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CIE 9709 2025 November Paper 31 Q4

A Level / CIE / P3

CIE 9709 2025 Nov Paper 31 Paper · Question 4

题目

Problem

The diagram shows the graph of y=esin⁡2xcos⁡4xy=\mathrm{e}^{\sin2x}\cos4x for 0≤x≤14π0\le x\le\frac14\pi, and its maximum point MM.

Find the xx-coordinate of MM.

(5)
题目中文翻译

图中显示了 y=esin⁡2xcos⁡4xy=\mathrm{e}^{\sin2x}\cos4x 在 0≤x≤14π0\le x\le\frac14\pi 上的图像及其极大值点 MM。

求 MM 的 xx 坐标。

(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

解法一

思路

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用乘积与链式法则求导。因指数因子恒正,驻点方程可约去它,再用二倍角化为关于 sin⁡2x\sin2x 的二次方程。

答题过程

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Differentiating,

dydx=esin⁡2x(2cos⁡2xcos⁡4x−4sin⁡4x).\frac{\mathrm{d}y}{\mathrm{d}x} =e^{\sin2x}\big(2\cos2x\cos4x-4\sin4x\big).

At the maximum, the derivative is zero. Since esin⁡2x>0e^{\sin2x}>0,

2cos⁡2xcos⁡4x−4sin⁡4x=0.2\cos2x\cos4x-4\sin4x=0.

Using sin⁡4x=2sin⁡2xcos⁡2x\sin4x=2\sin2x\cos2x,

2cos⁡2x(cos⁡4x−4sin⁡2x)=0.2\cos2x(\cos4x-4\sin2x)=0.

For an interior point 0<x<π/40<x<\pi/4, cos⁡2x>0\cos2x>0. Hence, using cos⁡4x=1−2sin⁡22x\cos4x=1-2\sin^22x,

2sin⁡22x+4sin⁡2x−1=0.2\sin^22x+4\sin2x-1=0.

Writing s=sin⁡2xs=\sin2x,

s=−4±244=−1±62.s=\frac{-4\pm\sqrt{24}}4 =-1\pm\frac{\sqrt6}{2}.

Only s=−1+6/2s=-1+\sqrt6/2 belongs to (0,1)(0,1). Therefore

x=12sin⁡−1(62−1)=0.113 radians.x=\frac12\sin^{-1}\bigg(\frac{\sqrt6}{2}-1\bigg) =\boxed{0.113\text{ radians}}.

As xx increases through this value, 1−2s2−4s1-2s^2-4s changes from positive to negative, so the stationary point is a maximum.