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CIE 9231 2023 June Paper 21 Q8

A Level / CIE / FP2

CIE 9231 2023 June Paper 21 Paper · Question 8

题目

Problem

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

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

The curve CC has parametric equations

x= 12tanh⁡2t+ln⁡sech⁡t,y= 1+tanh⁡4t,for t>0.\begin{align*} x=&\,\frac{1}{2}\tanh^2t+\ln\operatorname{sech}t,\\ y=&\,1+\tanh^4t,\qquad\text{for }t>0. \end{align*}

(b) Show that dydx=−4sech⁡2t\dfrac{\mathrm{d}y}{\mathrm{d}x}=-4\operatorname{sech}^2t.

(5)

(c) Find the coordinates of the point on CC with d2ydx2=−92\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\dfrac{9}{2}, giving your answer in the form (a+ln⁡b,c)(a+\ln b,c) where aa, bb and cc are rational numbers.

(6)
题目中文翻译

(a) 从 sech⁡\operatorname{sech} 和 tanh⁡\tanh 的指数形式定义出发，证明

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

曲线 CC 的参数方程为

x= 12tanh⁡2t+ln⁡sech⁡t,y= 1+tanh⁡4t,t>0.\begin{align*} x=&\,\frac{1}{2}\tanh^2t+\ln\operatorname{sech}t,\\ y=&\,1+\tanh^4t,\qquad t>0. \end{align*}

(b) 证明 dydx=−4sech⁡2t\dfrac{\mathrm{d}y}{\mathrm{d}x}=-4\operatorname{sech}^2t。

(5)

(c) 求曲线 CC 上满足 d2ydx2=−92\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\dfrac{9}{2} 的点的坐标，将答案写成 (a+ln⁡b,c)(a+\ln b,c) 的形式，其中 aa、bb 和 cc 为有理数。

(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

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用指数式通分。分子的平方减 44 可以写成两个指数之差的平方。

答题过程

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The exponential definitions are sech⁡t=2/(et+e−t)\operatorname{sech}t=2/(e^t+e^{-t}) and tanh⁡t=(et−e−t)/(et+e−t)\tanh t=(e^t-e^{-t})/(e^t+e^{-t}). Therefore,

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

(b)

解法一

思路

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分别求 x,yx,y 的参数导数。用 (a) 化简 x′x'，这样取比值时可约去 tanh⁡3t\tanh^3t。

答题过程

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Using (tanh⁡t)′=sech⁡2t(\tanh t)'=\operatorname{sech}^2t and (sech⁡t)′=−sech⁡ttanh⁡t(\operatorname{sech}t)'=-\operatorname{sech}t\tanh t, we have (ln⁡sech⁡t)′=−tanh⁡t(\ln\operatorname{sech}t)'=-\tanh t. Differentiating and using (a),

x′= tanh⁡t sech⁡2t−tanh⁡t= tanh⁡t(sech⁡2t−1)=−tanh⁡3t,y′= 4tanh⁡3t sech⁡2t.\begin{align*} x'=&\,\tanh t\,\operatorname{sech}^2t-\tanh t\\ =&\,\tanh t\big(\operatorname{sech}^2t-1\big)=-\tanh^3t,\\ y'=&\,4\tanh^3t\,\operatorname{sech}^2t. \end{align*}

Since t>0t>0, we have tanh⁡t>0\tanh t>0. Thus dydx=y′/x′=−4sech⁡2t\frac{\mathrm{d}y}{\mathrm{d}x}=y'/x'=-4\operatorname{sech}^2t.

(c)

解法一

思路

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先对 (b) 的斜率关于 tt 求导，再除以 x′x'。用 (a) 把二阶导数的条件改写成关于 tanh⁡2t\tanh^2t 的方程。

答题过程

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Differentiating the gradient gives ddt(−4sech⁡2t)=8sech⁡2ttanh⁡t\dfrac{\mathrm{d}}{\mathrm{d}t}(-4\operatorname{sech}^2t)=8\operatorname{sech}^2t\tanh t. The chain rule d2ydx2=ddt(dydx)/dxdt\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{\mathrm{d}}{\mathrm{d}t}\big(\dfrac{\mathrm{d}y}{\mathrm{d}x}\big)\big/\dfrac{\mathrm{d}x}{\mathrm{d}t} gives

d2ydx2=8sech⁡2t tanh⁡t−tanh⁡3t=−8sech⁡2ttanh⁡2t.\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =\frac{8\operatorname{sech}^2t\,\tanh t}{-\tanh^3t} =-8\frac{\operatorname{sech}^2t}{\tanh^2t}.

Setting this equal to −9/2-9/2 and using (a) yields 16(1−tanh⁡2t)=9tanh⁡2t16(1-\tanh^2t)=9\tanh^2t, hence tanh⁡2t=16/25\tanh^2t=16/25. Therefore sech⁡t=3/5\operatorname{sech}t=3/5, taking the positive square root. Substituting into the parametric equations gives

(x,y)=(825+ln⁡35, 881625).\boxed{(x,y)=\bigg(\frac8{25}+\ln\frac35,\ \frac{881}{625}\bigg)}.