题目

Problem

The curve $C$ has equation

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

The curve has a stationary point at the point $P$ .

(a) Show, using calculus, that the $x$ coordinate of $P$ is a solution of the equation

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

(4)

Using the iteration formula

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

(b) find the value of $x_2$ and the value of $x_6$ , giving your answers to 3 decimal places.

(3)

题目中文翻译

曲线 $C$ 的方程为

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

该曲线在点 $P$ 处有一个驻点。

(a) 用求导证明，点 $P$ 的 $x$ 坐标满足方程

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

使用迭代公式

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

(b) 求 $x_2$ 和 $x_6$ 的值，答案精确到小数点后 3 位。

解答

(a)

解法一

思路

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

答题过程

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The curve is

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

Differentiate using the product rule:

\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,

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

Since $0<x\leq\pi$ , we may divide by $x$ and rearrange:

\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 $0<\frac12x\leq\frac{\pi}{2}$ , so taking arctan gives

\frac12x=\arctan\left(\frac4x\right).

Hence

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

as required.

解法二

思路

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官方评分资料也接受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=2\arctan\left(\frac4x\right),

we obtain

\tan\left(\frac12x\right)=\frac4x.

Therefore

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

2x\cos\left(\frac12x\right) -\frac12x^2\sin\left(\frac12x\right)=0.

But

\frac{\mathrm{d}y}{\mathrm{d}x} =2x\cos\left(\frac12x\right) -\frac12x^2\sin\left(\frac12x\right),

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

\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

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

with $x_1=2$ .

For $x_2$ ,

x_2=2\arctan(2)=2.214297\ldots

Therefore

\boxed{x_2=2.214}

to 3 decimal places.

Continuing the iteration:

\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

\boxed{x_6=2.155}

to 3 decimal places.

IAL 2021 June Q1

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程