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IAL 2023 Jan Q1

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 1

题目

Problem

Given that y=ln⁡(5+3x)y = \ln(5 + 3x)

(a) determine, in simplest form, d3ydx3\frac{\mathrm{d}^3y}{\mathrm{d}x^3}

(3)

(b) Hence determine the Maclaurin series expansion of ln⁡(5+3x)\ln(5 + 3x), in ascending powers of xx up to and including the term in x3x^3, giving each coefficient in simplest form.

(2)

(c) Hence write down the Maclaurin series expansion of ln⁡(5−3x)\ln(5 - 3x), in ascending powers of xx up to and including the term in x3x^3, giving each coefficient in simplest form.

(1)

(d) Use the answers to parts (b) and (c) to determine the first 22 non-zero terms, in ascending powers of xx, of the Maclaurin series expansion of

ln⁡(5+3x5−3x)\begin{align*} \ln\left(\frac{5 + 3x}{5 - 3x}\right) \end{align*}
(2)

解答

(a)

解法一

思路

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直接连续求三次导。每次都要乘上内层 5+3x5+3x 的导数 33。

答题过程

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Given

y=ln⁡(5+3x).\begin{align*} y=\ln(5+3x). \end{align*}

Differentiate:

dydx= 35+3x= 3(5+3x)−1.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{3}{5+3x}\\[2mm] =&\,3(5+3x)^{-1}. \end{align*}

Differentiate again:

d2ydx2= 3(−1)(5+3x)−2⋅3= −9(5+3x)−2.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,3(-1)(5+3x)^{-2}\cdot 3\\[2mm] =&\,-9(5+3x)^{-2}. \end{align*}

So

d3ydx3= −9(−2)(5+3x)−3⋅3= 54(5+3x)−3= 54(5+3x)3.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,-9(-2)(5+3x)^{-3}\cdot 3\\[2mm] =&\,54(5+3x)^{-3}\\[2mm] =&\,\frac{54}{(5+3x)^3}. \end{align*}

Therefore

d3ydx3=54(5+3x)3.\begin{align*} \boxed{\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =\frac{54}{(5+3x)^3}}. \end{align*}

(b)

解法一

思路

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Maclaurin 展开需要 x=0x=0 处的函数值和导数值：

f(x)=f(0)+xf′(0)+x22!f′′(0)+x33!f′′′(0)+⋯ .\begin{align*} f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0) +\frac{x^3}{3!}f'''(0)+\cdots. \end{align*}

答题过程

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At x=0x=0,

y(0)= ln⁡5,y′(0)= 35,y′′(0)= −925,y′′′(0)= 54125.\begin{align*} y(0)=&\,\ln5,\\[2mm] y'(0)=&\,\frac{3}{5},\\[2mm] y''(0)=&\,-\frac{9}{25},\\[2mm] y'''(0)=&\,\frac{54}{125}. \end{align*}

Using the Maclaurin series,

ln⁡(5+3x)= ln⁡5+35x+12!(−925)x2 +13!(54125)x3+⋯= ln⁡5+35x−950x2+9125x3+⋯ .\begin{align*} \ln(5+3x) =&\,\ln5+\frac{3}{5}x +\frac{1}{2!}\left(-\frac{9}{25}\right)x^2\\[2mm] &\,\hspace{2pt}+\frac{1}{3!}\left(\frac{54}{125}\right)x^3+\cdots\\[2mm] =&\,\ln5+\frac{3}{5}x-\frac{9}{50}x^2 +\frac{9}{125}x^3+\cdots. \end{align*}

Thus

ln⁡(5+3x)=ln⁡5+35x−950x2+9125x3+⋯.\begin{align*} \boxed{\ln(5+3x)=\ln5+\frac{3}{5}x-\frac{9}{50}x^2 +\frac{9}{125}x^3+\cdots}. \end{align*}

(c)

解法一

思路

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ln⁡(5−3x)\ln(5-3x) 可以由 (b) 中的 ln⁡(5+3x)\ln(5+3x) 把 xx 换成 −x-x 得到。偶次项符号不变，奇次项符号改变。

答题过程

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Replace xx by −x-x in the expansion from part (b):

ln⁡(5−3x)= ln⁡5+35(−x)−950(−x)2+9125(−x)3+⋯= ln⁡5−35x−950x2−9125x3+⋯ .\begin{align*} \ln(5-3x) =&\,\ln5+\frac{3}{5}(-x)-\frac{9}{50}(-x)^2 +\frac{9}{125}(-x)^3+\cdots\\[2mm] =&\,\ln5-\frac{3}{5}x-\frac{9}{50}x^2 -\frac{9}{125}x^3+\cdots. \end{align*}

Therefore

ln⁡(5−3x)=ln⁡5−35x−950x2−9125x3+⋯.\begin{align*} \boxed{\ln(5-3x)=\ln5-\frac{3}{5}x-\frac{9}{50}x^2 -\frac{9}{125}x^3+\cdots}. \end{align*}

(d)

解法一

思路

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先用对数性质：

ln⁡(5+3x5−3x)=ln⁡(5+3x)−ln⁡(5−3x).\begin{align*} \ln\left(\frac{5+3x}{5-3x}\right) =\ln(5+3x)-\ln(5-3x). \end{align*}

然后把 (b) 和 (c) 的级数相减。常数项和 x2x^2 项会抵消，留下前两个非零项。

答题过程

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Using

ln⁡(5+3x5−3x)=ln⁡(5+3x)−ln⁡(5−3x),\begin{align*} \ln\left(\frac{5+3x}{5-3x}\right) =\ln(5+3x)-\ln(5-3x), \end{align*}

we get

ln⁡(5+3x5−3x)= (ln⁡5+35x−950x2+9125x3+⋯ ) −(ln⁡5−35x−950x2−9125x3+⋯ )= 65x+18125x3+⋯ .\begin{align*} \ln\left(\frac{5+3x}{5-3x}\right) =&\,\left(\ln5+\frac{3}{5}x-\frac{9}{50}x^2 +\frac{9}{125}x^3+\cdots\right)\\[2mm] &\,\hspace{2pt}-\left(\ln5-\frac{3}{5}x-\frac{9}{50}x^2 -\frac{9}{125}x^3+\cdots\right)\\[2mm] =&\,\frac{6}{5}x+\frac{18}{125}x^3+\cdots. \end{align*}

Therefore the first two non-zero terms are

65x+18125x3.\begin{align*} \boxed{\frac{6}{5}x+\frac{18}{125}x^3}. \end{align*}