题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

A curve $C$ has equation

y=2xe^{x^2+(3k-2)x}

where $k$ is a constant.

Given that $C$ has two distinct turning points, find the range of possible values of $k$ .

(7)

题目中文翻译

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不接受完全依赖计算器技术的解法。

曲线 $C$ 的方程为

y=2xe^{x^2+(3k-2)x}

其中 $k$ 为常数。

已知曲线 $C$ 有两个不同的驻点，求 $k$ 的可能取值范围。

解答

Let

y=2xe^{x^2+(3k-2)x}

For convenience, write

u=x^2+(3k-2)x

Then

\frac{\mathrm{d}u}{\mathrm{d}x}=2x+3k-2

Differentiate $y=2xe^u$ using the product rule:

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &=2e^u+2x e^u\frac{\mathrm{d}u}{\mathrm{d}x} \\ &=2e^u+2x e^u(2x+3k-2) \\ &=2e^u\left[1+x(2x+3k-2)\right] \\ &=2e^u\left[2x^2+(3k-2)x+1\right] \end{aligned}

Since $e^u>0$ for all real $x$ , the turning points occur when

2x^2+(3k-2)x+1=0

The curve has two distinct turning points when this quadratic has two distinct real roots.

So its discriminant must be positive:

(3k-2)^2-4(2)(1)>0

Hence

\begin{aligned} (3k-2)^2-8&>0 \\ 9k^2-12k+4-8&>0 \\ 9k^2-12k-4&>0 \end{aligned}

Now solve

9k^2-12k-4=0

Using the quadratic formula,

\begin{aligned} k&=\frac{12\pm\sqrt{(-12)^2-4(9)(-4)}}{2(9)} \\ &=\frac{12\pm\sqrt{144+144}}{18} \\ &=\frac{12\pm12\sqrt2}{18} \\ &=\frac{2\pm2\sqrt2}{3} \end{aligned}

Since $9k^2-12k-4>0$ is true outside the two roots,

\boxed{k<\frac{2-2\sqrt2}{3}\quad\text{or}\quad k>\frac{2+2\sqrt2}{3}}