CAIE P3 2008 June — Question 3 6 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2008
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeDerive equation from area/geometry
DifficultyStandard +0.8 Part (i) requires setting up geometric relationships (perimeter of sector equals half rectangle perimeter) and algebraic manipulation to derive the given equation—a multi-step problem requiring careful reasoning. Part (ii) is routine application of fixed point iteration with a given formula. The geometry setup and equation derivation elevate this above average difficulty, but it remains accessible with standard A-level techniques.
Spec1.05d 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

3 \includegraphics[max width=\textwidth, alt={}, center]{20893bfc-3300-4205-9d2c-729cc3243971-2_337_828_657_657} In the diagram, \(A B C D\) is a rectangle with \(A B = 3 a\) and \(A D = a\). A circular arc, with centre \(A\) and radius \(r\), joins points \(M\) and \(N\) on \(A B\) and \(C D\) respectively. The angle \(M A N\) is \(x\) radians. The perimeter of the sector \(A M N\) is equal to half the perimeter of the rectangle.
  1. Show that \(x\) satisfies the equation $$\sin x = \frac { 1 } { 4 } ( 2 + x ) \text {. }$$
  2. This equation has only one root in the interval \(0 < x < \frac { 1 } { 2 } \pi\). Use the iterative formula $$x _ { n + 1 } = \sin ^ { - 1 } \left( \frac { 2 + x _ { n } } { 4 } \right) ,$$ with initial value \(x _ { 1 } = 0.8\), to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

AnswerMarks Guidance
(i) State or imply \(r = a \cos ec\,x\), or equivalentB1
Using perimeters, obtain a correct equation in \(x\), e.g. \(2a \cos ec\,x + ax \cos ec\,x = 4a\), or \(2r + rx = 4a\)B1
Deduce the given form of equation correctlyB1 [3 marks]
(ii) Use the iterative formula correctly at least onceM1
Obtain final answer 0.76A1
Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show that there is a sign change in the value of \(\sin x - \frac{1}{4}(2+x)\) in the interval \((0.755, 0.765)\)A1 [3 marks]
**(i)** State or imply $r = a \cos ec\,x$, or equivalent | B1 |

Using perimeters, obtain a correct equation in $x$, e.g. $2a \cos ec\,x + ax \cos ec\,x = 4a$, or $2r + rx = 4a$ | B1 |

Deduce the given form of equation correctly | B1 | [3 marks]

**(ii)** Use the iterative formula correctly at least once | M1 |

Obtain final answer 0.76 | A1 |

Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show that there is a sign change in the value of $\sin x - \frac{1}{4}(2+x)$ in the interval $(0.755, 0.765)$ | A1 | [3 marks]

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\includegraphics[max width=\textwidth, alt={}, center]{20893bfc-3300-4205-9d2c-729cc3243971-2_337_828_657_657}

In the diagram, $A B C D$ is a rectangle with $A B = 3 a$ and $A D = a$. A circular arc, with centre $A$ and radius $r$, joins points $M$ and $N$ on $A B$ and $C D$ respectively. The angle $M A N$ is $x$ radians. The perimeter of the sector $A M N$ is equal to half the perimeter of the rectangle.\\
(i) Show that $x$ satisfies the equation

$$\sin x = \frac { 1 } { 4 } ( 2 + x ) \text {. }$$

(ii) This equation has only one root in the interval $0 < x < \frac { 1 } { 2 } \pi$. Use the iterative formula

$$x _ { n + 1 } = \sin ^ { - 1 } \left( \frac { 2 + x _ { n } } { 4 } \right) ,$$

with initial value $x _ { 1 } = 0.8$, to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

\hfill \mbox{\textit{CAIE P3 2008 Q3 [6]}}