| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Substitution into binomial expansion |
| Difficulty | Moderate -0.8 Part (i) is direct application of binomial expansion formula requiring only the first three terms of (1+x)^5, which is straightforward recall. Part (ii) requires substituting x→(px+x²) and equating coefficients, but the question explicitly tells students to use part (i), making this a guided, routine exercise with minimal problem-solving demand. This is easier than average A-level content. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((1+x)^5 = 1 + 5x + 10x^2\) | B2,1 [2] | Loses 1 for each error |
| Answer | Marks | Guidance |
|---|---|---|
| Coeff of \(x^2 = 5 + 10p^2 = 95 \rightarrow p = 3\) | M1, DM1, A1 [3] | Replace \(x\) by \((px+x^2)\) in their expansion; Considers 2 terms; co – no penalty for \(\pm 3\) |
**(i)** $(1+x)^5 = 1 + 5x + 10x^2$ | B2,1 [2] | Loses 1 for each error
**(ii)** $(1+px+x^2)^5$
$(1+) 5(px+x^2) + 10(px+x^2)^2$
Coeff of $x^2 = 5 + 10p^2 = 95 \rightarrow p = 3$ | M1, DM1, A1 [3] | Replace $x$ by $(px+x^2)$ in their expansion; Considers 2 terms; co – no penalty for $\pm 3$\begin{enumerate}[label=(\roman*)]
\item Find the first 3 terms, in ascending powers of $x$, in the expansion of $(1 + x)^5$. [2]
\end{enumerate}
The coefficient of $x^2$ in the expansion of $(1 + (px + x^2))^5$ is 95.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Use the answer to part (i) to find the value of the positive constant $p$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2014 Q3 [5]}}