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CIE 9231 2023 November Paper 23 Q6

A Level / CIE / FP2

CIE 9231 2023 Nov Paper 23 Paper · Question 6

题目

Problem

(a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that

sinh2x=2sinhxcoshx.\sinh 2x=2\sinh x\cosh x.
[3]

(b) Using the substitution u=sinhxu=\sinh x, find

sinh22xcoshxdx.\int\sinh^2 2x\cosh x\,\mathrm{d}x.
[4]

(c) Find the particular solution of the differential equation

dydx+ytanhx=sinh22x,\frac{\mathrm{d}y}{\mathrm{d}x}+y\tanh x=\sinh^2 2x,

given that y=4y=4 when x=0x=0. Give your answer in the form y=f(x)y=f(x).

[7]
题目中文翻译

(a) 从用指数表示的 cosh\coshsinh\sinh 的定义出发,证明

sinh2x=2sinhxcoshx\sinh 2x=2\sinh x\cosh x

(b) 使用代换 u=sinhxu=\sinh x,求

sinh22xcoshxdx\int\sinh^2 2x\cosh x\,\mathrm{d}x

(c) 求微分方程

dydx+ytanhx=sinh22x\frac{\mathrm{d}y}{\mathrm{d}x}+y\tanh x=\sinh^2 2x

的特解,已知当 x=0x=0y=4y=4。将答案写成 y=f(x)y=f(x) 的形式。

解答