曲线 $C$ 的方程为 $y=\dfrac{2x^2-x-1}{x^2+x+1}$。

(a) 证明 $C$ 没有竖直渐近线，并写出 $C$ 的水平渐近线方程。

(b) 求 $C$ 上驻点的坐标。

(c) 画出 $C$ 的草图，并写出曲线与坐标轴交点的坐标。

(d) 画出方程为 $y=\Big|\dfrac{2x^2-x-1}{x^2+x+1}\Big|$ 的曲线的草图，并写出使方程 $\Big|\dfrac{2x^2-x-1}{x^2+x+1}\Big|=k$ 有 $4$ 个不同实数解的 $k$ 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。

配方说明分母恒正；无穷远处比较分子、分母的最高次项。

For every real $x$,

The denominator never vanishes, so there are no vertical asymptotes. Also,

Thus the horizontal asymptote is $\boxed{y=2}$.

用商法则求导，把分子完全整理后解导数为零，并代回原函数求纵坐标。

By the quotient rule,

The denominator is positive. Thus stationary points occur at $x=-2$ and $x=0$. Substitution gives

The derivative is positive for $x<-2$, negative for $-2<x<0$, and positive for $x>0$. Therefore the stationary points are

另把曲线改写成关于 $x$ 的二次方程。驻点对应水平线与曲线相切，即二次方程有重根。

Rearrange the curve equation as

Here $F_y=x^2+x+1>0$. Implicit differentiation gives $F_x+F_y\,\mathrm{d}y/\mathrm{d}x=0$, so at a stationary point $F_x=0$ as well as $F=0$.

For $y\ne2$, this means the quadratic in $x$ has a repeated root. Its discriminant is

Thus $\Delta=0$ gives $y=-1$ or $y=3$. For $y=-1$, the equation is $-3x^2=0$, giving $x=0$. For $y=3$, it is $(x+2)^2=0$, giving $x=-2$.

The exceptional value $y=2$ gives $F=3x+3$ and $F_x=3\ne0$, so it supplies no stationary point. Hence the stationary points are

结合渐近线、驻点和截距作图，并检查从水平渐近线的哪一侧接近。

For the $x$-intercepts,

Hence the axis intersections are

Also,

The curve approaches $y=2$ from above as $x\to-\infty$, rises to $(-2,3)$, then falls through $(-1,2)$ and $(-1/2,0)$ to $(0,-1)$. It then rises through $(1,0)$ towards $y=2$ from below as $x\to\infty$.

绝对值只把横轴下方的一段反射到上方。数水平线与反射后曲线的交点，注意中央峰顶只能算一个交点。

Reflect the portion $-1/2<x<1$ of $C$ in the $x$-axis, leaving the rest unchanged. The reflected curve has cusps at $(-1/2,0)$ and $(1,0)$, a central maximum at $(0,1)$, and the unchanged maximum at $(-2,3)$. Its horizontal asymptote remains $y=2$.

For $0<k<1$, the line $y=k$ meets the central arch twice and the two outer portions once each, giving four distinct intersections. At $k=0$ there are only the two cusps, and at $k=1$ the two central intersections merge. For $k>1$ there are no central intersections, and the outer portions together give at most two. Negative $k$ gives no intersection.

Thus the required set is $\boxed{0<k<1}$.