A-Level Mathematics October/November 2024 Q7(b): Hence find the coefficient of x³ in the expansion of f(x).

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  From part (a), the partial fraction decomposition of $f(x)$ is: $f(x) = \frac{1}{1+2x} + \frac{2x+1}{2+x^2}$

  We need to find the coefficient of $x^3$ in the binomial expansion of each term.

  For the first term, $\frac{1}{1+2x} = (1+2x)^{-1}$: The general binomial expansion is $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + ...$ Here, $n=-1$ and $y=2x$. The term in $x^3$ is: $\frac{(-1)(-1-1)(-1-2)}{3!}(2x)^3 = \frac{(-1)(-2)(-3)}{6}(8x^3) = \frac{-6}{6}(8x^3) = -8x^3$. The coefficient of $x^3$ from this term is $-8$.

  For the second term, $\frac{2x+1}{2+x^2} = (2x+1)(2+x^2)^{-1}$: First, expand $(2+x^2)^{-1}$: $(2+x^2)^{-1} = 2^{-1}(1+\frac{x^2}{2})^{-1} = \frac{1}{2}(1+\frac{x^2}{2})^{-1}$

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

  Now, multiply by the numerator $(2x+1)$: $(2x+1) \left( \frac{1}{2} - \frac{x^2}{4} + ... \right)$ To find the term in $x^3$, we multiply the term in $x$ from the first bracket with the term in $x^2$ from the second bracket: $(2x) \times (-\frac{x^2}{4}) = -\frac{2x^3}{4} = -\frac{1}{2}x^3$. The coefficient of $x^3$ from this term is $-\frac{1}{2}$.

  Total coefficient of $x^3$: Sum the coefficients from both parts: Total coefficient = $(-8) + (-\frac{1}{2}) = -8.5$.

  Final Answer: The coefficient of $x^3$ is $-8.5$.

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  How the marks are awarded

  • B1 FT — The B1 mark is awarded for correctly finding the coefficient of $x^3$ from the expansion of $(1+2x)^{-1}$. This is shown by the calculation $\frac{(-1)(-2)(-3)}{3!}(2x)^3$, which gives the coefficient $-8$. This mark is 'Follow Through' (FT) based on the value of A found in part (a).
  • M1 — The M1 mark is for the correct method to expand the second fraction's denominator. This involves factorising out the 2 to get $(2+x^2)^{-1} = \frac{1}{2}(1+\frac{x^2}{2})^{-1}$ and then applying the binomial expansion to find the terms up to $x^2$.
  • A1 FT — The A1 FT mark is awarded for correctly multiplying the relevant terms to find the $x^3$ contribution from the second partial fraction. This is achieved by multiplying the $2x$ from the numerator with the $-\frac{x^2}{4}$ term from the expansion, resulting in a coefficient of $-\frac{1}{2}$. This mark follows through from the value of B found in part (a).
  • A1 — The final A1 mark is for combining the coefficients from both expansions ($-8$ and $-\frac{1}{2}$) to obtain the correct final answer of $-8.5$ or an equivalent fraction like $-\frac{17}{2}$.

  Common mistakes

  • Sign errors in the binomial expansion, for example calculating the coefficient from the first term as $+8$ instead of $-8$.
  • Forgetting to raise the entire term to the power, e.g., writing $2x^3$ instead of $(2x)^3$, which leads to an incorrect coefficient of $-2$ instead of $-8$.
  • Incorrectly identifying which terms to multiply. For example, trying to find an $x^3$ term in the expansion of $(2+x^2)^{-1}$ (which only has even powers of x) to multiply by the constant in the numerator.
  • Making an error when factoring out the constant from the second denominator, for example expanding $(2+x^2)^{-1}$ as $2^{-1}(1+x^2)^{-1}$ instead of $2^{-1}(1+x^2/2)^{-1}$.

  Examiner tip: When finding a specific coefficient from a product of expressions, such as $(Bx+C)(1+...)^{-1}$, systematically identify which term from the first expression multiplies with which term from the second to produce the required power of x.

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