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CIE 9709 2024 March Paper 32 Q5

A Level / CIE / P3

CIE 9709 2024 March Paper 32 Paper · Question 5

题目

Problem

(a) On a sketch of an Argand diagram, shade the region whose points represent complex numbers zz satisfying the inequalities ∣z−4−2i∣≤3|z-4-2\mathrm{i}|\le3 and ∣z∣≥∣10−z∣|z|\ge|10-z|.

(4)

(b) Find the greatest value of arg⁡z\arg z for points in this region.

(2)
题目中文翻译

(a) 在 Argand 图的草图中,用阴影标出满足不等式 ∣z−4−2i∣≤3|z-4-2\mathrm{i}|\le3 和 ∣z∣≥∣10−z∣|z|\ge|10-z| 的复数 zz 所对应的点组成的区域。

(4)

(b) 求此区域内各点所对应的 arg⁡z\arg z 的最大值。

(2)
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

展开

第一个条件是闭圆盘。第二个条件表示到原点的距离不小于到 (10,0)(10,0) 的距离,可平方化成半平面。

答题过程

展开

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

(x−4)2+(y−2)2≤9,(x-4)^2+(y-2)^2\le9,

the closed disc with centre (4,2)(4,2) and radius 33.

For the second inequality, both moduli are non-negative, so it is equivalent to

x2+y2≥ (10−x)2+y2,20x≥ 100,x≥ 5.\begin{align*} x^2+y^2\ge&\,(10-x)^2+y^2,\\[4mm] 20x\ge&\,100,\\[4mm] x\ge&\,5. \end{align*}

Draw the circle and the vertical line x=5x=5. Shade the part inside the circle and on or to the right of x=5x=5, including the boundaries.

(b)

解法一

思路

展开

涂色部分在 x≥5x\ge5,最大辐角出现在直线 x=5x=5 与圆的上方交点。须使用涂色区域,不能直接取整个圆的切线方向。

答题过程

展开

On x=5x=5, the circle gives

(5−4)2+(y−2)2=9,y=2±22.(5-4)^2+(y-2)^2=9, \qquad y=2\pm2\sqrt2.

The upper intersection is T=(5,2+22)T=(5,2+2\sqrt2).

For 5≤x≤75\le x\le7, the upper boundary is

y=2+9−(x−4)2.y=2+\sqrt{9-(x-4)^2}.

As xx increases, this positive yy decreases while xx increases, so y/xy/x decreases. Points below this boundary have smaller arguments. Thus the greatest argument occurs at TT:

max⁡arg⁡z=tan⁡−1(2+225)≈0.768 radians.\max\arg z=\tan^{-1}\bigg(\frac{2+2\sqrt2}{5}\bigg) \approx\boxed{0.768\text{ radians}}.