| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Two unknowns from two coefficient conditions |
| Difficulty | Standard +0.3 This is a straightforward application of the binomial expansion formula requiring students to find coefficients, set up an equation from given conditions, and solve. While it involves multiple steps and algebraic manipulation, each step follows standard procedures taught in P4 with no novel insight required. The constraint solving in part (b) is routine algebra, making this slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A = \frac{1}{9}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \("3^{-2}"(1+(-2)(\frac{kx}{3})+\frac{(-2)(-3)}{2}(\frac{kx}{3})^2+...)\) | B1 | Correct unsimplified coefficients for \(x\) and \(x^2\) either in expansion or separate. Accept \(B=-\frac{2k}{3}\) and \(C=\frac{k^2}{3}\) if \(3^{-2}\) missing. B0 if brackets on \((\frac{k}{3})^2\) missing unless implied by recovery |
| \(\frac{(-2)(-3)}{2}(\frac{k}{3})^2 = 3\times(-2)(\frac{k}{3}) \Rightarrow ...k^2=...k\) | M1 | Sets coefficient of \(x^2\) equal to 3 times coefficient of \(x\) to produce two-term quadratic |
| \(k^2+6k=0\) * | A1* | Achieves given answer from correct equation; condone if \(B\) and \(C\) both missed the \(3^{-2}\); may be scored if \(A\) was incorrect |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k=-6\) | B1 | \(k=-6\) only; \(k=0\) solution must be rejected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3^{-2}\frac{(-2)(-3)(-4)}{3!}(\frac{"-6"}{3})^3=\frac{32}{9}\) | M1A1 | M1: substitutes non-zero value for \(k\) into correct expression for coefficient of \(x^3\); must include \(3^{-2}\). A1: \(\frac{32}{9}\) oe |
# Question 1:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $A = \frac{1}{9}$ | B1 | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $"3^{-2}"(1+(-2)(\frac{kx}{3})+\frac{(-2)(-3)}{2}(\frac{kx}{3})^2+...)$ | B1 | Correct unsimplified coefficients for $x$ and $x^2$ either in expansion or separate. Accept $B=-\frac{2k}{3}$ and $C=\frac{k^2}{3}$ if $3^{-2}$ missing. B0 if brackets on $(\frac{k}{3})^2$ missing unless implied by recovery |
| $\frac{(-2)(-3)}{2}(\frac{k}{3})^2 = 3\times(-2)(\frac{k}{3}) \Rightarrow ...k^2=...k$ | M1 | Sets coefficient of $x^2$ equal to 3 times coefficient of $x$ to produce two-term quadratic |
| $k^2+6k=0$ * | A1* | Achieves given answer from correct equation; condone if $B$ and $C$ both missed the $3^{-2}$; may be scored if $A$ was incorrect |
## Part (c)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k=-6$ | B1 | $k=-6$ only; $k=0$ solution must be rejected |
## Part (c)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $3^{-2}\frac{(-2)(-3)(-4)}{3!}(\frac{"-6"}{3})^3=\frac{32}{9}$ | M1A1 | M1: substitutes non-zero value for $k$ into correct expression for coefficient of $x^3$; must include $3^{-2}$. A1: $\frac{32}{9}$ oe |
---\begin{enumerate}
\item The binomial expansion of
\end{enumerate}
$$( 3 + k x ) ^ { - 2 } \quad | k x | < 3$$
where $k$ is a non-zero constant, may be written in the form
$$A + B x + C x ^ { 2 } + D x ^ { 3 } + \ldots$$
where $A$, $B$, $C$ and $D$ are constants.\\
(a) Find the value of $A$
Given that $C = 3 B$\\
(b) show that
$$k ^ { 2 } + 6 k = 0$$
(c) Hence (i) find the value of $k$\\
(ii) find the value of $D$
\hfill \mbox{\textit{Edexcel P4 2022 Q1 [7]}}