曲线 $C$ 的方程为 $y=\dfrac{x^2+3}{x^2+1}$。

(a) 证明 $C$ 没有竖直渐近线，并写出水平渐近线的方程。

(b) 证明：对所有实数 $x$，$1<y\leqslant3$。

(c) 求 $C$ 上所有驻点的坐标。

(d) 画出 $C$ 的草图，写出它与坐标轴的所有交点的坐标，并标注渐近线。

(e) 画出方程为 $y=\dfrac{x^2+1}{x^2+3}$ 的曲线的草图，并求满足 $\dfrac{x^2+1}{x^2+3}<\dfrac12$ 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。

分母x平方加1恒正，因而没有竖直渐近线；无穷远处比值趋于1。

For all real $x$, $x^2+1\geqslant1>0$, so the denominator never vanishes. Hence there are no vertical asymptotes.

Also

so the horizontal asymptote is $\boxed{y=1}$.

移项得到关于x的二次方程，用实根判别式，再排除y=1的退化情形。

Rearranging,

The value $y=1$ is impossible, since it would give $2=0$. Otherwise the quadratic must have real roots, so

Thus $(1-y)(3-y)\leqslant0$, giving $1\leqslant y\leqslant3$. Excluding $y=1$ yields

改写成1加正分式，分母至少为1即可直接给出上下界。

Write

Since $x^2+1\geqslant1$,

Adding $1$ gives

The upper bound occurs at $x=0$; the lower bound is approached but never attained.

求导后分母恒正，导数只在x=0为零。

Differentiating,

This is zero only at $x=0$, where $y=3$. Thus the only stationary point is $\boxed{(0,3)}$, a maximum.

函数为偶函数，最高点在纵轴上，两侧向水平渐近线靠近，没有横轴交点。

The curve is symmetric about the $y$-axis. Its only axis intersection is $\boxed{(0,3)}$, because $x^2+3>0$ gives no $x$-intercepts.

Sketch a smooth maximum at $(0,3)$, increasing for $x<0$ and decreasing for $x>0$. Both tails approach the labelled asymptote $y=1$ from above.

倒数曲线仍为偶函数，原最大值3变成最小值三分之一。分母恒正，不等式交叉相乘不会反向。

For $g(x)=(x^2+1)/(x^2+3)$,

Sketch an even curve with smooth minimum $\boxed{(0,\frac13)}$, no $x$-intercepts and horizontal asymptote $\boxed{y=1}$ approached from below on both sides.

Since $x^2+3>0$,

Hence $\boxed{-1<x<1}$.