题目

Problem

The diagram shows the curve $y=\dfrac{1}{\sqrt{x^2+2x}}$ for $x\gt0$ , together with a set of $(n-1)$ rectangles of unit width.

By considering the sum of the areas of these rectangles, show that

\begin{align*} \sum_{r=1}^{n}\frac{1}{\sqrt{r^2+2r}}<&\,\ln\big(n+1+\sqrt{n^2+2n}\big)\\ &\,\hspace{2pt}+\frac{1}{3}\sqrt{3}-\ln\big(2+\sqrt{3}\big). \end{align*}

(10)

题目中文翻译

图中显示了曲线 $y=\dfrac{1}{\sqrt{x^2+2x}}$ （ $x\gt0$ ），以及一组宽均为一个单位的 $(n-1)$ 个矩形。

通过考虑这些矩形的面积之和，证明

\begin{align*} \sum_{r=1}^{n}\frac{1}{\sqrt{r^2+2r}}<&\,\ln\big(n+1+\sqrt{n^2+2n}\big)\\ &\,\hspace{2pt}+\frac{1}{3}\sqrt{3}-\ln\big(2+\sqrt{3}\big). \end{align*}

(10)

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

解法一

思路

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曲线递减，题图取右端点，面积和从 $r=2$ 开始。先与 $[1,n]$ 上的积分比较，配方后用反双曲余弦求积分，最后补回 $r=1$ 项。

答题过程

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Let $f(x)=(x^2+2x)^{-1/2}$ for $x>0$ . It is strictly decreasing because $x^2+2x$ is strictly increasing. The right-endpoint rectangles of width one on $[1,n]$ , with $n>1$ , give

\begin{align*} &\,f(2)+f(3)+f(4)+\cdots+f(n-1)+f(n)\\ =&\,\frac1{\sqrt8}+\frac1{\sqrt{15}}+\frac1{\sqrt{24}}+\cdots\\ &\,\hspace{2pt}+\frac1{\sqrt{(n-1)^2+2(n-1)}}+\frac1{\sqrt{n^2+2n}}\\ =&\,\sum_{r=2}^{n}f(r)<\int_1^n f(x)\,\mathrm{d}x. \end{align*}

The first and last terms in the sum are $f(2)$ and $f(n)$ . Since $x^2+2x=(x+1)^2-1$ , put $u=x+1$ , so $\mathrm{d}u=\mathrm{d}x$ and $u$ runs from $2$ to $n+1$ . The standard antiderivative is $\operatorname{arcosh}u=\ln(u+\sqrt{u^2-1})$ . Thus

\begin{align*} \int_1^n f(x)\,\mathrm{d}x =&\,[\operatorname{arcosh}(x+1)]_1^n\\ =&\,\big[\ln(x+1+\sqrt{x^2+2x})\big]_1^n\\ =&\,\ln(n+1+\sqrt{n^2+2n})-\ln(2+\sqrt3). \end{align*}

Adding $f(1)=1/\sqrt3=\sqrt3/3$ gives the required strict bound.

CIE 9231 2022 November Paper 22 Q6

题目

解答

解法一

思路

答题过程