A curve $C$ has equation $y = 2ax + \frac{b}{x}$ where $a$ and $b$ are non-zero real constants and $x \ne 0$.

1. Find the first derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$.

Given $y = 2ax + bx^{-1}$,
$\frac{\mathrm{d}y}{\mathrm{d}x} = 2a - bx^{-2} = 2a - \frac{b}{x^2}$

2. Set the derivative to zero to locate potential stationary points.

For stationary points, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$:
$2a - \frac{b}{x^2} = 0$
$x^2 = \frac{b}{2a}$

3. Use the condition $ab < 0$ to deduce if real roots exist.

We are given that $ab < 0$, which means $a$ and $b$ have opposite signs.
Thus, $\frac{b}{a} < 0$, which implies that $\frac{b}{2a} < 0$.
Since $x^2 \ge 0$ for all real $x$, the equation $x^2 = \frac{b}{2a}$ has no real solutions.
Therefore, the curve $C$ has no stationary points.

1. Substitute $b = \frac{1}{2}a$ into the equation and find the equations of the asymptotes.

$y = 2ax + \frac{a}{2x}$

As $x \to 0$, $y \to \pm\infty$. Thus, the vertical asymptote is $x = 0$ (the $y$-axis).

As $x \to \pm\infty$, $\frac{a}{2x} \to 0$. Thus, the oblique asymptote is $y = 2ax$.

2. Differentiate to find the coordinates of the stationary points.

$\frac{\mathrm{d}y}{\mathrm{d}x} = 2a - \frac{a}{2x^2}$

For stationary points, set $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$:
$2a - \frac{a}{2x^2} = 0$
$\frac{a}{2x^2} = 2a$
$x^2 = \frac{1}{4} \implies x = \frac{1}{2} \text{ or } x = -\frac{1}{2}$

When $x = \frac{1}{2}$, $y = 2a\left(\frac{1}{2}\right) + \frac{a}{2(1/2)} = a + a = 2a$.
When $x = -\frac{1}{2}$, $y = 2a\left(-\frac{1}{2}\right) + \frac{a}{2(-1/2)} = -a - a = -2a$.

The stationary points are $(\frac{1}{2}, 2a)$ and $(-\frac{1}{2}, -2a)$.

3. Check for any points where the curve crosses the axes.

$y$-intercept: Since $x = 0$ is an asymptote, the curve is undefined at $x=0$ and does not cross the $y$-axis.

$x$-intercept: Set $y = 0$
$2ax + \frac{a}{2x} = 0$
Multiply by $2x$:
$4ax^2 + a = 0$
$a(4x^2 + 1) = 0$
Since $a > 0$, $4x^2 + 1 \ne 0$ for any real $x$. Thus, there are no real roots and the curve does not cross the $x$-axis.

4. Sketch the curve, combining the asymptotes, stationary points, and general shape.

To sketch $C$:

1. Draw and label the vertical line $x = 0$ and the oblique line $y = 2ax$ (passing through the origin with a positive gradient since $a > 0$).
2. Plot the local minimum at $(\frac{1}{2}, 2a)$ in the first quadrant, lying above the oblique asymptote.
3. Plot the local maximum at $(-\frac{1}{2}, -2a)$ in the third quadrant, lying below the oblique asymptote.
4. Draw two hyperbola-like branches smoothly passing through the turning points and approaching the asymptotes.

1. Substitute $b$ and rearrange the equation into a quadratic form in terms of $x^2$.

Substitute $b = \frac{1}{2}a$ (given in part b) into the equation:
$2ax + \frac{a}{2x} = kx$
Multiply throughout by $x$ (since it is given that $x \ne 0$):
$2ax^2 + \frac{1}{2}a = kx^2$
Rearranging the terms to isolate $x^2$ gives:
$(2a - k)x^2 + \frac{1}{2}a = 0$
$x^2 = \frac{-a}{2(2a - k)} = \frac{a}{2(k - 2a)}$

2. Determine the range of $k$ for no real roots.

For the equation to have no real roots, there must be no valid real values for $x$.
Since $a > 0$, the numerator $a$ is positive. For $x^2$ to be negative (which yields no real roots), the denominator must be negative:
$k - 2a < 0 \implies k < 2a$
Additionally, we must check the boundary case where the denominator is zero, i.e., $k = 2a$. Substituting $k = 2a$ into our rearranged equation gives:
$(2a - 2a)x^2 + \frac{1}{2}a = 0 \implies \frac{1}{2}a = 0$
Since $a > 0$, this is a mathematical contradiction, meaning there are no solutions (and thus no real roots) when $k = 2a$.
Therefore, the complete range is $k \le 2a$.

