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CIE 9231 2025 November Paper 11 Q3

A Level / CIE / FP1

CIE 9231 2025 Nov Paper 11 Paper · Question 3

题目

Problem

Prove by mathematical induction that, for every positive integer nn,

d2n−1dx2n−1(xcos⁡x)=(−1)n(xsin⁡x−(2n−1)cos⁡x).\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}(x\cos x)=(-1)^n\big(x\sin x-(2n-1)\cos x\big).
(7)
题目中文翻译

利用数学归纳法证明：对每一个正整数 nn，都有

d2n−1dx2n−1(xcos⁡x)=(−1)n(xsin⁡x−(2n−1)cos⁡x).\frac{\mathrm{d}^{2n-1}}{\mathrm{d}x^{2n-1}}(x\cos x)=(-1)^n\big(x\sin x-(2n-1)\cos x\big).
(7)
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=1的直接求导开始。归纳步骤由2k−1阶到2k+1阶，需要连续求导两次，而不是只求一次。

答题过程

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Let f(x)=xcos⁡xf(x)=x\cos x. For n=1n=1, the product rule gives

f′(x)= cos⁡x−xsin⁡x= (−1)1(xsin⁡x−(2(1)−1)cos⁡x).\begin{align*} f'(x)=&\,\cos x-x\sin x\\[4mm] =&\,(-1)^1\big(x\sin x-(2(1)-1)\cos x\big). \end{align*}

Thus the base case holds.

Assume that for some positive integer kk,

f(2k−1)(x)=(−1)k(xsin⁡x−(2k−1)cos⁡x).f^{(2k-1)}(x) =(-1)^k\big(x\sin x-(2k-1)\cos x\big).

Differentiating once,

f(2k)(x)=(−1)k(xcos⁡x+2ksin⁡x).f^{(2k)}(x) =(-1)^k\big(x\cos x+2k\sin x\big).

Differentiating again,

f(2k+1)(x)= (−1)k(−xsin⁡x+(2k+1)cos⁡x)= (−1)k+1(xsin⁡x−(2(k+1)−1)cos⁡x).\begin{align*} f^{(2k+1)}(x) =&\,(-1)^k\big(-x\sin x+(2k+1)\cos x\big) \\[4mm] =&\,(-1)^{k+1}\big(x\sin x-(2(k+1)-1)\cos x\big). \end{align*}

This is the required statement for k+1k+1. Therefore, by mathematical induction, the stated identity holds for every positive integer nn.