题目

Problem

(i) Solve

\begin{align*} \frac3x>4. \end{align*}

(3)

(ii) Figure 1 shows a sketch of the curve $C$ and the straight line $l$ .

Figure 1

The infinite region $R$ , shown shaded in Figure 1, lies in quadrants 2 and 3 and is bounded by $C$ and $l$ only.

Given that

use inequalities to define the region $R$ .

(3)

解答

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分式不等式不能直接乘以 $x$ ，因为 $x$ 的正负未知。更稳的方法是移到一边，通分后看关键点 $x=0$ 和 $x=\frac34$ 。

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\begin{align*} \frac3x&>4\\ \frac3x-4&>0\\ \frac{3-4x}{x}&>0. \end{align*}

The critical values are

\begin{align*} x=0,\quad x=\frac34. \end{align*}

Test the intervals:

\begin{align*} x<0&:\quad \frac{3-4x}{x}<0,\\ 0<x<\frac34&:\quad \frac{3-4x}{x}>0,\\ x>\frac34&:\quad \frac{3-4x}{x}<0. \end{align*}

Therefore

\begin{align*} 0<x<\frac34. \end{align*}

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直线经过负 $x$ 轴上的交点。曲线与 $x$ 轴交于 $2x^2-50=0$ ，负的交点是 $x=-5$ ，所以直线经过 $(-5,0)$ ，斜率为 $3$ 。

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The curve meets the $x$ -axis when

\begin{align*} 2x^2-50=&\,0\\ x^2=&\,25\\ x=&\,\pm5. \end{align*}

The intersection on the negative $x$ -axis is $(-5,0)$ .

The line $l$ has gradient $3$ , so

\begin{align*} y-0=&\,3(x+5)\\ y=&\,3x+15. \end{align*}

From the diagram, $R$ is below the curve, above the line, and to the left of the intersection. Hence

\begin{align*} y&<2x^2-50,\\ y&>3x+15,\\ x&<-5. \end{align*}