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