(i) To express \(f(x)\) in the form \(\frac{A}{3-x} + \frac{Bx+C}{2+x^2}\), we equate:
\(\frac{x - 4x^2}{(3-x)(2+x^2)} = \frac{A}{3-x} + \frac{Bx+C}{2+x^2}\)
Multiply through by \((3-x)(2+x^2)\) to clear the denominators:
\(x - 4x^2 = A(2+x^2) + (Bx+C)(3-x)\)
Expand and collect like terms:
\(x - 4x^2 = 2A + Ax^2 + 3Bx + Cx - Bx^2 - Cx\)
\(x - 4x^2 = (A-B)x^2 + (3B+C)x + 2A\)
Equating coefficients, we get:
\(A - B = -4\)
\(3B + C = 1\)
\(2A = 0\) so \(A = 0\)
Substitute \(A = 0\) into \(A - B = -4\):
\(-B = -4\) so \(B = 4\)
Substitute \(B = 4\) into \(3B + C = 1\):
\(12 + C = 1\) so \(C = -11\)
Thus, \(A = 0\), \(B = 4\), \(C = -11\).
(ii) Expand \(\frac{1}{3-x}\) and \(\frac{4x-11}{2+x^2}\) using binomial series:
\(\frac{1}{3-x} = \frac{1}{3} \left(1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} \right)\)
\(\frac{4x-11}{2+x^2} = \frac{4x-11}{2} \left(1 - \frac{x^2}{2} + \frac{x^4}{4} \right)\)
Combine and simplify terms up to \(x^3\):
\(\frac{1}{6}x - \frac{11}{18}x^2 - \frac{31}{108}x^3\)
