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CIE 9231 2025 June Paper 22 Q4

A Level / CIE / FP2

CIE 9231 2025 June Paper 22 Paper · Question 4

题目

Problem

The diagram shows the curve with equation y=1xexy = \frac{1}{\sqrt{x}}e^{\sqrt{x}} for x≥1x \ge 1, together with a set of n−1n - 1 rectangles of unit width.

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

∑r=1n1rer< (2+1n)en−2e.\begin{align*} \sum_{r = 1}^{n} \frac{1}{\sqrt{r}}e^{\sqrt{r}} <&\, \bigg(2 + \frac{1}{\sqrt{n}}\bigg)e^{\sqrt{n}} - 2e. \end{align*}
[5]

(b) Use a similar method to find, in terms of nn, a lower bound for ∑r=1n1rer\sum_{r = 1}^{n} \frac{1}{\sqrt{r}}e^{\sqrt{r}}.

[4]
题目中文翻译

图中显示曲线 y=1xexy = \frac{1}{\sqrt{x}}e^{\sqrt{x}},其中 x≥1x \ge 1,以及一组宽度为 11 的 n−1n - 1 个矩形。

(a) 通过考虑这些矩形面积之和,证明

∑r=1n1rer< (2+1n)en−2e.\begin{align*} \sum_{r = 1}^{n} \frac{1}{\sqrt{r}}e^{\sqrt{r}} <&\, \bigg(2 + \frac{1}{\sqrt{n}}\bigg)e^{\sqrt{n}} - 2e. \end{align*}
(5)

(b) 使用类似方法,求用 nn 表示的 ∑r=1n1rer\sum_{r = 1}^{n} \frac{1}{\sqrt{r}}e^{\sqrt{r}} 的一个下界。

(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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题图的矩形取左端点。先确认函数递增,因此图中矩形面积和小于积分。这个和只到 n−1n-1,最后要补上第 nn 项。

答题过程

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Let f(x)=ex/xf(x)=e^{\sqrt x}/\sqrt x. For x>1x>1,

f′(x)=ex(x−1)2x3/2>0.f'(x)=\frac{e^{\sqrt x}(\sqrt x-1)}{2x^{3/2}}>0.

With u=xu=\sqrt x, we have dx=2u du\mathrm{d}x=2u\,\mathrm{d}u, so

∫1nf(x) dx=2∫1neu du=2en−2e.\int_1^n f(x)\,\mathrm{d}x=2\int_1^{\sqrt n}e^u\,\mathrm{d}u =2e^{\sqrt n}-2e.

The left-endpoint rectangles on [1,n][1,n], for n≥2n\ge2, have total area

f(1)+f(2)+f(3)+⋯+f(n−2)+f(n−1).f(1)+f(2)+f(3)+\cdots+f(n-2)+f(n-1).

Comparing with the integral gives

∑r=1n−1f(r)<∫1nf(x) dx=[2ex]1n=2en−2e.\sum_{r=1}^{n-1}f(r)<\int_1^n f(x)\,\mathrm{d}x =[2e^{\sqrt x}]_1^n=2e^{\sqrt n}-2e.

Adding f(n)f(n) to both sides gives

∑r=1nerr<(2+1n)en−2e.\sum_{r=1}^{n}\frac{e^{\sqrt r}}{\sqrt r} <\bigg(2+\frac1{\sqrt n}\bigg)e^{\sqrt n}-2e.

(b)

解法一

思路

展开

改用右端点,面积和大于同一个积分。右端点从 22 开始,加回第 11 项即可。

答题过程

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The right-endpoint rectangles have heights f(2),f(3),f(4),…,f(n−1),f(n)f(2),f(3),f(4),\ldots,f(n-1),f(n). They give ∑r=2nf(r)>∫1nf(x) dx\sum_{r=2}^n f(r)>\int_1^n f(x)\,\mathrm{d}x. Adding f(1)=ef(1)=e gives

∑r=1nf(r)>2en−2e+e=2en−e.\sum_{r=1}^n f(r)>2e^{\sqrt n}-2e+e =\boxed{2e^{\sqrt n}-e}.