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CIE 9231 2025 Nov Paper 24 Q4

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 24 Paper · Question 4

题目

Problem

The diagram shows the curve with equation y=12(3x)y = \frac{1}{2}(3^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

∫0112(3x) dx<UN,\int_0^1 \frac{1}{2}(3^x)\,\mathrm{d}x < U_N,

where

UN=31NN(31N−1).U_N = \frac{3^{\frac{1}{N}}}{N\big(3^{\frac{1}{N}} - 1\big)}.
[4]

(b) Use a similar method to find, in terms of NN, a lower bound LNL_N for ∫0112(3x) dx\int_0^1 \frac{1}{2}(3^x)\,\mathrm{d}x.

[4]

(c) By simplifying UN−LNU_N - L_N, show that lim⁡N→∞(UN−LN)=0\lim_{N \to \infty}(U_N - L_N) = 0.

[2]
题目中文翻译

图中显示曲线 y=12(3x)y = \frac{1}{2}(3^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

∫0112(3x) dx<UN,\int_0^1 \frac{1}{2}(3^x)\,\mathrm{d}x < U_N,

其中

UN=31NN(31N−1).U_N = \frac{3^{\frac{1}{N}}}{N\big(3^{\frac{1}{N}} - 1\big)}.
(4)

(b) 使用类似方法，求用 NN 表示的 ∫0112(3x) dx\int_0^1 \frac{1}{2}(3^x)\,\mathrm{d}x 的一个下界 LNL_N。

(4)

(c) 通过化简 UN−LNU_N - L_N，证明 lim⁡N→∞(UN−LN)=0\lim_{N \to \infty}(U_N - L_N) = 0。

(2)
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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题图按右端点取高度。指数函数递增，因此矩形面积和大于积分。相邻矩形的高度之比固定，用等比数列求和。

答题过程

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Let r=31/Nr=3^{1/N}. The curve is strictly increasing, so the area sum of the right-endpoint rectangles is a strict upper bound. It is

UN= 12N(r+r2+r3+⋯+rN−1+rN)= 12N∑j=1Nrj= r(rN−1)2N(r−1)= 31/NN(31/N−1),\begin{align*} U_N=&\,\frac1{2N}(r+r^2+r^3+\cdots+r^{N-1}+r^N)\\ =&\,\frac1{2N}\sum_{j=1}^N r^j\\ =&\,\frac{r(r^N-1)}{2N(r-1)}\\ =&\,\frac{3^{1/N}}{N(3^{1/N}-1)}, \end{align*}

using rN=3r^N=3.

(b)

解法一

思路

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改用左端点的高度，得到首项为 12N\frac1{2N} 的等比面积和。

答题过程

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The left-endpoint rectangles lie below the curve on the interior of every interval. Their area sum is

LN= 12N(1+r+r2+⋯+rN−2+rN−1)= 12N∑j=0N−1rj= rN−12N(r−1)= 1N(31/N−1).\begin{align*} L_N=&\,\frac1{2N}(1+r+r^2+\cdots+r^{N-2}+r^{N-1})\\ =&\,\frac1{2N}\sum_{j=0}^{N-1}r^j\\ =&\,\frac{r^N-1}{2N(r-1)}\\ =&\,\boxed{\frac{1}{N(3^{1/N}-1)}}. \end{align*}

(c)

解法一

思路

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两式通分后，分子正好消去公分母中的 31/N−13^{1/N}-1。

答题过程

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We obtain

UN−LN=31/N−1N(31/N−1)=1N,U_N-L_N=\frac{3^{1/N}-1}{N(3^{1/N}-1)}=\frac1N,

which tends to zero as N→∞N\to\infty.