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CIE 9709 2024 November Paper 33 Q2

A Level / CIE / P3

CIE 9709 2024 Nov Paper 33 Paper · Question 2

题目

Problem

Let f(x)=2x3−5x2+4\mathrm{f}(x)=2x^3-5x^2+4.

(a) Show that if a sequence of values given by the iterative formula

xn+1=45−2xnx_{n+1}=\sqrt{\frac4{5-2x_n}}

converges, then it converges to a root of the equation f(x)=0\mathrm{f}(x)=0.

(2)

(b) The equation has a root close to 1.21.2.

Use the iterative formula from part (a) and an initial value of 1.21.2 to determine the root correct to 22 decimal places. Give the result of each iteration to 44 decimal places.

(3)
题目中文翻译

设 f(x)=2x3−5x2+4\mathrm{f}(x)=2x^3-5x^2+4。

(a) 证明:如果由迭代公式

xn+1=45−2xnx_{n+1}=\sqrt{\frac4{5-2x_n}}

给出的数列收敛,那么它收敛于方程 f(x)=0\mathrm{f}(x)=0 的一个根。

(2)

(b) 此方程有一个接近 1.21.2 的根。

使用 (a) 问中的迭代公式,以 1.21.2 为初始值,求此根,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

收敛极限满足不动点方程,平方后去分母即可恢复原多项式方程。

答题过程

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Squaring the recurrence and multiplying by its denominator gives, for every iteration,

xn+12(5−2xn)=4.x_{n+1}^2(5-2x_n)=4.

If xn→Lx_n\to L, then also xn+1→Lx_{n+1}\to L. Taking limits in this polynomial relation gives

L2(5−2L)= 4,5L2−2L3= 4,2L3−5L2+4= 0.\begin{align*} L^2(5-2L)=&\,4,\\[4mm] 5L^2-2L^3=&\,4,\\[4mm] 2L^3-5L^2+4=&\,0. \end{align*}

Thus f(L)=0f(L)=0, so the sequence converges to a root of f(x)=0\boxed{f(x)=0}.

(b)

解法一

思路

展开

使用指定初值 1.21.2,计算保留完整精度,输出四位小数。

答题过程

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Starting with the required x0=1.2x_0=1.2, apply

xn+1=45−2xn.x_{n+1}=\sqrt{\frac4{5-2x_n}}.

The iterations are

nnxnx_n to 4 decimal places
01.2000
11.2403
21.2601
31.2700
41.2752
51.2778
61.2792
71.2800
81.2804

Hence the root is x=1.28\boxed{x=1.28} to 2 decimal places.