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CIE 9231 2023 June Paper 11 Q5

A Level / CIE / FP1

CIE 9231 2023 June Paper 11 Paper · Question 5

题目

Problem

The curve CC has polar equation r2=1θ2+1r^2=\dfrac{1}{\theta^2+1}, for 0≤θ≤π0\le\theta\le\pi.

(a) Sketch CC and state the polar coordinates of the point of CC furthest from the pole.

(3)

(b) Find the area of the region enclosed by CC, the initial line, and the half-line θ=π\theta=\pi.

(4)

(c) Show that, at the point of CC furthest from the initial line,

(θ+1θ)cot⁡θ−1=0\bigg(\theta+\frac1\theta\bigg)\cot\theta-1=0

and verify that this equation has a root between 1.11.1 and 1.21.2.

(5)
题目中文翻译

曲线 CC 的极坐标方程为 r2=1θ2+1r^2=\dfrac{1}{\theta^2+1}，其中 0≤θ≤π0\le\theta\le\pi。

(a) 画出 CC 的示意图，并写出 CC 上距极点最远的点的极坐标。

(3)

(b) 求由 CC、极轴及射线 θ=π\theta=\pi 围成的区域的面积。

(4)

(c) 证明，在 CC 上距极轴最远的点处，

(θ+1θ)cot⁡θ−1=0\bigg(\theta+\frac1\theta\bigg)\cot\theta-1=0

并验证此方程在 1.11.1 与 1.21.2 之间有一个根。

(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

(a)

解法一

思路

展开

取非负极径，随θ增加半径减小，但到π时并未到达极点。两端点的切线方向也要区分。

答题过程

展开

Here r=(1+θ2)−1/2r=(1+\theta^2)^{-1/2}, with

drdθ=−θ(1+θ2)3/2⩽0.\frac{\mathrm{d}r}{\mathrm{d}\theta} =-\frac\theta{(1+\theta^2)^{3/2}}\leqslant0.

Sketch an upper-half-plane arc running anticlockwise from (1,0)(1,0) at θ=0\theta=0 to (−1/1+π2,0)(-1/\sqrt{1+\pi^2},0) at θ=π\theta=\pi. The radius decreases strictly after the starting point and never reaches zero.

The tangent at (1,0)(1,0) is vertical. At the other endpoint its gradient is −(1+π2)/π-(1+\pi^2)/\pi, so it slopes downwards to the right.

The furthest point from the pole has polar coordinates (1,0)\boxed{(1,0)}.

(b)

解法一

思路

展开

两个边界射线给积分上下限0与π，代入r平方后是标准反正切积分。

答题过程

展开 A= 12∫0π11+θ2 dθ= 12[tan⁡−1θ]0π= 12tan⁡−1π.\begin{align*} A=&\,\frac12\int_0^\pi\frac1{1+\theta^2}\,\mathrm{d}\theta\\[4mm] =&\,\frac12\big[\tan^{-1}\theta\big]_0^\pi\\[4mm] =&\,\boxed{\frac12\tan^{-1}\pi}. \end{align*}

(c)

解法一

思路

展开

最大垂距点在上半平面的内部，对纵坐标求导。整理时只能除内部非零的θ与正弦，最后检查连续函数变号。

答题过程

展开

The distance from the initial line is

y=rsin⁡θ=sin⁡θ1+θ2.y=r\sin\theta=\frac{\sin\theta}{\sqrt{1+\theta^2}}.

It is zero at both endpoints and positive inside. Thus its maximum occurs at an interior point, where

0= dydθ= cos⁡θ1+θ2 −θsin⁡θ(1+θ2)3/2.\begin{align*} 0=&\,\frac{\mathrm{d}y}{\mathrm{d}\theta}\\[4mm] =&\,\frac{\cos\theta}{\sqrt{1+\theta^2}}\\[4mm] &\,-\frac{\theta\sin\theta}{(1+\theta^2)^{3/2}}. \end{align*}

Multiplying by (1+θ2)3/2(1+\theta^2)^{3/2} gives

(1+θ2)cos⁡θ−θsin⁡θ=0.(1+\theta^2)\cos\theta-\theta\sin\theta=0.

Since 0<θ<π0<\theta<\pi, divide by θsin⁡θ>0\theta\sin\theta>0 to obtain

(θ+1θ)cot⁡θ−1=0.\boxed{\bigg(\theta+\frac1\theta\bigg)\cot\theta-1=0}.

Let F(θ)F(\theta) be the left-hand side. In radians,

F(1.1)= 0.0225632…>0,F(1.2)= −0.2094815…<0.\begin{align*} F(1.1)=&\,0.0225632\ldots>0,\\[4mm] F(1.2)=&\,-0.2094815\ldots<0. \end{align*}

Since FF is continuous on [1.1,1.2][1.1,1.2], it has a root in (1.1,1.2)\boxed{(1.1,1.2)}.