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CIE 9709 2025 November Paper 33 Q4

A Level / CIE / P3

CIE 9709 2025 Nov Paper 33 Paper · Question 4

题目

Problem

On an Argand diagram shade the region whose points represent complex numbers zz which satisfy both the inequalities ∣z+2i∣≤3|z+2\mathrm{i}|\le3 and ∣z+2i∣≤∣z−2+4i∣|z+2\mathrm{i}|\le|z-2+4\mathrm{i}|.

(5)
题目中文翻译

在 Argand 图上用阴影标出一个区域,使其中的点表示同时满足不等式 ∣z+2i∣≤3|z+2\mathrm{i}|\le3 和 ∣z+2i∣≤∣z−2+4i∣|z+2\mathrm{i}|\le|z-2+4\mathrm{i}| 的复数 zz。

(5)
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

解法一

思路

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第一个条件选圆盘,第二个条件比较到两个定点的距离,选垂直平分线靠近圆心的一侧。两个边界都包括在内。

答题过程

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Write z=x+yiz=x+y\mathrm{i}. The first inequality is

x2+(y+2)2⩽9,x^2+(y+2)^2\leqslant9,

so draw the closed disk with centre (0,−2)(0,-2) and radius 33.

The second inequality means that the point is at least as close to A=(0,−2)A=(0,-2) as to B=(2,−4)B=(2,-4). Their midpoint is (1,−3)(1,-3), and the perpendicular bisector of ABAB is y=x−4y=x-4.

Algebraically,

x2+(y+2)2⩽ (x−2)2+(y+4)2,0⩽ −4x+4y+16,y⩾ x−4.\begin{align*} x^2+(y+2)^2\leqslant&\,(x-2)^2+(y+4)^2,\\[4mm] 0\leqslant&\,-4x+4y+16,\\[4mm] y\geqslant&\,x-4. \end{align*}

Shade the part of the disk on or above y=x−4y=x-4, including both boundaries. Mark the centre, radius, the points A,BA,B, and the midpoint or the corresponding scale clearly.