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CIE 9231 2023 November Paper 22 Q5

A Level / CIE / FP2

CIE 9231 2023 Nov Paper 22 Paper · Question 5

题目

Problem

The diagram shows part of the curve y=xsech⁡2xy=x\operatorname{sech}^2x and its maximum point MM.

(a) Show that, at MM,

2xtanh⁡x−1=02x\tanh x-1=0

and verify that this equation has a root between 0.70.7 and 0.80.8.

[4]

(b) By considering a suitable set of rectangles, use the diagram to show that

∑r=2nrsech⁡2r<ntanh⁡n+ln⁡(sech⁡n)−tanh⁡1−ln⁡(sech⁡1).\sum_{r=2}^{n}r\operatorname{sech}^2r<n\tanh n+\ln(\operatorname{sech}n)-\tanh 1-\ln(\operatorname{sech}1).
[6]
题目中文翻译

图中显示了曲线 y=xsech⁡2xy=x\operatorname{sech}^2x 的一部分及其最高点 MM。

(a) 证明在 MM 点,

2xtanh⁡x−1=02x\tanh x-1=0

并验证此方程在 0.70.7 与 0.80.8 之间有一个根。

(4)

(b) 考虑一组合适的矩形,利用图示证明

∑r=2nrsech⁡2r<ntanh⁡n+ln⁡(sech⁡n)−tanh⁡1−ln⁡(sech⁡1)\sum_{r=2}^{n}r\operatorname{sech}^2r<n\tanh n+\ln(\operatorname{sech}n)-\tanh 1-\ln(\operatorname{sech}1)
(6)
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

用乘积法则求导,提出始终为正的 sech⁡2x\operatorname{sech}^2x。再对剩下的方程检查两端的符号。

答题过程

展开

For f(x)=xsech⁡2xf(x)=x\operatorname{sech}^2x,

f′(x)= sech⁡2x−2xsech⁡2xtanh⁡x= sech⁡2x(1−2xtanh⁡x).\begin{align*} f'(x)=&\,\operatorname{sech}^2x-2x\operatorname{sech}^2x\tanh x\\ =&\,\operatorname{sech}^2x(1-2x\tanh x). \end{align*}

At MM the derivative vanishes. Since sech⁡2x>0\operatorname{sech}^2x>0, we obtain 2xtanh⁡x−1=02x\tanh x-1=0. The continuous function h(x)=2xtanh⁡x−1h(x)=2x\tanh x-1 satisfies

h(0.7)≈−0.154<0,h(0.8)≈0.062>0.h(0.7)\approx-0.154<0,\qquad h(0.8)\approx0.062>0.

So it has a root in (0.7,0.8)(0.7,0.8).

(b)

解法一

思路

展开

在 x≥1x\ge1 的下降段用宽为 11 的右端点矩形。它们给出积分的下界。积分时把 xx 保留作因子,对 sech⁡2x\operatorname{sech}^2x 积分。

答题过程

展开

For x>0x>0, h′(x)=2tanh⁡x+2xsech⁡2x>0h'(x)=2\tanh x+2x\operatorname{sech}^2x>0. Since h(0.8)>0h(0.8)>0, it follows that f′(x)<0f'(x)<0 for x≥1x\ge1. For r=2,…,nr=2,\ldots,n, the rectangle on [r−1,r][r-1,r] with height f(r)f(r) lies below the curve on the interval interior. The areas are 2sech⁡222\operatorname{sech}^22, 3sech⁡233\operatorname{sech}^23, 4sech⁡244\operatorname{sech}^24, …\ldots, (n−1)sech⁡2(n−1)(n-1)\operatorname{sech}^2(n-1), nsech⁡2nn\operatorname{sech}^2n. Thus

∑r=2nrsech⁡2r<∫1nxsech⁡2x dx.\sum_{r=2}^{n}r\operatorname{sech}^2r<\int_1^n x\operatorname{sech}^2x\,\mathrm{d}x.

Take u=xu=x and dv=sech⁡2x dx\mathrm{d}v=\operatorname{sech}^2x\,\mathrm{d}x, so du=dx\mathrm{d}u=\mathrm{d}x and v=tanh⁡xv=\tanh x. Integration by parts gives

∫1nxsech⁡2x dx= [xtanh⁡x]1n−∫1ntanh⁡x dx= [xtanh⁡x+ln⁡(sech⁡x)]1n= ntanh⁡n+ln⁡(sech⁡n) −tanh⁡1−ln⁡(sech⁡1).\begin{align*} \int_1^n x\operatorname{sech}^2x\,\mathrm{d}x =&\,[x\tanh x]_1^n-\int_1^n\tanh x\,\mathrm{d}x\\ =&\,[x\tanh x+\ln(\operatorname{sech}x)]_1^n\\ =&\,n\tanh n+\ln(\operatorname{sech}n)\\ &\,\hspace{2pt}-\tanh1-\ln(\operatorname{sech}1). \end{align*}

This proves the claimed bound.