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CIE 9709 2025 June Paper 32 Q1

A Level / CIE / P3

CIE 9709 2025 June Paper 32 Paper · Question 1

题目

Problem

Solve the equation ex+2e−xex−3=4\frac{\mathrm{e}^x+2\mathrm{e}^{-x}}{\mathrm{e}^x-3}=4. Give your answer correct to 33 decimal places.

(5)
题目中文翻译

解方程 ex+2e−xex−3=4\frac{\mathrm{e}^x+2\mathrm{e}^{-x}}{\mathrm{e}^x-3}=4。答案精确到小数点后 33 位。

(5)
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

解法一

思路

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令 t=ex>0t=e^x>0,把负指数换成倒数。原分母还要求 t≠3t\ne3,最后检查所选根。

答题过程

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Let t=ex>0t=e^x>0, with t≠3t\ne3. Then

t+2/tt−3=4.\frac{t+2/t}{t-3}=4.

Multiplying by t(t−3)t(t-3),

t2+2=4t2−12t⟹3t2−12t−2=0.t^2+2=4t^2-12t \quad\Longrightarrow\quad 3t^2-12t-2=0.

Hence

t=12±1686=6±423.t=\frac{12\pm\sqrt{168}}6 =\frac{6\pm\sqrt{42}}3.

The negative root is impossible since t>0t>0. The positive root is not 33, so it is valid. Thus

x= ln⁡(6+423)= 1.425574…=1.426.\begin{align*} x=&\,\ln\bigg(\frac{6+\sqrt{42}}3\bigg)\\[4mm] =&\,1.425574\ldots=\boxed{1.426}. \end{align*}