题目

Problem

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

Solutions relying on calculator technology are not acceptable.

Figure 3 shows

Figure 3

The line $l$ intersects the curve $C$ at the points $P$ and $Q$ , as shown in Figure 3.

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

(4)

The region $R$ , shown shaded in Figure 3, is bounded by $C$ , $l$ and the $x$ -axis.

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

(3)

解答

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交点满足直线和曲线的 $y$ 值相等。先解出 $x$ ，再代回直线方程求 $y$ 。

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The line is

\begin{align*} y=5x+75. \end{align*}

At intersections,

\begin{align*} 2x^2+x-21=&\,5x+75\\ 2x^2-4x-96=&\,0\\ x^2-2x-48=&\,0\\ (x-8)(x+6)=&\,0. \end{align*}

\begin{align*} x=8\quad\text{or}\quad x=-6. \end{align*}

Using $y=5x+75$ :

\begin{align*} x=-6&:\quad y=5(-6)+75=45,\\ x=8&:\quad y=5(8)+75=115. \end{align*}

Therefore

\begin{align*} P=(-6,45),\qquad Q=(8,115). \end{align*}

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图中阴影区域在 $x$ 轴上方，同时在直线下方、在抛物线下方。为了只取左边被三条边围住的那一块，还要限制 $x\le-\frac72$ ，因为抛物线和 $x$ 轴的左交点是 $x=-\frac72$ 。

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The region is above the $x$ -axis:

\begin{align*} y\ge0. \end{align*}

It is below the line:

\begin{align*} y\le5x+75. \end{align*}

It is also below the curve:

\begin{align*} y\le2x^2+x-21. \end{align*}

The left $x$ -intercept of the curve is found from

\begin{align*} 2x^2+x-21=&\,0\\ (2x+7)(x-3)=&\,0, \end{align*}

so it is $x=-\frac72$ .

Therefore $R$ is defined by

\begin{align*} y&\ge0,\\ y&\le5x+75,\\ y&\le2x^2+x-21,\\ x&\le-\frac72. \end{align*}