曲线 $C$ 的方程为
$y = \frac{x^2 + (\lambda - 1)x}{x - 1}, \quad x \ne 1,$
其中 $\lambda$ 为常数。求

1. Find the vertical asymptote

The vertical asymptote occurs when the denominator is zero:
$x - 1 = 0$
$x = 1$

2. Find the oblique asymptote by expressing the function in proper rational form

$y = \frac{x^2 + (\lambda - 1)x}{x - 1}$
$y = \frac{x(x - 1) + \lambda x}{x - 1}$
$y = \frac{x(x - 1) + \lambda(x - 1) + \lambda}{x - 1}$
$y = x + \lambda + \frac{\lambda}{x - 1}$
As $x \to \pm\infty$, $\frac{\lambda}{x - 1} \to 0$, so the oblique asymptote is:
$y = x + \lambda$

1. Express the equation of the curve in a form easier to differentiate (typically done in part (i) to find asymptotes) and find the first derivative.

By algebraic division, we rewrite the equation of $C$:
$y = \frac{x^2 + (\lambda - 1)x}{x - 1} = \frac{x(x - 1) + \lambda(x - 1) + \lambda}{x - 1}$
$y = x + \lambda + \frac{\lambda}{x - 1}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 1 - \frac{\lambda}{(x - 1)^2}$

2. Set the derivative to zero to find the condition for stationary points.

For stationary points, we set $\frac{dy}{dx} = 0$:
$1 - \frac{\lambda}{(x - 1)^2} = 0$
$(x - 1)^2 = \lambda$

3. Apply the condition that there are two distinct stationary points for $x > 0$ to find the range of $\lambda$.

For $C$ to have two distinct stationary points, the equation must have two distinct real roots, which requires:
$\lambda > 0$
Solving for $x$, we obtain the $x$-coordinates of the stationary points:
$x = 1 \pm \sqrt{\lambda}$
We are given that both stationary points occur for $x > 0$. Thus, both roots must be strictly positive. Since $1 + \sqrt{\lambda} > 0$ is inherently true for $\lambda > 0$, we must enforce that the smaller root is also positive:
$1 - \sqrt{\lambda} > 0$
$\sqrt{\lambda} < 1 \implies \lambda < 1$
Combining this inequality with $\lambda > 0$, the range of values for $\lambda$ is:
$0 < \lambda < 1$

1. Determine the axial intercepts of the curve.

Setting $y = 0$ in $y = \frac{x^2 + (\lambda - 1)x}{x - 1}$ gives:
$x^2 + (\lambda - 1)x = 0$
$x(x + \lambda - 1) = 0$
$x = 0 \quad \text{or} \quad x = 1 - \lambda$
Since $0 < \lambda < 1$ from part (ii), $1 - \lambda > 0$.
The intercepts are $(0,0)$ and $(1-\lambda, 0)$.

2. Determine the coordinates of the stationary points in terms of $\lambda$.

From part (ii), the stationary points occur at $x = 1 \pm \sqrt{\lambda}$.
Substituting these into the asymptotic form $y = x + \lambda + \frac{\lambda}{x - 1}$ from part (i):
For $x = 1 - \sqrt{\lambda}$:
$y = (1 - \sqrt{\lambda}) + \lambda + \frac{\lambda}{-\sqrt{\lambda}} = 1 - \sqrt{\lambda} + \lambda - \sqrt{\lambda} = (1 - \sqrt{\lambda})^2$
For $x = 1 + \sqrt{\lambda}$:
$y = (1 + \sqrt{\lambda}) + \lambda + \frac{\lambda}{\sqrt{\lambda}} = 1 + \sqrt{\lambda} + \lambda + \sqrt{\lambda} = (1 + \sqrt{\lambda})^2$
Thus, the stationary points are $(1-\sqrt{\lambda}, (1-\sqrt{\lambda})^2)$ and $(1+\sqrt{\lambda}, (1+\sqrt{\lambda})^2)$.

