| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show convergence to specific root |
| Difficulty | Standard +0.8 This is a multi-part fixed point iteration question requiring: (a) interval verification using cotangent evaluation, (b) algebraic proof that the iteration formula converges to the root by rearranging cot(x/2)=3x into the given form, and (c) numerical iteration to 2dp. The rearrangement in part (b) involves non-trivial manipulation of inverse trigonometric functions and the half-angle relationship, making it more challenging than standard iteration questions that simply verify g'(α)<1. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Calculate the values of a relevant expression or pair of expressions at \(x = 0.5\) and \(x = 1\) | M1 | Need to evaluate at both points, but M1 still available if one value incorrect. Use of degrees is M0. Correct use of a smaller interval is M1. If using \(g(x) - f(x)\), there needs to be a clear indication of the comparison being made e.g. by listing values in a table. Embedded values 0.5 and 1 are not sufficient. 3.92 and 1.83 alone are not sufficient |
| Complete the argument correctly with conclusion about change of sign or change of inequalities and with correct calculated values. Can all be in symbols – explanation in words is not required | A1 | e.g. \(3.92 > 1.5\), \(1.83 < 3\) or \(2.42 > 0\), \(-1.17 < 0\) |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(x = \frac{1}{3}\left(x + 4\tan^{-1}\frac{1}{3x}\right)\) | M1 | Or rearrange \(\cot\left(\frac{x}{2}\right) = 3x\) as far as \(2x = 4\tan^{-1}\left(\frac{1}{3x}\right)\) |
| Rearrange to the given equation \(\cot\left(\frac{x}{2}\right) = 3x\) | A1 | Or continue rearrangement to \(x = \frac{1}{3}\left(x + 4\tan^{-1}\frac{1}{3x}\right)\) and state iterative formula of \(x_{n+1} = \frac{1}{3}\left(x_n + 4\tan^{-1}\frac{1}{3x_n}\right)\). AG |
| Need intermediate step between \(\frac{x}{2} = \tan^{-1}\frac{1}{3x}\) and \(\cot\left(\frac{x}{2}\right) = 3x\) | ||
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the iterative process correctly at least once | M1 | Obtain one value and substitute back to obtain a second value. Working in degrees is M0. |
| Obtain final answer \(0.79\) | A1 | Must be to 2 d.p. |
| Show sufficient iterations to at least 4 d.p. to justify \(0.79\) to 2 d.p. or show there is a sign change in the interval \((0.785, 0.795)\) | A1 | e.g. \(1, 0.7623, 0.8037, 0.7921, 0.7951, 0.7943, 0.7945\) or \(0.5, 0.9506, 0.7665, 0.8024, 0.7924, 0.7950, 0.7944, 0.7945\) or \(0.75, 0.8076, 0.7911, 0.7954, 0.7943, 0.7946, 0.7945\). Condone truncation. Allow recovery. Condone minor differences in final d.p. |
| 3 | If they do the iteration in (b) but restate the conclusion here, no marks in (b) but could score 3/3 for (c). |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Calculate the values of a relevant expression or pair of expressions at $x = 0.5$ and $x = 1$ | M1 | Need to evaluate at both points, but M1 still available if one value incorrect. Use of degrees is M0. Correct use of a smaller interval is M1. If using $g(x) - f(x)$, there needs to be a clear indication of the comparison being made e.g. by listing values in a table. Embedded values 0.5 and 1 are not sufficient. 3.92 and 1.83 alone are not sufficient |
| Complete the argument correctly with conclusion about change of sign or change of inequalities and with correct calculated values. Can all be in symbols – explanation in words is not required | A1 | e.g. $3.92 > 1.5$, $1.83 < 3$ or $2.42 > 0$, $-1.17 < 0$ |
| **Total: 2** | | |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $x = \frac{1}{3}\left(x + 4\tan^{-1}\frac{1}{3x}\right)$ | M1 | Or rearrange $\cot\left(\frac{x}{2}\right) = 3x$ as far as $2x = 4\tan^{-1}\left(\frac{1}{3x}\right)$ |
| Rearrange to the given equation $\cot\left(\frac{x}{2}\right) = 3x$ | A1 | Or continue rearrangement to $x = \frac{1}{3}\left(x + 4\tan^{-1}\frac{1}{3x}\right)$ and state iterative formula of $x_{n+1} = \frac{1}{3}\left(x_n + 4\tan^{-1}\frac{1}{3x_n}\right)$. **AG** |
| Need intermediate step between $\frac{x}{2} = \tan^{-1}\frac{1}{3x}$ and $\cot\left(\frac{x}{2}\right) = 3x$ | | |
| **Total: 2** | | |
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the iterative process correctly at least once | M1 | Obtain one value and substitute back to obtain a second value. Working in degrees is M0. |
| Obtain final answer $0.79$ | A1 | Must be to 2 d.p. |
| Show sufficient iterations to at least 4 d.p. to justify $0.79$ to 2 d.p. or show there is a sign change in the interval $(0.785, 0.795)$ | A1 | e.g. $1, 0.7623, 0.8037, 0.7921, 0.7951, 0.7943, 0.7945$ or $0.5, 0.9506, 0.7665, 0.8024, 0.7924, 0.7950, 0.7944, 0.7945$ or $0.75, 0.8076, 0.7911, 0.7954, 0.7943, 0.7946, 0.7945$. Condone truncation. Allow recovery. Condone minor differences in final d.p. |
| | 3 | If they do the iteration in (b) but restate the conclusion here, no marks in (b) but could score 3/3 for (c). |
---6 The equation $\cot \frac { 1 } { 2 } x = 3 x$ has one root in the interval $0 < x < \pi$, denoted by $\alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show by calculation that $\alpha$ lies between 0.5 and 1 .
\item Show that, if a sequence of positive values given by the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 3 } \left( x _ { n } + 4 \tan ^ { - 1 } \left( \frac { 1 } { 3 x _ { n } } \right) \right)$$
converges, then it converges to $\alpha$.
\item Use this iterative formula to calculate $\alpha$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q6 [7]}}