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FP2 2024 Jan Q8

A Level / Edexcel / FP2

IAL 2024 Jan Paper · Question 8

题目

Problem

8.

(a) For all the values of xx where the identity is defined, prove that

cot⁡2x+tan⁡x≡cosec⁡2x\begin{align*} \cot 2x + \tan x \equiv \cosec 2x \end{align*}
(3)

(b) Show that the substitution y2=wsin⁡2xy^2 = w\sin 2x , where ww is a function of xx , transforms the differential equation

ydydx+y2tan⁡x= sin⁡x0<x<π2(I)\begin{align*} y\frac{\mathrm{d}y}{\mathrm{d}x} + y^2 \tan x =&\, \sin x \qquad 0 < x < \frac{\pi}{2} \qquad \text{(I)} \end{align*}

into the differential equation

dwdx+2wcosec⁡2x= sec⁡x0<x<π2(II)\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x} + 2w\cosec 2x =&\, \sec x \qquad 0 < x < \frac{\pi}{2} \qquad \text{(II)} \end{align*}
(4)

(c) By solving differential equation (II) , determine a general solution of differential equation (I) in the form y2=f(x)y^2 = f(x) , where f(x)f(x) is a function in terms of cos⁡x\cos x

You may use without proof

[∫cosec⁡2x dx=12ln⁡∣tan⁡x∣+(constant)]\Bigg[ \int \cosec 2x \,\mathrm{d}x = \frac{1}{2}\ln |\tan x| + (\text{constant})\Bigg]
(6)
题目中文翻译

(a) 对所有使恒等式有定义的 xx，证明

cot⁡2x+tan⁡x≡ cosec⁡2x. \begin{align*} \cot2x+\tan x\equiv&\,\cosec2x. \end{align*}
(3)

(b) 证明代换 y2=wsin⁡2xy^2=w\sin2x（其中 ww 是 xx 的函数）将微分方程

ydydx+y2tan⁡x= sin⁡x,0<x<π2(I) \begin{align*} y\frac{\mathrm{d}y}{\mathrm{d}x}+y^2\tan x =&\,\sin x,\qquad0<x<\frac{\pi}{2}\qquad\text{(I)} \end{align*}

变为微分方程

dwdx+2wcosec⁡2x= sec⁡x,0<x<π2(II). \begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x}+2w\cosec2x =&\,\sec x,\qquad0<x<\frac{\pi}{2}\qquad\text{(II)}. \end{align*}
(4)

(c) 通过求解微分方程 (II)，求微分方程 (I) 的一个通解，将答案写成 y2=f(x)y^2=f(x) 的形式，其中 f(x)f(x) 是只含 cos⁡x\cos x 的函数。

可以不加证明地使用：

∫cosec⁡2x dx= 12ln⁡∣tan⁡x∣+常数. \begin{align*} \int\cosec2x\,\mathrm{d}x=&\,\frac12\ln|\tan x|+\text{常数}. \end{align*}
(6)

解答

(a)

解法一

思路

展开

把 cot⁡2x\cot2x 和 tan⁡x\tan x 都写成 sin⁡,cos⁡\sin,\cos，通分后分子会出现 cos⁡2xcos⁡x+sin⁡2xsin⁡x\cos2x\cos x+\sin2x\sin x，可以用 cos⁡(A−B)\cos(A-B) 化简。

答题过程

展开 cot⁡2x+tan⁡x= cos⁡2xsin⁡2x+sin⁡xcos⁡x= cos⁡2xcos⁡x+sin⁡2xsin⁡xsin⁡2xcos⁡x= cos⁡(2x−x)sin⁡2xcos⁡x= cos⁡xsin⁡2xcos⁡x= 1sin⁡2x= cosec⁡2x\begin{align*} \cot2x+\tan x =&\, \frac{\cos2x}{\sin2x}+\frac{\sin x}{\cos x}\\[4mm] =&\, \frac{\cos2x\cos x+\sin2x\sin x} {\sin2x\cos x}\\[4mm] =&\, \frac{\cos(2x-x)}{\sin2x\cos x}\\[4mm] =&\, \frac{\cos x}{\sin2x\cos x}\\[4mm] =&\, \frac{1}{\sin2x}\\[4mm] =&\, \cosec2x \end{align*}

