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IAL 2022 Oct Q6

A Level / Edexcel / P3

IAL 2022 Oct Paper · Question 6

题目

Problem

y=2+3sin⁡xcos⁡x+sin⁡xy=\frac{2+3\sin x}{\cos x+\sin x}

Show that

dydx=atan⁡x+bsec⁡x+csec⁡x+2sin⁡x\frac{dy}{dx}=\frac{a\tan x+b\sec x+c}{\sec x+2\sin x}

where aa, bb and cc are integers to be found.

(6)
题目中文翻译 y=2+3sin⁡xcos⁡x+sin⁡xy=\frac{2+3\sin x}{\cos x+\sin x}

证明

dydx=atan⁡x+bsec⁡x+csec⁡x+2sin⁡x\frac{dy}{dx}=\frac{a\tan x+b\sec x+c}{\sec x+2\sin x}

其中 aa、bb 和 cc 是需要求出的整数。

解答

We have

y=2+3sin⁡xcos⁡x+sin⁡xy=\frac{2+3\sin x}{\cos x+\sin x}

Using the quotient rule,

dydx=(cos⁡x+sin⁡x)(3cos⁡x)−(2+3sin⁡x)(−sin⁡x+cos⁡x)(cos⁡x+sin⁡x)2\frac{dy}{dx} = \frac{(\cos x+\sin x)(3\cos x)-(2+3\sin x)(-\sin x+\cos x)} {(\cos x+\sin x)^2}

Expand the numerator:

(cos⁡x+sin⁡x)(3cos⁡x)−(2+3sin⁡x)(−sin⁡x+cos⁡x)=3cos⁡2x+3sin⁡xcos⁡x+2sin⁡x+3sin⁡2x−2cos⁡x−3sin⁡xcos⁡x=3cos⁡2x+3sin⁡2x+2sin⁡x−2cos⁡x\begin{aligned} &(\cos x+\sin x)(3\cos x)-(2+3\sin x)(-\sin x+\cos x) \\ &=3\cos^2x+3\sin x\cos x +2\sin x+3\sin^2x-2\cos x-3\sin x\cos x \\ &=3\cos^2x+3\sin^2x+2\sin x-2\cos x \end{aligned}

Using

sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1

this becomes

3+2sin⁡x−2cos⁡x3+2\sin x-2\cos x

The denominator is

(cos⁡x+sin⁡x)2=cos⁡2x+2sin⁡xcos⁡x+sin⁡2x(\cos x+\sin x)^2 =\cos^2x+2\sin x\cos x+\sin^2x

So

(cos⁡x+sin⁡x)2=1+2sin⁡xcos⁡x(\cos x+\sin x)^2=1+2\sin x\cos x

Therefore

dydx=3+2sin⁡x−2cos⁡x1+2sin⁡xcos⁡x\frac{dy}{dx} =\frac{3+2\sin x-2\cos x}{1+2\sin x\cos x}

Multiply numerator and denominator by sec⁡x\sec x:

dydx=(3+2sin⁡x−2cos⁡x)sec⁡x(1+2sin⁡xcos⁡x)sec⁡x\frac{dy}{dx} = \frac{(3+2\sin x-2\cos x)\sec x} {(1+2\sin x\cos x)\sec x}

So

dydx=3sec⁡x+2sin⁡xsec⁡x−2sec⁡x+2sin⁡x\frac{dy}{dx} = \frac{3\sec x+2\sin x\sec x-2} {\sec x+2\sin x}

Since

sin⁡xsec⁡x=tan⁡x\sin x\sec x=\tan x

we get

dydx=2tan⁡x+3sec⁡x−2sec⁡x+2sin⁡x\frac{dy}{dx} =\frac{2\tan x+3\sec x-2}{\sec x+2\sin x}

Therefore

a=2,b=3,c=−2\boxed{a=2,\qquad b=3,\qquad c=-2}