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CIE 9709 2022 November Paper 32 Q3

A Level / CIE / P3

CIE 9709 2022 Nov Paper 32 Paper · Question 3

题目

Problem

The equation of a curve is y=sin⁡xsin⁡2xy=\sin x\sin2x. The curve has a stationary point in the interval 0<x<12π0<x<\frac12\pi.

Find the xx-coordinate of this point, giving your answer correct to 33 significant figures.

(6)
题目中文翻译

一条曲线的方程为 y=sin⁡xsin⁡2xy=\sin x\sin2x。曲线在区间 0<x<12π0<x<\frac12\pi 内有一个驻点。

求该点的 xx 坐标,答案保留 33 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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用乘积法则求导,再用二倍角公式把所有项改成 sin⁡x,cos⁡x\sin x,\cos x。在第一象限中 sin⁡x>0\sin x>0,可约去它。

答题过程

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Differentiating,

dydx=cos⁡xsin⁡2x+2sin⁡xcos⁡2x.\frac{\mathrm{d}y}{\mathrm{d}x} =\cos x\sin2x+2\sin x\cos2x.

Using sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x and cos⁡2x=cos⁡2x−sin⁡2x\cos2x=\cos^2x-\sin^2x,

dydx= 2sin⁡xcos⁡2x+2sin⁡x(cos⁡2x−sin⁡2x)= 2sin⁡x(2cos⁡2x−sin⁡2x)= 2sin⁡x(2−3sin⁡2x).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,2\sin x\cos^2x +2\sin x(\cos^2x-\sin^2x)\\[4mm] =&\,2\sin x(2\cos^2x-\sin^2x)\\[4mm] =&\,2\sin x(2-3\sin^2x). \end{align*}

Since 0<x<π/20<x<\pi/2, sin⁡x>0\sin x>0. At the stationary point,

sin⁡2x=23,sin⁡x=23.\sin^2x=\frac23,\qquad \sin x=\sqrt{\frac23}.

Thus

x=sin⁡−123≈0.955x=\sin^{-1}\sqrt{\frac23}\approx\boxed{0.955}

to 3 significant figures.

解法二

思路

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先用二倍角公式把函数写成 cos⁡x\cos x 的三次多项式,再用链式法则求导。

答题过程

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Using sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x,

y=2sin⁡2xcos⁡x=2cos⁡x−2cos⁡3x.y=2\sin^2x\cos x=2\cos x-2\cos^3x.

Hence

dydx= −2sin⁡x+6cos⁡2xsin⁡x= 2sin⁡x(3cos⁡2x−1).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,-2\sin x+6\cos^2x\sin x\\[4mm] =&\,2\sin x(3\cos^2x-1). \end{align*}

In 0<x<π/20<x<\pi/2, sin⁡x>0\sin x>0, so the stationary condition is

cos⁡2x=13,cos⁡x=13.\cos^2x=\frac13,\qquad \cos x=\frac1{\sqrt3}.

Therefore

x=cos⁡−113≈0.955x=\cos^{-1}\frac1{\sqrt3}\approx\boxed{0.955}

to 3 significant figures.