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CIE 9709 2025 November Paper 35 Q10

A Level / CIE / P3

CIE 9709 2025 Nov Paper 35 Paper · Question 10

题目

Problem

The constant aa is such that ∫0axe12x dx=6\displaystyle\int_0^ax\mathrm{e}^{\frac12x}\,\mathrm{d}x=6.

(a) Show that a=2+e−12aa=2+\mathrm{e}^{-\frac12a}.

(5)

(b) Verify by calculation that aa lies between 2.22.2 and 2.42.4.

(2)

(c) Use an iterative formula based on the equation in part (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

常数 aa 满足 ∫0axe12x dx=6\displaystyle\int_0^ax\mathrm{e}^{\frac12x}\,\mathrm{d}x=6。

(a) 证明 a=2+e−12aa=2+\mathrm{e}^{-\frac12a}。

(5)

(b) 通过计算验证 aa 位于 2.22.2 与 2.42.4 之间。

(2)

(c) 使用基于 (a) 问中方程的迭代公式,求 aa 的值,精确到小数点后 22 位。每次迭代的结果保留小数点后 44 位。

(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

分部积分先求原函数,再代入上下限。注意零下限处指数项的值不为零。

答题过程

展开

Using u=xu=x and dv=ex/2 dx\mathrm{d}v=e^{x/2}\,\mathrm{d}x, integration by parts gives

∫xex/2 dx= 2xex/2−2∫ex/2 dx= 2xex/2−4ex/2+C.\begin{align*} \int xe^{x/2}\,\mathrm{d}x =&\,2xe^{x/2}-2\int e^{x/2}\,\mathrm{d}x\\ =&\,2xe^{x/2}-4e^{x/2}+C. \end{align*}

Thus the given condition becomes

(2aea/2−4ea/2)−(−4)=6.(2ae^{a/2}-4e^{a/2})-(-4)=6.

Hence

2(a−2)ea/2=2⟹a−2=e−a/2,2(a-2)e^{a/2}=2 \quad\Longrightarrow\quad a-2=e^{-a/2},

giving

a=2+e−a/2.\boxed{a=2+e^{-a/2}}.

(b)

解法一

思路

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把重排方程两边之差设成连续函数,计算两端并检查变号。

答题过程

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Let f(a)=a−2−e−a/2f(a)=a-2-e^{-a/2}. Then

f(2.2)=−0.132871…,f(2.4)=0.0988058….f(2.2)=-0.132871\ldots,\qquad f(2.4)=0.0988058\ldots.

Since ff is continuous and changes sign, a root lies in (2.2,2.4)(2.2,2.4). Moreover,

f′(a)=1+12e−a/2>0,f'(a)=1+\frac12e^{-a/2}>0,

so it is the unique root. Hence 2.2<a<2.4\boxed{2.2<a<2.4}.

(c)

解法一

思路

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按(a)的重排建立迭代,取区间中的值作为初值。输出四位小数,内部计算保留完整精度。

答题过程

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Use

an+1=2+e−an/2a_{n+1}=2+e^{-a_n/2}

with a0=2.3a_0=2.3. The iterations are

nnana_n to 4 decimal places
02.3000
12.3166
22.3140
32.3144
42.3144
52.3144

Therefore a=2.31\boxed{a=2.31} to 2 decimal places.