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CIE 9709 2023 June Paper 33 Q5

A Level / CIE / P3

CIE 9709 2023 June Paper 33 Paper · Question 5

题目

Problem

The diagram shows the part of the curve y=x2cos⁡3xy=x^2\cos3x for 0≤x≤16π0\le x\le\frac16\pi, and its maximum point MM, where x=ax=a.

(a) Show that aa satisfies the equation a=13tan⁡−1(23a)a=\frac13\tan^{-1}\big(\frac{2}{3a}\big).

(3)

(b) Use an iterative formula based on the equation in (a) to determine aa correct to 22 decimal places. Give the result of each iteration to 44 decimal places.

(3)
题目中文翻译

图中显示了曲线 y=x2cos⁡3xy=x^2\cos3x 在 0≤x≤16π0\le x\le\frac16\pi 范围内的部分,以及其最大值点 MM,该点处 x=ax=a。

(a) 证明 aa 满足方程 a=13tan⁡−1(23a)a=\frac13\tan^{-1}\big(\frac{2}{3a}\big)。

(3)

(b) 使用基于 (a) 问方程的迭代公式求 aa,结果保留 22 位小数。每次迭代的结果保留 44 位小数。

(3)
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

最高点位于区间内部,导数为零。先约去非零的 aa 与 cos⁡3a\cos3a,再取反正切;用 0<3a<π/20<3a<\pi/2 确定分支。

答题过程

展开

Differentiating,

dydx=2xcos⁡3x−3x2sin⁡3x.\frac{\mathrm{d}y}{\mathrm{d}x} =2x\cos3x-3x^2\sin3x.

At the interior maximum MM,

2acos⁡3a−3a2sin⁡3a=0.2a\cos3a-3a^2\sin3a=0.

Since 0<a<π/60<a<\pi/6, both aa and cos⁡3a\cos3a are non-zero. Dividing gives

2= 3atan⁡3a,tan⁡3a= 23a.\begin{align*} 2=&\,3a\tan3a,\\[4mm] \tan3a=&\,\frac2{3a}. \end{align*}

Also 0<3a<π/20<3a<\pi/2, so

a=13tan⁡−1(23a).\boxed{a=\frac13\tan^{-1}\bigg(\frac2{3a}\bigg)}.

(b)

解法一

思路

展开

用(a)给出的式子迭代,角度用弧度,内部保留精度,列表到四位小数。

答题过程

展开

Use

an+1= 13tan⁡−1(23an),a0= 0.4.\begin{align*} a_{n+1}=&\,\frac13\tan^{-1}\bigg(\frac2{3a_n}\bigg),\\[4mm] a_0=&\,0.4. \end{align*}

The successive values are:

nnana_n to 4 decimal places
00.4000
10.3435
20.3650
30.3566
40.3599
50.3586
60.3591
70.3589
80.3590
90.3590

Hence a=0.36\boxed{a=0.36} to 2 decimal places.