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CIE 9231 2024 November Paper 21 Q6

A Level / CIE / FP2

CIE 9231 2024 Nov Paper 21 Paper · Question 6

题目

Problem

The diagram shows the curve with equation y=(12)xy = \left(\frac{1}{2}\right)^x for 0≤x≤10 \le x \le 1, together with a set of NN rectangles each of width 1N\frac{1}{N}.

(a) By considering the sum of the areas of these rectangles, show that

∫01(12)x dx>LN,\int_0^1 \left(\frac{1}{2}\right)^x\,\mathrm{d}x > L_N,

where

LN=12N(21N−1).L_N = \frac{1}{2N\left(2^{\frac{1}{N}} - 1\right)}.
[4]

(b) Use a similar method to find, in terms of NN, an upper bound UNU_N for

∫01(12)x dx.\int_0^1 \left(\frac{1}{2}\right)^x\,\mathrm{d}x.
[4]

(c) Find the least value of NN such that UN−LN≤10−3U_N - L_N \le 10^{-3}.

[2]

(d) Given that

∫01(12)x dx=12ln⁡2,\int_0^1 \left(\frac{1}{2}\right)^x\,\mathrm{d}x = \frac{1}{2\ln 2},

use the value of NN found in part (c) to find upper and lower bounds for ln⁡2\ln 2.

[4]
题目中文翻译

图中显示曲线 y=(12)xy = \left(\frac{1}{2}\right)^x，其中 0≤x≤10 \le x \le 1，以及一组宽度均为 1N\frac{1}{N} 的 NN 个矩形。

(a) 通过考虑这些矩形的面积之和，证明

∫01(12)x dx>LN\int_0^1 \left(\frac{1}{2}\right)^x\,\mathrm{d}x > L_N

其中

LN=12N(21N−1)L_N = \frac{1}{2N\left(2^{\frac{1}{N}} - 1\right)}
(4)

(b) 使用类似方法，求

∫01(12)x dx\int_0^1 \left(\frac{1}{2}\right)^x\,\mathrm{d}x

的一个关于 NN 的上界 UNU_N。

(4)

(c) 求使 UN−LN≤10−3U_N - L_N \le 10^{-3} 的最小 NN 值。

(2)

(d) 已知

∫01(12)x dx=12ln⁡2\int_0^1 \left(\frac{1}{2}\right)^x\,\mathrm{d}x = \frac{1}{2\ln 2}

使用 (c) 部分求得的 NN 值，求 ln⁡2\ln 2 的上下界。

(4)
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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题图取右端点的高度。曲线递减，所以这些矩形的面积和严格小于积分。面积构成等比数列，可用求和公式。

答题过程

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Let r=2−1/Nr=2^{-1/N}. The right-endpoint rectangles lie below the decreasing curve on the interior of each interval. Hence their area sum satisfies

LN= 1N(r+r2+r3+⋯+rN−1+rN)= r(1−rN)N(1−r)= 12N(r−1−1)= 12N(21/N−1)< ∫01(1/2)x dx,\begin{align*} L_N=&\,\frac1N(r+r^2+r^3+\cdots+r^{N-1}+r^N)\\[4mm] =&\,\frac{r(1-r^N)}{N(1-r)}\\[4mm] =&\,\frac{1}{2N(r^{-1}-1)}\\[4mm] =&\,\frac{1}{2N(2^{1/N}-1)}\\[4mm] <&\,\int_0^1(1/2)^x\,\mathrm{d}x, \end{align*}

where rN=12r^N=\frac12. The displayed terms indicate the pattern of NN rectangles; for small NN, the sum contains each of its NN terms once.

(b)

解法一

思路

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改用左端点的高度。与 (a) 的等比和相比，每一项都除以公比 rr。

答题过程

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The left-endpoint rectangles lie above the decreasing curve, so their area sum gives the upper bound

UN= 1N(1+r+r2+⋯+rN−2+rN−1)= 1−rNN(1−r)= 21/N2N(21/N−1).\begin{align*} U_N=&\,\frac1N(1+r+r^2+\cdots+r^{N-2}+r^{N-1})\\[4mm] =&\,\frac{1-r^N}{N(1-r)}\\[4mm] =&\,\boxed{\frac{2^{1/N}}{2N(2^{1/N}-1)}}. \end{align*}

(c)

解法一

思路

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上下界相减后，分子消去同一个指数差。再解不等式并取最小正整数。

答题过程

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We have

UN−LN=21/N−12N(21/N−1)=12N.U_N-L_N=\frac{2^{1/N}-1}{2N(2^{1/N}-1)} =\frac1{2N}.

The condition 12N≤10−3\frac1{2N}\le10^{-3} gives N≥500N\ge500, so the least value is 500\boxed{500}.

(d)

解法一

思路

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将给定积分夹在 LNL_N 和 UNU_N 之间。三项都为正，取倒数时不等号反向。

答题过程

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From LN<1/(2ln⁡2)<UNL_N<1/(2\ln2)<U_N we obtain

12UN<ln⁡2<12LN.\frac1{2U_N}<\ln2<\frac1{2L_N}.

Putting N=500N=500 gives the exact bounds

500(1−2−1/500)<ln⁡2<500(21/500−1).500(1-2^{-1/500})<\ln2<500(2^{1/500}-1).

The bounds are approximately 0.69266694950.6926669495 and 0.69362785570.6936278557, respectively. Rounding outward gives 0.69266<ln⁡2<0.693630.69266<\ln2<0.69363.