(i) Express \(f(x)\) in partial fractions:

Assume \(\frac{10x - 2x^2}{(x+3)(x-1)^2} = \frac{A}{x+3} + \frac{B}{x-1} + \frac{C}{(x-1)^2}\).

Multiply through by \((x+3)(x-1)^2\) to clear the denominators:

\(10x - 2x^2 = A(x-1)^2 + B(x+3)(x-1) + C(x+3)\).

Expand and equate coefficients to solve for \(A\), \(B\), and \(C\):

\(A = -3, \ B = 1, \ C = 2\).

(ii) Expand in ascending powers of \(x\):

Use the expansions:

\((x+3)^{-1} = \frac{1}{3} - \frac{x}{9} + \frac{x^2}{27} + \ldots\)

\((1-x)^{-1} = 1 + x + x^2 + \ldots\)

\((1-x)^{-2} = 1 + 2x + 3x^2 + \ldots\)

Substitute and simplify:

\(\frac{-3}{x+3} = -\frac{1}{3} + \frac{x}{9} - \frac{x^2}{27} + \ldots\)

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

\(\frac{2}{(x-1)^2} = 2 + 4x + 6x^2 + \ldots\)

Add the series up to \(x^2\):

\(f(x) = \left(-\frac{1}{3} + \frac{x}{9} - \frac{x^2}{27}\right) + (1 + x + x^2) + (2 + 4x + 6x^2)\)

Combine terms:

\(f(x) = \frac{10}{3}x + \frac{44}{9}x^2 + \ldots\)

(a) Express \(f(x)\) in partial fractions:

Assume \(\frac{24x + 13}{(1 - 2x)(2 + x)^2} = \frac{A}{1 - 2x} + \frac{B}{2 + x} + \frac{C}{(2 + x)^2}\).

Multiply through by the denominator \((1 - 2x)(2 + x)^2\) to clear the fractions:

\(24x + 13 = A(2 + x)^2 + B(1 - 2x)(2 + x) + C(1 - 2x)\).

Expand and equate coefficients to find \(A = 4\), \(B = 2\), \(C = -7\).

(b) Expand each partial fraction:

\(\frac{A}{1 - 2x} = 4(1 - 2x)^{-1} \approx 4(1 + 2x + 4x^2)\).

\(\frac{B}{2 + x} = 2(2 + x)^{-1} \approx 2(1 - \frac{x}{2} + \frac{x^2}{4})\).

\(\frac{C}{(2 + x)^2} = -7(2 + x)^{-2} \approx -7(\frac{1}{4} - \frac{x}{4} + \frac{3x^2}{16})\).

Combine these to get:

\(f(x) \approx \frac{13}{4} + \frac{37}{4}x + \frac{239}{16}x^2\).

(c) The expansion is valid for \(|x| < \frac{1}{2}\).

(i) Express \(f(x)\) in partial fractions:

Assume \(\frac{4x^2 + 12}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}\).

Multiply through by \((x+1)(x-3)^2\) to clear the denominators:

\(4x^2 + 12 = A(x-3)^2 + B(x+1)(x-3) + C(x+1)\).

Expand and equate coefficients to solve for \(A, B,\) and \(C\):

\(A = 1, B = 3, C = 12\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the partial fraction decomposition:

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

\(\frac{1}{x-3} = -\frac{1}{3} - \frac{1}{9}x - \ldots\)

\(\frac{1}{(x-3)^2} = \frac{1}{9} + \frac{2}{27}x + \ldots\)

Combine these expansions:

\(f(x) = \frac{1}{x+1} + 3\left(\frac{1}{x-3}\right) + 12\left(\frac{1}{(x-3)^2}\right)\)

\(= \left(1 - x + x^2\right) + 3\left(-\frac{1}{3} - \frac{1}{9}x\right) + 12\left(\frac{1}{9} + \frac{2}{27}x\right)\)

\(= \frac{4}{3} - \frac{4}{9}x + \frac{4}{3}x^2\).

(i) Express \(f(x)\) in partial fractions:

Assume \(\frac{5x^2 + x + 6}{(3 - 2x)(x^2 + 4)} = \frac{A}{3 - 2x} + \frac{Bx + C}{x^2 + 4}\).

Multiply through by the denominator \((3 - 2x)(x^2 + 4)\) to get:

\(5x^2 + x + 6 = A(x^2 + 4) + (Bx + C)(3 - 2x)\).

Expand and equate coefficients to solve for \(A, B, C\):

\(A = 3, B = -1, C = -2\).

Thus, \(f(x) = \frac{3}{3 - 2x} + \frac{-x - 2}{x^2 + 4}\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

For \(\frac{3}{3 - 2x}\), use the expansion \((3 - 2x)^{-1} = \frac{1}{3}(1 - \frac{2}{3}x)^{-1}\).

First two terms: \(\frac{1}{3}(1 + \frac{2}{3}x + (\frac{2}{3}x)^2) = \frac{1}{3} + \frac{2}{9}x + \frac{4}{27}x^2\).

For \(\frac{-x - 2}{x^2 + 4}\), use the expansion \((x^2 + 4)^{-1} = \frac{1}{4}(1 - \frac{1}{4}x^2)\).

First two terms: \(\frac{-x - 2}{4}(1 - \frac{1}{4}x^2) = \frac{-x}{4} - \frac{1}{2} + \frac{x^3}{16}\).

Combine terms up to \(x^2\):

\(\frac{1}{3} + \frac{2}{9}x + \frac{4}{27}x^2 - \frac{x}{4} - \frac{1}{2}\).

Simplify to get the final answer:

\(\frac{1}{2} + \frac{5}{12}x + \frac{41}{72}x^2\).

(i) Express \(f(x)\) in partial fractions:

Assume \(\frac{x^2 - 8x + 9}{(1-x)(2-x)^2} = \frac{A}{1-x} + \frac{B}{2-x} + \frac{C}{(2-x)^2}\).

Multiply through by \((1-x)(2-x)^2\) to clear the denominators:

\(x^2 - 8x + 9 = A(2-x)^2 + B(1-x)(2-x) + C(1-x)\).

Expand and equate coefficients to solve for \(A, B, C\):

\(A = 2, B = -1, C = 3\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the partial fractions: \(\frac{2}{1-x} - \frac{1}{2-x} + \frac{3}{(2-x)^2}\).

Expand each term using binomial series:

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

\(\frac{-1}{2-x} = -\frac{1}{2}(1 + \frac{x}{2} + \frac{x^2}{4} + \ldots)\)

\(\frac{3}{(2-x)^2} = \frac{3}{4}(1 + \frac{x}{2} + \frac{3x^2}{8} + \ldots)\)

Add the expansions up to \(x^2\):

\(\frac{9}{4} + \frac{5}{2}x + \frac{39}{16}x^2\).