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 \(x = -3\) (a) (i) Explain why you can deduce \(q = 4\) (ii) Show that \(p = 15\) (b) Show the exact value of the area of R is \(a \ln 2 + b \ln 3\), where a and b are rational constants to be found. - Edexcel - A-Level Maths Pure - Question 14 - 2019 - Paper 1

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

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

This simplifies to:

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

Cross-multiplying yields:

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

To find the area of region R, we will integrate with respect to (x) from 3 to the vertical asymptote at (x = -3). The integral to calculate the area A is given by:

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

Using partial fractions, we express:

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

Solving for A and B provides values to evaluate the definite integrals leading to the area, which results in:

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

Final evaluation gives specific rational constants for a and b.