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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 0x10 \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)xdx>LN,\int_0^1 \left(\frac{1}{2}\right)^x\,\mathrm{d}x > L_N,

where

LN=12N(21N1).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)xdx.\int_0^1 \left(\frac{1}{2}\right)^x\,\mathrm{d}x.
[4]

(c) Find the least value of NN such that UNLN103U_N - L_N \le 10^{-3}.

[2]

(d) Given that

01(12)xdx=12ln2,\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 ln2\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 0x10 \le x \le 1,以及一组宽度均为 1N\frac{1}{N}NN 个矩形。

(a) 通过考虑这些矩形的面积之和,证明

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

其中

LN=12N(21N1)L_N = \frac{1}{2N\left(2^{\frac{1}{N}} - 1\right)}

(b) 使用类似方法,求

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

的一个关于 NN 的上界 UNU_N

(c) 求使 UNLN103U_N - L_N \le 10^{-3} 的最小 NN 值。

(d) 已知

01(12)xdx=12ln2\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 ln2\ln 2 的上下界。

解答