A-Level Mathematics May/June 2025 Q7(b): Hence obtain the expansion of f(x) in ascending powers of x, up to and including the te…

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  From part (a), we have the partial fraction decomposition of $f(x)$:

  $f(x) = \frac{3}{3a+2x} - \frac{1}{2a-x}$

  First, we prepare each term for binomial expansion by writing it in the form $k(1+y)^n$.

  For the first term: $\frac{3}{3a+2x} = 3(3a+2x)^{-1} = 3(3a(1+\frac{2x}{3a}))^{-1} = 3(3a)^{-1}(1+\frac{2x}{3a})^{-1} = \frac{1}{a}(1+\frac{2x}{3a})^{-1}$

  Using the binomial expansion $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots$, with $n=-1$ and $y=\frac{2x}{3a}$: $\frac{1}{a}(1+\frac{2x}{3a})^{-1} = \frac{1}{a} \left[ 1 + (-1)(\frac{2x}{3a}) + \frac{(-1)(-2)}{2}(\frac{2x}{3a})^2 + \dots \right]$ $= \frac{1}{a} \left[ 1 - \frac{2x}{3a} + \frac{4x^2}{9a^2} + \dots \right]$ $= \frac{1}{a} - \frac{2x}{3a^2} + \frac{4x^2}{9a^3} + \dots$

  For the second term: $-\frac{1}{2a-x} = -(2a-x)^{-1} = -(2a(1-\frac{x}{2a}))^{-1} = -(2a)^{-1}(1-\frac{x}{2a})^{-1} = -\frac{1}{2a}(1-\frac{x}{2a})^{-1}$

  Using the binomial expansion with $n=-1$ and $y=-\frac{x}{2a}$: $-\frac{1}{2a}(1-\frac{x}{2a})^{-1} = -\frac{1}{2a} \left[ 1 + (-1)(-\frac{x}{2a}) + \frac{(-1)(-2)}{2}(-\frac{x}{2a})^2 + \dots \right]$ $= -\frac{1}{2a} \left[ 1 + \frac{x}{2a} + \frac{x^2}{4a^2} + \dots \right]$ $= -\frac{1}{2a} - \frac{x}{4a^2} - \frac{x^2}{8a^3} + \dots$

  Now, we combine the two expansions: $f(x) = \left( \frac{1}{a} - \frac{2x}{3a^2} + \frac{4x^2}{9a^3} \right) + \left( -\frac{1}{2a} - \frac{x}{4a^2} - \frac{x^2}{8a^3} \right) + \dots$

  Collecting like terms: Constant term: $\frac{1}{a} - \frac{1}{2a} = \frac{1}{2a}$ Term in $x$: $-\frac{2x}{3a^2} - \frac{x}{4a^2} = (-\frac{2}{3} - \frac{1}{4})\frac{x}{a^2} = (-\frac{8}{12} - \frac{3}{12})\frac{x}{a^2} = -\frac{11x}{12a^2}$ Term in $x^2$: $\frac{4x^2}{9a^3} - \frac{x^2}{8a^3} = (\frac{4}{9} - \frac{1}{8})\frac{x^2}{a^3} = (\frac{32}{72} - \frac{9}{72})\frac{x^2}{a^3} = \frac{23x^2}{72a^3}$

  So, the expansion of $f(x)$ is: $f(x) = \frac{1}{2a} - \frac{11x}{12a^2} + \frac{23x^2}{72a^3} + \dots$

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  How the marks are awarded

  • M1 — Correctly rewriting either partial fraction into the form $k(1+y)^n$ and applying the binomial formula to find the first two terms. For example, expanding $\frac{1}{a}(1+\frac{2x}{3a})^{-1}$ to get $\frac{1}{a}(1 - \frac{2x}{3a} + ...)$.
  • A2 — Obtaining the correct, unsimplified expansions for BOTH partial fractions up to the $x^2$ term. This corresponds to the lines $\frac{1}{a} \left[ 1 - \frac{2x}{3a} + \frac{4x^2}{9a^2} + \dots \right]$ and $-\frac{1}{2a} \left[ 1 + \frac{x}{2a} + \frac{x^2}{4a^2} + \dots \right]$. One mark (A1) is awarded if only one of these is correct.
  • A1 — Correctly combining the terms from both expansions and simplifying to obtain the final answer $\frac{1}{2a} - \frac{11x}{12a^2} + \frac{23x^2}{72a^3}$.

  Common mistakes

  • Incorrectly factoring out the constant term, for example writing $3(3a+2x)^{-1}$ as $3a(1+2x)^{-1}$ instead of $\frac{1}{a}(1+\frac{2x}{3a})^{-1}$.
  • Errors in the binomial formula, particularly with signs. Forgetting that in $(1-\frac{x}{2a})^{-1}$, the term for expansion is $y = -\frac{x}{2a}$, leading to all positive terms in the bracket.
  • Forgetting to square all parts of the term in the $x^2$ coefficient, e.g., writing $(\frac{2x}{3a})^2$ as $\frac{2x^2}{9a^2}$ instead of the correct $\frac{4x^2}{9a^2}$.
  • Arithmetic errors when combining the simplified fractions to find the final coefficients, especially with finding a common denominator for the $x$ and $x^2$ terms.

  Examiner tip: Master the two-stage process of first correctly factoring out the constant to get the required $(1+y)^n$ form, and then carefully applying the binomial expansion formula.

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