曲线 $C$ 的方程为 $y=\dfrac{ax^2+x-1}{x-1}$，其中 $a$ 是正常数。

(a) 求 $C$ 的渐近线方程。

(b) 证明 $C$ 上不存在满足 $1<y<1+4a$ 的点。

(c) 画出 $C$ 的草图。你不需要求出曲线与坐标轴交点的坐标。

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。

多项式除法给出斜渐近线；竖直渐近线处还要确认分子不为零。

At $x=1$, the numerator is $a>0$, so $\boxed{x=1}$ is a vertical asymptote.

Polynomial division gives

Since $a/(x-1)\to0$ as $x\to\pm\infty$, the oblique asymptote is $\boxed{y=ax+a+1}$.

把曲线方程视为关于 $x$ 的二次方程。若存在实数横坐标，判别式必须非负。

Rearranging the curve equation gives

Since $a>0$, a real point requires

This holds only when $y\le1$ or $y\ge1+4a$. For $1<y<1+4a$, the two factors have opposite signs and $\Delta<0$, so no real $x$ exists. Therefore there is no point on $C$ with

另可分别找出两条分支的极值，并用渐近线与单调性说明它们的完整纵坐标范围。

From part (a),

The stationary points are $x=0$ and $x=2$, with ordinates $1$ and $1+4a$ respectively.

On $x<1$, the derivative is positive for $x<0$ and negative for $0<x<1$. This branch tends to $-\infty$ at both ends ($x\to-\infty$ and $x\to1^-$), so its maximum is $(0,1)$ and its range is $y\le1$.

On $x>1$, the derivative is negative for $1<x<2$ and positive for $x>2$. This branch tends to $+\infty$ at both ends, so its minimum is $(2,1+4a)$ and its range is $y\ge1+4a$.

Hence neither branch contains a point with $\boxed{1<y<1+4a}$.

除渐近线外，图形还要体现左支穿过四个象限、左支在斜线下方以及右支在斜线上方。

Draw the asymptotes $x=1$ and $y=ax+a+1$. Since

the left branch lies below the oblique asymptote and the right branch lies above it.

The left branch rises from $-\infty$ to the maximum $(0,1)$ and then falls to $-\infty$ as $x\to1^-$. It crosses both halves of the $x$-axis on opposite sides of the $y$-axis and therefore passes through all four quadrants. The right branch falls from $+\infty$ to the minimum $(2,1+4a)$ and rises again towards the oblique asymptote. Label these two stationary points and both asymptotes.