题目

Problem

The curve $C$ has polar equation $r=\cos\frac12\theta$ , for $0\leqslant\theta\leqslant\pi$ .

(a) Sketch $C$ .

(2)

In parts (b) and (c) you may use the identities $\sin\theta=2\sin\frac12\theta\cos\frac12\theta$ and $\cos\theta=2\cos^2\frac12\theta-1$ .

(b) Find the exact value of the area of the region enclosed by $C$ and the initial line.

(4)

(c) Find the maximum distance of a point on $C$ from the initial line. Give your answer in the form $p\sqrt q$ , where $p$ and $q$ are rational numbers to be determined.

(6)

题目中文翻译

曲线 $C$ 的极坐标方程为 $r=\cos\frac12\theta$ ，其中 $0\leqslant\theta\leqslant\pi$ 。

(a) 画出 $C$ 的草图。

(2)

在 (b) 和 (c) 中，你可以使用恒等式 $\sin\theta=2\sin\frac12\theta\cos\frac12\theta$ 和 $\cos\theta=2\cos^2\frac12\theta-1$ 。

(b) 求由 $C$ 与始边围成的区域面积的精确值。

(4)

(6)

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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半径从1连续降到0，曲线依次经过第一、第二象限；不能只画第一象限。

答题过程

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The radius decreases from $r=1$ at $\theta=0$ to $r=0$ at $\theta=\pi$ . Sketch one smooth arc in the upper half-plane, starting at $(1,0)$ with a vertical tangent, passing through $(0,1/\sqrt2)$ at $\theta=\pi/2$ , then entering the second quadrant and ending at the pole with a horizontal tangent.

Mark the pole, initial line and endpoint $(1,0)$ .

(b)

解法一

思路

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极坐标面积中对cos平方半角使用题目给定恒等式，保留面积公式外面的二分之一。

答题过程

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Using $\cos^2(\theta/2)=(1+\cos\theta)/2$ ,

\begin{align*} A=&\,\frac12\int_0^\pi\cos^2\frac\theta2\,\mathrm{d}\theta \\[4mm] =&\,\frac14\int_0^\pi(1+\cos\theta)\,\mathrm{d}\theta \\[4mm] =&\,\frac14\big[\theta+\sin\theta\big]_0^\pi \\[4mm] =&\,\boxed{\frac\pi4}. \end{align*}

(c)

解法一

思路

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最大距离对应最大纵坐标。求导后用半角恒等式把条件化为cos平方半角的方程，并检查极大值。

答题过程

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The distance from the initial line is

y=r\sin\theta=\cos\frac\theta2\sin\theta.

Differentiating,

\frac{\mathrm{d}y}{\mathrm{d}\theta} =\cos\frac\theta2\cos\theta -\frac12\sin\frac\theta2\sin\theta.

Using the given identities,

\frac{\mathrm{d}y}{\mathrm{d}\theta} =\cos\frac\theta2\bigg(3\cos^2\frac\theta2-2\bigg).

For $0<\theta<\pi$ , the first factor is positive. Thus the interior stationary point has $\cos^2(\theta/2)=2/3$ and $\sin(\theta/2)=1/\sqrt3$ .

The derivative changes from positive to negative there, while the endpoint distances are zero. Therefore the greatest distance is

\begin{align*} y_{\max}=&\,2\cos^2\frac\theta2\sin\frac\theta2\\[4mm] =&\,2\frac23\frac1{\sqrt3}\\[4mm] =&\,\boxed{\frac49\sqrt3}. \end{align*}

Thus $p=4/9$ and $q=3$ .

CIE 9231 2025 November Paper 14 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程