题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Figure 1 shows a sketch of part of the curve with equation $y=f(x)$ where

f(x)=(x-2)^2e^{3x}\qquad x\in\mathbb{R}

The curve has a maximum turning point at $A$ and a minimum turning point at $(2,0)$ .

(a) Use calculus to find the exact coordinates of $A$ .

(5)

Given that the equation $f(x)=k$ , where $k$ is a constant, has at least two distinct roots,

(b) state the range of possible values for $k$ .

(2)

题目中文翻译

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

图 1 给出了部分曲线 $y=f(x)$ 的示意图，其中

f(x)=(x-2)^2e^{3x}\qquad x\in\mathbb{R}

该曲线在点 $A$ 处有一个极大转折点，在 $(2,0)$ 处有一个极小转折点。

(a) 用微积分求点 $A$ 的精确坐标。

已知方程 $f(x)=k$ （其中 $k$ 是常数）至少有两个不同的根。

(b) 写出 $k$ 的可能取值范围。

解答

We have

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

求导得

f'(x)=2(x-2)e^{3x}+(x-2)^2\cdot3e^{3x}

Factorise:

f'(x)=e^{3x}(x-2)\left[2+3(x-2)\right]

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

At a turning point,

f'(x)=0

Since

e^{3x}>0

we have

x-2=0

3x-4=0

Thus

x=2 \qquad\text{or}\qquad x=\frac43

The point $(2,0)$ is the minimum point, so $A$ has

x=\frac43

Now find the $y$ coordinate:

y=\left(\frac43-2\right)^2e^{3(4/3)}

y=\left(-\frac23\right)^2e^4 =\frac49e^4

Therefore

\boxed{A=\left(\frac43,\frac49e^4\right)}

The curve has a local minimum at

(2,0)

and a local maximum at

\left(\frac43,\frac49e^4\right)

For the equation

f(x)=k

to have at least two distinct roots, the horizontal line $y=k$ must meet the curve at least twice.

Since

f(x)=(x-2)^2e^{3x}\ge0

we need

k>0

Also, if

k=\frac49e^4

the line passes through the local maximum and also meets the right branch, so there are still at least two distinct roots.

Hence

\boxed{0<k\le \frac49e^4}