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CIE 9709 2025 March Paper 32 Q2

A Level / CIE / P3

CIE 9709 2025 March Paper 32 Paper · Question 2

题目

Problem

The equation of a curve is xy2+ln⁡(x+2y)=1xy^2+\ln(x+2y)=1.

Find the gradient of the curve at the point where x=0x=0.

(5)
题目中文翻译

一条曲线的方程为 xy2+ln⁡(x+2y)=1xy^2+\ln(x+2y)=1。

求曲线上 x=0x=0 处的斜率。

(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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先用 x=0x=0 求对应的 yy,再做隐式微分。对数内部的导数必须包括 2y′2y'。

答题过程

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When x=0x=0,

ln⁡(2y)=1⟹y=e2.\ln(2y)=1 \quad\Longrightarrow\quad y=\frac e2.

Differentiating the curve equation implicitly,

y2+2xydydx+1+2 dy/dxx+2y=0.y^2+2xy\frac{\mathrm{d}y}{\mathrm{d}x} +\frac{1+2\,\mathrm{d}y/\mathrm{d}x}{x+2y}=0.

At (0,e/2)(0,e/2), let the gradient be mm. Then

e24+1+2me=0.\frac{e^2}{4}+\frac{1+2m}{e}=0.

Multiplying by ee,

e34+1+2m=0,\frac{e^3}{4}+1+2m=0,

so the gradient is

m=−e3+48.\boxed{m=-\frac{e^3+4}{8}}.