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CIE 9231 2023 November Paper 12 Q6

A Level / CIE / FP1

CIE 9231 2023 Nov Paper 12 Paper · Question 6

题目

Problem

The curve CC has polar equation r=e−θ−e−12πr=\mathrm{e}^{-\theta}-\mathrm{e}^{-\frac12\pi}, where 0≤θ≤12π0\le\theta\le\frac12\pi.

(a) Sketch CC and state, in exact form, the greatest distance of a point on CC from the pole.

(3)

(b) Find the exact value of the area of the region bounded by CC and the initial line.

(5)

(c) Show that, at the point on CC farthest from the initial line,

1−eθ−12π−tan⁡θ=01-\mathrm{e}^{\theta-\frac12\pi}-\tan\theta=0

and verify that this equation has a root between 0.560.56 and 0.570.57.

(5)
题目中文翻译

曲线 CC 的极坐标方程为 r=e−θ−e−12πr=\mathrm{e}^{-\theta}-\mathrm{e}^{-\frac12\pi}，其中 0≤θ≤12π0\le\theta\le\frac12\pi。

(a) 画出 CC 的示意图，并写出 CC 上的点到极点的最大距离的精确值。

(3)

(b) 求由 CC 与极轴围成的区域面积的精确值。

(5)

(c) 证明，在 CC 上距极轴最远的点处，

1−eθ−12π−tan⁡θ=01-\mathrm{e}^{\theta-\frac12\pi}-\tan\theta=0

并验证此方程在 0.560.56 与 0.570.57 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

极径随角度严格减小，曲线在第一象限由正极轴走到极点。极点处为竖直切线，起点处斜率为负。

答题过程

展开

Since

drdθ=−e−θ<0,\frac{\mathrm{d}r}{\mathrm{d}\theta}=-\mathrm e^{-\theta}<0,

the radius decreases from 1−e−π/21-\mathrm e^{-\pi/2} at θ=0\theta=0 to 00 at θ=π/2\theta=\pi/2.

Sketch a first-quadrant arc from (1−e−π/2,0)(1-\mathrm e^{-\pi/2},0) to the pole. At the starting point its Cartesian gradient is −(1−e−π/2)-(1-\mathrm e^{-\pi/2}); at the pole its tangent is vertical.

The greatest distance from the pole is 1−e−π/2\boxed{1-\mathrm e^{-\pi/2}}.

(b)

解法一

思路

展开

用极坐标面积，先展开平方再逐项积分，保留指数精确形式。

答题过程

展开 A= 12∫0π/2(e−θ−e−π/2)2 dθ= 12∫0π/2e−2θ dθ −∫0π/2e−θ−π/2 dθ +12∫0π/2e−π dθ= 12[−12e−2θ+2e−θ−π/2+θe−π]0π/2= 12[32e−π+π2e−π+12−2e−π/2]= 14+3+π4e−π−e−π/2.\begin{align*} A=&\,\frac12\int_0^{\pi/2} \big(\mathrm e^{-\theta}-\mathrm e^{-\pi/2}\big)^2\,\mathrm{d}\theta\\[4mm] =&\,\frac12\int_0^{\pi/2}\mathrm e^{-2\theta}\,\mathrm{d}\theta\\[4mm] &\,-\int_0^{\pi/2}\mathrm e^{-\theta-\pi/2}\,\mathrm{d}\theta\\[4mm] &\,+\frac12\int_0^{\pi/2}\mathrm e^{-\pi}\,\mathrm{d}\theta\\[4mm] =&\,\frac12\big[-\tfrac12\mathrm e^{-2\theta} +2\mathrm e^{-\theta-\pi/2}+\theta\mathrm e^{-\pi}\big]_0^{\pi/2}\\[4mm] =&\,\frac12\big[\tfrac32\mathrm e^{-\pi} +\tfrac\pi2\mathrm e^{-\pi}+\tfrac12-2\mathrm e^{-\pi/2}\big]\\[4mm] =&\,\boxed{\frac14+\frac{3+\pi}{4}\mathrm e^{-\pi}-\mathrm e^{-\pi/2}}. \end{align*}

(c)

解法一

思路

展开

距离是纵坐标，端点都为零，最大点在内部。求导后除以内部恒正的指数与余弦，得到所给方程，再验证变号。

答题过程

展开

The distance from the initial line is

y=(e−θ−e−π/2)sin⁡θ.y=\big(\mathrm e^{-\theta}-\mathrm e^{-\pi/2}\big)\sin\theta.

It is zero at both endpoints and positive inside, so its maximum is attained at an interior stationary point. There,

0=dydθ= (e−θ−e−π/2)cos⁡θ −e−θsin⁡θ.\begin{align*} 0=\frac{\mathrm{d}y}{\mathrm{d}\theta} =&\,\big(\mathrm e^{-\theta}-\mathrm e^{-\pi/2}\big)\cos\theta\\[4mm] &\,-\mathrm e^{-\theta}\sin\theta. \end{align*}

Since 0<θ<π/20<\theta<\pi/2, division by e−θcos⁡θ>0\mathrm e^{-\theta}\cos\theta>0 gives

1−eθ−π/2−tan⁡θ=0.\boxed{1-\mathrm e^{\theta-\pi/2}-\tan\theta=0}.

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

F(0.56)= 0.0091214…>0,F(0.57)= −0.0085552…<0.\begin{align*} F(0.56)=&\,0.0091214\ldots>0,\\[4mm] F(0.57)=&\,-0.0085552\ldots<0. \end{align*}

Since FF is continuous on [0.56,0.57][0.56,0.57], it has a root in (0.56,0.57)\boxed{(0.56,0.57)}.