A-Level Mathematics May/June 2020 Q2: The coefficient of $\frac{1}{x}$ in the expansion of $(kx + \frac{1}{x})^5 + (1 - \frac…

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  The problem requires us to find the coefficient of $\frac{1}{x}$ (or $x^{-1}$) in the expansion of $(kx + \frac{1}{x})^5 + (1 - \frac{2}{x})^8$ and use it to find the value of the positive constant $k$.

  First, consider the expansion of $(kx + \frac{1}{x})^5$. The general term is given by $\binom{5}{r}(kx)^{5-r}(\frac{1}{x})^r = \binom{5}{r}k^{5-r}x^{5-r}x^{-r} = \binom{5}{r}k^{5-r}x^{5-2r}$. To find the term in $x^{-1}$, we set the power of $x$ to $-1$: $5 - 2r = -1$ $2r = 6$ $r = 3$

  The term is $\binom{5}{3}k^{5-3}x^{-1} = \binom{5}{3}k^2x^{-1}$. The coefficient is $\binom{5}{3}k^2 = 10k^2$.

  Next, consider the expansion of $(1 - \frac{2}{x})^8$. The general term is given by $\binom{8}{r}(1)^{8-r}(-\frac{2}{x})^r = \binom{8}{r}(-2)^r x^{-r}$. To find the term in $x^{-1}$, we set the power of $x$ to $-1$: $-r = -1$ $r = 1$

  The term is $\binom{8}{1}(-2)^1 x^{-1}$. The coefficient is $8 \times (-2) = -16$.

  The total coefficient of $x^{-1}$ is the sum of the coefficients from both expansions: Total coefficient = $10k^2 - 16$.

  We are given that this coefficient is 74. $10k^2 - 16 = 74$ $10k^2 = 90$ $k^2 = 9$ $k = \pm 3$

  Since $k$ is a positive constant, we take the positive root. $k = 3$

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  How the marks are awarded

  • B1 — Correctly identifying the binomial coefficient for the first expansion as $\binom{5}{3}$ which evaluates to 10.
  • B1 — Correctly identifying the power of $k$ as $k^2$, leading to the full coefficient of the term being $10k^2$.
  • B2 — Correctly calculating the coefficient of $x^{-1}$ in the expansion of $(1 - \frac{2}{x})^8$ as $-16$. This is worth two marks.
  • B1 — Forming the correct equation $10k^2 - 16 = 74$ and solving it to find the final answer $k=3$, having correctly rejected the negative solution.

  Common mistakes

  • Forgetting the negative sign in the second expansion, $(1 - \frac{2}{x})^8$, and calculating the coefficient as $+16$ instead of $-16$. This leads to an incorrect final equation.
  • Incorrectly determining the power of $k$ in the first expansion. For example, using $r=2$ instead of $r=3$, which would lead to a term in $k^3$ instead of $k^2$.
  • Making a sign error when solving the equation for the power of $x$. For example, in $5-2r=-1$, incorrectly rearranging to get $2r=4$ and thus $r=2$.
  • Forgetting that the question specifies $k$ is a positive constant and either giving both solutions ($k=\pm 3$) or failing to select the positive root as the final answer.

  Examiner tip: When finding a specific term in a binomial expansion, first use the general term formula $\binom{n}{r}a^{n-r}b^r$ to set up an equation for the required power of the variable and solve for $r$.

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