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CIE 9709 2022 November Paper 13 Q10

A Level / CIE / P1

CIE 9709 2022 Nov Paper 13 Paper · Question 10

题目

Problem

The diagram shows the circle x2+y2=2x^2+y^2=2 and the straight line y=2x−1y=2x-1 intersecting at the points AA and BB. The point DD on the xx-axis is such that ADAD is perpendicular to the xx-axis.

(a) Find the coordinates of AA.

(4)

(b) Find the volume of revolution when the shaded region is rotated through 360∘360^\circ about the xx-axis. Give your answer in the form πa(bc−d)\dfrac{\pi}{a}\big(b\sqrt{c}-d\big), where aa, bb, cc and dd are integers.

(4)

(c) Find an exact expression for the perimeter of the shaded region.

(2)
题目中文翻译

图中圆 x2+y2=2x^2+y^2=2 与直线 y=2x−1y=2x-1 相交于点 AA 和 BB。点 DD 位于 xx 轴上,且 ADAD 垂直于 xx 轴。

(a) 求 AA 的坐标。

(4)

(b) 将阴影区域绕 xx 轴旋转 360∘360^\circ,求所得旋转体的体积。答案写成 πa(bc−d)\dfrac{\pi}{a}\big(b\sqrt{c}-d\big) 的形式,其中 aa、bb、cc 和 dd 为整数。

(4)

(c) 求阴影区域周长的精确表达式。

(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

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将直线代入圆得到两个交点。按原图选择第一象限的交点 AA。

答题过程

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Substituting y=2x−1y=2x-1 gives

x2+(2x−1)2=2⟺(5x+1)(x−1)=0.x^2+(2x-1)^2=2\quad\Longleftrightarrow\quad(5x+1)(x-1)=0.

The first-quadrant point uses x=1x=1, so y=1y=1 and A=(1,1)A=(1,1).

(b)

解法一

思路

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从图中读出阴影的横向范围。绕横轴旋转时,每个圆截面的半径是上半圆的纵坐标。

答题过程

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The region extends from x=1x=1 to x=2x=\sqrt2. The squared radius is y2=2−x2y^2=2-x^2, so the volume is

 π∫12(2−x2) dx= π[2x−x33]12= π3(42−5).\begin{align*} &\,\pi\int_1^{\sqrt2}(2-x^2)\,\mathrm{d}x\\ =&\,\pi\Big[2x-\frac{x^3}{3}\Big]_1^{\sqrt2}\\ =&\,\frac\pi3(4\sqrt2-5). \end{align*}

(c)

解法一

思路

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周长包括竖直线段、横轴上的线段和圆弧。点 AA 的两个坐标相同,所以圆弧所对的圆心角是 π/4\pi/4。

答题过程

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Let E=(2,0)E=(\sqrt2,0) be the right intercept of the circle. The straight lengths are AD=1AD=1 and DE=2−1DE=\sqrt2-1. The arc AEAE has radius 2\sqrt2 and angle π/4\pi/4, so the perimeter is

1+(2−1)+π24=2(1+π4).1+(\sqrt2-1)+\frac{\pi\sqrt2}{4}=\sqrt2\Big(1+\frac\pi4\Big).