A-Level Mathematics October/November 2024 Q1: Expand (9-3x) in ascending powers of x, up to and including the term in x², simplifying…

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  The expression is $(9-3x)^{\frac{1}{2}}$.

  First, rewrite the expression in the form $a(1+y)^n$: $(9-3x)^{\frac{1}{2}} = \left(9\left(1 - \frac{3x}{9}\right)\right)^{\frac{1}{2}} = 9^{\frac{1}{2}}\left(1 - \frac{x}{3}\right)^{\frac{1}{2}} = 3\left(1 - \frac{x}{3}\right)^{\frac{1}{2}}$

  The binomial expansion formula for $(1+y)^n$ is $1 + ny + \frac{n(n-1)}{2!}y^2 + \dots$

  Here, $n = \frac{1}{2}$ and $y = -\frac{x}{3}$.

  Substitute these values into the expansion: $3\left(1 - \frac{x}{3}\right)^{\frac{1}{2}} = 3\left[ 1 + \left(\frac{1}{2}\right)\left(-\frac{x}{3}\right) + \frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)}{2!}\left(-\frac{x}{3}\right)^2 + \dots \right]$ $= 3\left[ 1 + \left(\frac{1}{2}\right)\left(-\frac{x}{3}\right) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2}\left(\frac{x^2}{9}\right) + \dots \right]$

  Now, simplify the terms: $= 3\left[ 1 - \frac{x}{6} - \frac{1}{8}\left(\frac{x^2}{9}\right) + \dots \right]$ $= 3\left[ 1 - \frac{x}{6} - \frac{x^2}{72} + \dots \right]$

  Finally, multiply by the 3 outside the bracket: $= 3(1) + 3\left(-\frac{x}{6}\right) + 3\left(-\frac{x^2}{72}\right) + \dots$ $= 3 - \frac{1}{2}x - \frac{1}{24}x^2 + \dots$

  So, the expansion up to the term in $x^2$ is: $3 - \frac{1}{2}x - \frac{1}{24}x^2$

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  How the marks are awarded

  • B1 — Stating the correct first term, 3. This is achieved by correctly calculating $9^{\frac{1}{2}}$ after factoring it out.
  • M1 — Demonstrating the correct method for the $x$ or $x^2$ term. This is shown by substituting $n=\frac{1}{2}$ and $y=-\frac{x}{3}$ into the binomial formula, for example in the line $3\left[ 1 + (\frac{1}{2})(-\frac{x}{3}) + \dots \right]$ or the subsequent line showing the unsimplified $x^2$ term.
  • A1 — Obtaining the correct simplified second term, $-\frac{1}{2}x$. This comes from simplifying $3 \times (\frac{1}{2})(-\frac{x}{3})$.
  • A1 — Obtaining the correct simplified third term, $-\frac{1}{24}x^2$. This is the result of simplifying $3 \times \frac{(\frac{1}{2})(-\frac{1}{2})}{2} \times (\frac{x^2}{9})$.

  Common mistakes

  • Forgetting to apply the power of $\frac{1}{2}$ to the factored-out 9, leading to an incorrect multiplier of 9 instead of 3.
  • Making a sign error when substituting for $y$, using $y = \frac{x}{3}$ instead of $y = -\frac{x}{3}$, which results in an incorrect positive sign for the $x$ term.
  • Errors in the $x^2$ term, such as forgetting to square the denominator of $(-\frac{x}{3})$ to get $\frac{x^2}{9}$, or making an arithmetic slip in the coefficient $\frac{n(n-1)}{2!}$.
  • Not multiplying all the terms of the expansion by the factor of 3 at the end.

  Examiner tip: For binomial expansions of the form $(a+bx)^n$, always factor out the constant 'a' first to get $a^n(1+\frac{b}{a}x)^n$, as this greatly simplifies the subsequent calculations.

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