Given that $f(x)$ can be expressed in the form $$f(x) = \frac{A}{(3x + 2)} + \frac{B}{(3x + 2)} + \frac{C}{(1 - x)}, \quad |x| < \frac{2}{3}$$ (a) find the values of $B$ and $C$ and show that $A = 0$ - Edexcel - A-Level Maths Pure - Question 5 - 2009 - Paper 3

Start with the function:

$f(x) = \frac{4}{(3x + 2)} + \frac{C}{(1 - x)}$

Using the formula for geometric series expansion:

1. For the first part:

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

2. For the second part:

   $\frac{C}{(1 - x)} = C\sum_{n=0}^{\infty} x^n$

Now, combine terms, simplifying up to $x^2$ gives:

$= 4(0) + x^2( ... ) + ...$

To estimate $f(0.2)$, we use the series expansion we derived:

1. Calculate the actual value of $f(0.2)$ using the original function:

   $f(0.2) = \frac{4}{(3(0.2) + 2)} + \frac{C}{(1 - 0.2)}$

2. For the estimate using series:

   $Estimate = ...$

3. Calculate the percentage error:

   $\text{Percentage Error} = \frac{\text{Actual} - \text{Estimate}}{\text{Actual}} \times 100$

4. Finally, present the answer to 2 significant figures.