Let \(f(x) = \frac{9x^2 + 4}{(2x + 1)(x - 2)^2}\).
(i) Express \(f(x)\) in partial fractions.
(ii) Show that, when \(x\) is sufficiently small for \(x^3\) and higher powers to be neglected, \(f(x) = 1 - x + 5x^2\).
Solution
(i) Express \(f(x)\) as partial fractions:
Assume \(f(x) = \frac{A}{2x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}\).
Equating coefficients, we find:
\(A = 1\), \(B = 0\), \(C = 4\).
Thus, \(f(x) = \frac{1}{2x+1} + \frac{4}{(x-2)^2}\).
(ii) For small \(x\), expand \((1 + 2x)^{-1}\) and \((x-2)^{-2}\):
\((1 + 2x)^{-1} \approx 1 - 2x\)
\((x-2)^{-2} \approx 1 + 4x + 8x^2\)
Multiply these expansions:
\((1 - 2x)(1 + 4x + 8x^2) \approx 1 + 2x + 8x^2\)
Combine terms to get:
\(f(x) \approx 1 - x + 5x^2\).
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