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CIE 9231 2025 June Paper 23 Q2

A Level / CIE / FP2

CIE 9231 2025 June Paper 23 Paper · Question 2

题目

Problem

(a) Starting from the definitions of tanh⁡\tanh and sech⁡\operatorname{sech} in terms of exponentials, prove that

tanh⁡2t+sech⁡2t=1.\tanh^2 t + \operatorname{sech}^2 t = 1.
[3]

(b) The curve CC has parametric equations

x= ln⁡(cosh⁡t),y=tan⁡−1(sinh⁡t),for 0≤t≤1.\begin{align*} x =&\, \ln(\cosh t), \quad y = \tan^{-1}(\sinh t), \quad \text{for } 0 \le t \le 1. \end{align*}

Find the length of CC.

[5]
题目中文翻译

(a) 从 tanh⁡\tanh 和 sech⁡\operatorname{sech} 用指数表示的定义出发，证明

tanh⁡2t+sech⁡2t=1.\tanh^2 t + \operatorname{sech}^2 t = 1.
(3)

(b) 曲线 CC 的参数方程为

x= ln⁡(cosh⁡t),y=tan⁡−1(sinh⁡t),for 0≤t≤1.\begin{align*} x =&\, \ln(\cosh t), \quad y = \tan^{-1}(\sinh t), \quad \text{for } 0 \le t \le 1. \end{align*}

求 CC 的长度。

(5)
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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代入两个双曲函数的指数定义。通分后，分子正好等于分母。

答题过程

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The exponential definitions are

tanh⁡t=et−e−tet+e−t,sech⁡t=2et+e−t.\tanh t=\frac{e^t-e^{-t}}{e^t+e^{-t}},\qquad \operatorname{sech}t=\frac2{e^t+e^{-t}}.

Therefore

 tanh⁡2t+sech⁡2t= (et−e−t)2+4(et+e−t)2= e2t+2+e−2t(et+e−t)2=1.\begin{align*} &\,\tanh^2t+\operatorname{sech}^2t\\[4mm] =&\,\frac{(e^t-e^{-t})^2+4}{(e^t+e^{-t})^2}\\[4mm] =&\,\frac{e^{2t}+2+e^{-2t}}{(e^t+e^{-t})^2}=1. \end{align*}

(b)

解法一

思路

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分别对 x,yx,y 关于 tt 求导。两个导数的平方和正是 (a) 的左边，所以速度恒为 11。

答题过程

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The derivatives are

dxdt= tanh⁡t,dydt= cosh⁡t1+sinh⁡2t= sech⁡t.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t}=&\,\tanh t,\\[4mm] \frac{\mathrm{d}y}{\mathrm{d}t}=&\,\frac{\cosh t}{1+\sinh^2t}\\[4mm] =&\,\operatorname{sech}t. \end{align*}

The parametric arc-length formula gives

L=∫01(dxdt)2+(dydt)2 dt.L=\int_0^1\sqrt{\bigg(\frac{\mathrm{d}x}{\mathrm{d}t}\bigg)^2+\bigg(\frac{\mathrm{d}y}{\mathrm{d}t}\bigg)^2}\,\mathrm{d}t.

By (a), the length is

∫01tanh⁡2t+sech⁡2t dt=∫011 dt=1.\int_0^1\sqrt{\tanh^2t+\operatorname{sech}^2t}\,\mathrm{d}t =\int_0^1 1\,\mathrm{d}t=\boxed{1}.