The curves $C_1$ and $C_2$ have equations $y=\dfrac{x-b}{x-a}$ and $y=\dfrac{x-b}{b}$ respectively, where $a$ and $b$ are constants with $1<a<b$.

1. Set the equations of $C_1$ and $C_2$ equal to find the $x$-coordinates of their intersections.

$\frac{x-b}{x-a} = \frac{x-b}{b}$
$\frac{x-b}{x-a} - \frac{x-b}{b} = 0$
$(x-b) \left( \frac{1}{x-a} - \frac{1}{b} \right) = 0$

2. Solve for $x$ to show the coordinates given in the question.

$x - b = 0 \implies x = b$
$\text{or} \quad \frac{1}{x-a} = \frac{1}{b} \implies x - a = b \implies x = a + b$
Hence, the $x$-coordinates of intersection are $b$ and $a+b$ as required.

3. Determine the asymptotes of $C_1$.

For $C_1$ with equation $y = \frac{x-b}{x-a}$:
The vertical asymptote occurs where the denominator is zero:
$x - a = 0 \implies x = a$
The horizontal asymptote occurs as $x \to \pm\infty$:
$y = \lim_{x \to \pm\infty} \frac{x(1-b/x)}{x(1-a/x)} = 1 \implies y = 1$

4. Determine the axes intercepts for both curves.

For $C_1$ ($y = \frac{x-b}{x-a}$):
When $x=0$, $y = \frac{-b}{-a} = \frac{b}{a}$. ($y$-intercept)
When $y=0$, $x-b=0 \implies x=b$. ($x$-intercept)

For $C_2$ ($y = \frac{x-b}{b}$):
When $x=0$, $y = \frac{-b}{b} = -1$. ($y$-intercept)
When $y=0$, $x-b=0 \implies x=b$. ($x$-intercept)

5. Describe the sketching of $C_1$ and $C_2$ on a single diagram.

• Draw a dashed vertical line for $x=a$ and a dashed horizontal line for $y=1$.
• Since $1<a<b$, note that $\frac{b}{a} > 1$. Plot the $y$-intercept $(0, \frac{b}{a})$ above $y=1$.
• Draw the two smooth branches of $C_1$ approaching its asymptotes.
• Draw $C_2$ as a straight line passing through $(0, -1)$ and $(b, 0)$.
• Mark the intersections points $(b,0)$ and $(a+b, \frac{a}{b})$ clearly.

1. Relate the inequality to the given graphs.

The inequality $\dfrac{x-b}{x-a} \ge \dfrac{x-b}{b}$ is equivalent to $C_1 \ge C_2$. From the diagram, we need to find the range of $x$-values where the curve $C_1$ lies above or intersects the line $C_2$.

2. Identify the regions from the graph where $C_1$ is above or meets $C_2$.

From the sketch in part (i):

• For $x < a$ (to the left of the vertical asymptote), $C_1$ lies above $C_2$.
• For $a < x < b$, $C_1$ lies below $C_2$.
• For $b \le x \le a+b$ (between the two points of intersection), $C_1$ lies on or above $C_2$.
• For $x > a+b$, $C_1$ lies below $C_2$.

Thus, the solution is $x < a$ or $b \le x \le a+b$.

1. Find the $y$-limits of the region bounded by the curves.

From part (i), the $x$-coordinates of the intersections are $x = b = 3$ and $x = a + b = 5$.
Substitute $x=3$ into $C_2$: $y = \frac{3-3}{3} = 0$.
Substitute $x=5$ into $C_2$: $y = \frac{5-3}{3} = \frac{2}{3}$.
Thus, the region is bounded between $y=0$ and $y=\frac{2}{3}$.

2. Express $x$ in terms of $y$ for both curves.

For $C_2$ ($y = \frac{x-3}{3}$):
$x = 3y + 3$

For $C_1$ ($y = \frac{x-3}{x-2}$):
$y(x-2) = x-3$
$xy - 2y = x - 3$
$x(y-1) = 2y - 3$
$x = \frac{2y-3}{y-1}$

3. Set up the definite integral for the volume of revolution about the $y$-axis.

Testing a point in the interval $0 < y < \frac{2}{3}$ (e.g., $y = 0.5$):
$x_{C_2} = 3(0.5) + 3 = 4.5$
$x_{C_1} = \frac{2(0.5)-3}{0.5-1} = 4$
Since $x_{C_2} > x_{C_1}$ within this interval, $C_2$ forms the outer radius and $C_1$ forms the inner radius.
The volume $V$, formed by rotating the region through 4 right angles ($360^\circ$) about the $y$-axis, is given by:
$V = \pi \int_{0}^{\frac{2}{3}} \left[ (3y+3)^2 - \left( \frac{2y-3}{y-1} \right)^2 \right] \text{d}y$

4. Evaluate the integral using a graphing calculator.

Using a graphing calculator to evaluate the definite integral directly:
$V \approx 5.74211$
$V = 5.742 \text{ (to 3 d.p.)}$