题目

Problem

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

Solutions relying on calculator technology are not acceptable.

A curve has equation

\begin{align*} y=3x^5+4x^3-x+5. \end{align*}

The points $P$ and $Q$ lie on the curve.

The gradient of the curve at both point $P$ and point $Q$ is $2$ .

Find the $x$ coordinates of $P$ and $Q$ .

(5)

解答

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曲线的梯度由导数给出。题目说梯度为 $2$ ，所以先求 $\frac{dy}{dx}$ ，再令它等于 $2$ 。整理后会得到关于 $x^2$ 的二次方程。

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\begin{align*} \frac{dy}{dx} =&\,15x^4+12x^2-1. \end{align*}

The gradient is $2$ , so

\begin{align*} 15x^4+12x^2-1=&\,2\\ 15x^4+12x^2-3=&\,0\\ 5x^4+4x^2-1=&\,0. \end{align*}

Factorise as a quadratic in $x^2$ :

\begin{align*} 5x^4+4x^2-1 =&\,(5x^2-1)(x^2+1). \end{align*}

Therefore

\begin{align*} (5x^2-1)(x^2+1)=&\,0. \end{align*}

Since $x^2+1=0$ has no real solutions,

\begin{align*} 5x^2-1=&\,0\\ x^2=&\,\frac15\\ x=&\,\pm\frac1{\sqrt5}. \end{align*}

So the $x$ coordinates are

\begin{align*} -\frac1{\sqrt5}\quad\text{and}\quad \frac1{\sqrt5}. \end{align*}