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IAL 2021 Oct Q5

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 5

题目

Problem

Given that y=tan⁡2xy = \tan^2 x

(a) show that

d3ydx3=8tan⁡xsec⁡2x(psec⁡2x+q)\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = 8\tan x \sec^2 x (p \sec^2 x + q) \end{align*}

where pp and qq are integers to be determined.

(5)

(b) Hence determine the Taylor series expansion about π3\frac{\pi}{3} of tan⁡2x\tan^2 x in ascending powers of (x−π3)\left( x - \frac{\pi}{3} \right) up to and including the term in (x−π3)3\left( x - \frac{\pi}{3} \right)^3, giving each coefficient in simplest form.

(3)

解答

(a)

解法一

思路

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先求一阶、二阶、三阶导数。二阶导数可以先写成含 tan⁡x\tan x 和 sec⁡x\sec x 的形式，三阶导数再整理成题目要求的

8tan⁡xsec⁡2x(psec⁡2x+q).\begin{align*} 8\tan x\sec^2x(p\sec^2x+q). \end{align*}

答题过程

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Given

y=tan⁡2x.\begin{align*} y=\tan^2x. \end{align*}

Differentiate:

dydx= 2tan⁡xsec⁡2x.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,2\tan x\sec^2x. \end{align*}

Differentiate again using the product rule:

d2ydx2= 2sec⁡2xsec⁡2x+2tan⁡x(2sec⁡2xtan⁡x)= 2sec⁡4x+4sec⁡2xtan⁡2x.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,2\sec^2x\sec^2x +2\tan x\left(2\sec^2x\tan x\right)\\[2mm] =&\,2\sec^4x+4\sec^2x\tan^2x. \end{align*}

Now differentiate once more:

d3ydx3= 2(4sec⁡4xtan⁡x) +4[(2sec⁡2xtan⁡x)tan⁡2x+sec⁡2x(2tan⁡xsec⁡2x)]= 8sec⁡4xtan⁡x+8sec⁡2xtan⁡3x +8sec⁡4xtan⁡x.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,2\left(4\sec^4x\tan x\right)\\[2mm] &\,\hspace{2pt}+4\left[ \left(2\sec^2x\tan x\right)\tan^2x +\sec^2x\left(2\tan x\sec^2x\right) \right]\\[2mm] =&\,8\sec^4x\tan x +8\sec^2x\tan^3x\\[2mm] &\,\hspace{2pt}+8\sec^4x\tan x. \end{align*}

So

d3ydx3= 16sec⁡4xtan⁡x+8sec⁡2xtan⁡3x.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,16\sec^4x\tan x +8\sec^2x\tan^3x. \end{align*}

Use

tan⁡2x=sec⁡2x−1.\begin{align*} \tan^2x=\sec^2x-1. \end{align*}

Then

8sec⁡2xtan⁡3x= 8sec⁡2xtan⁡x(sec⁡2x−1).\begin{align*} 8\sec^2x\tan^3x =&\,8\sec^2x\tan x(\sec^2x-1). \end{align*}

Therefore

d3ydx3= 16sec⁡4xtan⁡x+8sec⁡2xtan⁡x(sec⁡2x−1)= 24sec⁡4xtan⁡x−8sec⁡2xtan⁡x= 8tan⁡xsec⁡2x(3sec⁡2x−1).\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,16\sec^4x\tan x +8\sec^2x\tan x(\sec^2x-1)\\[2mm] =&\,24\sec^4x\tan x -8\sec^2x\tan x\\[2mm] =&\,8\tan x\sec^2x(3\sec^2x-1). \end{align*}

Hence

p=3,q=−1.\begin{align*} p=3,\qquad q=-1. \end{align*}

解法二

思路

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二阶导数也可以先用 tan⁡2x=sec⁡2x−1\tan^2x=\sec^2x-1 化成只含 sec⁡x\sec x 的形式：

d2ydx2=6sec⁡4x−4sec⁡2x.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =6\sec^4x-4\sec^2x. \end{align*}

这样求三阶导数会更短。

答题过程

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From

y=tan⁡2x,\begin{align*} y=\tan^2x, \end{align*}

we get

dydx= 2tan⁡xsec⁡2x.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,2\tan x\sec^2x. \end{align*}

