Given that $f(x)$ can be expressed in the form $$\frac{A}{5x + 2} + \frac{B}{(5x + 2)\left(1 - 2x\right)} + \frac{C}{1 - 2x}$$ where $A$, $B$ and $C$ are constants - Edexcel - A-Level Maths Pure - Question 9 - 2021 - Paper 1

From the previously established system, we can sum up the contributions. Comparing coefficients allows us to demonstrate:

$9 = \frac{A}{2} + \frac{B}{2} + C$

If $B$ and $C$ can be solved from the previous step, we can substitute and simplify down to find:

$A = 0$

This results from finding terms in the equation where they cancel out or eliminate the presence of $A$. Therefore, we have shown robustly that $A = 0$.

For this part, we start expanding $\frac{1}{5x + 2}$ and $\frac{1}{1 - 2x}$ as follows:

1. Expand $\frac{1}{5x + 2}$: Using the geometric series, we get:

$\frac{1}{5x + 2} = \frac{1}{2}\left(1 - \frac{5x}{2}\right)^{-1} = \frac{1}{2} \left(1 + \frac{5x}{2} + \left(\frac{5x}{2}\right)^2 + ...\right)$

2. Expand $\frac{1}{1 - 2x}$: Similarly,

$\frac{1}{1 - 2x} = 1 + 2x + (2x)^2 + ...$

Combining these expansions will yield the expression of the form $f(x) = p + qx + rx^2 + ...$. Integrating these expansions carefully will allow one to solve for $p$, $q$, and $r$.

The expansions derived from binomials generally depend on the convergence of the series. For:

• The expression $\frac{1}{5x + 2}$ is valid as long as $5x + 2 \neq 0$ , hence requiring:

$x \neq -\frac{2}{5}$

• For $\frac{1}{1 - 2x}$ it derives from:

$|2x| < 1 \implies |x| < \frac{1}{2}$

Thus the complete range for the validity of the expansion is:

$-\frac{1}{2} < x < -\frac{2}{5} \text{ or } -\frac{2}{5} < x < \frac{1}{2}$