| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 6 |
| Paper | 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 by equating coefficients. The algebra is routine and the problem-solving element (recognizing which terms contribute to the constant and x² terms after multiplication) is minimal, making it slightly easier than average. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(2+\frac{1}{3}kx\right)^6 = 2^6 + {}^6C_1 2^5\left(\frac{1}{3}kx\right) + {}^6C_2 2^4\left(\frac{1}{3}kx\right)^2 + \ldots\) | M1 | Attempt at least 2 terms – products of binomial coefficients and correct powers of 2 and \(\frac{1}{3}kx\). Using \(kx\) rather than \(\frac{1}{3}kx\) mark as MR -2 |
| \(64 + 64kx\) | A1 | |
| \(+\frac{80}{3}k^2x^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3-4x)\left(64 + 64kx + \frac{80}{3}k^2x^2 + \ldots\right) = 192 + \ldots + \left(80k^2 - 256k\right)x^2\) | M1* | Using two terms from expansion in (a) to find coefficient of \(x^2\) |
| \(5k^2 - 16k - 12 = 0 \Rightarrow k = \ldots\) | M1dep* | Forming a 3TQ in \(k\). Using \(3\times\) their constant term from (a) |
| \(k = \frac{8+2\sqrt{31}}{5}\) | A1 | BC must be positive root only |
## Question 4:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(2+\frac{1}{3}kx\right)^6 = 2^6 + {}^6C_1 2^5\left(\frac{1}{3}kx\right) + {}^6C_2 2^4\left(\frac{1}{3}kx\right)^2 + \ldots$ | M1 | Attempt at least 2 terms – products of binomial coefficients and correct powers of 2 and $\frac{1}{3}kx$. Using $kx$ rather than $\frac{1}{3}kx$ mark as MR -2 |
| $64 + 64kx$ | A1 | |
| $+\frac{80}{3}k^2x^2$ | A1 | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3-4x)\left(64 + 64kx + \frac{80}{3}k^2x^2 + \ldots\right) = 192 + \ldots + \left(80k^2 - 256k\right)x^2$ | M1* | Using two terms from expansion in (a) to find coefficient of $x^2$ |
| $5k^2 - 16k - 12 = 0 \Rightarrow k = \ldots$ | M1dep* | Forming a 3TQ in $k$. Using $3\times$ their constant term from (a) |
| $k = \frac{8+2\sqrt{31}}{5}$ | A1 | BC must be positive root only |
---4
\begin{enumerate}[label=(\alph*)]
\item Find and simplify the first three terms in the expansion, in ascending powers of $x$, of $\left( 2 + \frac { 1 } { 3 } k x \right) ^ { 6 }$, where $k$ is a constant.
\item In the expansion of $( 3 - 4 x ) \left( 2 + \frac { 1 } { 3 } k x \right) ^ { 6 }$, the constant term is equal to the coefficient of $x ^ { 2 }$.
Determine the exact value of $k$, given that $k$ is positive.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q4 [6]}}