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CIE 9231 2025 Nov Paper 23 Q7

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 23 Paper · Question 7

题目

Problem

(a) Show that

ddx(tanh⁡−1x)=11−x2.\frac{\mathrm{d}}{\mathrm{d}x}\big(\tanh^{-1}x\big) = \frac{1}{1 - x^2}.
[3]

(b) Find the solution of the differential equation

xdydx−y=x2tanh⁡−1x,x\frac{\mathrm{d}y}{\mathrm{d}x} - y = x^2\tanh^{-1}x,

for 0<x<10 < x < 1, given that y=0y = 0 when x=12x = \frac{1}{2}. Give your answer in an exact form.

[9]
题目中文翻译

(a) 证明

ddx(tanh⁡−1x)=11−x2.\frac{\mathrm{d}}{\mathrm{d}x}\big(\tanh^{-1}x\big) = \frac{1}{1 - x^2}.
(3)

(b) 求微分方程

xdydx−y=x2tanh⁡−1xx\frac{\mathrm{d}y}{\mathrm{d}x} - y = x^2\tanh^{-1}x

在 0<x<10 < x < 1 上的解，已知当 x=12x = \frac{1}{2} 时 y=0y = 0。答案需为精确形式。

(9)
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

(a)

解法一

思路

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把反双曲正切写成两个对数之差，求导后通分。

答题过程

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For −1<x<1-1<x<1,

tanh⁡−1x=12[ln⁡(1+x)−ln⁡(1−x)].\tanh^{-1}x=\frac12[\ln(1+x)-\ln(1-x)].

Differentiating gives

ddx(tanh⁡−1x)= 12(11+x+11−x)= 11−x2.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}(\tanh^{-1}x)=&\,\frac12\bigg(\frac1{1+x}+\frac1{1-x}\bigg)\\ =&\,\frac1{1-x^2}. \end{align*}

解法二

思路

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也可先写成 x=tanh⁡ux=\tanh u，再用隐式求导和 sech⁡2u=1−tanh⁡2u\operatorname{sech}^2u=1-\tanh^2u。

答题过程

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Put u=tanh⁡−1xu=\tanh^{-1}x, so tanh⁡u=x\tanh u=x. Differentiating implicitly gives sech⁡2u dudx=1\operatorname{sech}^2u\,\frac{\mathrm{d}u}{\mathrm{d}x}=1. Therefore

dudx= 1sech⁡2u= 11−tanh⁡2u= 11−x2.\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x}=&\,\frac1{\operatorname{sech}^2u}\\ =&\,\frac1{1-\tanh^2u}\\ =&\,\frac1{1-x^2}. \end{align*}

(b)

解法一

思路

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先除以 xx 化为一阶线性方程，用积分因子 x−1x^{-1}。右边的反双曲正切用分部积分，再由给定点求积分常数。

答题过程

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On 0<x<10<x<1, the equation is y′−yx=xtanh⁡−1xy'-\frac yx=x\tanh^{-1}x. Its integrating factor is

e∫−1/x dx=e−ln⁡x=x−1,e^{\int-1/x\,\mathrm{d}x}=e^{-\ln x}=x^{-1},

giving

ddx(yx)=tanh⁡−1x.\frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac yx\bigg)=\tanh^{-1}x.

Integrating by parts with u=tanh⁡−1xu=\tanh^{-1}x and dv=dx\mathrm{d}v=\mathrm{d}x, we have v=xv=x and, by (a), du=dx/(1−x2)\mathrm{d}u=\mathrm{d}x/(1-x^2). Hence

yx= xtanh⁡−1x−∫x1−x2 dx= xtanh⁡−1x+12ln⁡(1−x2)+C.\begin{align*} \frac yx=&\,x\tanh^{-1}x-\int\frac{x}{1-x^2}\,\mathrm{d}x\\ =&\,x\tanh^{-1}x+\frac12\ln(1-x^2)+C. \end{align*}

At x=12x=\frac12, tanh⁡−1(12)=12ln⁡3\tanh^{-1}(\frac12)=\frac12\ln 3, so

C= −14ln⁡3−12ln⁡34= ln⁡2−34ln⁡3.\begin{align*} C=&\,-\frac14\ln3-\frac12\ln\frac34\\ =&\,\ln2-\frac34\ln3. \end{align*}

Thus

y=x[xtanh⁡−1x+12ln⁡(1−x2)+ln⁡2−34ln⁡3].y=x\bigg[x\tanh^{-1}x+\frac12\ln(1-x^2)+\ln2-\frac34\ln3\bigg].