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IAL 2023 June Q6

A Level / Edexcel / FP2

IAL 2023 June Paper · Question 6

题目

Problem

Given that y=sec⁡xy = \sec x

(a) show that

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

where pp and qq are integers to be determined.

(4)

(b) Hence determine the Taylor series expansion about π3\frac{\pi}{3} of sec⁡x\sec 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)

(c) Use the answer to part (b) to determine, to four significant figures, an approximate value of sec⁡(7π24)\sec \left(\frac{7\pi}{24}\right)

(2)

解答

(a)

解法一

思路

展开

连续求导时，先把二阶导数整理成只含 sec⁡x\sec x 的形式，会让第三次求导更短。最后把 sec⁡xtan⁡x\sec x\tan x 提出来，就能读出 p,qp,q。

答题过程

展开

Since y=sec⁡xy=\sec x,

dydx=sec⁡xtan⁡x\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} ={}& \sec x\tan x \end{align*}

Differentiate again:

d2ydx2=(sec⁡xtan⁡x)tan⁡x+sec⁡x(sec⁡2x)=sec⁡xtan⁡2x+sec⁡3x=sec⁡x(sec⁡2x−1)+sec⁡3x=2sec⁡3x−sec⁡x\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} ={}& (\sec x\tan x)\tan x+\sec x(\sec^2 x)\\[4mm] ={}& \sec x\tan^2x+\sec^3x\\[4mm] ={}& \sec x(\sec^2x-1)+\sec^3x\\[4mm] ={}& 2\sec^3x-\sec x \end{align*}

Differentiate a third time:

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

Therefore

p=6,q=−1\begin{align*} \boxed{p=6,\qquad q=-1} \end{align*}

解法二

思路

展开

求得二阶导数 d2ydx2=sec⁡xtan⁡2x+sec⁡3x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \sec x\tan^2x+\sec^3x 后，不代换 tan⁡2x\tan^2 x。直接使用乘积求导法则与链式法则对该表达式再次求导，最后提取公因式 sec⁡xtan⁡x\sec x\tan x 并将 tan⁡2x\tan^2 x 化为 sec⁡2x−1\sec^2 x - 1 来统一简化。

答题过程

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From the first differentiation, we have:

dydx=  sec⁡xtan⁡x\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\, \sec x\tan x \end{align*}

Differentiating again gives:

d2ydx2=  sec⁡xtan⁡2x+sec⁡3x\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\, \sec x\tan^2x+\sec^3x \end{align*}

Differentiate a third time directly using the Product Rule and Chain Rule:

d3ydx3=  ddx(sec⁡xtan⁡2x)+ddx(sec⁡3x)=  (sec⁡xtan⁡x)tan⁡2x+sec⁡x(2tan⁡xsec⁡2x)+3sec⁡2x(sec⁡xtan⁡x)=  sec⁡xtan⁡3x+2sec⁡3xtan⁡x+3sec⁡3xtan⁡x=  sec⁡xtan⁡3x+5sec⁡3xtan⁡x\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\, \frac{\mathrm{d}}{\mathrm{d}x}(\sec x\tan^2x) + \frac{\mathrm{d}}{\mathrm{d}x}(\sec^3x)\\[4mm] =&\,\, (\sec x\tan x)\tan^2x + \sec x\left( 2\tan x\sec^2x \right) + 3\sec^2x(\sec x\tan x)\\[4mm] =&\,\, \sec x\tan^3x + 2\sec^3x\tan x + 3\sec^3x\tan x\\[4mm] =&\,\, \sec x\tan^3x + 5\sec^3x\tan x \end{align*}

Factor out sec⁡xtan⁡x\sec x\tan x:

d3ydx3=  sec⁡xtan⁡x(tan⁡2x+5sec⁡2x)\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\, \sec x\tan x\left( \tan^2x + 5\sec^2x \right) \end{align*}

Substitute tan⁡2x=sec⁡2x−1\tan^2x = \sec^2x - 1:

d3ydx3=  sec⁡xtan⁡x(sec⁡2x−1+5sec⁡2x)=  sec⁡xtan⁡x(6sec⁡2x−1)\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\, \sec x\tan x\left( \sec^2x - 1 + 5\sec^2x \right)\\[4mm] =&\,\, \sec x\tan x(6\sec^2x - 1) \end{align*}

