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CIE 9709 2022 March Paper 12 Q1

A Level / CIE / P1

CIE 9709 2022 March Paper 12 Paper · Question 1

题目

Problem

A curve with equation y=f(x)y=\mathrm{f}(x) is such that f′(x)=2x−13−x13\mathrm{f}'(x)=2x^{-\frac13}-x^{\frac13}. It is given that f(8)=5\mathrm{f}(8)=5.

Find f(x)\mathrm{f}(x).

(4)
题目中文翻译

一条方程为 y=f(x)y=\mathrm{f}(x) 的曲线满足 f′(x)=2x−13−x13\mathrm{f}'(x)=2x^{-\frac13}-x^{\frac13}。已知 f(8)=5\mathrm{f}(8)=5。

求 f(x)\mathrm{f}(x)。

(4)
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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逐项积分后用 f(8)=5\mathrm f(8)=5 求常数。计算分数幂时先取立方根。

答题过程

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On the interval containing 88, integration gives

f(x)=3x2/3−34x4/3+C.\mathrm f(x)=3x^{2/3}-\frac34x^{4/3}+C.

At x=8x=8, the first two terms give 3(4)−34(16)=03(4)-\tfrac34(16)=0, so C=5C=5. Hence f(x)=3x2/3−34x4/3+5\mathrm f(x)=3x^{2/3}-\tfrac34x^{4/3}+5 on x>0x>0.