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CIE 9231 2023 November Paper 12 Q2

A Level / CIE / FP1

CIE 9231 2023 Nov Paper 12 Paper · Question 2

题目

Problem

Prove by mathematical induction that, for all positive integers nn,

dndxn(x2ex)=(x2+2nx+n(n−1))ex.\frac{\mathrm{d}^n}{\mathrm{d}x^n}(x^2\mathrm{e}^x)=\bigl(x^2+2nx+n(n-1)\bigr)\mathrm{e}^x.
(6)
题目中文翻译

用数学归纳法证明,对于所有正整数 nn,

dndxn(x2ex)=(x2+2nx+n(n−1))ex.\frac{\mathrm{d}^n}{\mathrm{d}x^n}(x^2\mathrm{e}^x)=\bigl(x^2+2nx+n(n-1)\bigr)\mathrm{e}^x.
(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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用乘积法则验证第一阶导数,再在归纳步骤把指数函数提出,合并多项式系数。

答题过程

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For n=1n=1,

ddx(x2ex)=(x2+2x)ex,\frac{\mathrm{d}}{\mathrm{d}x}(x^2\mathrm e^x) =(x^2+2x)\mathrm e^x,

which agrees with the given formula, so the result is true.

Assume that for some positive integer kk,

dkdxk(x2ex)=(x2+2kx+k(k−1))ex.\frac{\mathrm{d}^k}{\mathrm{d}x^k}(x^2\mathrm e^x) =\big(x^2+2kx+k(k-1)\big)\mathrm e^x.

Differentiating using the product rule,

 dk+1dxk+1(x2ex)= [x2+2kx+k(k−1)+2x+2k]ex= [x2+2(k+1)x+k(k+1)]ex.\begin{align*} &\,\frac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}}(x^2\mathrm e^x)\\[4mm] =&\,\big[x^2+2kx+k(k-1)+2x+2k\big]\mathrm e^x\\[4mm] =&\,\big[x^2+2(k+1)x+k(k+1)\big]\mathrm e^x. \end{align*}

This is the required expression for n=k+1n=k+1. Since it holds for n=1n=1 and truth for n=kn=k implies truth for n=k+1n=k+1, the statement follows by mathematical induction for all positive integers nn.