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CIE 9709 2022 June Paper 32 Q4

A Level / CIE / P3

CIE 9709 2022 June Paper 32 Paper · Question 4

题目

Problem

The equation of a curve is y=cos⁡3xsin⁡xy=\cos^3x\sqrt{\sin x}. It is given that the curve has one stationary point in the interval 0<x<12π0<x<\frac12\pi.

Find the xx-coordinate of this stationary point, giving your answer correct to 3 significant figures.

(6)
题目中文翻译

一条曲线的方程为 y=cos⁡3xsin⁡xy=\cos^3x\sqrt{\sin x}。已知该曲线在区间 0<x<12π0<x<\frac12\pi 内有一个驻点。

求该驻点的 xx 坐标,答案保留 3 位有效数字。

(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

解法一

思路

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同时使用乘积法则与链式法则。区间内正弦、余弦均为正,清除根式分母后可约去非零的 cos⁡2x\cos^2x。

答题过程

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Differentiating,

dydx= −3cos⁡2xsin⁡xsin⁡x +cos⁡4x2sin⁡x.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,-3\cos^2x\sin x\sqrt{\sin x}\\[4mm] &\,\hspace{2pt}+\frac{\cos^4x}{2\sqrt{\sin x}}. \end{align*}

At a stationary point, multiplication by 2sin⁡x>02\sqrt{\sin x}>0 gives

−6cos⁡2xsin⁡2x+cos⁡4x=0.-6\cos^2x\sin^2x+\cos^4x=0.

Since cos⁡x>0\cos x>0 in the interval,

cos⁡2x=6sin⁡2x.\cos^2x=6\sin^2x.

Using cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x,

7sin⁡2x=1,sin⁡x=17.7\sin^2x=1,\qquad \sin x=\frac1{\sqrt7}.

Hence

x=sin⁡−117≈0.388x=\sin^{-1}\frac1{\sqrt7}\approx\boxed{0.388}

to 3 significant figures.