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