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CIE 9231 2021 November Paper 21 Q4

A Level / CIE / FP2

CIE 9231 2021 Nov Paper 21 Paper · Question 4

题目

Problem

The diagram shows the curve with equation y=lnxx2y=\dfrac{\ln x}{x^2} for x2x\ge2, together with a set of (N2)(N-2) rectangles of unit width.

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

r=1Nlnrr2<2+3ln241+lnNN.\begin{aligned} \sum_{r=1}^{N}\frac{\ln r}{r^2}<&\,\frac{2+3\ln2}{4}\\ &\,-\frac{1+\ln N}{N}. \end{aligned}
[7]

(b) Use a similar method to find, in terms of NN, a lower bound for r=1Nlnrr2\displaystyle\sum_{r=1}^{N}\frac{\ln r}{r^2}.

[3]
题目中文翻译

图中显示了方程为 y=lnxx2y=\dfrac{\ln x}{x^2} 的曲线,其中 x2x\ge2,以及一组宽度为 1 的 (N2)(N-2) 个矩形。

(a) 考虑这些矩形的面积之和,证明

r=1Nlnrr2<2+3ln241+lnNN\begin{aligned} \sum_{r=1}^{N}\frac{\ln r}{r^2}<&\,\frac{2+3\ln2}{4}\\ &\,-\frac{1+\ln N}{N} \end{aligned}

(b) 使用类似方法,求出一个关于 NNr=1Nlnrr2\displaystyle\sum_{r=1}^{N}\frac{\ln r}{r^2} 的下界。

解答