题目

Problem

Figure 3 shows a sketch of part of the curve with equation

y=2\cos 3x-3x+4\qquad x>0

where $x$ is measured in radians.

The curve crosses the $x$ -axis at the point $P$ , as shown in Figure 3.

Given that the $x$ coordinate of $P$ is $\alpha$ ,

(a) show that $\alpha$ lies between $0.8$ and $0.9$ .

(2)

The iteration formula

x_{n+1}=\frac{1}{3}\arccos(1.5x_n-2)

can be used to find an approximate value for $\alpha$ .

(b) Using this iteration formula with $x_1=0.8$ find, to 4 decimal places, the value of

(i) $x_2$

(ii) $x_5$

(3)

The point $Q$ and the point $R$ are local minimum points on the curve, as shown in Figure 3.

Given that the $x$ coordinates of $Q$ and $R$ are $\beta$ and $\lambda$ respectively, and that they are the two smallest values of $x$ at which local minima occur,

(6)

题目中文翻译

图 3 给出了曲线

y=2\cos 3x-3x+4\qquad x>0

的一部分草图，其中 $x$ 用弧度表示。

如图所示，曲线在点 $P$ 处与 $x$ 轴相交。

已知点 $P$ 的 $x$ 坐标为 $\alpha$ ，

(a) 证明 $\alpha$ 位于 $0.8$ 与 $0.9$ 之间。

迭代公式

x_{n+1}=\frac{1}{3}\arccos(1.5x_n-2)

可用于求 $\alpha$ 的近似值。

(b) 用该迭代公式并取 $x_1=0.8$ ，求下列各值，答案精确到小数点后 4 位：

(i) $x_2$

(ii) $x_5$

点 $Q$ 与点 $R$ 为曲线上的局部极小点，如图 3 所示。

已知点 $Q$ 与点 $R$ 的 $x$ 坐标分别为 $\beta$ 与 $\lambda$ ，并且它们是曲线上取得局部极小值时最小的两个 $x$ 值，

解答

(a)

The $x$ -coordinate of $P$ is a root of

2\cos3x-3x+4=0

Let

F(x)=2\cos3x-3x+4

At $x=0.8$ ,

F(0.8)=2\cos(2.4)-3(0.8)+4

Using a calculator,

F(0.8)=0.125\ldots>0

At $x=0.9$ ,

F(0.9)=2\cos(2.7)-3(0.9)+4

Using a calculator,

F(0.9)=-0.508\ldots<0

Since $F(x)$ changes sign between $0.8$ and $0.9$ , there is a root between $0.8$ and $0.9$ .

Therefore

\boxed{0.8<\alpha<0.9}

(b)

The iteration formula is

x_{n+1}=\frac13\arccos(1.5x_n-2)

with

x_1=0.8

(i)

\begin{aligned} x_2 &=\frac13\arccos(1.5(0.8)-2)\\ &=\frac13\arccos(-0.8)\\ &=0.832697\ldots \end{aligned}

\boxed{x_2=0.8327}

to 4 decimal places.

(ii)

Continue the iteration:

\begin{aligned} x_2&=0.832697\ldots\\ x_3&=0.806767\ldots\\ x_4&=0.827120\ldots\\ x_5&=0.811022\ldots \end{aligned}

Therefore

\boxed{x_5=0.8110}

to 4 decimal places.

(c)

The curve is

y=2\cos3x-3x+4

Differentiate:

\frac{\mathrm{d}y}{\mathrm{d}x}=-6\sin3x-3

At a stationary point,

-6\sin3x-3=0

\sin3x=-\frac12

For local minima, we need to classify the stationary points.

Differentiate again:

\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-18\cos3x

A local minimum occurs when

\frac{\mathrm{d}^2y}{\mathrm{d}x^2}>0

-18\cos3x>0

which means

\cos3x<0

Now solve

\sin3x=-\frac12

with $\cos3x<0$ .

The relevant angles are in the third quadrant:

3x=\frac{7\pi}{6}+2n\pi

where $n$ is an integer.

Therefore

x=\frac{7\pi}{18}+\frac{2n\pi}{3}

The two smallest positive values occur when $n=0$ and $n=1$ .

\beta=\frac{7\pi}{18}

and

\lambda=\frac{7\pi}{18}+\frac{2\pi}{3}

Hence

\lambda=\frac{7\pi}{18}+\frac{12\pi}{18} =\frac{19\pi}{18}

Therefore

\boxed{\beta=\frac{7\pi}{18},\qquad \lambda=\frac{19\pi}{18}}

IAL 2020 Jan Q7

题目

解答

(a)

(b)

(i)

(ii)

(c)