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IAL 2021 Oct Q4

A Level / Edexcel / P4

IAL 2021 Oct Paper · Question 4

题目

Problem

f(x)=1−4x2∣x∣<12\mathrm{f}(x)=\sqrt{1-4x^2}\qquad |x|<\frac{1}{2}

(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\mathrm{f}(x)=\sqrt{1-4x^2}\qquad |x|<\frac{1}{2}

(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。

(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。

解答

(a)

解法一

思路

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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

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Using the binomial expansion,

(1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯ .(1+u)^n =1+nu+\frac{n(n-1)}{2!}u^2 +\frac{n(n-1)(n-2)}{3!}u^3+\cdots.

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!} (-4x^2)^3+\cdots\\ =&\,\boxed{1-2x^2-2x^4-4x^6+\cdots}. \end{align*}

解法二

思路

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官方替代路线先因式分解 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.\sqrt{1-4x^2} =(1-2x)^{1/2}(1+2x)^{1/2}.

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-\frac12x^3\\ &\,-\frac58x^4-\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+\frac12x^3\\ &\,-\frac58x^4+\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{ \sqrt{1-4x^2} =1-2x^2-2x^4-4x^6+\cdots }.

(b)

解法一

思路

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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

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When x=14x=\frac14,

f(14)=1−4(14)2=32.\mathrm{f}\bigg(\frac14\bigg) =\sqrt{1-4\bigg(\frac14\bigg)^2} =\frac{\sqrt3}{2}.

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[ 1-\frac18-\frac{1}{128}-\frac{1}{1024} \bigg]\\ =&\,1.732421875. \end{align*}

Hence, to four decimal places,

3≈1.7324.\boxed{\sqrt3\approx1.7324}.