| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Coefficient zero after multiplying binomial |
| Difficulty | Moderate -0.8 This is a straightforward binomial expansion question requiring routine application of the binomial theorem formula to find specific terms, followed by a simple algebraic manipulation to find k by setting the x³ coefficient to zero. The calculations are mechanical with no conceptual challenges beyond standard AS-level binomial theorem knowledge. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Term in \(x^3\): \(^6C_1 \times \left(\frac{3x}{2}\right)^3 = \frac{-540x^3}{8}\) | M1 A1 | For either unsimplified term co |
| A1 [3] | co (omission or error with "\(-\)" can still gain 2 out of 3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\to k = 1\) | M1 A1 [2] | considers exactly 2 terms in \(x^3\) co |
**(i)** Term in $x^2$: $^6C_3 \times \left(\frac{3x}{2}\right)^3 = \frac{135x^2}{4}$
Term in $x^3$: $^6C_1 \times \left(\frac{3x}{2}\right)^3 = \frac{-540x^3}{8}$ | M1 A1 | For either unsimplified term co
| | A1 [3] | co (omission or error with "$-$" can still gain 2 out of 3)
**(ii)** Term in $x^3 = \frac{270x^3}{4} - \frac{135x^3}{2}$
$\to k = 1$ | M1 A1 [2] | considers exactly 2 terms in $x^3$ co\begin{enumerate}[label=(\roman*)]
\item Find the terms in $x^2$ and $x^3$ in the expansion of $\left(1 - \frac{3}{2}x\right)^6$. [3]
\item Given that there is no term in $x^3$ in the expansion of $(k + 2x)\left(1 - \frac{3}{2}x\right)^6$, find the value of the constant $k$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q2 [5]}}