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

Assume \(f(x) = \frac{A}{2-x} + \frac{Bx+C}{x^2+1}\).

Equating coefficients, solve for \(A, B,\) and \(C\):

\(A = 4, \ B = 4, \ C = 1\).

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

(ii) Show the expansion:

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

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

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

Multiply the expansions: \((4x+1)(1-x^2)^{-1}\) and simplify.

Combine terms to obtain: \(f(x) = 3 + 5x - \frac{1}{2}x^2 - \frac{15}{4}x^3\).

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

Assume \(\frac{2x^2 + 7x + 8}{(1+x)(2+x)^2} = \frac{A}{1+x} + \frac{B}{2+x} + \frac{C}{(2+x)^2}\).

By equating coefficients, we find \(A = 3\), \(B = -1\), \(C = -2\).

(b) Expand each term:

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

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

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

Combine and simplify to get \(2 - \frac{9}{4}x + \frac{5}{2}x^2\).

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

Assume \(\frac{5x^2 + 8x - 3}{(x-2)(2x^2 + 3)} = \frac{A}{x-2} + \frac{Bx + C}{2x^2 + 3}\).

Multiply through by the denominator: \(5x^2 + 8x - 3 = A(2x^2 + 3) + (Bx + C)(x-2)\).

Equate coefficients to find \(A = 3\), \(B = -1\), \(C = 6\).

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

Use the expansions:

\((x-2)^{-1} = \frac{1}{2} \left[ 1 - \left( \frac{x}{2} \right) + \left( \frac{x}{2} \right)^2 + \ldots \right]\)

\((2x^2 + 3)^{-1} = \frac{1}{3} \left[ 1 - \frac{2x^2}{3} + \ldots \right]\)

Combine these with the partial fractions:

\(\frac{A}{x-2} = \frac{3}{2} \left[ 1 - \frac{x}{2} + \left( \frac{x}{2} \right)^2 + \ldots \right]\)

\(\frac{Bx + C}{2x^2 + 3} = \frac{-x + 6}{3} \left[ 1 - \frac{2x^2}{3} + \ldots \right]\)

Combine and simplify to get the expansion:

\(\frac{1}{2} - \frac{13}{12}x - \frac{41}{24}x^2\).

(a) Express \(f(x)\) in the form \(\frac{A}{2+x} + \frac{B+Cx}{3+x^2}\).

Equating coefficients, solve for \(A, B,\) and \(C\):

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

(b) Expand \(\left(1 + \frac{1}{2}x\right)^{-1}\) and \(\left(1 + \frac{1}{3}x^2\right)^{-1}\) up to \(x^2\):

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

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

Substitute back into the partial fractions and simplify:

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

Multiply out and collect terms up to \(x^2\):

\(\frac{7}{3} - \frac{5}{3}x + \frac{7}{18}x^2\)

(a) To express \(f(x)\) in partial fractions, assume \(f(x) = \frac{A}{1+2x} + \frac{B}{1-2x} + \frac{C}{2+x}\).

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

\(2 + 11x - 10x^2 = A(1-2x)(2+x) + B(1+2x)(2+x) + C(1+2x)(1-2x)\).

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

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

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

(b) Expand each partial fraction up to \(x^2\):

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

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

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

Add the expansions:

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

Combine like terms:

\(f(x) = 1 + 5x - \frac{7}{2}x^2\).