(Note: Graphically, the curve $y = 2ax + \frac{a}{2x}$ has an oblique asymptote $y = 2ax$. A line $y = kx$ passing through the origin will strictly not intersect the curve if its gradient $k$ is less than or equal to $2a$.)

1. Express $x$ in terms of $y$ and select the correct branch for the given region.

From $y = 2ax + \frac{a}{2x}$, we rearrange to form a quadratic equation in $x$:
$4ax^2 - 2xy + a = 0$
Using the quadratic formula to solve for $x$:
$x = \frac{2y \pm \sqrt{4y^2 - 16a^2}}{8a} = \frac{y \pm \sqrt{y^2 - 4a^2}}{4a}$
From part (b), the minimum point is $(1/2, 2a)$. Since the region is bounded by $x = \frac{1}{2}$ and the $y$-axis, we must consider the branch of the curve where $x \le \frac{1}{2}$. Thus, we reject the positive root and take:
$x = \frac{y - \sqrt{y^2 - 4a^2}}{4a}$

2. Set up the integral expression for the volume generated.

The volume consists of a cylindrical region below the minimum point (from $y = 0$ to $y = 2a$) and the solid generated by revolving the curve $C$ (from $y = 2a$ to $y = 4a$).
The cylinder has a constant radius $x = \frac{1}{2}$ and height $2a$.
$\text{Volume} = \pi \left(\frac{1}{2}\right)^2(2a) + \int_{2a}^{4a} \pi x^2 \; \mathrm{d}y$
$= \frac{1}{2}a\pi + \int_{2a}^{4a} \pi \left( \frac{y - \sqrt{y^2 - 4a^2}}{4a} \right)^2 \; \mathrm{d}y$

3. Expand the integrand to derive the required volume formula.

Expanding $x^2$:
$x^2 = \frac{y^2 - 2y\sqrt{y^2 - 4a^2} + (y^2 - 4a^2)}{16a^2}$
$= \frac{2y^2 - 4a^2 - 2y\sqrt{y^2 - 4a^2}}{16a^2}$
$= \frac{y^2 - 2a^2 - y\sqrt{y^2 - 4a^2}}{8a^2}$
Substituting this into the volume integral yields the required expression:
$\text{Volume} = \frac{1}{2}a\pi + \frac{\pi}{8a^2} \int_{2a}^{4a} \left(y^2 - 2a^2 - y\sqrt{y^2 - 4a^2}\right) \; \mathrm{d}y \quad \text{(Shown)}$

4. Evaluate the polynomial part of the definite integral.

$\int_{2a}^{4a} (y^2 - 2a^2) \; \mathrm{d}y = \left[ \frac{y^3}{3} - 2a^2y \right]_{2a}^{4a}$
$= \left( \frac{64a^3}{3} - 8a^3 \right) - \left( \frac{8a^3}{3} - 4a^3 \right)$
$= \frac{40a^3}{3} - \left( -\frac{4a^3}{3} \right) = \frac{44a^3}{3}$

5. Evaluate the radical part of the definite integral using substitution.

Using substitution or the reverse chain rule, recognizing the form $\int f'(y)[f(y)]^n \; \mathrm{d}y$ where $f(y) = y^2 - 4a^2$:
$\int_{2a}^{4a} -y(y^2 - 4a^2)^{1/2} \; \mathrm{d}y = \left[ -\frac{1}{2} \cdot \frac{2}{3}(y^2 - 4a^2)^{3/2} \right]_{2a}^{4a}$
$= \left[ -\frac{1}{3}(y^2 - 4a^2)^{3/2} \right]_{2a}^{4a}$
$= -\frac{1}{3} \left( (16a^2 - 4a^2)^{3/2} - 0 \right)$
$= -\frac{1}{3}(12a^2)^{3/2} = -\frac{1}{3}(12\sqrt{12} a^3) = -8\sqrt{3}a^3$

6. Combine the evaluated parts to find the exact total volume.

Substitute the evaluated integrals back into the volume formula:
$\text{Volume} = \frac{1}{2}a\pi + \frac{\pi}{8a^2} \left( \frac{44a^3}{3} - 8\sqrt{3}a^3 \right)$
$= \frac{1}{2}a\pi + \pi a \left( \frac{11}{6} - \sqrt{3} \right)$
$= \pi a \left( \frac{3}{6} + \frac{11}{6} - \sqrt{3} \right) = \pi a \left( \frac{7}{3} - \sqrt{3} \right)$