题目

Problem

Figure 3 shows a sketch of part of the curve with equation $y=f(x)$ , where

f(x)=x(x^2-4)e^{-x/2}

(a) Find $f'(x)$ .

(2)

The line $l$ is the normal to the curve at $O$ and meets the curve again at the point $P$ .

The point $P$ lies in the 3rd quadrant, as shown in Figure 3.

(b) Show that the $x$ coordinate of $P$ is a solution of the equation

x=-\frac12\sqrt{16+e^{x/2}}

(4)

x_{n+1}=-\frac12\sqrt{16+e^{x_n/2}}\qquad \text{with }x_1=-2

find, to 4 decimal places,

(i) the value of $x_2$

(ii) the $x$ coordinate of $P$ .

(3)

题目中文翻译

图 3 给出了部分曲线的示意图，其方程为 $y=f(x)$ ，其中

f(x)=x(x^2-4)e^{-x/2}

(a) 求 $f'(x)$ 。

直线 $l$ 是曲线在点 $O$ 处的法线，并与曲线再次相交于点 $P$ 。

点 $P$ 位于第三象限，如图 3 所示。

(b) 证明点 $P$ 的 $x$ 坐标是方程

x=-\frac12\sqrt{16+e^{x/2}}

的一个解。

x_{n+1}=-\frac12\sqrt{16+e^{x_n/2}}\qquad \text{其中 }x_1=-2

求下列各值（精确到小数点后 4 位）：

(i) $x_2$

(ii) 点 $P$ 的 $x$ 坐标。

解答

(a)

First write

f(x)=(x^3-4x)e^{-x/2}

Using the product rule,

\begin{aligned} f'(x) &=(3x^2-4)e^{-x/2} +(x^3-4x)\left(-\frac12e^{-x/2}\right) \\ &=e^{-x/2}\left(3x^2-4-\frac12x^3+2x\right) \end{aligned}

\boxed{f'(x)=\left(-\frac12x^3+3x^2+2x-4\right)e^{-x/2}}

(b)

At $O$ , we have $x=0$ .

From part (a),

f'(0)=-4

So the gradient of the tangent at $O$ is $-4$ .

Therefore the gradient of the normal is

\frac14

Hence the normal $l$ has equation

y=\frac14x

At the second intersection point $P$ ,

x(x^2-4)e^{-x/2}=\frac14x

Since $P$ is not $O$ , $x\ne0$ , so we can divide by $x$ :

(x^2-4)e^{-x/2}=\frac14

Thus

x^2-4=\frac14e^{x/2}

4x^2=16+e^{x/2}

and hence

x^2=\frac14(16+e^{x/2})

Since $P$ lies in the third quadrant, its $x$ coordinate is negative. Therefore

\boxed{x=-\frac12\sqrt{16+e^{x/2}}}

(c)

(i)

Using $x_1=-2$ ,

x_2=-\frac12\sqrt{16+e^{-1}}

\boxed{x_2=-2.0229}

to 4 decimal places.

(ii)

Continuing the iteration,

\begin{aligned} x_1&=-2,\\ x_2&=-2.0229,\\ x_3&=-2.0226,\\ x_4&=-2.0226 \end{aligned}

The values converge to

\boxed{x=-2.0226}

So the $x$ coordinate of $P$ is

\boxed{-2.0226}

IAL 2021 Oct Q9

题目

解答

(a)

(b)

(c)

(i)

(ii)