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CIE 9231 2023 June Paper 23 Q1

A Level / CIE / FP2

CIE 9231 2023 June Paper 23 Paper · Question 1

题目

Problem

(a) Find the Maclaurin series for sin⁡−1x\sin^{-1}x up to and including the term in x3x^3.

(5)

(b) Deduce an approximation to ∫01511−u2 du\displaystyle\int_0^{\frac{1}{5}}\frac{1}{\sqrt{1-u^2}}\,\mathrm{d}u, giving your answer as a fraction.

(1)
题目中文翻译

(a) 求 sin⁡−1x\sin^{-1}x 的麦克劳林级数,展开至并包括 x3x^3 项。

(5)

(b) 由此推导出 ∫01511−u2 du\displaystyle\int_0^{\frac{1}{5}}\frac{1}{\sqrt{1-u^2}}\,\mathrm{d}u 的近似值,将答案写成分数。

(1)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

从 arcsin⁡x\arcsin x 的导数出发,用乘积法则求到三阶。代入 x=0x=0 后,一次项和三次项留下。

答题过程

展开

Let f(x)=arcsin⁡xf(x)=\arcsin x. Then

f′(x)= (1−x2)−1/2,f′′(x)= x(1−x2)−3/2,f′′′(x)= (1−x2)−3/2+3x2(1−x2)−5/2.\begin{align*} f'(x)=&\,(1-x^2)^{-1/2},\\ f''(x)=&\,x(1-x^2)^{-3/2},\\ f'''(x)=&\,(1-x^2)^{-3/2}+3x^2(1-x^2)^{-5/2}. \end{align*}

At x=0x=0, f(0)=0f(0)=0, f′(0)=1f'(0)=1, f′′(0)=0f''(0)=0 and f′′′(0)=1f'''(0)=1. Using

f(x)=f(0)+f′(0)x+f′′(0)2!x2+f(3)(0)3!x3+⋯ ,f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\cdots,

the terms through degree three are x+x36\boxed{x+\frac{x^3}{6}}.

解法二

思路

展开

也可令 y=arcsin⁡xy=\arcsin x,用 sin⁡y=x\sin y=x 隐式求导。这样各阶导数都可先用 yy 表示,再在 y=0y=0 处取值。

答题过程

展开

From sin⁡y=x\sin y=x, implicit differentiation gives y′=sec⁡yy'=\sec y. Using the chain rule twice gives

y′′= (sec⁡ytan⁡y)y′=sec⁡2ytan⁡y,y′′′= (2sec⁡2ytan⁡2y+sec⁡4y)y′= 2sec⁡3ytan⁡2y+sec⁡5y.\begin{align*} y''=&\,(\sec y\tan y)y'=\sec^2y\tan y,\\ y'''=&\,\big(2\sec^2y\tan^2y+\sec^4y\big)y'\\ =&\,2\sec^3y\tan^2y+\sec^5y. \end{align*}

At x=0x=0, y=0y=0, so these derivatives have values 1,0,11,0,1 respectively. The expansion through the term in x3x^3 is x+x3/6x+x^3/6.

(b)

解法一

思路

展开

被积函数是 arcsin⁡u\arcsin u 的导数。用 (a) 在上限 u=1/5u=1/5 处近似,减去零点的值。

答题过程

展开

Since arcsin⁡0=0\arcsin0=0, the integral equals arcsin⁡(1/5)\arcsin(1/5). Using (a), its approximation is

15+16(15)3=151750.\frac15+\frac16\bigg(\frac15\bigg)^3=\boxed{\frac{151}{750}}.