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CIE 9709 2026 March Paper 12 Q1

A Level / CIE / P1

CIE 9709 2026 March Paper 12 Paper · Question 1

题目

Problem

The equation of a curve is

y=x3−3x2+9.y=x^3-3x^2+9.

Find the coordinates of the stationary points of the curve and determine their nature.

(5)
题目中文翻译

一条曲线的方程为

y=x3−3x2+9.y=x^3-3x^2+9.

求该曲线驻点的坐标,并确定这些驻点的性质。

(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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求导并因式分解以找到驻点,再用二阶导数的符号判断极大或极小。

答题过程

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The derivative is 3x2−6x=3x(x−2)3x^2-6x=3x(x-2), so x=0x=0 or x=2x=2. Substitution gives the points (0,9)(0,9) and (2,5)(2,5). The second derivative is 6x−66x-6. At x=0x=0 it equals −6<0-6<0, and at x=2x=2 it equals 6>06>0. Thus (0,9)(0,9) is a local maximum and (2,5)(2,5) is a local minimum.