CalcGospel 國際數學圖譜
返回

CIE 9231 2026 June Paper 23 Q5

A Level / CIE / FP2

CIE 9231 2026 June Paper 23 Paper · Question 5

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)

It is given that

x=tsinh⁡tandy=tcosh⁡t.x=t\sinh t\qquad\text{and}\qquad y=t\cosh t.

(b) Show that dydx=ttanh⁡t+1t+tanh⁡t\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{t\tanh t+1}{t+\tanh t}.

(3)

(c) Show that d2ydx2=(t2−2)sech⁡nt(t+tanh⁡t)n\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{\big(t^2-2\big)\operatorname{sech}^n t}{\big(t+\tanh t\big)^n}, where nn is an integer to be determined.

(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)

已知

x=tsinh⁡t且y=tcosh⁡t.x=t\sinh t\qquad\text{且}\qquad y=t\cosh t.

(b) 证明 dydx=ttanh⁡t+1t+tanh⁡t\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{t\tanh t+1}{t+\tanh t}。

(3)

(c) 证明 d2ydx2=(t2−2)sech⁡nt(t+tanh⁡t)n\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{\big(t^2-2\big)\operatorname{sech}^n t}{\big(t+\tanh t\big)^n}，其中 nn 是待求的整数。

(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

展开

代入 sech⁡t\operatorname{sech}t 的指数定义，通分。分子的平方减 44 后，正好等于 (et−e−t)2(e^t-e^{-t})^2。

答题过程

展开

The exponential definitions are

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

Thus

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

展开

用乘积法则求两个参数导数。取比值后，分子分母同除以 cosh⁡t\cosh t。

答题过程

展开

Differentiating gives x′=sinh⁡t+tcosh⁡tx'=\sinh t+t\cosh t and y′=cosh⁡t+tsinh⁡ty'=\cosh t+t\sinh t. Therefore, where x′≠0x'\ne0,

dydx= cosh⁡t+tsinh⁡tsinh⁡t+tcosh⁡t= 1+ttanh⁡tt+tanh⁡t.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{\cosh t+t\sinh t}{\sinh t+t\cosh t}\\ =&\,\frac{1+t\tanh t}{t+\tanh t}. \end{align*}

(c)

解法一

思路

展开

把 tanh⁡t\tanh t 和它的导数简记为两个量。用商法则整理分子，再用 (a) 合并；最后乘 dtdx\frac{\mathrm{d}t}{\mathrm{d}x}。

答题过程

展开

Put u=tanh⁡tu=\tanh t and h=sech⁡2th=\operatorname{sech}^2t, so u′=hu'=h and u2=1−hu^2=1-h. The quotient rule gives

ddt(dydx)= (u+th)(t+u)−(1+tu)(1+h)(t+u)2= tu+u2+t2h+tuh−1−h−tu−tuh(t+u)2= u2+t2h−1−h(t+u)2= (t2−2)h(t+u)2.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg) =&\,\frac{(u+th)(t+u)-(1+tu)(1+h)}{(t+u)^2}\\ =&\,\frac{tu+u^2+t^2h+tuh-1-h-tu-tuh}{(t+u)^2}\\ =&\,\frac{u^2+t^2h-1-h}{(t+u)^2}\\ =&\,\frac{(t^2-2)h}{(t+u)^2}. \end{align*}

Since dxdt=cosh⁡t (t+u)\frac{\mathrm{d}x}{\mathrm{d}t}=\cosh t\,(t+u), the chain rule gives

d2ydx2=(t2−2)sech⁡3t(t+tanh⁡t)3.\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =\frac{(t^2-2)\operatorname{sech}^3t}{(t+\tanh t)^3}.

Thus n=3n=3. The expression applies for t≠0t\ne0, since t+tanh⁡tt+\tanh t has the same sign as tt when t≠0t\ne0.