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CIE 9709 2025 June Paper 33 Q5

A Level / CIE / P3

CIE 9709 2025 June Paper 33 Paper · Question 5

题目

Problem

The equation of a curve is xy+y2e−x=4xy+y^2\mathrm{e}^{-x}=4.

(a) Show that dydx=y2−yexxex+2y\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y^2-y\mathrm{e}^x}{x\mathrm{e}^x+2y}.

(4)

(b) Find the gradients of the tangents to the curve when x=0x=0.

(2)
题目中文翻译

一条曲线的方程为 xy+y2e−x=4xy+y^2\mathrm{e}^{-x}=4。

(a) 证明 dydx=y2−yexxex+2y\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y^2-y\mathrm{e}^x}{x\mathrm{e}^x+2y}。

(4)

(b) 求当 x=0x=0 时该曲线的切线的斜率。

(2)
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

两个乘积都隐式求导,再收集导数项。最后同乘 exe^x 得到目标形式。

答题过程

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Differentiating implicitly,

 y+xdydx+2ye−xdydx −y2e−x=0.\begin{align*} &\,y+x\frac{\mathrm{d}y}{\mathrm{d}x} +2ye^{-x}\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}-y^2e^{-x}=0. \end{align*}

Thus

(x+2ye−x)dydx=y2e−x−y.(x+2ye^{-x})\frac{\mathrm{d}y}{\mathrm{d}x} =y^2e^{-x}-y.

Multiplying numerator and denominator by exe^x gives

dydx=y2−yexxex+2y.\boxed{\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{y^2-ye^x}{xe^x+2y}}.

解法二

思路

展开

也可先把原式乘以 exe^x 再求导,最后必须利用原方程消去额外项。

答题过程

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Multiplying the curve equation by exe^x,

xyex+y2=4ex.xye^x+y^2=4e^x.

Differentiating,

 xexdydx+y(x+1)ex +2ydydx=4ex.\begin{align*} &\,xe^x\frac{\mathrm{d}y}{\mathrm{d}x} +y(x+1)e^x\\[4mm] &\,\hspace{2pt}+2y\frac{\mathrm{d}y}{\mathrm{d}x}=4e^x. \end{align*}

Hence

(xex+2y)dydx=4ex−y(x+1)ex.(xe^x+2y)\frac{\mathrm{d}y}{\mathrm{d}x} =4e^x-y(x+1)e^x.

The original equation gives 4ex−xyex=y24e^x-xye^x=y^2. Substituting,

dydx=y2−yexxex+2y.\boxed{\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{y^2-ye^x}{xe^x+2y}}.

(b)

解法一

思路

展开

先从原曲线求 x=0x=0 对应的两个点,再分别代入导数。

答题过程

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When x=0x=0, the curve equation gives y2=4y^2=4, so y=±2y=\pm2.

Using (a),

dydx=y2−y2y=y−12.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{y^2-y}{2y} =\frac{y-1}{2}.

Thus the tangent gradients are

12 at (0,2),−32 at (0,−2).\boxed{\frac12\text{ at }(0,2),\qquad -\frac32\text{ at }(0,-2)}.