3. Sketch the curve using the identified key features.

Using the asymptotes $x = 1$ and $y = x + \lambda$:

• Since $0 < \lambda < \sqrt{\lambda} < 1$, the $x$-coordinates follow the order: $0 < 1-\sqrt{\lambda} < 1-\lambda < 1$.
• The left branch passes through $(0,0)$, goes up to reach a local maximum at $(1-\sqrt{\lambda}, (1-\sqrt{\lambda})^2)$, cuts the $x$-axis again at $(1-\lambda, 0)$, and approaches the vertical asymptote $x = 1$ downwards.
• The right branch approaches $x = 1$ from positive infinity, goes down to reach a local minimum at $(1+\sqrt{\lambda}, (1+\sqrt{\lambda})^2)$, and approaches $y = x + \lambda$ from above.
  (Student draws the graph indicating these key features.)

1. Find the expression for $\mathrm{f}'(x)$ and determine the range of $\lambda$ from part (ii).

From part (i), we can rewrite $\mathrm{f}(x)$:
$\mathrm{f}(x) = \frac{x^2 - x + \lambda x}{x - 1} = x + \lambda + \frac{\lambda}{x - 1}$
Differentiating with respect to $x$:
$\mathrm{f}'(x) = 1 - \frac{\lambda}{(x - 1)^2}$
For part (ii), setting $\mathrm{f}'(x) = 0$ gives the $x$-coordinates of the stationary points:
$(x - 1)^2 = \lambda \implies x = 1 \pm \sqrt{\lambda}$
For 2 stationary points for $x > 0$, we require $\lambda > 0$ (for distinct real roots) and $1 - \sqrt{\lambda} > 0$ (for both roots to be positive).
This gives the range $0 < \lambda < 1$.

2. Find the equations of the asymptotes for the graph of $y = \mathrm{f}'(x)$.

The expression $\mathrm{f}'(x) = 1 - \frac{\lambda}{(x - 1)^2}$ is undefined at $x = 1$.
As $x \to 1$, $(x - 1)^2 \to 0^+$, so $\mathrm{f}'(x) \to -\infty$. Thus, the vertical asymptote is $x = 1$.
As $x \to \pm\infty$, $\frac{\lambda}{(x - 1)^2} \to 0$, meaning $\mathrm{f}'(x) \to 1$. Thus, the horizontal asymptote is $y = 1$.

3. Determine the coordinates where the graph cuts the axes.

Setting $\mathrm{f}'(x) = 0$, the $x$-intercepts are $x = 1 \pm \sqrt{\lambda}$.
The coordinates are $(1 - \sqrt{\lambda}, 0)$ and $(1 + \sqrt{\lambda}, 0)$.
Since $0 < \lambda < 1$, we have $1 - \sqrt{\lambda} > 0$, meaning both intercepts firmly lie on the positive $x$-axis.
Setting $x = 0$, the $y$-intercept is $\mathrm{f}'(0) = 1 - \frac{\lambda}{(0 - 1)^2} = 1 - \lambda$. Since $\lambda < 1$, the $y$-intercept $(0, 1-\lambda)$ is strictly positive.

4. Summarise the features for the sketch of the graph.

The graph of $y = 1 - \frac{\lambda}{(x - 1)^2}$ consists of two symmetrical branches separated by the vertical asymptote $x = 1$.

• Both branches approach $y = 1$ from below as $x \to \pm\infty$.
• Both branches turn downwards, approaching $x = 1$ asymptotically towards $-\infty$.
• The left branch cuts the positive $y$-axis at $1-\lambda$ and the positive $x$-axis at $1 - \sqrt{\lambda}$.
• The right branch cuts the positive $x$-axis at $1 + \sqrt{\lambda}$.
  (The final response is a sketch drawn on a coordinate plane incorporating these features.)