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IAL 2021 Jan FP3 Q2

A Level / Edexcel / FP3

IAL 2021 Jan Paper · Question 2

题目

Problem

y=ln(tanh2x)x>0y = \ln(\tanh 2x) \qquad x > 0

(a) Show that

dydx=pcosech4x\frac{dy}{dx} = p\,\operatorname{cosech} 4x

where pp is a constant to be determined.

(b) Hence determine, in simplest form, the exact value of xx for which

dydx=1.\frac{dy}{dx} = 1.
(6)
题目中文翻译

y=ln(tanh2x)x>0y = \ln(\tanh 2x) \qquad x > 0

(a) 证明

dydx=pcosech4x\frac{dy}{dx} = p\,\operatorname{cosech} 4x

其中 pp 为待定常数。

(b) 因此求使

dydx=1\frac{dy}{dx} = 1

成立的 xx 的精确值,并写成最简形式。

解答