(a) 画出方程为 $y=\dfrac{x+1}{x-1}$ 的曲线的草图。

(b) 画出方程为 $y=\dfrac{|x|+1}{|x|-1}$ 的曲线的草图，并求满足 $\dfrac{|x|+1}{|x|-1}<-2$ 的 $x$ 的取值集合。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

先写成常数加分式，直接读出两条渐近线，再用截距与单调性确定分支位置。

Write

Here $x\ne1$.

The asymptotes are $x=1$ and $y=1$. The curve passes through $(-1,0)$ and $(0,-1)$. Since

both branches decrease. The branch $x>1$ lies above $y=1$, tending to $+\infty$ as $x\to1^+$. The branch $x<1$ lies below $y=1$, tending to $-\infty$ as $x\to1^-$.

将(a)的右半边关于纵轴反射；不等式令 $t=|x|$，先检查分母符号，再解临界点。

Keep the part of the graph in (a) with $x\ge0$ and reflect it in the $y$-axis. The new graph is even, with vertical asymptotes $x=\pm1$, horizontal asymptote $y=1$, and a cusp at $(0,-1)$. The two outer branches lie above $y=1$; the central portion tends to $-\infty$ as $x\to\pm1$ from inside.

Put $t=|x|\ge0$, with $t\ne1$. Then

For $t>1$, both numerator and denominator are positive, so there is no solution. For $0\le t<1$, the denominator is negative, and the inequality requires $3t-1>0$. Thus $1/3<t<1$, giving