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IAL 2024 Jan Q4

A Level / Edexcel / FP2

IAL 2024 Jan Paper · Question 4

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(a) Determine, in ascending powers of (x−π6)\left( x - \frac{\pi}{6} \right) up to and including the term in (x−π6)3\left( x - \frac{\pi}{6} \right)^3 , the Taylor series expansion about π6\frac{\pi}{6} of

y=tan⁡(3x2)\begin{align*} y = \tan \left( \frac{3x}{2} \right) \end{align*}

giving each coefficient in simplest form.

(7)

(b) Hence show that

tan⁡3π8≈1+π4+π2A+π3B\begin{align*} \tan \frac{3\pi}{8} \approx{}& 1+\frac{\pi}{4}+\frac{\pi^2}{A}+\frac{\pi^3}{B} \end{align*}

where AA and BB are integers to be determined.

(2)

解答

(a)

解法一

思路

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Taylor series 到三次项需要 y,y′,y′′,y′′′y,y',y'',y''' 在 x=π6x=\frac{\pi}{6} 的值。这里 3x2\frac{3x}{2} 在 x=π6x=\frac{\pi}{6} 时等于 π4\frac{\pi}{4}，所以 tan⁡π4=1\tan\frac{\pi}{4}=1，sec⁡2π4=2\sec^2\frac{\pi}{4}=2，数值会比较整齐。

答题过程

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Let

y=tan⁡(3x2)\begin{align*} y=\tan\left(\frac{3x}{2}\right) \end{align*}

Then

y′=32sec⁡2(3x2)\begin{align*} y' ={}& \frac32\sec^2\left(\frac{3x}{2}\right) \end{align*}

Differentiate again:

y′′=32⋅2sec⁡(3x2)(sec⁡(3x2)tan⁡(3x2)⋅32)=92sec⁡2(3x2)tan⁡(3x2)\begin{align*} y'' ={}& \frac32\cdot 2\sec\left(\frac{3x}{2}\right) \left(\sec\left(\frac{3x}{2}\right) \tan\left(\frac{3x}{2}\right)\cdot\frac32\right)\\[4mm] ={}& \frac92\sec^2\left(\frac{3x}{2}\right) \tan\left(\frac{3x}{2}\right) \end{align*}

For the third derivative, use the product rule on sec⁡2(3x2)tan⁡(3x2)\sec^2\left(\frac{3x}{2}\right)\tan\left(\frac{3x}{2}\right):

y′′′=92[3sec⁡2(3x2)tan⁡2(3x2)+32sec⁡4(3x2)]=272sec⁡2(3x2)tan⁡2(3x2)+274sec⁡4(3x2)\begin{align*} y''' ={}& \frac92\left[ 3\sec^2\left(\frac{3x}{2}\right) \tan^2\left(\frac{3x}{2}\right) +\frac32\sec^4\left(\frac{3x}{2}\right) \right]\\[4mm] ={}& \frac{27}{2}\sec^2\left(\frac{3x}{2}\right) \tan^2\left(\frac{3x}{2}\right) +\frac{27}{4}\sec^4\left(\frac{3x}{2}\right) \end{align*}

At x=π6x=\frac{\pi}{6}, we have 3x2=π4\frac{3x}{2}=\frac{\pi}{4}. Therefore

y(π6)=1y′(π6)=32⋅2=3y′′(π6)=92⋅2⋅1=9y′′′(π6)=272⋅2⋅12+274⋅22=54\begin{align*} y\left(\frac{\pi}{6}\right) ={}&1\\[2mm] y'\left(\frac{\pi}{6}\right) ={}& \frac32\cdot 2=3\\[2mm] y''\left(\frac{\pi}{6}\right) ={}& \frac92\cdot 2\cdot 1=9\\[2mm] y'''\left(\frac{\pi}{6}\right) ={}& \frac{27}{2}\cdot 2\cdot 1^2 +\frac{27}{4}\cdot 2^2\\[2mm] ={}&54 \end{align*}

Using Taylor’s formula about x=π6x=\frac{\pi}{6},

y=1+3(x−π6)+92!(x−π6)2 +543!(x−π6)3+⋯=1+3(x−π6)+92(x−π6)2+9(x−π6)3+⋯\begin{align*} y ={}& 1+3\left(x-\frac{\pi}{6}\right) +\frac{9}{2!}\left(x-\frac{\pi}{6}\right)^2\\[4mm] &\,\hspace{2pt}+\frac{54}{3!}\left(x-\frac{\pi}{6}\right)^3+\cdots\\[4mm] ={}& \boxed{ 1+3\left(x-\frac{\pi}{6}\right) +\frac92\left(x-\frac{\pi}{6}\right)^2 +9\left(x-\frac{\pi}{6}\right)^3+\cdots } \end{align*}

(b)

解法一

思路

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因为 y=tan⁡(3x2)y=\tan\left(\frac{3x}{2}\right)，要得到 tan⁡3π8\tan\frac{3\pi}{8}，令 x=π4x=\frac{\pi}{4}。这时 x−π6=π12x-\frac{\pi}{6}=\frac{\pi}{12}。

答题过程

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Put x=π4x=\frac{\pi}{4} in the expansion from part (a). Then

x−π6=π4−π6=π12\begin{align*} x-\frac{\pi}{6} ={}& \frac{\pi}{4}-\frac{\pi}{6}\\[4mm] ={}& \frac{\pi}{12} \end{align*}

So

tan⁡3π8≈1+3(π12)+92(π12)2+9(π12)3=1+π4+9π2288+9π31728=1+π4+π232+π3192\begin{align*} \tan\frac{3\pi}{8} \approx{}& 1+3\left(\frac{\pi}{12}\right) +\frac92\left(\frac{\pi}{12}\right)^2 +9\left(\frac{\pi}{12}\right)^3\\[4mm] ={}& 1+\frac{\pi}{4} +\frac{9\pi^2}{288} +\frac{9\pi^3}{1728}\\[4mm] ={}& 1+\frac{\pi}{4} +\frac{\pi^2}{32} +\frac{\pi^3}{192} \end{align*}

Hence

A=32,B=192\begin{align*} \boxed{A=32,\qquad B=192} \end{align*}