| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Product with unknown constant to determine |
| Difficulty | Standard +0.3 This is a straightforward binomial expansion question requiring standard application of the binomial theorem to find three terms, then solving a linear equation after multiplying polynomials. The algebra is routine and the method is direct, making it slightly easier than average for A-level. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((2+px)^6 = (2^6+)\ 6\times2^5(px)+\frac{6\times5}{2}\times2^4(px)^2+...\) | M1 | Correct binomial coefficients linked with correct powers of \(x\); condone invisible brackets and \(p\) not squared on third term |
| \(2^6\) or \(64\) | B1 | Sight of \(2^6\) or \(64\) as first term |
| \(+192px\) or \(+240p^2x^2\) | A1 | Having gained the M; allow also for \(2^6\left(1+3px+\frac{15}{4}p^2x^2\right)\) |
| \((64+)192px+240p^2x^2\) | A1 | Both correct \(x\) and \(x^2\) terms; ISW after correct solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(3-\frac{1}{2}x\right)(2+px)^6\Rightarrow\left(3-\frac{1}{2}x\right)(64+192px+240p^2x^2)\) | — | Use of part (a) result |
| Attempts \(3\times\)"\(240p^2\)" and \(\left(-\frac{1}{2}\right)\times\)"\(192p\)" | M1 | Identifies two \(x^2\) terms |
| \(3\times\)"\(240p^2\)"\(+\left(-\frac{1}{2}\right)\times\)"\(192p\)"\(=-\frac{3}{4}\) | dM1 | Complete method; sets coefficient equal to \(-\frac{3}{4}\); dependent on previous M |
| \(2880p^2-384p+3=0\Rightarrow p=...\) | ddM1 | Attempts to solve 3TQ in \(p\); dependent on both previous M marks |
| \((p=)\ \frac{1}{8},\frac{1}{120}\) | A1 | Accept decimal equivalents \(0.125\) and awrt \(0.00833\) |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2+px)^6 = (2^6+)\ 6\times2^5(px)+\frac{6\times5}{2}\times2^4(px)^2+...$ | M1 | Correct binomial coefficients linked with correct powers of $x$; condone invisible brackets and $p$ not squared on third term |
| $2^6$ or $64$ | B1 | Sight of $2^6$ or $64$ as first term |
| $+192px$ or $+240p^2x^2$ | A1 | Having gained the M; allow also for $2^6\left(1+3px+\frac{15}{4}p^2x^2\right)$ |
| $(64+)192px+240p^2x^2$ | A1 | Both correct $x$ and $x^2$ terms; ISW after correct solution |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(3-\frac{1}{2}x\right)(2+px)^6\Rightarrow\left(3-\frac{1}{2}x\right)(64+192px+240p^2x^2)$ | — | Use of part (a) result |
| Attempts $3\times$"$240p^2$" and $\left(-\frac{1}{2}\right)\times$"$192p$" | M1 | Identifies two $x^2$ terms |
| $3\times$"$240p^2$"$+\left(-\frac{1}{2}\right)\times$"$192p$"$=-\frac{3}{4}$ | dM1 | Complete method; sets coefficient equal to $-\frac{3}{4}$; dependent on previous M |
| $2880p^2-384p+3=0\Rightarrow p=...$ | ddM1 | Attempts to solve 3TQ in $p$; dependent on both previous M marks |
| $(p=)\ \frac{1}{8},\frac{1}{120}$ | A1 | Accept decimal equivalents $0.125$ and awrt $0.00833$ |
---4. (a) Find the first three terms, in ascending powers of $x$, of the binomial expansion of
$$( 2 + p x ) ^ { 6 }$$
where $p$ is a constant. Give each term in simplest form.
Given that in the expansion of
$$\left( 3 - \frac { 1 } { 2 } x \right) ( 2 + p x ) ^ { 6 }$$
the coefficient of $x ^ { 2 }$ is $- \frac { 3 } { 4 }$\\
(b) find the possible values of $p$.\\
\includegraphics[max width=\textwidth, alt={}, center]{52c90d0e-a5e4-45fa-95a4-9523287e7588-11_2255_50_314_34}\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
\hline
\end{tabular}
\end{center}
\hfill \mbox{\textit{Edexcel P2 2021 Q4 [8]}}