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CIE 9709 2025 June Paper 35 Q4

A Level / CIE / P3

CIE 9709 2025 June Paper 35 Paper · Question 4

题目

Problem

Find the exact coordinates of the stationary point of the curve with equation y=3x3ln⁡x4y=3x^3\ln x^4, for x>0x>0.

(5)
题目中文翻译

求方程为 y=3x3ln⁡x4y=3x^3\ln x^4 的曲线的驻点的精确坐标,其中 x>0x>0。

(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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因 x>0x>0,先把 ln⁡(x4)\ln(x^4) 写成 4ln⁡x4\ln x。乘积求导后令导数为零,并代回求纵坐标。

答题过程

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For x>0x>0, ln⁡(x4)=4ln⁡x\ln(x^4)=4\ln x, so y=12x3ln⁡xy=12x^3\ln x.

Differentiating,

dydx= 36x2ln⁡x+12x2= 12x2(3ln⁡x+1).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,36x^2\ln x+12x^2\\ =&\,12x^2(3\ln x+1). \end{align*}

Since x>0x>0, a stationary point requires

3ln⁡x+1=0⟹x=e−1/3.3\ln x+1=0 \quad\Longrightarrow\quad x=e^{-1/3}.

Substituting,

y=12e−1(−13)=−4e.y=12e^{-1}\bigg(-\frac13\bigg)=-\frac4e.

The stationary point is

(e−1/3,−4e).\boxed{\bigg(e^{-1/3},-\frac4e\bigg)}.