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P4 2024 Oct Q1

A Level / Edexcel / P4

IAL 2024 Oct Paper · Question 1

题目

Problem

(a) Find the first 4 terms of the binomial expansion, in ascending powers of xx, of

(8−3x)−13∣x∣<83(8-3x)^{-\frac{1}{3}}\qquad |x|<\frac{8}{3}

giving each coefficient as a simplified fraction.

(4)

(b) Use the answer from part (a) with x=23x=\dfrac{2}{3} to find a rational approximation to 63\sqrt[3]{6}.

(2)
题目中文翻译

(a) 求

(8−3x)−13∣x∣<83(8-3x)^{-\frac{1}{3}}\qquad |x|<\frac{8}{3}

按 xx 的升幂展开时的前 4 项,并将每个系数化为最简分数。

(4)

(b) 利用 (a) 的结果并取 x=23x=\dfrac{2}{3},求 63\sqrt[3]{6} 的一个有理近似值。

(2)

解答

(a)

解法一

思路

展开

先提出常数 88,将表达式写成标准二项式形式。计算并简化系数后代入,最后乘回 1/21/2。

答题过程

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First factor out 88:

(8−3x)−1/3=12(1+u)−1/3(8-3x)^{-1/3}=\frac12(1+u)^{-1/3}

where u=−3x/8u=-3x/8. The binomial formula is

 (1+u)n=1+nu +n(n−1)2u2 +n(n−1)(n−2)6u3+⋯\begin{align*} &\,(1+u)^n=1+nu\\ &\,\hspace{2pt}+\frac{n(n-1)}2u^2\\ &\,\hspace{4pt}+\frac{n(n-1)(n-2)}6u^3+\cdots \end{align*}

For n=−1/3n=-1/3, the second- and third-power coefficients are

n(n−1)2=29\frac{n(n-1)}2=\frac29

and

n(n−1)(n−2)6= (−1)(−4)(−7)27×6= −1481.\begin{align*} \frac{n(n-1)(n-2)}6 =&\,\frac{(-1)(-4)(-7)}{27\times6}\\ =&\,-\frac{14}{81}. \end{align*}

Hence

 (1+u)−1/3=1−13u +29u2−1481u3+⋯\begin{align*} &\,(1+u)^{-1/3}=1-\frac13u\\ &\,\hspace{2pt}+\frac29u^2-\frac{14}{81}u^3+\cdots \end{align*}

Substituting u=−3x/8u=-3x/8 gives

 (1−3x8)−1/3 =1+x8+x232 +7x3768+⋯\begin{align*} &\,\bigg(1-\frac{3x}{8}\bigg)^{-1/3}\\ &\,=1+\frac{x}{8}+\frac{x^2}{32}\\ &\,\hspace{2pt}+\frac{7x^3}{768}+\cdots \end{align*}

Multiply each term by 1/21/2. The first four terms are

 12+x16 +x264+7x31536.\boxed{ \begin{align*} &\,\frac12+\frac{x}{16}\\ &\,\hspace{2pt}+\frac{x^2}{64}+\frac{7x^3}{1536} \end{align*} }.

(b)

解法一

思路

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代入 x=2/3x=2/3 得到的是 6−1/36^{-1/3} 的近似值,因此通分求和后还要取倒数。

答题过程

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When x=2/3x=2/3, 8−3x=68-3x=6. Thus part (a) approximates 1/631/\sqrt[3]{6}, not 63\sqrt[3]{6} itself.

The last three terms become

x16= 124x264= 11447x31536= 75184.\begin{align*} \frac{x}{16}=&\,\frac1{24}\\ \frac{x^2}{64}=&\,\frac1{144}\\ \frac{7x^3}{1536}=&\,\frac7{5184}. \end{align*}

Therefore

163≈ 12+124 +1144+75184= 2592+216+36+75184= 28515184.\begin{align*} \frac1{\sqrt[3]{6}} \approx&\,\frac12+\frac1{24}\\ &\,\hspace{2pt}+\frac1{144}+\frac7{5184}\\ =&\,\frac{2592+216+36+7}{5184}\\ =&\,\frac{2851}{5184}. \end{align*}

Taking the reciprocal gives the rational approximation

63≈51842851.\boxed{\sqrt[3]{6}\approx\frac{5184}{2851}}.