The curve $C$ has equation $y=\dfrac{x^2+(\lambda-1)x}{x-1},\quad x\ne 1,$ where $\lambda$ is a constant. Find

1. Identify the vertical asymptote from the denominator.

The function is undefined when the denominator is zero. Setting the denominator to zero gives:
$x - 1 = 0$
$x = 1$
This gives the vertical asymptote.

2. Perform algebraic manipulation to find the oblique asymptote.

We express the rational function in the form $y = mx + c + \frac{k}{x-1}$ by rewriting the numerator:

As $x \to \pm\infty$, $\frac{\lambda}{x - 1} \to 0$.
Therefore, the equation of the oblique asymptote is $y = x + \lambda$.

1. Rewrite the equation of the curve and differentiate to find the first derivative.

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

2. Equate the derivative to zero to find the $x$-coordinates of the stationary points.

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

3. Determine the condition for the curve to have two distinct stationary points.

\text{For the curve to have 2 distinct stationary points, the roots for } x \text{ must be real and distinct. Therefore:}
$\lambda > 0$

4. Apply the condition that both stationary points occur for $x > 0$ to find the final range for $\lambda$.

\text{We require both stationary points to occur at } x > 0. \text{ Thus, the smaller root must be strictly positive:}
$1 - \sqrt{\lambda} > 0$
$\sqrt{\lambda} < 1$
\text{Since } \lambda > 0, \text{ this implies }
$\lambda < 1$
\text{Combining these conditions, the required range of values is:}
$0 < \lambda < 1$

1. Recall the equations of the asymptotes from part (i).

From part (i), by performing algebraic division, the equation of the curve can be written as $y = x + \lambda + \dfrac{\lambda}{x-1}$.
Thus, the vertical asymptote is $x = 1$ and the oblique asymptote is $y = x + \lambda$.

2. Find the coordinates of the $x$-intercepts.

Set $y = 0$:
$\frac{x^2 + (\lambda - 1)x}{x - 1} = 0$
$x^2 + (\lambda - 1)x = 0$
$x(x + \lambda - 1) = 0$
$x = 0$ or $x = 1 - \lambda$
The coordinates of the points where the graph cuts the $x$-axis are $(0, 0)$ and $(1 - \lambda, 0)$.

3. Deduce the relative positions of the intercepts and asymptotes using the condition from part (ii).

From part (ii), we require $0 < \lambda < 1$.
Therefore, $0 < 1 - \lambda < 1$.
This means both $x$-intercepts $(0,0)$ and $(1 - \lambda, 0)$ lie to the left of the vertical asymptote $x = 1$. The $y$-intercept of the oblique asymptote, $(0, \lambda)$, is strictly positive.

4. Determine the shape and sketch the graph.

Draw the asymptotes $x = 1$ and $y = x + \lambda$ (noting its positive $y$-intercept).
Plot the intercepts $(0,0)$ and $(1-\lambda, 0)$.

• Left branch ($x < 1$): Starts from the third quadrant just below $y = x + \lambda$, passes up through $(0,0)$, reaches a local maximum in the first quadrant, passes down through $(1-\lambda, 0)$, and approaches $-\infty$ as $x \to 1^-$.
• Right branch ($x > 1$): Approaches $+\infty$ as $x \to 1^+$, falls to a local minimum in the first quadrant, and goes to $+\infty$ approaching $y = x + \lambda$ from above.

1. Differentiate $f(x)$ to find the expression for $f'(x)$.

Using the quotient rule for $f(x) = \frac{x^2+(\lambda-1)x}{x-1}$:
Let $u = x^2+(\lambda-1)x$ and $v = x-1$.
$u' = 2x+\lambda-1$ and $v' = 1$.

2. Find the equations of the asymptotes of $y=f'(x)$.

For the vertical asymptote, the denominator is zero at $x = 1$.
As $x \to 1$, $f'(x) \to -\infty$. Thus, the vertical asymptote is $x = 1$.

For the horizontal asymptote, evaluate the limit as $x \to \pm\infty$:
$\lim_{x \to \pm\infty} \left( 1 - \frac{\lambda}{(x-1)^2} \right) = 1 - 0 = 1$
Thus, the horizontal asymptote is $y = 1$.

3. Determine the coordinates of the $x$-intercepts.

Set $f'(x) = 0$ to find where the graph cuts the $x$-axis:
$1 - \frac{\lambda}{(x-1)^2} = 0$
$(x-1)^2 = \lambda$
$x - 1 = \pm\sqrt{\lambda}$
$x = 1 \pm \sqrt{\lambda}$
Since $0 < \lambda < 1$ from part (ii), $1-\sqrt{\lambda} > 0$.
The coordinates of the points where the graph cuts the $x$-axis are $(1-\sqrt{\lambda}, 0)$ and $(1+\sqrt{\lambda}, 0)$.

4. Deduce the overall shape to sketch the graph of $y=f'(x)$.

The function $f'(x) = 1 - \frac{\lambda}{(x-1)^2}$ is an inverted, shifted transformation of $y = \frac{1}{x^2}$.

• Sketch the asymptotes $x=1$ and $y=1$ using dashed lines.
• Plot the intercepts $(1-\sqrt{\lambda}, 0)$ and $(1+\sqrt{\lambda}, 0)$ on the positive $x$-axis.
• Draw two distinct curve branches: both bounded above by $y=1$ for large $|x|$ and plunging downwards to negative infinity as they approach the vertical line $x=1$.