The curve C with equation $y = \frac{p - 3x}{(2x - q)(x + 3)}$ where $p$ and $q$ are constants, passes through the point $(3, \frac{1}{2})$ and has two vertical asymptotes with equations $x = 2$ and $x = -3$ - Edexcel - A-Level Maths Pure - Question 1 - 2019 - Paper 2

To find $p$, we substitute the point $(3, \frac{1}{2})$ into the curve equation:

$\frac{1}{2} = \frac{p - 3(3)}{(2(3) - 4)(3 + 3)}$.

This simplifies to:

$\frac{1}{2} = \frac{p - 9}{(6 - 4)(6)} = \frac{p - 9}{2 \cdot 6} = \frac{p - 9}{12}.$

Cross-multiplying gives:

$1 \cdot 12 = 2(p - 9) \Rightarrow 12 = 2p - 18 \Rightarrow 2p = 30 \Rightarrow p = 15.$

To find the area under the curve $C$, we need to evaluate the integral of the function over the specified interval. The area is given by:

$A = \int_{3}^{5} \left( \frac{15 - 3x}{(2x - 4)(x + 3)} \right) dx.$

Using partial fractions, we can express this as:

$A = \int (\frac{A}{(2x-4)} + \frac{B}{(x+3)}) dx.$

Evaluating the integrals leads to the results involving $\ln$ functions, which can be combined to find constants $a$ and $b$. Finally, by evaluating from $3$ to $5$, we deduce:

$A = a \ln 2 + b \ln 3.$