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CIE 9709 2023 March Paper 32 Q7

A Level / CIE / P3

CIE 9709 2023 March Paper 32 Paper · Question 7

题目

Problem

The diagram shows a circle with centre OO and radius rr. The angle of the minor sector AOBAOB of the circle is xx radians. The area of the major sector of the circle is 33 times the area of the shaded region.

(a) Show that x=34sin⁡x+12πx=\frac34\sin x+\frac12\pi.

(4)

(b) Show by calculation that the root of the equation in (a) lies between 22 and 2.52.5.

(2)

(c) Use an iterative formula based on the equation in (a) to calculate this root correct to 22 decimal places. Give the result of each iteration to 44 decimal places.

(3)
题目中文翻译

图中显示了一个圆心为 OO、半径为 rr 的圆。圆的小扇形 AOBAOB 的圆心角为 xx 弧度。圆的大扇形的面积是阴影区域面积的 33 倍。

(a) 证明 x=34sin⁡x+12πx=\frac34\sin x+\frac12\pi。

(4)

(b) 通过计算证明 (a) 问方程的根位于 22 与 2.52.5 之间。

(2)

(c) 使用基于 (a) 问方程的迭代公式求此根,结果保留 22 位小数。每次迭代的结果保留 44 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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阴影是小扇形减去三角形,不能把阴影当作整个小扇形。写出大扇形与阴影的面积,利用三倍关系消去 r2r^2。

答题过程

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The major sector has angle 2π−x2\pi-x, so its area is

12r2(2π−x).\frac12r^2(2\pi-x).

The shaded segment is the minor sector minus triangle AOBAOB, with area

12r2x−12r2sin⁡x.\frac12r^2x-\frac12r^2\sin x.

Using the given relationship,

12r2(2π−x)=3(12r2x−12r2sin⁡x).\frac12r^2(2\pi-x) =3\bigg(\frac12r^2x-\frac12r^2\sin x\bigg).

Cancelling r2/2>0r^2/2>0,

2π−x= 3x−3sin⁡x,4x= 2π+3sin⁡x.\begin{align*} 2\pi-x=&\,3x-3\sin x,\\[4mm] 4x=&\,2\pi+3\sin x. \end{align*}

Hence

x=34sin⁡x+π2.\boxed{x=\frac34\sin x+\frac\pi2}.

(b)

解法一

思路

展开

把(a)移项成连续函数为零,在两个端点计算函数值并比较符号。

答题过程

展开

Let f(x)=x−34sin⁡x−π2f(x)=x-\tfrac34\sin x-\tfrac\pi2. In radians,

f(2)= −0.252769…<0,f(2.5)= 0.480349…>0.\begin{align*} f(2)=&\,-0.252769\ldots<0,\\[4mm] f(2.5)=&\,0.480349\ldots>0. \end{align*}

Since ff is continuous, it has a zero between 22 and 2.52.5. Also f′(x)=1−34cos⁡x≥1/4>0f'(x)=1-\tfrac34\cos x\ge1/4>0, so the root is unique.

(c)

解法一

思路

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使用(a)的右侧作为迭代函数。计算时保留精度,显示四位小数,直到结果稳定到两位小数。

答题过程

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Use

xn+1= 34sin⁡xn+π2,x0= 2,\begin{align*} x_{n+1}=&\,\frac34\sin x_n+\frac\pi2,\\[4mm] x_0=&\,2, \end{align*}

with angles in radians.

nnxnx_n to 4 decimal places
02.0000
12.2528
22.1530
32.1972
42.1784
52.1866
62.1830
72.1846
82.1839
92.1842
102.1841
112.1841

Therefore x=2.18\boxed{x=2.18} to 2 decimal places.