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

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

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

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

Expand and equate coefficients:

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

Equating coefficients gives:

\(3A + B = 12\)

\(2A - B = 4\)

Solving these equations, we find \(A = 4\) and \(B = 3\).

Thus, \(f(x) = \frac{4}{x-1} + \frac{3}{3x+2}\).

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

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

Expand \(\frac{1}{3x+2}\) as \(\frac{1}{2} - \frac{3}{4}x + \frac{9}{8}x^2 - \ldots\)

Combine these expansions:

\(f(x) = 4(-1 - x - x^2) + 3(\frac{1}{2} - \frac{3}{4}x + \frac{9}{8}x^2)\)

\(= -4 - 4x - 4x^2 + \frac{3}{2} - \frac{9}{4}x + \frac{27}{8}x^2\)

Combine like terms:

\(= \frac{1}{2} - \frac{9}{4}x - \frac{33}{8}x^2\)

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

Assume \(\frac{8x^2 + 9x + 8}{(1-x)(2x+3)^2} = \frac{A}{1-x} + \frac{B}{2x+3} + \frac{C}{(2x+3)^2}\).

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

\(8x^2 + 9x + 8 = A(2x+3)^2 + B(1-x)(2x+3) + C(1-x)\).

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

\(A = 1, B = -2, C = 5\).

Thus, \(f(x) = \frac{1}{1-x} - \frac{2}{2x+3} + \frac{5}{(2x+3)^2}\).

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

Use the binomial expansion for each term:

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

\(\frac{1}{2x+3} = \frac{1}{3} - \frac{2}{9}x + \cdots\)

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

Combine the expansions:

\(f(x) = \left(1 + x + x^2 + \cdots\right) - 2\left(\frac{1}{3} - \frac{2}{9}x + \cdots\right) + 5\left(\frac{1}{9} - \frac{4}{27}x + \cdots\right)\).

Simplify to obtain the expansion up to \(x^2\):

\(f(x) = \frac{8}{9} + \frac{19}{27}x + \frac{13}{9}x^2\).

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

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

Multiply through by the denominator \((3x+2)(x^2+5)\) to clear the fractions:

\(5x^2 - 7x + 4 = A(x^2 + 5) + (Bx + C)(3x + 2)\).

Expand and equate coefficients:

\(5x^2 - 7x + 4 = Ax^2 + 5A + 3Bx^2 + 2Bx + 3Cx + 2C\).

Combine like terms:

\((A + 3B)x^2 + (2B + 3C)x + (5A + 2C)\).

Equate coefficients with \(5x^2 - 7x + 4\):

\(A + 3B = 5\)

\(2B + 3C = -7\)

\(5A + 2C = 4\)

Solve these equations to find \(A = 2\), \(B = 1\), \(C = -3\).

Thus, \(f(x) = \frac{2}{3x+2} + \frac{x-3}{x^2+5}\).

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

Expand \(\frac{1}{3x+2}\) using geometric series:

\(\frac{1}{3x+2} = \frac{1}{2} \cdot \frac{1}{1 - (-\frac{3}{2}x)} \approx \frac{1}{2} (1 + \frac{3}{2}x + (\frac{3}{2}x)^2)\)

\(= \frac{1}{2} + \frac{3}{4}x + \frac{9}{8}x^2\).

Expand \(\frac{x-3}{x^2+5}\):

\(\frac{x-3}{x^2+5} = (x-3)(\frac{1}{5} - \frac{1}{5^2}x^2)\)

\(= \frac{1}{5}x - \frac{3}{5} - \frac{1}{25}x^3 + \frac{3}{25}x^2\).

Combine the expansions:

\(f(x) = \frac{2}{5} - \frac{13}{10}x + \frac{237}{100}x^2\).

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

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

Multiply through by the denominator \((2+x)(4+x^2)\) to clear fractions:

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

Expand and equate coefficients:

\(x(6-x) = 4A + Ax^2 + 2Bx + Bx^2 + Cx + Cx^2\).

\(x(6-x) = (A+B+C)x^2 + (2B+C)x + 4A\).

Equate coefficients:

\(A + B + C = 0\)

\(2B + C = 6\)

\(4A = 0\)

From \(4A = 0\), \(A = -2\).

Substitute \(A = -2\) into \(A + B + C = 0\):

\(-2 + B + C = 0 \Rightarrow B + C = 2\).

Substitute \(B + C = 2\) into \(2B + C = 6\):

\(2B + C = 6 \Rightarrow 2B + (2-B) = 6 \Rightarrow B = 1\).

Then \(C = 2 - B = 1\).

Thus, \(A = -2\), \(B = 1\), \(C = 4\).

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

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

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

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

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

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

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

\(f(x) \approx -2 + x + x + 4 - \frac{x^2}{4}\).

Final answer: \(\frac{3}{4}x - \frac{1}{2}x^2\).

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

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

Multiply through by \((x+2)(x^2+4)\) to clear the denominators:

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

Expand and collect like terms:

\(3x^2 + x + 6 = Ax^2 + 4A + Bx^2 + 2Bx + Cx + 2C\).

Combine terms:

\(3x^2 + x + 6 = (A+B)x^2 + (2B+C)x + (4A+2C)\).

Equate coefficients:

\(A + B = 3\)

\(2B + C = 1\)

\(4A + 2C = 6\)

Solve these equations to find \(A = 2\), \(B = 1\), \(C = -1\).

Thus, \(f(x) = \frac{2}{x+2} + \frac{x-1}{x^2+4}\).

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

Use the binomial expansion for \((x+2)^{-1}\) and \((x^2+4)^{-1}\):

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

\((x^2+4)^{-1} = \frac{1}{4} - \frac{1}{16}x^2 + \cdots\)

Substitute back into the partial fractions:

\(\frac{2}{x+2} = 2 \left( \frac{1}{2} - \frac{1}{4}x \right) = 1 - \frac{1}{2}x\)

\(\frac{x-1}{x^2+4} = (x-1) \left( \frac{1}{4} - \frac{1}{16}x^2 \right) = \frac{1}{4}x - \frac{1}{4} - \frac{1}{16}x^3 + \cdots\)

Add the expansions:

\(f(x) = 1 - \frac{1}{2}x + \frac{1}{4}x - \frac{1}{4} + \cdots\)

Combine like terms:

\(f(x) = \frac{3}{4} - \frac{1}{4}x + \frac{5}{16}x^2\).