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IAL 2021 Jan Q6

A Level / Edexcel / P3

IAL 2021 Jan Paper · Question 6

题目

Problem

f(x)=xcos⁡(x3)x>0f(x)=x\cos\left(\frac{x}{3}\right)\qquad x>0

(a) Find f′(x)f'(x).

(2)

(b) Show that the equation f′(x)=0f'(x)=0 can be written as

x=karctan⁡(kx)x=k\arctan\left(\frac{k}{x}\right)

where kk is an integer to be found.

(2)

(c) Starting with x1=2.5x_1=2.5 use the iteration formula

xn+1=karctan⁡(kxn)x_{n+1}=k\arctan\left(\frac{k}{x_n}\right)

with the value of kk found in part (b), to calculate the values of x2x_2 and x6x_6 giving your answers to 3 decimal places.

(2)

(d) Using a suitable interval and a suitable function that should be stated, show that a root of f′(x)=0f'(x)=0 is 2.5812.581 correct to 3 decimal places.

(2)
题目中文翻译 f(x)=xcos⁡(x3)x>0f(x)=x\cos\left(\frac{x}{3}\right)\qquad x>0

(a) 求 f′(x)f'(x)。

(b) 证明方程 f′(x)=0f'(x)=0 可写为

x=karctan⁡(kx)x=k\arctan\left(\frac{k}{x}\right)

其中 kk 为待求整数。

(c) 取 x1=2.5x_1=2.5，并使用迭代公式

xn+1=karctan⁡(kxn)x_{n+1}=k\arctan\left(\frac{k}{x_n}\right)

其中 kk 取自(b)中的结果，计算 x2x_2 和 x6x_6 的值，答案精确到小数点后 3 位。

(d) 选取合适区间并写出合适函数，证明 f′(x)=0f'(x)=0 的一个根为 2.5812.581（精确到小数点后 3 位）。

解答

(a)

We have

f(x)=xcos⁡(x3)f(x)=x\cos\left(\frac{x}{3}\right)

Differentiate using the product rule:

f′(x)=cos⁡(x3)+x[−sin⁡(x3)]⋅13=cos⁡(x3)−x3sin⁡(x3)\begin{aligned} f'(x) &=\cos\left(\frac{x}{3}\right) +x\left[-\sin\left(\frac{x}{3}\right)\right]\cdot \frac13\\ &=\cos\left(\frac{x}{3}\right)-\frac{x}{3}\sin\left(\frac{x}{3}\right) \end{aligned}

Therefore

f′(x)=cos⁡(x3)−x3sin⁡(x3)\boxed{f'(x)=\cos\left(\frac{x}{3}\right)-\frac{x}{3}\sin\left(\frac{x}{3}\right)}

(b)

Set

f′(x)=0f'(x)=0

So

cos⁡(x3)−x3sin⁡(x3)=0\cos\left(\frac{x}{3}\right)-\frac{x}{3}\sin\left(\frac{x}{3}\right)=0

Rearrange:

cos⁡(x3)=x3sin⁡(x3)\cos\left(\frac{x}{3}\right)=\frac{x}{3}\sin\left(\frac{x}{3}\right)

Divide by

cos⁡(x3)\cos\left(\frac{x}{3}\right)

and by x3\dfrac{x}{3}:

tan⁡(x3)=3x\tan\left(\frac{x}{3}\right)=\frac{3}{x}

Taking arctan,

x3=arctan⁡(3x)\frac{x}{3}=\arctan\left(\frac{3}{x}\right)

Therefore

x=3arctan⁡(3x)x=3\arctan\left(\frac{3}{x}\right)

So

k=3\boxed{k=3}

(c)

Using

k=3k=3

the iteration formula is

xn+1=3arctan⁡(3xn)x_{n+1}=3\arctan\left(\frac{3}{x_n}\right)

with

x1=2.5x_1=2.5

Now

x2=3arctan⁡(32.5)x_2=3\arctan\left(\frac{3}{2.5}\right)

So

x2=2.628174…x_2=2.628174\ldots

Continuing the iteration:

x2=2.628174…x3=2.554102…x4=2.596527…x5=2.572101…x6=2.586122…\begin{aligned} x_2&=2.628174\ldots\\ x_3&=2.554102\ldots\\ x_4&=2.596527\ldots\\ x_5&=2.572101\ldots\\ x_6&=2.586122\ldots \end{aligned}

Therefore

x2=2.628,x6=2.586\boxed{x_2=2.628,\qquad x_6=2.586}

to 3 decimal places.

(d)

Use the function

F(x)=f′(x)=cos⁡(x3)−x3sin⁡(x3)F(x)=f'(x)=\cos\left(\frac{x}{3}\right)-\frac{x}{3}\sin\left(\frac{x}{3}\right)

To show that the root is 2.5812.581 to 3 decimal places, check the interval

2.5805<x<2.58152.5805<x<2.5815

Now

F(2.5805)=0.000346…>0F(2.5805)=0.000346\ldots>0

and

F(2.5815)=−0.000346…<0F(2.5815)=-0.000346\ldots<0

Since F(x)F(x) is continuous and changes sign in this interval, there is a root between 2.58052.5805 and 2.58152.5815.

Therefore the root is

2.581\boxed{2.581}

to 3 decimal places.