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CIE 9231 2023 November Paper 21 Q3

A Level / CIE / FP2

CIE 9231 2023 Nov Paper 21 Paper · Question 3

题目

Problem

Find the first three terms in the Maclaurin’s series for

tanh⁡−1(12ex)\tanh^{-1}\left(\frac{1}{2}e^x\right)

in the form 12ln⁡a+bx+cx2\frac{1}{2}\ln a+bx+cx^2, giving the exact values of the constants aa, bb and cc.

[6]
题目中文翻译

求

tanh⁡−1(12ex)\tanh^{-1}\left(\frac{1}{2}e^x\right)

的 Maclaurin 级数的前三项,并将其写成 12ln⁡a+bx+cx2\frac{1}{2}\ln a+bx+cx^2 的形式,求出常数 aa、bb 和 cc 的精确值。

(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

展开

令 u=ex/2u=e^x/2,利用 u′=uu'=u 求前两阶导数。常数项用反双曲正切的对数形式求出。

答题过程

展开

Let f(x)=tanh⁡−1(u)f(x)=\tanh^{-1}(u) with u=ex/2u=e^x/2. Since u′=uu'=u,

f′(x)= u1−u2,f′′(x)= u(1−u2)−u(−2u2)(1−u2)2= u(1+u2)(1−u2)2.\begin{align*} f'(x)=&\,\frac{u}{1-u^2},\\ f''(x)=&\,\frac{u(1-u^2)-u(-2u^2)}{(1-u^2)^2}\\ =&\,\frac{u(1+u^2)}{(1-u^2)^2}. \end{align*}

At x=0x=0, u=12u=\frac12. The logarithmic form gives

f(0)=12ln⁡1+1/21−1/2=12ln⁡3.f(0)=\frac12\ln\frac{1+1/2}{1-1/2}=\frac12\ln3.

Also f′(0)=23f'(0)=\frac23 and f′′(0)=109f''(0)=\frac{10}{9}. Substituting into f(0)+xf′(0)+x22f′′(0)f(0)+xf'(0)+\frac{x^2}{2}f''(0) gives

f(x)=12ln⁡3+23x+59x2+O(x3).f(x)=\frac12\ln3+\frac23x+\frac59x^2+O(x^3).

Therefore (a,b,c)=(3,23,59)(a,b,c)=\big(3,\frac23,\frac59\big).

解法二

思路

展开

也可先写成两个对数之差,再分别代入 exe^x 和 ln⁡(1+w)\ln(1+w) 的展开。

答题过程

展开

Use ex=1+x+12x2+O(x3)e^x=1+x+\frac12x^2+O(x^3) and ln⁡(1+w)=w−12w2+O(w3)\ln(1+w)=w-\frac12w^2+O(w^3). For u=ex/2=12+x2+x24+O(x3)u=e^x/2=\frac12+\frac x2+\frac{x^2}{4}+O(x^3),

ln⁡(1+u)= ln⁡32+ln⁡(1+x3+x26+O(x3))= ln⁡32+x3+x29+O(x3),ln⁡(1−u)= ln⁡12+ln⁡(1−x−x22+O(x3))= ln⁡12−x−x2+O(x3).\begin{align*} \ln(1+u)=&\,\ln\frac32+\ln\bigg(1+\frac x3+\frac{x^2}{6}+O(x^3)\bigg)\\ =&\,\ln\frac32+\frac x3+\frac{x^2}{9}+O(x^3),\\ \ln(1-u)=&\,\ln\frac12+\ln\bigg(1-x-\frac{x^2}{2}+O(x^3)\bigg)\\ =&\,\ln\frac12-x-x^2+O(x^3). \end{align*}

Since tanh⁡−1u=12(ln⁡(1+u)−ln⁡(1−u))\tanh^{-1}u=\frac12\big(\ln(1+u)-\ln(1-u)\big), taking half the difference gives

f(x)= 12ln⁡3+12(13+1)x +12(19+1)x2+O(x3)= 12ln⁡3+23x+59x2+O(x3).\begin{align*} f(x)=&\,\frac12\ln3+\frac12\bigg(\frac13+1\bigg)x\\ &\,\hspace{2pt}+\frac12\bigg(\frac19+1\bigg)x^2+O(x^3)\\ =&\,\frac12\ln3+\frac23x+\frac59x^2+O(x^3). \end{align*}

Thus a=3a=3, b=23b=\frac23, and c=59c=\frac59.