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CIE 9709 2022 November Paper 32 Q9

A Level / CIE / P3

CIE 9709 2022 Nov Paper 32 Paper · Question 9

题目

Problem

The diagram shows a semicircle with diameter ABAB, centre OO and radius rr. The shaded region is the minor segment on the chord ACAC and its area is one third of the area of the semicircle. The angle CABCAB is θ\theta radians.

(a) Show that θ=13(π−1.5sin⁡2θ)\theta=\frac13(\pi-1.5\sin2\theta).

(4)

(b) Verify by calculation that 0.5<θ<0.70.5<\theta<0.7.

(2)

(c) Use an iterative formula based on the equation in part (a) to determine θ\theta correct to 33 decimal places. Give the result of each iteration to 55 decimal places.

(3)
题目中文翻译

图中显示了一个直径为 ABAB、圆心为 OO、半径为 rr 的半圆。阴影区域是弦 ACAC 所对应的小弓形,其面积为半圆面积的三分之一。角 CABCAB 为 θ\theta 弧度。

(a) 证明 θ=13(π−1.5sin⁡2θ)\theta=\frac13(\pi-1.5\sin2\theta)。

(4)

(b) 通过计算验证 0.5<θ<0.70.5<\theta<0.7。

(2)

(c) 使用基于 (a) 问方程的迭代公式求 θ\theta,结果保留 33 位小数。每次迭代的结果保留 55 位小数。

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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在等腰三角形 AOCAOC 中,底角均为 θ\theta,所以圆心角为 π−2θ\pi-2\theta。阴影是相应扇形减三角形,并等于半圆面积的三分之一。

答题过程

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Since OA=OC=rOA=OC=r and ∠OAC=θ\angle OAC=\theta,

∠AOC=π−2θ.\angle AOC=\pi-2\theta.

The shaded segment has area

12r2(π−2θ)−12r2sin⁡(π−2θ).\frac12r^2(\pi-2\theta) -\frac12r^2\sin(\pi-2\theta).

It equals one third of the semicircle area, so

12r2(π−2θ)−12r2sin⁡(π−2θ)=16πr2.\frac12r^2(\pi-2\theta) -\frac12r^2\sin(\pi-2\theta) =\frac16\pi r^2.

Cancelling r2/2r^2/2 and using sin⁡(π−2θ)=sin⁡2θ\sin(\pi-2\theta)=\sin2\theta,

π−2θ−sin⁡2θ=π3.\pi-2\theta-\sin2\theta=\frac\pi3.

Thus

3θ=π−32sin⁡2θ,3\theta=\pi-\frac32\sin2\theta,

giving

θ=13(π−1.5sin⁡2θ).\boxed{\theta=\frac13(\pi-1.5\sin2\theta)}.

(b)

解法一

思路

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将(a)右侧减去左侧,计算端点的符号。两端都在几何上允许的角度范围内。

答题过程

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Let f(θ)=13(π−1.5sin⁡2θ)−θf(\theta)=\tfrac13(\pi-1.5\sin2\theta)-\theta. Working in radians,

f(0.5)= 0.126462…>0,f(0.7)= −0.145527…<0.\begin{align*} f(0.5)=&\,0.126462\ldots>0,\\[4mm] f(0.7)=&\,-0.145527\ldots<0. \end{align*}

Since ff is continuous, there is a root between 0.50.5 and 0.70.7. Also

f′(θ)=−cos⁡2θ−1=−2cos⁡2θ<0f'(\theta)=-\cos2\theta-1=-2\cos^2\theta<0

for 0<θ<π/20<\theta<\pi/2, so the geometric root is unique. Hence 0.5<θ<0.7\boxed{0.5<\theta<0.7}.

(c)

解法一

思路

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按(a)迭代并列到五位小数,不能只列四位;继续到相邻数值支持同一个三位小数答案。

答题过程

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Use

θn+1= 13(π−1.5sin⁡2θn),θ0= 0.6,\begin{align*} \theta_{n+1}=&\,\frac13(\pi-1.5\sin2\theta_n),\\[4mm] \theta_0=&\,0.6, \end{align*}

in radians.

nnθn\theta_n to 5 decimal places
00.60000
10.58118
20.58833
30.58553
40.58661
50.58619
60.58636
70.58629
80.58632
90.58631
100.58631

Therefore θ=0.586\boxed{\theta=0.586} to 3 decimal places.