The curve C with equation $$y = \frac{p - 3x}{(2x - q)(x + 3)}$$ where p and q are constants, passes through the point \( \left( 3, \frac{1}{2} \right) \) and has two vertical asymptotes with equations \( x = 2 \) and \( x = -3 \) - Edexcel - A-Level Maths Pure - Question 14 - 2019 - Paper 2

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

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

Calculating the denominator gives:

$2(3) - 4 = 6 - 4 = 2 \quad \text{and} \quad 3 + 3 = 6 \Rightarrow 2 \times 6 = 12.$

Thus, the equation simplifies to:

$\frac{1}{2} = \frac{p - 9}{12}.$

Cross-multiplying results in:

$12 \times \frac{1}{2} = p - 9 \Rightarrow 6 = p - 9 \Rightarrow p = 15.$

Therefore, we have shown that ( p = 15 ).

To determine the area of region R, we calculate:

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

Using partial fractions, we can express:

$\frac{15 - 3x}{(2x - 4)(x + 3)} = \frac{A}{2x - 4} + \frac{B}{x + 3}.$

Finding A and B involves equating coefficients and matching terms. On solving, we get:

$A = 0.9, \quad B = -2.4.$

This allows us to integrate and find the area described by integrating from x=3 to x=5, yielding:

$Area = 0.9 \ln(2x - 4) \bigg|_3^5 - 2.4 \ln(x + 3) \bigg|_3^5.$

Calculating at the boundaries gives:

$\Rightarrow Area = \ln(2) + b \ln(3) \text{ with a = 1 and b = }-2.4.$

Thus, the exact value for the area of R is expressed as ( a \ln 2 + b \ln 3 ).