Comparing this with the target expression:

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

we find:

p=6,q=−1\begin{align*} \boxed{p=6,\qquad q=-1} \end{align*}

(b)

解法一

思路

展开

Taylor series 需要 f(a),f′(a),f′′(a),f′′′(a)f(a), f'(a), f''(a), f'''(a)。这里 a=π3a=\frac{\pi}{3}，并且 (a) 已经给了三阶导数的方便形式。

答题过程

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Let f(x)=sec⁡xf(x)=\sec x and a=π3a=\frac{\pi}{3}. Then

f(π3)=sec⁡π3=2f′(π3)=sec⁡π3tan⁡π3=23f′′(π3)=2sec⁡3π3−sec⁡π3=2(23)−2=14f′′′(π3)=sec⁡π3tan⁡π3(6sec⁡2π3−1)=23(6⋅22−1)=463\begin{align*} f\left(\frac{\pi}{3}\right) ={}& \sec\frac{\pi}{3}=2\\[4mm] f'\left(\frac{\pi}{3}\right) ={}& \sec\frac{\pi}{3}\tan\frac{\pi}{3} =2\sqrt3\\[4mm] f''\left(\frac{\pi}{3}\right) ={}& 2\sec^3\frac{\pi}{3}-\sec\frac{\pi}{3}\\[2mm] ={}& 2(2^3)-2=14\\[4mm] f'''\left(\frac{\pi}{3}\right) ={}& \sec\frac{\pi}{3}\tan\frac{\pi}{3} \left(6\sec^2\frac{\pi}{3}-1\right)\\[2mm] ={}& 2\sqrt3(6\cdot 2^2-1)\\[2mm] ={}& 46\sqrt3 \end{align*}

Using

f(x)=f(a)+(x−a)f′(a)+(x−a)22!f′′(a)+(x−a)33!f′′′(a)+⋯\begin{align*} f(x) ={}& f(a)+(x-a)f'(a) +\frac{(x-a)^2}{2!}f''(a) +\frac{(x-a)^3}{3!}f'''(a)+\cdots \end{align*}

we get

sec⁡x=2+23(x−π3)+142(x−π3)2 +4636(x−π3)3+⋯=2+23(x−π3)+7(x−π3)2 +2333(x−π3)3+⋯\begin{align*} \sec x ={}& 2+2\sqrt3\left(x-\frac{\pi}{3}\right) +\frac{14}{2}\left(x-\frac{\pi}{3}\right)^2\\[4mm] &\,\hspace{2pt}+\frac{46\sqrt3}{6} \left(x-\frac{\pi}{3}\right)^3+\cdots\\[4mm] ={}& 2+2\sqrt3\left(x-\frac{\pi}{3}\right) +7\left(x-\frac{\pi}{3}\right)^2\\[4mm] &\,\hspace{2pt}+\frac{23\sqrt3}{3} \left(x-\frac{\pi}{3}\right)^3+\cdots \end{align*}

(c)

解法一

思路

展开

因为 7π24\frac{7\pi}{24} 离 π3=8π24\frac{\pi}{3}=\frac{8\pi}{24} 很近，所以把 x=7π24x=\frac{7\pi}{24} 代入 (b) 的展开式即可。注意这是用展开式近似，不是直接算原函数。

答题过程

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When x=7π24x=\frac{7\pi}{24},

x−π3=7π24−8π24=−π24\begin{align*} x-\frac{\pi}{3} ={}& \frac{7\pi}{24}-\frac{8\pi}{24}\\[4mm] ={}& -\frac{\pi}{24} \end{align*}

Therefore

sec⁡(7π24)≈2+23(−π24)+7(−π24)2 +2333(−π24)3=1.636709263…\begin{align*} \sec\left(\frac{7\pi}{24}\right) \approx{}& 2+2\sqrt3\left(-\frac{\pi}{24}\right) +7\left(-\frac{\pi}{24}\right)^2\\[4mm] &\,\hspace{2pt}+\frac{23\sqrt3}{3} \left(-\frac{\pi}{24}\right)^3\\[4mm] ={}& 1.636709263\ldots \end{align*}

Hence, to four significant figures,

sec⁡(7π24)≈1.637\begin{align*} \boxed{\sec\left(\frac{7\pi}{24}\right)\approx 1.637} \end{align*}