曲线 $C$ 的方程为 $y=\dfrac{x^2+ax+1}{x+2}$，其中 $a>\frac52$。

(a) 求 $C$ 的渐近线方程。

(b) 证明 $C$ 没有驻点。

(c) 画出 $C$ 的草图，写出它与 $y$ 轴交点的坐标，并标明渐近线。

(d) (i) 画出方程为 $y=\left|\dfrac{x^2+ax+1}{x+2}\right|$ 的曲线的草图。

(ii) 在 (i) 的草图上画出直线 $y=a$。

(iii) 已知当 $-5-\sqrt{14}<x<-3$ 以及 $-5+\sqrt{14}<x<3$ 时，$\left|\dfrac{x^2+ax+1}{x+2}\right|<a$。

求 $a$ 的值。

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。

多项式除法分出斜渐近线与余项。参数条件确保x=−2处分子不为零。

Division gives

Since $a>\frac52$, the remainder $5-2a$ is nonzero. Therefore the asymptotes are

从除法结果求导，参数条件使额外的分式为正，导数恒大于1。

Differentiating the expression in (a),

For $x\ne-2$, the denominator is positive and $2a-5>0$. Hence $\mathrm{d}y/\mathrm{d}x>1>0$, so $C$ has no stationary points.

也可从商法则得到导数分子的二次式，证明其判别式为负，因此没有导数为零的实数点。

The quotient rule gives

A stationary point would require $x^2+4x+2a-1=0$. Its discriminant is

since $a>\frac52$. Thus there are no real solutions and no stationary points.

两支都严格上升。余项符号决定左支在斜渐近线上方、右支在下方；横轴交点分居竖直渐近线两侧。

At $x=0$, $y=\frac12$, so the required $y$-intercept is $\boxed{(0,\frac12)}$.

Label the asymptotes $x=-2$ and $y=x+a-2$. Sketch two increasing branches:

• For $x<-2$, the curve lies above the oblique asymptote and rises from $-\infty$ to $+\infty$ as $x\to-2^-$.
• For $x>-2$, it lies below the oblique asymptote, rising from $-\infty$ as $x\to-2^+$, through $(0,\frac12)$, and onwards to $+\infty$.

The $x$-intercepts are $\xi_\pm=(-a\pm\sqrt{a^2-4})/2$, with $\xi_-<-2<\xi_+<0$. Both tails approach the oblique asymptote. Its intersection with the vertical asymptote is $(-2,a-4)$, whose height depends on $a$.

把原曲线横轴以下的部分向上反射，横轴以上不变。两个原横轴交点成为尖点，竖直渐近线两侧都趋于正无穷。

Reflect the negative parts of $C$ in the $x$-axis, retaining the nonnegative parts.

The graph has sharp minima at $(\xi_-,0)$ and $(\xi_+,0)$, where $\xi_\pm=(-a\pm\sqrt{a^2-4})/2$. The $y$-intercept remains $(0,\frac12)$, and both sides tend to $+\infty$ at $x=-2$.

At the far left the graph approaches $y=-x-a+2$; at the far right it approaches $y=x+a-2$.

a大于二分之五，所以水平线在纵轴截距上方，并与两个尖点的两侧各交一次。

On the sketch from (i), draw the horizontal line $\boxed{y=a}$. Since $a>\frac52>\frac12$, it lies above $(0,\frac12)$ and intersects the four arms once each.

利用已知右端点3对应曲线与水平线的交点，先求a，再核对其余三个端点，保证没有误用正负分支。

At the rightmost boundary $x=3$, the original function is positive, so

Thus $10+3a=5a$, giving $\boxed{a=5}$.

To check all endpoints, when $a=5$ the equations $y=5$ and $y=-5$ become, respectively,

Their roots are $x=\pm3$ and $x=-5\pm\sqrt{14}$. They are ordered

Using the two reflected minima from (i), the regions below $y=5$ are exactly

matching the given intervals; $a=5>\frac52$ is valid.