Mathematics 9709/12 — May/June 2017

Walkthrough

We are asked for the coefficient of $x$ in the expansion of $\left(2x - \frac{1}{x}\right)^5$. The binomial expansion has terms of the form $\binom{5}{r}(2x)^{5-r}\left(-\frac{1}{x}\right)^r$. Each such term contributes a power of $x$ equal to $5-r$ from $x^{5-r}$ and $-r$ from $x^{-r}$, so the combined power is $5-2r$. To get an $x$ term we need this power to be $1$, which gives $r=2$. We then substitute $r=2$ into the coefficient formula: $\binom{5}{2}2^{3}(-1)^2 = 10 \cdot 8 = 80$. This is the coefficient of $x$.

Key Takeaways

The general term of $(a+b)^n$ is $\binom{n}{r}a^{n-r}b^r$. To find the coefficient of a particular power, set the exponent of $x$ equal to that power and solve for $r$. Remember to include signs and constants from both factors.

Common Mistakes

• Forgetting the $(-1)^r$ factor, which would give the wrong sign for odd values of $r$.
• Setting the exponent incorrectly: the $x$ and $\frac{1}{x}$ powers combine as $5-2r$, not $5-r$.
• Choosing $r=3$ instead of $r=2$; $r=3$ gives the $\frac{1}{x}$ term, not the $x$ term.
• The mark scheme notes that the correct value must be selected; if only $+80$ or $-80$ appears in an expansion, it may earn partial credit.

Things to Be Careful About

• Check the exponent equation carefully: $5-2r=1$ gives $r=2$.
• The sign is positive because $(-1)^2=1$.
• In the mark scheme, the correct value must be selected for both marks; an unsupported answer may not receive full credit.

Walkthrough

We already know the coefficient of $x$ in $\left(2x - \frac{1}{x}\right)^5$ is $80$ from part (i). When we multiply the whole expansion by $1+3x^2$, the only ways to get an $x$ term are:

• $1$ times the $x$ term in the expansion, contributing $1 \times 80x = 80x$.
• $3x^2$ times the $\frac{1}{x}$ term in the expansion, because $x^2 \cdot x^{-1} = x$.

So we need the coefficient of $x^{-1}$ in the expansion. Using the general term, the exponent is $5-2r$. Set $5-2r=-1$, so $r=3$. The coefficient is $\binom{5}{3}2^2(-1)^3 = 10 \cdot 4 \cdot (-1) = -40$. Then the second contribution is $3 \times (-40) = -120$. Adding gives $80 - 120 = -40$.

Key Takeaways

When multiplying an expansion by another polynomial, each term in the multiplier can combine with a different term of the expansion to produce the required power. The phrase 'hence' means the previous result should be reused.

Common Mistakes

• Forgetting the factor $3$ from $3x^2$ when combining.
• Using the wrong power: the term needed is $x^{-1}$, not $x$.
• Sign errors: the coefficient of $x^{-1}$ is negative because $(-1)^3 = -1$.
• The mark scheme gives B2 for the correct $\frac{1}{x}$ coefficient seen or implied, and M1 for linking the appropriate two terms only; a final answer without showing the combination may lose marks.

Things to Be Careful About

• Check that $x^2 \cdot x^{-1} = x^1$, so the $\frac{1}{x}$ term is the one that matters.
• Keep the sign of $-40$ when combining: $80 + 3(-40) = -40$.
• If powers are left unsimplified, the mark scheme only allows credit if the correct term is selected.