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CIE 9231 2025 June Paper 14 Q2

A Level / CIE / FP1

CIE 9231 2025 June Paper 14 Paper · Question 2

题目

Problem

Prove by mathematical induction that, for every integer n⩾2n\geqslant2,

dndxn(xln⁡x)=(−1)n(n−2)!x1−n.\frac{\mathrm{d}^n}{\mathrm{d}x^n}(x\ln x)=(-1)^n(n-2)!x^{1-n}.
(6)
题目中文翻译

用数学归纳法证明:对每一个整数 n⩾2n\geqslant2,

dndxn(xln⁡x)=(−1)n(n−2)!x1−n.\frac{\mathrm{d}^n}{\mathrm{d}x^n}(x\ln x)=(-1)^n(n-2)!x^{1-n}.
(6)
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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从n=2验证,不从n=1开始。归纳步骤再求一次导数,指数带出的负因子同时改变符号并补全阶乘。

答题过程

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Let f(x)=xln⁡xf(x)=x\ln x, where x>0x>0.

For n=2n=2,

f′(x)=1+ln⁡x,f′′(x)=x−1.f'(x)=1+\ln x,\qquad f''(x)=x^{-1}.

This agrees with (−1)2(2−2)!x1−2=x−1(-1)^2(2-2)!x^{1-2}=x^{-1}, so the result is true for n=2n=2.

Assume that, for some integer k⩾2k\geqslant2,

f(k)(x)=(−1)k(k−2)!x1−k.f^{(k)}(x)=(-1)^k(k-2)!x^{1-k}.

Differentiating both sides,

f(k+1)(x)= (−1)k(k−2)!(1−k)x−k= (−1)k+1(k−1)!x−k= (−1)k+1((k+1)−2)!x1−(k+1).\begin{align*} f^{(k+1)}(x)=&\,(-1)^k(k-2)!(1-k)x^{-k}\\[4mm] =&\,(-1)^{k+1}(k-1)!x^{-k}\\[4mm] =&\,(-1)^{k+1}((k+1)-2)!x^{1-(k+1)}. \end{align*}

Thus truth for n=kn=k implies truth for n=k+1n=k+1. Since the result holds for n=2n=2, by mathematical induction it holds for every integer n⩾2n\geqslant2.