The curve C has equation $y=\dfrac{1}{4x - x^2}$.

1. Find the first derivative $\dfrac{dy}{dx}$ using the chain rule.

$y = \dfrac{1}{4x - x^2} = (4x - x^2)^{-1}$

Using the chain rule:
$\dfrac{dy}{dx} = -(4x - x^2)^{-2} \cdot \dfrac{d}{dx}(4x - x^2)$
$\dfrac{dy}{dx} = -(4x - x^2)^{-2}(4 - 2x)$
$\dfrac{dy}{dx} = \dfrac{2x - 4}{(4x - x^2)^2}$

2. Equate the derivative to zero to find the $x$-coordinate of the turning point.

At a turning point, $\dfrac{dy}{dx} = 0$.

$\dfrac{2x - 4}{(4x - x^2)^2} = 0$
$2x - 4 = 0$
$2x = 4$
$x = 2$

Therefore, $x_1 = 2$.

3. Determine the nature of the turning point using the first derivative test.

Using the first derivative test around $x = 2$:

• When $x = 1.9$, $\dfrac{dy}{dx} = \dfrac{2(1.9) - 4}{(4(1.9) - 1.9^2)^2} = \dfrac{-0.2}{(+)} < 0$
• When $x = 2$, $\dfrac{dy}{dx} = 0$
• When $x = 2.1$, $\dfrac{dy}{dx} = \dfrac{2(2.1) - 4}{(4(2.1) - 2.1^2)^2} = \dfrac{0.2}{(+)} > 0$

Since the gradient changes from negative to positive as $x$ increases through $x=2$, the turning point is a minimum point.

1. Find the $x$-coordinate of the turning point, $x_1$, from part (i).

$\frac{dy}{dx} = -\frac{4 - 2x}{(4x - x^2)^2} = \frac{2x - 4}{(4x - x^2)^2}

At the turning point, $\frac{dy}{dx} = 0 \implies 2x - 4 = 0 \implies x_1 = 2$.

2. Set up the definite integral for the area and complete the square for the denominator.

$\text{Area} = \int_1^2 \frac{1}{4x - x^2} \, dx$

$\phantom{\text{Area}} = \int_1^2 \frac{1}{4 - (x^2 - 4x + 4)} \, dx$

$\phantom{\text{Area}} = \int_1^2 \frac{1}{2^2 - (x-2)^2} \, dx$

3. Integrate using the standard formula for $\int \frac{1}{a^2-x^2} \, dx$.

$= \left[ \frac{1}{2(2)} \ln \left| \frac{2 + (x-2)}{2 - (x-2)} \right| \right]_1^2$

$= \left[ \frac{1}{4} \ln \left| \frac{x}{4 - x} \right| \right]_1^2$

4. Evaluate the definite integral to find the exact area.

$= \frac{1}{4} \ln \left( \frac{2}{4-2} \right) - \frac{1}{4} \ln \left( \frac{1}{4-1} \right)$

$= \frac{1}{4} \ln 1 - \frac{1}{4} \ln \left( \frac{1}{3} \right)$

$= 0 - \frac{1}{4} (-\ln 3)$

$= \frac{1}{4} \ln 3$