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07r 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 |
|---|---|---|
| Answer | Mark | Guidance |
| Use correct product rule | M1 | \(\frac{d}{dx}(x^2)\cos(3x) + x^2\frac{d}{dx}(\cos 3x)\) |
| Obtain correct derivative in any form | A1 | e.g. \(2x\cos 3x - 3x^2\sin 3x\) |
| Equate derivative to zero and obtain \(a = \frac{1}{3}\tan^{-1}\!\left(\frac{2}{3a}\right)\) | A1 | AG; must at least reach \(2x = 3x^2\tan(3x)\) or better before final answer; final answer must be in terms of \(a\); look for \(\frac{2}{3}a\) or \(\frac{2}{3}x\) in working not immediately corrected as penultimate line A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use iterative process \(a_{n+1} = \frac{1}{3}\tan^{-1}\!\left(\frac{2}{3a_n}\right)\) correctly at least twice | M1 | Degrees 0/3 |
| Obtain final answer \(0.36\) | A1 | Must be 2 d.p. |
| Show sufficient iterations to 4 or more d.p. to justify \(0.36\) to 2 d.p. or show sign change in interval \((0.355, 0.365)\) | A1 | Allow small errors in 4th d.p.; allow errors at start if self corrects later |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct product rule | M1 | $\frac{d}{dx}(x^2)\cos(3x) + x^2\frac{d}{dx}(\cos 3x)$ |
| Obtain correct derivative in any form | A1 | e.g. $2x\cos 3x - 3x^2\sin 3x$ |
| Equate derivative to zero and obtain $a = \frac{1}{3}\tan^{-1}\!\left(\frac{2}{3a}\right)$ | A1 | AG; must at least reach $2x = 3x^2\tan(3x)$ or better before final answer; final answer must be in terms of $a$; look for $\frac{2}{3}a$ or $\frac{2}{3}x$ in working not immediately corrected as penultimate line A0 |
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## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use iterative process $a_{n+1} = \frac{1}{3}\tan^{-1}\!\left(\frac{2}{3a_n}\right)$ correctly at least twice | M1 | Degrees 0/3 |
| Obtain final answer $0.36$ | A1 | Must be 2 d.p. |
| Show sufficient iterations to 4 or more d.p. to justify $0.36$ to 2 d.p. or show sign change in interval $(0.355, 0.365)$ | A1 | Allow small errors in 4th d.p.; allow errors at start if self corrects later |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{72042f09-3495-42e9-bee9-96ec5ac0bf0c-06_352_643_274_744}
The diagram shows the part of the curve $y = x ^ { 2 } \cos 3 x$ for $0 \leqslant x \leqslant \frac { 1 } { 6 } \pi$, and its maximum point $M$, where $x = a$.
\begin{enumerate}[label=(\alph*)]
\item Show that $a$ satisfies the equation $a = \frac { 1 } { 3 } \tan ^ { - 1 } \left( \frac { 2 } { 3 a } \right)$.
\item Use an iterative formula based on the equation in (a) to determine $a$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q5 [6]}}