Mathematics 9709/13 — May/June 2022

Walkthrough

We need the coefficient of $x^3$ in $\left(p + \frac{1}{p}x\right)^4$. Each bracket contributes either $p$ or $\frac{x}{p}$. To get $x^3$, three brackets must contribute $\frac{x}{p}$ and one bracket must contribute $p$. The single bracket that contributes $p$ can be chosen in ${}^4C_1 = 4$ ways.

The numerical part from three $\frac{x}{p}$ factors is $\left(\frac{1}{p}\right)^3 = \frac{1}{p^3}$", and the one $p$ factor multiplies this, giving $p \cdot \frac{1}{p^3} = \frac{1}{p^2}$. Multipying by the 4 ways gives the coefficient $\frac{4}{p^2}$.

We are told the coefficient is 144, so we set $\frac{4}{p^2} = 144$. Dividing both sides by 4 gives $p^2 = \frac{1}{36}$. Taking the square root of both sides gives two solutions, $p = \frac{1}{6}$ and $p = -\frac{1}{6}$, because squaring removes the sign.

Key Takeaways

This question tests the binomial expansion of $(a + b)^n$ when the two terms contain powers of the same parameter $p$. The key skill is to identify the correct term and then simplify the powers of $p$ correctly. It also tests that solving $p^2 = k$ produces two possible signs.

Common Mistakes

• A common error is to use $r=1$ or $r=2$ instead of $r=3$ when looking for the $x^3$ term.
• Some students stop at $p^2 = \frac{1}{36}$ and forget to state both $p = \frac{1}{6}$ and $p = -\frac{1}{6}$.
• Another common error is to write the coefficient as $4p \cdot \frac{1}{p^3}$ and then simplify incorrectly, or to leave $p$ in both numerator and denominator without cancelling.
• The mark scheme awards a special case of just one mark for writing $\pm \sqrt{\frac{1}{36}}$ without simplifying fully; it is better to give the simplified $\pm \frac{1}{6}$.

Things to Be Careful About

• The coefficient of a term is the complete numerical and algebraic factor multiplying $x^3$; the $x^3$ factor itself is not part of the coefficient.
• Since $p$ appears in a denominator, $p = 0$ is not possible, though it would also make the original expression undefined.
• Remember that both positive and negative $p$ give the same value of $p^2$ and therefore both satisfy the condition.
• Check your simplification: $p \cdot \frac{1}{p^3} = \frac{1}{p^2}$, not $\frac{1}{p^4}$.