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P4 2021 Oct Q4

A Level / Edexcel / P4

IAL 2021 Oct Paper · Question 4

题目

Problem

 f(x)=1−4x2 ∣x∣<12\begin{align*} &\,\mathrm{f}(x)=\sqrt{1-4x^2}\\ &\,|x|<\frac{1}{2} \end{align*}

(a) Find, in ascending powers of xx, the first four non-zero terms of the binomial expansion of f(x)\mathrm{f}(x). Give each coefficient in simplest form.

(4)

(b) By substituting x=14x=\dfrac{1}{4} into the binomial expansion of f(x)\mathrm{f}(x), obtain an approximation for 3\sqrt{3}

Give your answer to 4 decimal places.

(2)
题目中文翻译  f(x)=1−4x2 ∣x∣<12\begin{align*} &\,\mathrm{f}(x)=\sqrt{1-4x^2}\\ &\,|x|<\frac{1}{2} \end{align*}

(a) 求 f(x)\mathrm{f}(x) 的二项展开式中按 xx 的升幂排列的前四个非零项，并将每个系数写成最简形式。

(4)

(b) 将 x=14x=\dfrac{1}{4} 代入 f(x)\mathrm{f}(x) 的二项展开式，求 3\sqrt{3} 的近似值。

答案保留 4 位小数。

(2)

解答

(a)

解法一

思路

展开

把函数写成 (1+u)1/2(1+u)^{1/2}，其中 u=−4x2u=-4x^2，再使用二项展开至 u3u^3。由于 uu 含 x2x^2，前三次幂会依次产生 x2,x4,x6x^2,x^4,x^6。

答题过程

展开

Using the binomial expansion,

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

Taking n=12n=\frac12 and u=−4x2u=-4x^2,

 (1−4x2)1/2= 1+12(−4x2) +12(−12)2!(−4x2)2 +12(−12)(−32)3! ⋅(−4x2)3+⋯=  1−2x2 −2x4−4x6+⋯.\begin{align*} &\,\big(1-4x^2\big)^{1/2}\\ =&\,1+\frac12(-4x^2)\\ &\,+\frac{\frac12(-\frac12)}{2!} (-4x^2)^2\\ &\,+\frac{\frac12(-\frac12)(-\frac32)}{3!} \\ &\,\hspace{2pt}\cdot(-4x^2)^3+\cdots\\ =&\,\boxed{\begin{aligned} &\,1-2x^2\\ &\,-2x^4-4x^6+\cdots \end{aligned}}. \end{align*}

解法二

思路

展开

先因式分解 1−4x2=(1−2x)(1+2x)1-4x^2=(1-2x)(1+2x)，分别展开两个平方根，再相乘并合并同次幂项。奇次幂会互相抵消。

答题过程

展开

We may write

 1−4x2= (1−2x)1/2(1+2x)1/2.\begin{align*} &\,\sqrt{1-4x^2}\\ =&\,(1-2x)^{1/2}(1+2x)^{1/2}. \end{align*}

Expanding each factor to the required order gives

 (1−2x)1/2= 1−x−12x2 −12x3−58x4 −78x5−2116x6+⋯\begin{align*} &\,(1-2x)^{1/2}\\ =&\,1-x-\frac12x^2\\ &\,\hspace{2pt}-\frac12x^3-\frac58x^4\\ &\,\hspace{4pt}-\frac78x^5-\frac{21}{16}x^6+\cdots \end{align*}

and

 (1+2x)1/2= 1+x−12x2 +12x3−58x4 +78x5−2116x6+⋯ .\begin{align*} &\,(1+2x)^{1/2}\\ =&\,1+x-\frac12x^2\\ &\,\hspace{2pt}+\frac12x^3-\frac58x^4\\ &\,\hspace{4pt}+\frac78x^5-\frac{21}{16}x^6+\cdots. \end{align*}

Multiplying the two series and collecting like powers, all odd-power terms cancel, giving

 1−4x2= 1−2x2 −2x4−4x6+⋯.\boxed{ \begin{aligned} &\,\sqrt{1-4x^2}\\ =&\,1-2x^2\\ &\,\hspace{2pt}-2x^4-4x^6+\cdots \end{aligned} }.

(b)

解法一

思路

展开

承接 (a)，代入 x=14x=\frac14。此时 f(14)=32\mathrm{f}(\frac14)=\frac{\sqrt3}{2}，所以将展开式的近似值乘以 22 才是 3\sqrt3 的近似值。

答题过程

展开

When x=14x=\frac14,

 f(14)= 1−4(14)2= 32.\begin{align*} &\,\mathrm{f}\bigg(\frac14\bigg)\\ =&\,\sqrt{1-4\bigg(\frac14\bigg)^2}\\ =&\,\frac{\sqrt3}{2}. \end{align*}

Using the expansion from part (a),

 32≈ 1−2(14)2 −2(14)4−4(14)6.\begin{align*} &\,\frac{\sqrt3}{2}\\ \approx&\,1-2\bigg(\frac14\bigg)^2\\ &\,-2\bigg(\frac14\bigg)^4 -4\bigg(\frac14\bigg)^6. \end{align*}

Therefore,

 3≈ 2[ 1−18 −1128−11024]= 1.732421875.\begin{align*} &\,\sqrt3\\ \approx&\,2\bigg[ \begin{aligned} &\,1-\frac18\\ &\,-\frac{1}{128}-\frac{1}{1024} \end{aligned} \bigg]\\ =&\,1.732421875. \end{align*}

Hence, to four decimal places,

3≈1.7324.\boxed{\sqrt3\approx1.7324}.