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CIE 9709 2022 November Paper 33 Q1

A Level / CIE / P3

CIE 9709 2022 Nov Paper 33 Paper · Question 1

题目

Problem

Solve the equation ln⁡(2x−1)=2ln⁡(x+1)−ln⁡x\ln(2x-1)=2\ln(x+1)-\ln x. Give your answer correct to 33 decimal places.

(4)
题目中文翻译

解方程 ln⁡(2x−1)=2ln⁡(x+1)−ln⁡x\ln(2x-1)=2\ln(x+1)-\ln x。答案保留 33 位小数。

(4)
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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先检查三个对数的定义域,再将右侧合并为商的对数。指数化得到二次方程后,排除不满足定义域的根。

答题过程

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The logarithms require 2x−1>02x-1>0, x+1>0x+1>0 and x>0x>0, so x>1/2x>1/2.

Using logarithm laws,

ln⁡(2x−1)=ln⁡(x+1)2x.\ln(2x-1)=\ln\frac{(x+1)^2}{x}.

Hence

x(2x−1)=(x+1)2,x(2x-1)=(x+1)^2,

giving

x2−3x−1= 0,x= 3±132.\begin{align*} x^2-3x-1=&\,0,\\[4mm] x=&\,\frac{3\pm\sqrt{13}}2. \end{align*}

The minus sign gives a negative value and is rejected. Therefore

x=3+132≈3.303x=\frac{3+\sqrt{13}}2\approx\boxed{3.303}

to 3 decimal places.