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IAL 2021 June Q1

A Level / Edexcel / P4

IAL 2021 June Paper · Question 1

题目

Problem

Given that kk is a constant and the binomial expansion of

1+kxkx<1\sqrt{1+kx}\qquad |kx|<1

in ascending powers of xx up to the term in x3x^3 is

1+18x+Ax2+Bx31+\frac{1}{8}x+Ax^2+Bx^3

(a) (i) find the value of kk,

(ii) find the value of the constant AA and the constant BB.

(5)

(b) Use the expansion to find an approximate value to 1.15\sqrt{1.15}

Show your working and give your answer to 6 decimal places.

(2)
题目中文翻译

已知 kk 是一个常数,并且

1+kxkx<1\sqrt{1+kx}\qquad |kx|<1

的二项展开式按 xx 的升幂排列,到 x3x^3 项为止可写成

1+18x+Ax2+Bx31+\frac{1}{8}x+Ax^2+Bx^3

的形式。

(a) (i) 求 kk 的值;

(ii) 求常数 AA 和常数 BB 的值。

(b) 利用该展开式求 1.15\sqrt{1.15} 的近似值。

写出计算过程,并将答案保留到小数点后 6 位。

解答

(a)(i)

解法一

思路

展开

写出 (1+kx)1/2(1+kx)^{1/2} 的二项展开式,并把 xx 项系数与题目给出的 18\frac18 比较,即可求出 kk

答题过程

展开

Using the binomial expansion,

(1+kx)1/2=1+12(kx)+12(12)2!(kx)2+12(12)(32)3!(kx)3+.\begin{align*} (1+kx)^{1/2} =&\,1+\frac12(kx)\\ &\,\hspace{2pt} +\frac{\frac12\big(-\frac12\big)}{2!} (kx)^2\\ &\,\hspace{4pt} +\frac{\frac12\big(-\frac12\big) \big(-\frac32\big)}{3!}(kx)^3\\ &\,\hspace{6pt}+\cdots. \end{align*}

The coefficient of xx is therefore k2\frac{k}{2}. Comparing this with the given coefficient,

k2=18,\frac{k}{2}=\frac18,

so

k=14.\boxed{k=\frac14}.

(a)(ii)

解法一

思路

展开

由二项展开式分别读出 x2x^2x3x^3 的系数,再代入 (a)(i) 求得的 k=14k=\frac14,即可得到 AABB

答题过程

展开

From the binomial expansion,

A=12(12)2!k2=18k2.\begin{align*} A =&\,\frac{\frac12\big(-\frac12\big)}{2!}k^2\\ =&\,-\frac18k^2. \end{align*}

Using k=14k=\frac14,

A=18(14)2=1128.\begin{align*} A =&\,-\frac18\bigg(\frac14\bigg)^2\\ =&\,\boxed{-\frac{1}{128}}. \end{align*}

Also,

B=12(12)(32)3!k3=116k3.\begin{align*} B =&\,\frac{\frac12\big(-\frac12\big) \big(-\frac32\big)}{3!}k^3\\ =&\,\frac{1}{16}k^3. \end{align*}

Hence,

B=116(14)3=11024.\begin{align*} B =&\,\frac{1}{16}\bigg(\frac14\bigg)^3\\ =&\,\boxed{\frac{1}{1024}}. \end{align*}

(b)

解法一

思路

展开

要使 1+kx=1.151+kx=1.15,代入 k=14k=\frac14 后可得 x=0.6x=0.6。把这个值代入题目给出的关于 xx 的展开式,并使用 (a)(ii) 的 A,BA,B,最后按要求保留六位小数。

答题过程

展开

Since k=14k=\frac14, set

1+14x=1.15,1+\frac14x=1.15,

which gives x=0.6x=0.6. Therefore, using the given expansion,

1.151+18(0.6)1128(0.6)2+11024(0.6)3=1.0723984375.\begin{align*} \sqrt{1.15} \approx&\,1+\frac18(0.6)\\ &\,\hspace{2pt} -\frac{1}{128}(0.6)^2\\ &\,\hspace{4pt} +\frac{1}{1024}(0.6)^3\\ =&\,1.0723984375. \end{align*}

Thus, to 6 decimal places,

1.151.072398.\boxed{\sqrt{1.15}\approx1.072398}.

解法二

思路

展开

也可以保留以 kxkx 为整体的标准二项展开式。因为 1.15=1+0.151.15=1+0.15,直接代入 kx=0.15kx=0.15,不必先求 xx

答题过程

展开

Using the expansion directly in powers of kxkx,

(1+kx)1/2=1+12(kx)18(kx)2+116(kx)3+.\begin{align*} (1+kx)^{1/2} =&\,1+\frac12(kx)\\ &\,\hspace{2pt}-\frac18(kx)^2\\ &\,\hspace{4pt}+\frac{1}{16}(kx)^3\\ &\,\hspace{6pt}+\cdots. \end{align*}

For 1.15\sqrt{1.15}, take kx=0.15kx=0.15. Since 0.15<1|0.15|<1, the expansion is valid. Hence,

1.151+12(0.15)18(0.15)2+116(0.15)3=1.07239843751.072398.\begin{align*} \sqrt{1.15} \approx&\,1+\frac12(0.15)\\ &\,\hspace{2pt}-\frac18(0.15)^2\\ &\,\hspace{4pt}+\frac{1}{16}(0.15)^3\\ =&\,1.0723984375\\ \approx&\,\boxed{1.072398}. \end{align*}