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P3 2021 June Q1

A Level / Edexcel / P3

IAL 2021 June Paper · Question 1

题目

Problem

The curve CC has equation

y=x2cos⁡(12x)0<x≤πy=x^2\cos\left(\frac{1}{2}x\right)\qquad 0<x\le \pi

The curve has a stationary point at the point PP.

(a) Show, using calculus, that the xx coordinate of PP is a solution of the equation

x=2arctan⁡(4x)x=2\arctan\left(\frac{4}{x}\right)
(4)

Using the iteration formula

xn+1=2arctan⁡(4xn)x1=2x_{n+1}=2\arctan\left(\frac{4}{x_n}\right)\qquad x_1=2

(b) find the value of x2x_2 and the value of x6x_6, giving your answers to 3 decimal places.

(3)
题目中文翻译

曲线 CC 的方程为

y=x2cos⁡(12x)0<x≤πy=x^2\cos\left(\frac{1}{2}x\right)\qquad 0<x\le \pi

该曲线在点 PP 处有一个驻点。

(a) 用求导证明,点 PP 的 xx 坐标满足方程

x=2arctan⁡(4x)x=2\arctan\left(\frac{4}{x}\right)

使用迭代公式

xn+1=2arctan⁡(4xn)x1=2x_{n+1}=2\arctan\left(\frac{4}{x_n}\right)\qquad x_1=2

(b) 求 x2x_2 和 x6x_6 的值,答案精确到小数点后 3 位。

解答

(a)

解法一

思路

展开

先用乘积法则求导,再令导数等于零。利用题目给出的范围 0<x≤π0<x\leq\pi 排除 x=0x=0,便可除以 xx 并整理成指定的迭代公式。

答题过程

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The curve is

y=x2cos⁡(12x).y=x^2\cos\left(\frac12x\right).

Differentiate using the product rule:

dydx= 2xcos⁡(12x)−12x2sin⁡(12x).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,2x\cos\left(\frac12x\right) -\frac12x^2\sin\left(\frac12x\right). \end{align*}

At a stationary point,

dydx=0.\frac{\mathrm{d}y}{\mathrm{d}x}=0.

Since 0<x≤π0<x\leq\pi, we may divide by xx and rearrange:

2cos⁡(12x)−12xsin⁡(12x)= 0,4cos⁡(12x)= xsin⁡(12x),tan⁡(12x)= 4x.\begin{align*} 2\cos\left(\frac12x\right) -\frac12x\sin\left(\frac12x\right)=&\,0,\\[2mm] 4\cos\left(\frac12x\right) =&\,x\sin\left(\frac12x\right),\\[2mm] \tan\left(\frac12x\right)=&\,\frac4x. \end{align*}

Here 0<12x≤π20<\frac12x\leq\frac{\pi}{2}, so taking arctan gives

12x=arctan⁡(4x).\frac12x=\arctan\left(\frac4x\right).

Hence

x=2arctan⁡(4x)\boxed{x=2\arctan\left(\frac4x\right)}

as required.

解法二

思路

展开

也可以从题目要求证明的等式反向整理。把它化回导数为零的条件,即可说明该等式确实刻画曲线的驻点。

答题过程

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Starting from

x=2arctan⁡(4x),x=2\arctan\left(\frac4x\right),

we obtain

tan⁡(12x)=4x.\tan\left(\frac12x\right)=\frac4x.

Therefore

xsin⁡(12x)=4cos⁡(12x),x\sin\left(\frac12x\right) =4\cos\left(\frac12x\right),

so

2xcos⁡(12x)−12x2sin⁡(12x)=0.2x\cos\left(\frac12x\right) -\frac12x^2\sin\left(\frac12x\right)=0.

But

dydx=2xcos⁡(12x)−12x2sin⁡(12x),\frac{\mathrm{d}y}{\mathrm{d}x} =2x\cos\left(\frac12x\right) -\frac12x^2\sin\left(\frac12x\right),

hence dydx=0\frac{\mathrm{d}y}{\mathrm{d}x}=0. All the steps above are reversible for 0<x≤π0<x\leq\pi, so the stationary-point condition is equivalent to

x=2arctan⁡(4x).\boxed{x=2\arctan\left(\frac4x\right)}.

(b)

解法一

思路

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把初值逐次代入题目给出的迭代式。计算器须使用弧度制,并保留足够多的中间位数,最后才按要求取三位小数。

答题过程

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The iteration formula is

xn+1=2arctan⁡(4xn),x_{n+1}=2\arctan\left(\frac{4}{x_n}\right),

with x1=2x_1=2.

For x2x_2,

x2=2arctan⁡(2)=2.214297…x_2=2\arctan(2)=2.214297\ldots

Therefore

x2=2.214\boxed{x_2=2.214}

to 3 decimal places.

Continuing the iteration:

x2= 2.214297…x3= 2.130426…x4= 2.162810…x5= 2.150239…x6= 2.155109…\begin{align*} x_2=&\,2.214297\ldots\\ x_3=&\,2.130426\ldots\\ x_4=&\,2.162810\ldots\\ x_5=&\,2.150239\ldots\\ x_6=&\,2.155109\ldots \end{align*}

Hence

x6=2.155\boxed{x_6=2.155}

to 3 decimal places.