| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive equation from area/geometry |
| Difficulty | Challenging +1.2 This is a multi-part question involving geometric reasoning with circles and sectors, followed by standard iterative methods. Part (i) requires basic trigonometry in a triangle, part (ii) involves equating areas (sector formulas) and algebraic manipulation, and parts (iii)-(iv) are routine numerical work with iteration. While it requires careful setup and multiple steps, the techniques are all standard A-level material with no novel insights required. The geometric visualization and area equation setup elevate it slightly above average difficulty. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct use of trigonometry to obtain \(AB = 2r\cos x\) | B1 | AG |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use correct method for finding the area of the sector and the semicircle and form an equation in \(x\): \(\frac{1}{2} \times \frac{1}{2}\pi r^2 = \frac{1}{2}(2r\cos x)^2 \cdot 2x\) | M1 | |
| Obtain \(x = \cos^{-1}\sqrt{\frac{\pi}{16x}}\) correctly | A1 | AG. Via correct simplification e.g. from \(\cos^2 x = \frac{\pi}{16x}\) |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Calculate values of a relevant expression or pair of expressions at \(x=1\) and \(x=1.5\). Must be working in radians | M1 | e.g. \(x=1: 1\to 1.11\); \(x=1.5: 1.5\to 1.20\). Accept \(f(1)=1.11\), \(f(1.5)=1.20\). \(f(x)=x-\cos^{-1}\sqrt{\frac{\pi}{16x}}\): \(f(1)=-0.111\), \(f(1.5)=0.3..\) \(f(x)=\cos x - \sqrt{\frac{\pi}{16x}}\): \(f(1)=0.097..\), \(f(1.5)=-0.291\). For \(16x\cos^2 x - \pi\): \(f(1)=1.529..\), \(f(1.5)=-3.02..\) Must find values. M1 if at least one value correct |
| Correct values and complete the argument correctly | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(x_{n+1} = \cos^{-1}\sqrt{\dfrac{\pi}{16x_n}}\) correctly at least twice; must be working in radians | M1 | 1, 1.11173, 1.13707, 1.14225, 1.14329, 1.14349, 1.14354, 1.14354; 1.25, 1.16328, 1.14742, 1.14432, 1.14370; 1.5, 1.20060, 1.15447, 1.14570, 1.14397, 1.14363 |
| Obtain final answer 1.144 | A1 | |
| Show sufficient iterations to at least 5 d.p. to justify 1.144 to 3 d.p., or show sign change in interval (1.1435, 1.1445) | A1 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct use of trigonometry to obtain $AB = 2r\cos x$ | B1 | AG |
| **Total** | **1** | |
---
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct method for finding the area of the sector and the semicircle and form an equation in $x$: $\frac{1}{2} \times \frac{1}{2}\pi r^2 = \frac{1}{2}(2r\cos x)^2 \cdot 2x$ | M1 | |
| Obtain $x = \cos^{-1}\sqrt{\frac{\pi}{16x}}$ correctly | A1 | AG. Via correct simplification e.g. from $\cos^2 x = \frac{\pi}{16x}$ |
| **Total** | **2** | |
---
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Calculate values of a relevant expression or pair of expressions at $x=1$ and $x=1.5$. Must be working in radians | M1 | e.g. $x=1: 1\to 1.11$; $x=1.5: 1.5\to 1.20$. Accept $f(1)=1.11$, $f(1.5)=1.20$. $f(x)=x-\cos^{-1}\sqrt{\frac{\pi}{16x}}$: $f(1)=-0.111$, $f(1.5)=0.3..$ $f(x)=\cos x - \sqrt{\frac{\pi}{16x}}$: $f(1)=0.097..$, $f(1.5)=-0.291$. For $16x\cos^2 x - \pi$: $f(1)=1.529..$, $f(1.5)=-3.02..$ Must find values. M1 if at least one value correct |
| Correct values and complete the argument correctly | A1 | |
| **Total** | **2** | |
## Question 6(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $x_{n+1} = \cos^{-1}\sqrt{\dfrac{\pi}{16x_n}}$ correctly at least twice; must be working in radians | M1 | 1, 1.11173, 1.13707, 1.14225, 1.14329, 1.14349, 1.14354, 1.14354; 1.25, 1.16328, 1.14742, 1.14432, 1.14370; 1.5, 1.20060, 1.15447, 1.14570, 1.14397, 1.14363 |
| Obtain final answer 1.144 | A1 | |
| Show sufficient iterations to at least 5 d.p. to justify 1.144 to 3 d.p., or show sign change in interval (1.1435, 1.1445) | A1 | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{772393d7-6e81-4b99-913a-63c9f87d1af2-08_492_812_260_664}
In the diagram, $A$ is the mid-point of the semicircle with centre $O$ and radius $r$. A circular arc with centre $A$ meets the semicircle at $B$ and $C$. The angle $O A B$ is equal to $x$ radians. The area of the shaded region bounded by $A B , A C$ and the arc with centre $A$ is equal to half the area of the semicircle.\\
(i) Use triangle $O A B$ to show that $A B = 2 r \cos x$.\\
(ii) Hence show that $x = \cos ^ { - 1 } \sqrt { } \left( \frac { \pi } { 16 x } \right)$.\\
(iii) Verify by calculation that $x$ lies between 1 and 1.5.\\
(iv) Use an iterative formula based on the equation in part (ii) to determine $x$ correct to 3 decimal places. Give the result of each iteration to 5 decimal places.\\
\hfill \mbox{\textit{CAIE P3 2019 Q6 [8]}}