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CIE 9231 2024 November Paper 22 Q6

A Level / CIE / FP2

CIE 9231 2024 Nov Paper 22 Paper · Question 6

题目

Problem

The diagram shows the curve with equation y=e1−xy = e^{1 - x} for 0≤x≤10 \le x \le 1, together with a set of nn rectangles of width 1n\frac{1}{n}.

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

∫01e1−x dx<Un,\int_0^1 e^{1 - x}\,\mathrm{d}x < U_n,

where

Un=e−1n(1−e−1n).U_n = \frac{e - 1}{n\left(1 - e^{-\frac{1}{n}}\right)}.
[4]

(b) Use a similar method to find, in terms of nn, a lower bound LnL_n for

∫01e1−x dx.\int_0^1 e^{1 - x}\,\mathrm{d}x.
[4]

(c) Show that

lim⁡n→∞(Un−Ln)=0.\lim_{n\to\infty}(U_n - L_n) = 0.
[2]

(d) Use the Maclaurin’s series for exe^x given in the list of formulae (MF19) to find the first three terms of the series expansion of z(1−e−1z)z\left(1 - e^{-\frac{1}{z}}\right), in ascending powers of 1z\frac{1}{z}, and deduce the value of

lim⁡n→∞(Un).\lim_{n\to\infty}(U_n).
[3]
题目中文翻译

图中显示曲线 y=e1−xy = e^{1 - 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) 通过考虑这些矩形面积之和，证明

∫01e1−x dx<Un\int_0^1 e^{1 - x}\,\mathrm{d}x < U_n

其中

Un=e−1n(1−e−1n)U_n = \frac{e - 1}{n\left(1 - e^{-\frac{1}{n}}\right)}
(4)

(b) 使用类似方法，求关于 nn 的积分

∫01e1−x dx\int_0^1 e^{1 - x}\,\mathrm{d}x

的下界 LnL_n。

(4)

(c) 证明

lim⁡n→∞(Un−Ln)=0\lim_{n\to\infty}(U_n - L_n) = 0
(2)

(d) 使用公式表（MF19）中给出的 exe^x 的 Maclaurin 级数，求 z(1−e−1z)z\left(1 - e^{-\frac{1}{z}}\right) 的级数展开的前三项，按 1z\frac{1}{z} 的升幂排列，并据此推出

lim⁡n→∞(Un)\lim_{n\to\infty}(U_n)

的值。

(3)
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

展开

递减曲线的左端点矩形位于曲线上方，题图正是这一组矩形。把面积和写成等比和。

答题过程

展开

Let r=e−1/nr=e^{-1/n}. The diagram uses left endpoints, and e1−xe^{1-x} is strictly decreasing, so the rectangle sum is an upper bound. It is

Un= en(1+r+r2+⋯+rn−2+rn−1)= e(1−rn)n(1−r)= e−1n(1−e−1/n),\begin{align*} U_n=&\,\frac en(1+r+r^2+\cdots+r^{n-2}+r^{n-1})\\[4mm] =&\,\frac{e(1-r^n)}{n(1-r)}\\[4mm] =&\,\frac{e-1}{n(1-e^{-1/n})}, \end{align*}

Here rn=e−1r^n=e^{-1}, and the displayed terms indicate the pattern of nn rectangles, with each term counted once. Thus ∫01e1−x dx<Un\int_0^1 e^{1-x}\,\mathrm{d}x<U_n.

(b)

解法一

思路

展开

右端点比对应的左端点多出 1n\frac1n，所以每个矩形的高度都乘 rr。这些矩形给出下界。

答题过程

展开

The right-endpoint rectangles lie below the curve on the interior of each interval. Their total area is

Ln= en(r+r2+r3+⋯+rn−1+rn)= rUn= e−1n(e1/n−1).\begin{align*} L_n=&\,\frac en(r+r^2+r^3+\cdots+r^{n-1}+r^n)\\[4mm] =&\,rU_n\\[4mm] =&\,\boxed{\frac{e-1}{n(e^{1/n}-1)}}. \end{align*}

(c)

解法一

思路

展开

由 Ln=rUnL_n=rU_n，上下界之差正好消去分母中的 1−r1-r。

答题过程

展开

Since Ln=rUnL_n=rU_n,

Un−Ln=(1−r)Un=e−1n⟶0.U_n-L_n=(1-r)U_n=\frac{e-1}{n}\longrightarrow0.

(d)

解法一

思路

展开

将 −1/z-1/z 代入 exe^x 的 Maclaurin 级数。因为最后要乘 zz，所以指数展开必须保留到 z−3z^{-3}。

答题过程

展开

The standard Maclaurin series is

ev=1+v+v22!+v33!+O(v4).e^v=1+v+\frac{v^2}{2!}+\frac{v^3}{3!}+O(v^4).

Putting v=−1/zv=-1/z, for large positive zz,

e−1/z=1−1z+12z2−16z3+O(z−4).e^{-1/z}=1-\frac1z+\frac1{2z^2}-\frac1{6z^3}+O(z^{-4}).

Hence

z(1−e−1/z)=1−12z+16z2+O(z−3).z(1-e^{-1/z})=1-\frac1{2z}+\frac1{6z^2}+O(z^{-3}).

The denominator of UnU_n is this expression with z=nz=n, so it tends to 11. Therefore lim⁡n→∞Un=e−1\boxed{\lim_{n\to\infty}U_n=e-1}.