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CIE 9231 2022 June Paper 22 Q4

A Level / CIE / FP2

CIE 9231 2022 June Paper 22 Paper · Question 4

题目

Problem

The diagram shows the curve with equation y=2xy=2^x for 0≤x≤10\le x\le1, together with a set of NN rectangles each of width 1N\dfrac{1}{N}.

(a) By considering the sum of the areas of these rectangles, show that ∫012x dx<UN\displaystyle\int_0^1 2^x\,\mathrm{d}x<U_N, where

UN=21NN(21N−1).U_N=\frac{2^{\frac{1}{N}}}{N\big(2^{\frac{1}{N}}-1\big)}.
(4)

(b) Use a similar method to find, in terms of NN, a lower bound LNL_N for ∫012x dx\displaystyle\int_0^1 2^x\,\mathrm{d}x.

(4)

(c) Find the least value of NN such that UN−LN<10−4U_N-L_N<10^{-4}.

(2)
题目中文翻译

图中显示了方程为 y=2xy=2^x 的曲线(0≤x≤10\le x\le1),以及一组 NN 个矩形,每个矩形的宽均为 1N\dfrac{1}{N}。

(a) 通过考虑这些矩形的面积之和,证明 ∫012x dx<UN\displaystyle\int_0^1 2^x\,\mathrm{d}x<U_N,其中

UN=21NN(21N−1).U_N=\frac{2^{\frac{1}{N}}}{N\big(2^{\frac{1}{N}}-1\big)}.
(4)

(b) 用类似的方法求 ∫012x dx\displaystyle\int_0^1 2^x\,\mathrm{d}x 的一个下界 LNL_N,用含 NN 的式子表示。

(4)

(c) 求满足 UN−LN<10−4U_N-L_N<10^{-4} 的 NN 的最小值。

(2)
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

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指数函数递增,题图取右端点,所以面积和给出严格上界。相邻矩形高度的比是 21/N2^{1/N}。

答题过程

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Let r=21/Nr=2^{1/N}, so rN=2r^N=2. The right-endpoint rectangle sum is

UN= 1N(r+r2+r3+⋯+rN−1+rN)= 1N⋅r(rN−1)r−1= 21/NN(21/N−1).\begin{align*} U_N=&\,\frac1N\big(r+r^2+r^3+\cdots+r^{N-1}+r^N\big)\\ =&\,\frac1N\cdot\frac{r(r^N-1)}{r-1}\\ =&\,\frac{2^{1/N}}{N(2^{1/N}-1)}. \end{align*}

Since 2x2^x is strictly increasing, the curve lies below the top of each rectangle on the interval interior, so ∫012x dx<UN\int_0^1 2^x\,\mathrm{d}x<U_N.

(b)

解法一

思路

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改用左端点。矩形高度从 11 开始,共 NN 项,仍可用等比求和。

答题过程

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The left-endpoint rectangles give a lower bound since 2x2^x is increasing. Their area sum is

LN= 1N(1+r+r2+⋯+rN−2+rN−1)= 1N⋅rN−1r−1= 1N(21/N−1).\begin{align*} L_N=&\,\frac1N\big(1+r+r^2+\cdots+r^{N-2}+r^{N-1}\big)\\ =&\,\frac1N\cdot\frac{r^N-1}{r-1}\\ =&\,\boxed{\frac{1}{N(2^{1/N}-1)}}. \end{align*}

The rectangles lie strictly below the curve on each interval interior, so LN<∫012x dxL_N<\int_0^1 2^x\,\mathrm{d}x.

(c)

解法一

思路

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上下界相减后约掉 21/N−12^{1/N}-1。题设是不含等号的严格不等式,取整数时需再进一位。

答题过程

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We have

UN−LN=21/N−1N(21/N−1)=1N.U_N-L_N=\frac{2^{1/N}-1}{N(2^{1/N}-1)}=\frac1N.

The inequality 1/N<10−41/N<10^{-4} gives N>10000N>10000, so the least integer is 10001\boxed{10001}.