题目

Problem

Figure 1 shows a line $l$ with equation $x+y=6$ and a curve $C$ with equation

\begin{align*} y=6x-2x^2+1. \end{align*}

Figure 1

The line $l$ intersects the curve $C$ at the points $P$ and $Q$ as shown in Figure 1.

(a) Find, using algebra, the coordinates of $P$ and the coordinates of $Q$ .

(4)

The region $R$ , shown shaded in Figure 1, is bounded by $C$ , $l$ and the $x$ -axis.

(b) Use inequalities to define the region $R$ .

(3)

解答

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交点同时满足直线和曲线。先由直线写出 $y=6-x$ ，再代入曲线方程，得到一个二次方程。题目要求用代数，所以要把解方程过程写出来。

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From the line $x+y=6$ ,

\begin{align*} y=6-x. \end{align*}

At the points of intersection,

\begin{align*} 6-x=&\,6x-2x^2+1\\ 0=&\,7x-2x^2-5\\ 2x^2-7x+5=&\,0\\ (2x-5)(x-1)=&\,0. \end{align*}

\begin{align*} x=1\quad\text{or}\quad x=\frac52. \end{align*}

Using $y=6-x$ :

\begin{align*} x=1&\implies y=5,\\ x=\frac52&\implies y=\frac72. \end{align*}

Therefore the points of intersection are

\begin{align*} (1,5)\quad\text{and}\quad \left(\frac52,\frac72\right). \end{align*}

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区域由三条边界围成：曲线、直线和 $x$ 轴。图中区域在曲线的上方、直线的下方、 $x$ 轴的上方，并且在两个交点中较右侧交点的右边开始。

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The region is above the curve $C$ , so

\begin{align*} y\ge 6x-2x^2+1. \end{align*}

It is below the line $l$ , so

\begin{align*} x+y\le 6. \end{align*}

It is above the $x$ -axis, so

\begin{align*} y\ge 0. \end{align*}

From part (a), the region begins at the right-hand intersection, where $x=\frac52$ . Hence

\begin{align*} x\ge \frac52. \end{align*}

Therefore $R$ is defined by

\begin{align*} y&\ge 6x-2x^2+1,\\ x+y&\le 6,\\ y&\ge 0,\\ x&\ge \frac52. \end{align*}