Therefore

cot⁡2x+tan⁡x≡cosec⁡2x\begin{align*} \cot2x+\tan x\equiv \cosec2x \end{align*}

解法二

思路

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将两项通分到 sin⁡2x\sin2x，再用 sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x 和 cos⁡2x=1−2sin⁡2x\cos2x=1-2\sin^2x，使分子化为 11。

答题过程

展开

For all xx for which the expressions are defined,

cot⁡2x+tan⁡x= cos⁡2x+tan⁡xsin⁡2xsin⁡2x= cos⁡2x+sin⁡xcos⁡x⋅2sin⁡xcos⁡xsin⁡2x= cos⁡2x+2sin⁡2xsin⁡2x= 1−2sin⁡2x+2sin⁡2xsin⁡2x= 1sin⁡2x= cosec⁡2x.\begin{align*} \cot2x+\tan x =&\,\frac{\cos2x+\tan x\sin2x}{\sin2x}\\[2mm] =&\,\frac{\cos2x+\frac{\sin x}{\cos x}\cdot2\sin x\cos x}{\sin2x}\\[2mm] =&\,\frac{\cos2x+2\sin^2x}{\sin2x}\\[2mm] =&\,\frac{1-2\sin^2x+2\sin^2x}{\sin2x}\\[2mm] =&\,\frac1{\sin2x}\\[2mm] =&\,\cosec2x. \end{align*}

解法三

思路

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利用 cot⁡2x=1−tan⁡2x2tan⁡x\cot2x=\frac{1-\tan^2x}{2\tan x}，先把左边化成只含 tan⁡x\tan x 的分式，再用 1+tan⁡2x=sec⁡2x1+\tan^2x=\sec^2x 化简。

答题过程

展开

Using the double-angle identity for cot⁡2x\cot2x,

cot⁡2x+tan⁡x= 1−tan⁡2x2tan⁡x+tan⁡x= 1−tan⁡2x+2tan⁡2x2tan⁡x= 1+tan⁡2x2tan⁡x= sec⁡2x2tan⁡x= 1/cos⁡2x2sin⁡x/cos⁡x= 12sin⁡xcos⁡x= cosec⁡2x.\begin{align*} \cot2x+\tan x =&\,\frac{1-\tan^2x}{2\tan x}+\tan x\\[2mm] =&\,\frac{1-\tan^2x+2\tan^2x}{2\tan x}\\[2mm] =&\,\frac{1+\tan^2x}{2\tan x}\\[2mm] =&\,\frac{\sec^2x}{2\tan x}\\[2mm] =&\,\frac{1/\cos^2x}{2\sin x/\cos x}\\[2mm] =&\,\frac1{2\sin x\cos x}\\[2mm] =&\,\cosec2x. \end{align*}

(b)

解法一

思路

展开

先对 y2=wsin⁡2xy^2=w\sin2x 两边求导，得到 2ydydx2y\frac{\mathrm{d}y}{\mathrm{d}x} 的表达式。再把原方程中的 ydydxy\frac{\mathrm{d}y}{\mathrm{d}x} 和 y2y^2 都换成 ww，最后用 (a) 的恒等式化简。

答题过程

展开

Given

y2=wsin⁡2x\begin{align*} y^2=w\sin2x \end{align*}

differentiate both sides with respect to xx:

2ydydx= dwdxsin⁡2x+2wcos⁡2x\begin{align*} 2y\frac{\mathrm{d}y}{\mathrm{d}x} =&\, \frac{\mathrm{d}w}{\mathrm{d}x}\sin2x+2w\cos2x \end{align*}

Hence

ydydx= 12dwdxsin⁡2x+wcos⁡2x\begin{align*} y\frac{\mathrm{d}y}{\mathrm{d}x} =&\, \frac12\frac{\mathrm{d}w}{\mathrm{d}x}\sin2x+w\cos2x \end{align*}

Substitute this and y2=wsin⁡2xy^2=w\sin2x into (I):

12dwdxsin⁡2x+wcos⁡2x+wsin⁡2xtan⁡x= sin⁡x\begin{align*} \frac12\frac{\mathrm{d}w}{\mathrm{d}x}\sin2x +w\cos2x +w\sin2x\tan x =&\, \sin x \end{align*}

