CAIE P3 2023 March — Question 7 9 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2023
SessionMarch
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeDerive equation from area/geometry
DifficultyChallenging +1.2 This question requires deriving a relationship from geometric areas (sector and triangle), then applying standard iterative methods. Part (a) involves setting up equations for areas and algebraic manipulation, which is moderately challenging but follows a clear path. Parts (b) and (c) are routine A-level iteration techniques. The geometric setup and algebraic derivation elevate this slightly above average difficulty, but it remains a standard P3 question type with no novel insights required.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

7 \includegraphics[max width=\textwidth, alt={}, center]{8c26235b-c78c-40d8-9e8e-213dc1311186-10_627_611_255_767} The diagram shows a circle with centre \(O\) and radius \(r\). The angle of the minor sector \(A O B\) of the circle is \(x\) radians. The area of the major sector of the circle is 3 times the area of the shaded region.
  1. Show that \(x = \frac { 3 } { 4 } \sin x + \frac { 1 } { 2 } \pi\).
  2. Show by calculation that the root of the equation in (a) lies between 2 and 2.5.
  3. Use an iterative formula based on the equation in (a) to calculate this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

Question 7(a):
AnswerMarks Guidance
AnswerMark Guidance
Area of major sector \(= \frac{1}{2}r^2(2\pi - x)\)B1 OE
Area of shaded segment \(= \frac{1}{2}r^2x - \frac{1}{2}r^2\sin x\)B1 OE; \(r^2\sin(x/2)\cos(x/2)\) B0 until changed to \(\frac{1}{2}r^2\sin x\)
State \(\frac{1}{2}r^2(2\pi-x) = 3\left(\frac{1}{2}r^2x - \frac{1}{2}r^2\sin x\right)\)M1 OE; Area of major sector = 3 times (area of minor sector – area of triangle). Allow \(r^2\sin(x/2)\cos(x/2)\)
Obtain \(x = \frac{3}{4}\sin x + \frac{1}{2}\pi\) after full and correct workingA1 AG Allow rectified slip if before penultimate line
Total: 4
Question 7(b):
AnswerMarks Guidance
AnswerMark Guidance
Calculate values of a relevant expression or pair of expressions at \(x=2\) and \(x=2.5\)M1 \(x=2\): \((3/4)\sin x + (1/2)\pi \approx 2.2\), so \(2 < 2.2\); \(x-(3/4)\sin x-(1/2)\pi \approx -0.2(5277)<0\). \(x=2.5\): \(\approx 2.0(197)\), \(2.5>2.0\); \(+0.4(803)>0\). Attempt both values and one correct for M1
Complete the argument correctly with correct calculated valuesA1 Degrees award 0/2
Total: 2
Question 7(c):
AnswerMarks Guidance
AnswerMark Guidance
Use the iterative formula correctly at least twiceM1
Obtain final answer 2.18A1
Show sufficient iterations to 4 d.p. to justify 2.18 to 2 d.p. or show sign change in interval \((2.175, 2.185)\)A1 Starting from 2: 2.2528, 2.1530, 2.1972, 2.1784, 2.1866, 2.1830, 2.1846... Degrees award 0/3
Total: 3
## Question 7(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| Area of major sector $= \frac{1}{2}r^2(2\pi - x)$ | B1 | OE |
| Area of shaded segment $= \frac{1}{2}r^2x - \frac{1}{2}r^2\sin x$ | B1 | OE; $r^2\sin(x/2)\cos(x/2)$ B0 until changed to $\frac{1}{2}r^2\sin x$ |
| State $\frac{1}{2}r^2(2\pi-x) = 3\left(\frac{1}{2}r^2x - \frac{1}{2}r^2\sin x\right)$ | M1 | OE; Area of major sector = 3 times (area of minor sector – area of triangle). Allow $r^2\sin(x/2)\cos(x/2)$ |
| Obtain $x = \frac{3}{4}\sin x + \frac{1}{2}\pi$ after full and correct working | A1 | AG Allow rectified slip if before penultimate line |
| **Total: 4** | | |

## Question 7(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| Calculate values of a relevant expression or pair of expressions at $x=2$ and $x=2.5$ | M1 | $x=2$: $(3/4)\sin x + (1/2)\pi \approx 2.2$, so $2 < 2.2$; $x-(3/4)\sin x-(1/2)\pi \approx -0.2(5277)<0$. $x=2.5$: $\approx 2.0(197)$, $2.5>2.0$; $+0.4(803)>0$. Attempt both values and one correct for M1 |
| Complete the argument correctly with correct calculated values | A1 | Degrees award 0/2 |
| **Total: 2** | | |

## Question 7(c):

| Answer | Mark | Guidance |
|--------|------|----------|
| Use the iterative formula correctly at least twice | M1 | |
| Obtain final answer 2.18 | A1 | |
| Show sufficient iterations to 4 d.p. to justify 2.18 to 2 d.p. or show sign change in interval $(2.175, 2.185)$ | A1 | Starting from 2: 2.2528, 2.1530, 2.1972, 2.1784, 2.1866, 2.1830, 2.1846... Degrees award 0/3 |
| **Total: 3** | | |
7\\
\includegraphics[max width=\textwidth, alt={}, center]{8c26235b-c78c-40d8-9e8e-213dc1311186-10_627_611_255_767}

The diagram shows a circle with centre $O$ and radius $r$. The angle of the minor sector $A O B$ of the circle is $x$ radians. The area of the major sector of the circle is 3 times the area of the shaded region.
\begin{enumerate}[label=(\alph*)]
\item Show that $x = \frac { 3 } { 4 } \sin x + \frac { 1 } { 2 } \pi$.
\item Show by calculation that the root of the equation in (a) lies between 2 and 2.5.
\item Use an iterative formula based on the equation in (a) to calculate this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2023 Q7 [9]}}