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CIE 9709 2023 Nov Paper 33 Q10

A Level / CIE / P3

CIE 9709 2023 Nov Paper 33 Paper · Question 10

题目

Problem

The diagram shows the curve y=xcos⁡2xy=x\cos2x, for x≥0x\ge0.

(a) Find the equation of the tangent to the curve at the point where x=12πx=\frac12\pi.

(4)

(b) Find the exact area of the shaded region shown in the diagram, bounded by the curve and the xx-axis.

(5)
题目中文翻译

图中显示了曲线 y=xcos⁡2xy=x\cos2x,其中 x≥0x\ge0。

(a) 求该曲线上 x=12πx=\frac12\pi 处的切线方程。

(4)

(b) 求图中由该曲线与 xx 轴围成的阴影区域的精确面积。

(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

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用乘积法则求导,在指定横坐标计算纵坐标与切线斜率,再用点斜式。

答题过程

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For y=xcos⁡2xy=x\cos2x,

dydx=cos⁡2x−2xsin⁡2x.\frac{\mathrm{d}y}{\mathrm{d}x}=\cos2x-2x\sin2x.

At x=π/2x=\pi/2,

y= π2cos⁡π=−π2,dydx= cos⁡π−πsin⁡π=−1.\begin{align*} y=&\,\frac\pi2\cos\pi=-\frac\pi2,\\[4mm] \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\cos\pi-\pi\sin\pi=-1. \end{align*}

Thus the tangent is

y+π2=−(x−π2),y+\frac\pi2=-\bigg(x-\frac\pi2\bigg),

or x+y=0\boxed{x+y=0}.

(b)

解法一

思路

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阴影是原点右侧第一块在横轴上方的区域,右端为 cos⁡2x=0\cos2x=0 的第一个正根 π/4\pi/4。用分部积分处理 xcos⁡2xx\cos2x。

答题过程

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The first positive zero is x=π/4x=\pi/4, and xcos⁡2x≥0x\cos2x\ge0 on 0≤x≤π/40\le x\le\pi/4. Thus

A=∫0π/4xcos⁡2x dx.A=\int_0^{\pi/4}x\cos2x\,\mathrm{d}x.

Integrating by parts with u=xu=x and dv=cos⁡2x dx\mathrm{d}v=\cos2x\,\mathrm{d}x,

∫xcos⁡2x dx= x2sin⁡2x−12∫sin⁡2x dx= x2sin⁡2x+14cos⁡2x+C.\begin{align*} \int x\cos2x\,\mathrm{d}x =&\,\frac x2\sin2x-\frac12\int\sin2x\,\mathrm{d}x\\[4mm] =&\,\frac x2\sin2x+\frac14\cos2x+C. \end{align*}

Therefore

A= [x2sin⁡2x+14cos⁡2x]0π/4= π8−14= π−28.\begin{align*} A=&\,\bigg[\frac x2\sin2x+\frac14\cos2x\bigg]_0^{\pi/4}\\[4mm] =&\,\frac\pi8-\frac14\\[4mm] =&\,\boxed{\frac{\pi-2}{8}}. \end{align*}