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CIE 9709 2025 June Paper 13 Q4

A Level / CIE / P1

CIE 9709 2025 June Paper 13 Paper · Question 4

题目

Problem

(a) Find the first three terms in the expansion of (2−32x)5\left(2 - \frac{3}{2}x\right)^5 in ascending powers of xx.

[3]

(b) Use your answer to part (a), with a suitable value of xx, to find an approximation to 1.98551.985^5.

[3]
题目中文翻译

(a) 求 (2−32x)5\left(2 - \frac{3}{2}x\right)^5 按 xx 的升幂展开式中的前三项。

[3]

(b) 利用你在 (a) 中的答案,并选取合适的 xx 值,求 1.98551.985^5 的近似值。

[3]

解答

(a)

解法一

思路

展开

按二项式定理取 xx 的零次、一次和二次项。每个 xx 都来自一个 −3x/2-3x/2。

答题过程

展开

The first three terms are

 25+(51)24(−32x) +(52)23(−32x)2= 32−120x+180x2.\begin{align*} &\,2^5+\binom51 2^4\Big(-\frac32x\Big)\\ &\,\hspace{2pt}+\binom52 2^3\Big(-\frac32x\Big)^2\\ =&\,32-120x+180x^2. \end{align*}

(b)

解法一

思路

展开

让括号里的 2−3x/22-3x/2 等于 1.9851.985。所得的 xx 很小,可用 (a) 的三项作近似。

答题过程

展开

Solving 2−32x=1.9852-\frac32x=1.985 gives x=0.01x=0.01. Substitution into the first three terms gives

1.9855≈32−120(0.01)+180(0.01)2=30.818.1.985^5\approx32-120(0.01)+180(0.01)^2=30.818.