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CIE 9709 2024 November Paper 33 Q11

A Level / CIE / P3

CIE 9709 2024 Nov Paper 33 Paper · Question 11

题目

Problem

The diagram shows the curve y=2sin⁡x2+cos⁡xy=2\sin x\sqrt{2+\cos x}, for 0≤x≤2π0\le x\le2\pi, and its minimum point MM, where x=ax=a.

(a) Find the value of aa correct to 22 decimal places.

(5)

(b) Use the substitution u=2+cos⁡xu=2+\cos x to find the exact area of the shaded region RR.

(6)
题目中文翻译

图中显示了曲线 y=2sin⁡x2+cos⁡xy=2\sin x\sqrt{2+\cos x} 及其极小值点 MM,其中 0≤x≤2π0\le x\le2\pi,且 MM 处的 x=ax=a。

(a) 求 aa 的值,结果保留 22 位小数。

(5)

(b) 使用代换 u=2+cos⁡xu=2+\cos x,求阴影区域 RR 的精确面积。

(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

(a)

解法一

思路

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乘积与链式法则求导后,把驻点条件化成关于 cos⁡x\cos x 的二次方程。选择下半周期对应的极小点。

答题过程

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Differentiating,

dydx= 2cos⁡x2+cos⁡x −sin⁡2x2+cos⁡x.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,2\cos x\sqrt{2+\cos x}\\[4mm] &\,\hspace{2pt}-\frac{\sin^2x}{\sqrt{2+\cos x}}. \end{align*}

Since 2+cos⁡x⩾12+\cos x\geqslant1, setting the derivative to zero gives

2cos⁡x(2+cos⁡x)−sin⁡2x=0.2\cos x(2+\cos x)-\sin^2x=0.

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

3cos⁡2x+4cos⁡x−1=0.3\cos^2x+4\cos x-1=0.

Thus

cos⁡x=−2±73.\cos x=\frac{-2\pm\sqrt7}{3}.

Only (−2+7)/3(-2+\sqrt7)/3 is in [−1,1][-1,1]. The corresponding point in (0,π)(0,\pi) has positive yy, while the one in (π,2π)(\pi,2\pi) has negative yy and is the minimum.

Therefore

a= 2π−cos⁡−1(−2+73)= 4.929337…,\begin{align*} a=&\,2\pi-\cos^{-1}\bigg(\frac{-2+\sqrt7}{3}\bigg)\\[4mm] =&\,4.929337\ldots, \end{align*}

so a=4.93\boxed{a=4.93}.

(b)

解法一

思路

展开

阴影是零到 π\pi 的正面积。指定换元后,负微分与倒转的上下限相抵。

答题过程

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The shaded region is above the horizontal axis for 0⩽x⩽π0\leqslant x\leqslant\pi. Its area is

A=∫0π2sin⁡x2+cos⁡x dx.A=\int_0^\pi2\sin x\sqrt{2+\cos x}\,\mathrm{d}x.

Let u=2+cos⁡xu=2+\cos x. Then du=−sin⁡x dx\mathrm{d}u=-\sin x\,\mathrm{d}x, and

x=0 ⇒u=3,x=π ⇒u=1.\begin{align*} x=0&\,\Rightarrow u=3,\\[4mm] x=\pi&\,\Rightarrow u=1. \end{align*}

Therefore

A= −2∫31u1/2 du= 2∫13u1/2 du= [43u3/2]13= 43(33−1).\begin{align*} A=&\,-2\int_3^1u^{1/2}\,\mathrm{d}u\\ =&\,2\int_1^3u^{1/2}\,\mathrm{d}u\\ =&\,\bigg[\frac43u^{3/2}\bigg]_1^3\\ =&\,\boxed{\frac43(3\sqrt3-1)}. \end{align*}