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CIE 9709 2024 November Paper 31 Q6

A Level / CIE / P3

CIE 9709 2024 Nov Paper 31 Paper · Question 6

题目

Problem

The diagram shows the curve y=sin⁡2x(1+sin⁡2x)y=\sin2x(1+\sin2x), for 0≤x≤34π0\le x\le\frac34\pi, and its minimum point MM. The shaded region bounded by the curve that lies above the xx-axis and the xx-axis itself is denoted by RR.

(a) Given that the xx-coordinate of MM lies in the interval 12π<x<34π\frac12\pi<x<\frac34\pi, find the exact coordinates of MM.

(4)

(b) Find the exact area of the region RR.

(4)
题目中文翻译

图中显示了曲线 y=sin⁡2x(1+sin⁡2x)y=\sin2x(1+\sin2x) 及其极小值点 MM,其中 0≤x≤34π0\le x\le\frac34\pi。由位于 xx 轴上方的曲线部分与 xx 轴围成的阴影区域记为 RR。

(a) 已知 MM 的 xx 坐标位于区间 12π<x<34π\frac12\pi<x<\frac34\pi 内,求 MM 的精确坐标。

(4)

(b) 求区域 RR 的精确面积。

(4)
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

展开

乘积求导后因式分解,在指定开区间选驻点。再代回函数求纵坐标,并检查是最小值。

答题过程

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Differentiating,

dydx= 2cos⁡2x(1+sin⁡2x)+2sin⁡2xcos⁡2x= 2cos⁡2x(1+2sin⁡2x).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,2\cos2x(1+\sin2x)+2\sin2x\cos2x\\ =&\,2\cos2x(1+2\sin2x). \end{align*}

For π/2<x<3π/4\pi/2<x<3\pi/4, we have π<2x<3π/2\pi<2x<3\pi/2, so cos⁡2x\cos2x is nonzero. Thus the stationary point satisfies

sin⁡2x=−12.\sin2x=-\frac12.

In this interval, 2x=7π/62x=7\pi/6, giving x=7π/12x=7\pi/12. Then

y=−12(1−12)=−14.y=-\frac12\bigg(1-\frac12\bigg)=-\frac14.

Also y=s+s2=(s+1/2)2−1/4y=s+s^2=(s+1/2)^2-1/4, with s=sin⁡2xs=\sin2x, so −1/4-1/4 is the minimum value. Therefore

M=(7π12,−14).\boxed{M=\bigg(\frac{7\pi}{12},-\frac14\bigg)}.

(b)

解法一

思路

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阴影在横轴上方,对应 00 到 π/2\pi/2。平方正弦用降幂公式积分。

答题过程

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The curve is above the horizontal axis for 0<x<π/20<x<\pi/2. For π/2<x<3π/4\pi/2<x<3\pi/4, it is below it, so only the first interval contributes to RR.

Thus

A= ∫0π/2(sin⁡2x+sin⁡22x) dx= ∫0π/2(sin⁡2x+12−12cos⁡4x) dx= [−12cos⁡2x+x2−18sin⁡4x]0π/2= 12+π4+12= 1+π4.\begin{align*} A=&\,\int_0^{\pi/2}(\sin2x+\sin^22x)\,\mathrm{d}x\\ =&\,\int_0^{\pi/2}\bigg(\sin2x+\frac12-\frac12\cos4x\bigg)\,\mathrm{d}x\\ =&\,\bigg[-\frac12\cos2x+\frac x2-\frac18\sin4x\bigg]_0^{\pi/2}\\ =&\,\frac12+\frac\pi4+\frac12\\ =&\,\boxed{1+\frac\pi4}. \end{align*}

解法二

思路

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也可把原来的两个因式直接用于分部积分,再把剩下的平方余弦降幂。

答题过程

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As in the first method, the area is the integral over 0⩽x⩽π/20\leqslant x\leqslant\pi/2. Take u=1+sin⁡2xu=1+\sin2x and dv=sin⁡2x dx\mathrm{d}v=\sin2x\,\mathrm{d}x. Then du=2cos⁡2x dx\mathrm{d}u=2\cos2x\,\mathrm{d}x and v=−12cos⁡2xv=-\frac12\cos2x. Thus

A= [−12cos⁡2x(1+sin⁡2x)]0π/2 +∫0π/2cos⁡22x dx= 1+∫0π/21+cos⁡4x2 dx= 1+[x2+18sin⁡4x]0π/2= 1+π4.\begin{align*} A=&\,\bigg[-\frac12\cos2x(1+\sin2x)\bigg]_0^{\pi/2}\\[4mm] &\,\hspace{2pt}+\int_0^{\pi/2}\cos^22x\,\mathrm{d}x\\[4mm] =&\,1+\int_0^{\pi/2}\frac{1+\cos4x}{2}\,\mathrm{d}x\\[4mm] =&\,1+\bigg[\frac x2+\frac18\sin4x\bigg]_0^{\pi/2}\\[4mm] =&\,\boxed{1+\frac\pi4}. \end{align*}