9709 P31 - Nov 2023 - Q10
2060
Let \(f(x) = \frac{24x + 13}{(1 - 2x)(2 + x)^2}\).
(a) Express \(f(x)\) in partial fractions.
(b) Hence obtain the expansion of \(f(x)\) in ascending powers of \(x\), up to and including the term in \(x^2\).
(c) State the set of values of \(x\) for which the expansion in (b) is valid.
Solution
(a) Express \(f(x)\) in partial fractions:
Assume \(\frac{24x + 13}{(1 - 2x)(2 + x)^2} = \frac{A}{1 - 2x} + \frac{B}{2 + x} + \frac{C}{(2 + x)^2}\).
Multiply through by the denominator \((1 - 2x)(2 + x)^2\) to clear the fractions:
\(24x + 13 = A(2 + x)^2 + B(1 - 2x)(2 + x) + C(1 - 2x)\).
Expand and equate coefficients to find \(A = 4\), \(B = 2\), \(C = -7\).
(b) Expand each partial fraction:
\(\frac{A}{1 - 2x} = 4(1 - 2x)^{-1} \approx 4(1 + 2x + 4x^2)\).
\(\frac{B}{2 + x} = 2(2 + x)^{-1} \approx 2(1 - \frac{x}{2} + \frac{x^2}{4})\).
\(\frac{C}{(2 + x)^2} = -7(2 + x)^{-2} \approx -7(\frac{1}{4} - \frac{x}{4} + \frac{3x^2}{16})\).
Combine these to get:
\(f(x) \approx \frac{13}{4} + \frac{37}{4}x + \frac{239}{16}x^2\).
(c) The expansion is valid for \(|x| < \frac{1}{2}\).