Then

d2ydx2= 2sec⁡4x+4sec⁡2xtan⁡2x.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,2\sec^4x+4\sec^2x\tan^2x. \end{align*}

Use tan⁡2x=sec⁡2x−1\tan^2x=\sec^2x-1:

d2ydx2= 2sec⁡4x+4sec⁡2x(sec⁡2x−1)= 6sec⁡4x−4sec⁡2x.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,2\sec^4x +4\sec^2x(\sec^2x-1)\\[2mm] =&\,6\sec^4x-4\sec^2x. \end{align*}

Differentiate this:

d3ydx3= 6(4sec⁡4xtan⁡x)−4(2sec⁡2xtan⁡x)= 24sec⁡4xtan⁡x−8sec⁡2xtan⁡x= 8tan⁡xsec⁡2x(3sec⁡2x−1).\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,6\left(4\sec^4x\tan x\right) -4\left(2\sec^2x\tan x\right)\\[2mm] =&\,24\sec^4x\tan x -8\sec^2x\tan x\\[2mm] =&\,8\tan x\sec^2x(3\sec^2x-1). \end{align*}

Therefore

p=3,q=−1.\begin{align*} p=3,\qquad q=-1. \end{align*}

(b)

解法一

思路

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Taylor series about x=π3x=\dfrac{\pi}{3} 需要

y,y′,y′′,y′′′\begin{align*} y,\quad y',\quad y'',\quad y''' \end{align*}

在 x=π3x=\dfrac{\pi}{3} 的值。这里

tan⁡π3=3,sec⁡2π3=4.\begin{align*} \tan\frac{\pi}{3}=\sqrt3, \qquad \sec^2\frac{\pi}{3}=4. \end{align*}

答题过程

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At

x=π3,\begin{align*} x=\frac{\pi}{3}, \end{align*}

we have

tan⁡x=3,sec⁡2x=4.\begin{align*} \tan x=\sqrt3, \qquad \sec^2x=4. \end{align*}

So

y= tan⁡2x=3.\begin{align*} y =&\,\tan^2x=3. \end{align*}

Also

y′= 2tan⁡xsec⁡2x= 2(3)(4)= 83.\begin{align*} y' =&\,2\tan x\sec^2x\\[2mm] =&\,2(\sqrt3)(4)\\[2mm] =&\,8\sqrt3. \end{align*}

From part (a),

y′′′= 8tan⁡xsec⁡2x(3sec⁡2x−1).\begin{align*} y''' =&\,8\tan x\sec^2x(3\sec^2x-1). \end{align*}

At x=π3x=\dfrac{\pi}{3},

y′′′= 8(3)(4)(3⋅4−1)= 3523.\begin{align*} y''' =&\,8(\sqrt3)(4)(3\cdot4-1)\\[2mm] =&\,352\sqrt3. \end{align*}

We also need y′′y''. From part (a),

y′′= 6sec⁡4x−4sec⁡2x.\begin{align*} y'' =&\,6\sec^4x-4\sec^2x. \end{align*}

Therefore

y′′= 6(42)−4(4)= 96−16= 80.\begin{align*} y'' =&\,6(4^2)-4(4)\\[2mm] =&\,96-16\\[2mm] =&\,80. \end{align*}

The Taylor series is

y= y0+y0′(x−π3)+y0′′2!(x−π3)2 +y0′′′3!(x−π3)3+⋯ .\begin{align*} y =&\,y_0+y_0'\left(x-\frac{\pi}{3}\right) +\frac{y_0''}{2!}\left(x-\frac{\pi}{3}\right)^2\\[2mm] &\,\hspace{2pt}+\frac{y_0'''}{3!} \left(x-\frac{\pi}{3}\right)^3+\cdots. \end{align*}

Hence

tan⁡2x= 3+83(x−π3)+40(x−π3)2 +17633(x−π3)3+⋯ .\begin{align*} \tan^2x =&\,3+8\sqrt3\left(x-\frac{\pi}{3}\right) +40\left(x-\frac{\pi}{3}\right)^2\\[2mm] &\,\hspace{2pt}+\frac{176\sqrt3}{3} \left(x-\frac{\pi}{3}\right)^3+\cdots. \end{align*}