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CIE 9709 2025 June Paper 31 Q2

A Level / CIE / P3

CIE 9709 2025 June Paper 31 Paper · Question 2

题目

Problem

It is given that 2ln⁡p+ln⁡(p−1)−12ln⁡(q+1)=32\ln p+\ln(p-1)-\frac12\ln(q+1)=3.

Find qq in terms of pp.

(3)
题目中文翻译

已知 2ln⁡p+ln⁡(p−1)−12ln⁡(q+1)=32\ln p+\ln(p-1)-\frac12\ln(q+1)=3。

用 pp 表示 qq。

(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

解法一

思路

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合并对数后指数化,把平方根移到一边再平方。原对数同时给出 pp 与 qq 的范围。

答题过程

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The original logarithms require p>1p>1 and q>−1q>-1. Combining them,

ln⁡p2(p−1)q+1=3.\ln\frac{p^2(p-1)}{\sqrt{q+1}}=3.

Exponentiating,

p2(p−1)q+1= e3,q+1= p2(p−1)e3.\begin{align*} \frac{p^2(p-1)}{\sqrt{q+1}}=&\,e^3,\\[4mm] \sqrt{q+1}=&\,\frac{p^2(p-1)}{e^3}. \end{align*}

Both sides are positive. Squaring and rearranging,

q=p4(p−1)2e6−1,p>1.\boxed{q=\frac{p^4(p-1)^2}{e^6}-1,\qquad p>1}.