Multiply by 22:

dwdxsin⁡2x+2wcos⁡2x+2wsin⁡2xtan⁡x= 2sin⁡x\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x}\sin2x +2w\cos2x +2w\sin2x\tan x =&\, 2\sin x \end{align*}

Since 0<x<π20<x<\frac{\pi}{2}, sin⁡2x≠0\sin2x\ne0, so divide by sin⁡2x\sin2x:

dwdx+2wcos⁡2xsin⁡2x+2wtan⁡x= 2sin⁡xsin⁡2xdwdx+2w(cot⁡2x+tan⁡x)= 2sin⁡x2sin⁡xcos⁡xdwdx+2w(cot⁡2x+tan⁡x)= sec⁡x\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x} +2w\frac{\cos2x}{\sin2x} +2w\tan x =&\, \frac{2\sin x}{\sin2x}\\[4mm] \frac{\mathrm{d}w}{\mathrm{d}x} +2w(\cot2x+\tan x) =&\, \frac{2\sin x}{2\sin x\cos x}\\[4mm] \frac{\mathrm{d}w}{\mathrm{d}x} +2w(\cot2x+\tan x) =&\, \sec x \end{align*}

Using the identity from part (a),

dwdx+2wcosec⁡2x= sec⁡x\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x} +2w\cosec2x =&\, \sec x \end{align*}

This is differential equation (II).

(c)

解法一

思路

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(II) 是一阶线性微分方程。积分因子是 e∫2cosec⁡2x dx\mathrm{e}^{\int 2\cosec2x\,\mathrm{d}x}，题目已经给出所需积分。最后要把 ww 换回 y2y^2，并整理成只含 cos⁡x\cos x 的形式。

答题过程

展开

The differential equation is

dwdx+2wcosec⁡2x= sec⁡x\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x} +2w\cosec2x =&\, \sec x \end{align*}

The integrating factor is

I.F.= e∫2cosec⁡2x dx= eln⁡∣tan⁡x∣= tan⁡x\begin{align*} \mathrm{I.F.} =&\, \mathrm{e}^{\int 2\cosec2x\,\mathrm{d}x}\\[4mm] =&\, \mathrm{e}^{\ln|\tan x|}\\[4mm] =&\, \tan x \end{align*}

For 0<x<π20<x<\frac{\pi}{2}, tan⁡x>0\tan x>0, so the absolute value causes no difficulty.

Multiply (II) by tan⁡x\tan x:

tan⁡xdwdx+2wtan⁡xcosec⁡2x= sec⁡xtan⁡xddx(wtan⁡x)= sec⁡xtan⁡x\begin{align*} \tan x\frac{\mathrm{d}w}{\mathrm{d}x} +2w\tan x\cosec2x =&\, \sec x\tan x\\[4mm] \frac{\mathrm{d}}{\mathrm{d}x}(w\tan x) =&\, \sec x\tan x \end{align*}

Integrate:

wtan⁡x= ∫sec⁡xtan⁡x dx= sec⁡x+C\begin{align*} w\tan x =&\, \int \sec x\tan x\,\mathrm{d}x\\[4mm] =&\, \sec x+C \end{align*}

Since y2=wsin⁡2xy^2=w\sin2x,

w= y2sin⁡2x\begin{align*} w=&\,\frac{y^2}{\sin2x} \end{align*}

Substitute into wtan⁡x=sec⁡x+Cw\tan x=\sec x+C:

y2sin⁡2xtan⁡x= sec⁡x+Cy22sin⁡xcos⁡x⋅sin⁡xcos⁡x= 1cos⁡x+Cy22cos⁡2x= 1cos⁡x+C\begin{align*} \frac{y^2}{\sin2x}\tan x =&\, \sec x+C\\[4mm] \frac{y^2}{2\sin x\cos x}\cdot\frac{\sin x}{\cos x} =&\, \frac{1}{\cos x}+C\\[4mm] \frac{y^2}{2\cos^2x} =&\, \frac{1}{\cos x}+C \end{align*}

Multiply by 2cos⁡2x2\cos^2x:

y2= 2cos⁡x+2Ccos⁡2x\begin{align*} y^2 =&\, 2\cos x+2C\cos^2x \end{align*}

Writing the arbitrary constant as AA, the general solution is

y2=2cos⁡x+Acos⁡2x\begin{align*} \boxed{y^2=2\cos x+A\cos^2x} \end{align*}