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CIE 9231 2024 June Paper 11 Q7

A Level / CIE / FP1

CIE 9231 2024 June Paper 11 Paper · Question 7

题目

Problem

The curve CC has polar equation r2=(π−θ)tan⁡−1(π−θ)r^2=(\pi-\theta)\tan^{-1}(\pi-\theta), for 0⩽θ⩽π0\leqslant\theta\leqslant\pi.

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

(3)

(b) Using the substitution u=π−θu=\pi-\theta, or otherwise, find the area of the region enclosed by CC and the initial line.

(7)

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

 2(π−θ)tan⁡−1(π−θ)cot⁡θ −π−θ1+(π−θ)2−tan⁡−1(π−θ)=0\begin{align*} &\,2(\pi-\theta)\tan^{-1}(\pi-\theta)\cot\theta\\ &\,\hspace{2pt}-\frac{\pi-\theta}{1+(\pi-\theta)^2}-\tan^{-1}(\pi-\theta)=0 \end{align*}

and verify that this equation has a root for θ\theta between 1.21.2 and 1.31.3.

(5)
题目中文翻译

曲线 CC 的极坐标方程为 r2=(π−θ)tan⁡−1(π−θ)r^2=(\pi-\theta)\tan^{-1}(\pi-\theta)，其中 0⩽θ⩽π0\leqslant\theta\leqslant\pi。

(a) 画出 CC 的草图，并写出 CC 上距离极点最远的点的极坐标。

(3)

(b) 使用代换 u=π−θu=\pi-\theta 或其他方法，求 CC 与极轴所围区域的面积。

(7)

(c) 证明：在 CC 上距离极轴最远的点处，

 2(π−θ)tan⁡−1(π−θ)cot⁡θ −π−θ1+(π−θ)2−tan⁡−1(π−θ)=0\begin{align*} &\,2(\pi-\theta)\tan^{-1}(\pi-\theta)\cot\theta\\ &\,\hspace{2pt}-\frac{\pi-\theta}{1+(\pi-\theta)^2}-\tan^{-1}(\pi-\theta)=0 \end{align*}

并验证这个方程在 θ\theta 为 1.21.2 与 1.31.3 之间时有一个根。

(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

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令u=π−θ，u乘反正切u随u增加，所以极径随θ减小。草图从正极轴上最大极径点逆时针绕到极点，后半段必须在第二象限。

答题过程

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Put u=π−θu=\pi-\theta. For u⩾0u\geqslant0,

ddu(utan⁡−1u)=tan⁡−1u+u1+u2>0(u>0).\frac{\mathrm{d}}{\mathrm{d}u}\big(u\tan^{-1}u\big) =\tan^{-1}u+\frac{u}{1+u^2}>0\quad(u>0).

Thus rr decreases as θ\theta increases from 00 to π\pi. Sketch an upper-half-plane curve starting on the positive initial line at r=πtan⁡−1πr=\sqrt{\pi\tan^{-1}\pi}, passing across the positive yy-axis and into the second quadrant, then reaching the pole at θ=π\theta=\pi with tangent along the initial line.

The furthest point from the pole has polar coordinates

(πtan⁡−1π,0)(≈(1.99,0)).\boxed{\big(\sqrt{\pi\tan^{-1}\pi},0\big)} \quad\big(\approx(1.99,0)\big).

(b)

解法一

思路

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极坐标面积先换元为u乘反正切u的积分，再用分部积分。剩余的有理式先拆成1减去标准反正切导数。

答题过程

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The required area is

A=12∫0π(π−θ)tan⁡−1(π−θ) dθ.A=\frac12\int_0^\pi(\pi-\theta)\tan^{-1}(\pi-\theta)\,\mathrm{d}\theta.

Using u=π−θu=\pi-\theta, dθ=−du\mathrm{d}\theta=-\mathrm{d}u, the limits change from (0,π)(0,\pi) to (π,0)(\pi,0), giving

A=12∫0πutan⁡−1u du.A=\frac12\int_0^\pi u\tan^{-1}u\,\mathrm{d}u.

Integrating by parts with v=u2/2v=u^2/2 and d(tan⁡−1u)/du=1/(1+u2)\mathrm{d}(\tan^{-1}u)/\mathrm{d}u=1/(1+u^2),

A= 14[u2tan⁡−1u]0π −14∫0πu21+u2 du.\begin{align*} A=&\,\frac14\big[u^2\tan^{-1}u\big]_0^\pi\\[4mm] &\,-\frac14\int_0^\pi\frac{u^2}{1+u^2}\,\mathrm{d}u. \end{align*}

Now

u21+u2=1−11+u2,\frac{u^2}{1+u^2}=1-\frac1{1+u^2},

so

A= 14[u2tan⁡−1u]0π −14[u−tan⁡−1u]0π= 14((π2+1)tan⁡−1π−π).\begin{align*} A=&\,\frac14\big[u^2\tan^{-1}u\big]_0^\pi\\[4mm] &\,-\frac14\big[u-\tan^{-1}u\big]_0^\pi\\[4mm] =&\,\boxed{\frac14\big((\pi^2+1)\tan^{-1}\pi-\pi\big)}. \end{align*}

(c)

解法一

思路

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上半平面到极轴的距离就是纵坐标。对r乘正弦求导，再由r平方的导数代入，避免直接展开根式导数；内部极值可除以正弦。

答题过程

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Let u=π−θu=\pi-\theta, so r2=utan⁡−1ur^2=u\tan^{-1}u. Differentiation gives

2rdrdθ=−tan⁡−1u−u1+u2.2r\frac{\mathrm{d}r}{\mathrm{d}\theta} =-\tan^{-1}u-\frac{u}{1+u^2}.

The perpendicular distance from the initial line is y=rsin⁡θy=r\sin\theta. At both endpoints it is zero, and it is positive for 0<θ<π0<\theta<\pi. Hence the furthest point lies in the interior, where r>0r>0 and sin⁡θ>0\sin\theta>0.

Setting its derivative to zero,

drdθsin⁡θ+rcos⁡θ=0.\frac{\mathrm{d}r}{\mathrm{d}\theta}\sin\theta+r\cos\theta=0.

Multiply by 2r/sin⁡θ2r/\sin\theta and substitute the expressions above:

2utan⁡−1ucot⁡θ−u1+u2−tan⁡−1u=0.2u\tan^{-1}u\cot\theta-\frac{u}{1+u^2}-\tan^{-1}u=0.

Thus, with u=π−θu=\pi-\theta,

 2(π−θ)tan⁡−1(π−θ)cot⁡θ −π−θ1+(π−θ)2−tan⁡−1(π−θ)=0,\begin{align*} &\,2(\pi-\theta)\tan^{-1}(\pi-\theta)\cot\theta\\[4mm] &\,\hspace{2pt}-\frac{\pi-\theta}{1+(\pi-\theta)^2} -\tan^{-1}(\pi-\theta)=0, \end{align*}

as required.

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

F(1.2)= 0.1511598…>0,F(1.3)= −0.3951967…<0.\begin{align*} F(1.2)=&\,0.1511598\ldots>0,\\[4mm] F(1.3)=&\,-0.3951967\ldots<0. \end{align*}

The function is continuous on [1.2,1.3][1.2,1.3], so there is a root in (1.2,1.3)\boxed{(1.2,1.3)}.