| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Coefficient zero after multiplying binomial |
| Difficulty | Standard +0.3 This is a straightforward binomial expansion question requiring students to expand (3+ax)^5 to three terms, multiply by (a-x), and set the coefficient of x to zero. While it involves multiple steps and algebraic manipulation, it follows a standard template with no novel insight required—slightly easier than average due to being a routine C2-level question. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.04b Binomial probabilities: link to binomial expansion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3+ax)^5 = 3^5 + \binom{5}{1}3^4(ax) + \binom{5}{2}3^3(ax)^2 + ...\) | M1 | Attempt at binomial expansion; correct binomial coefficient with correct power of 3 and correct power of \(x\); ignore bracketing errors |
| \(= 243 + 405ax + 270a^2x^2 + ...\) | B1, A1, A1 [4] | B1: simplified to 243 (just \(3^5\) is B0); A1: one correct from \(405ax\) and \(270a^2x^2\); A1: both correct. Allow \(270(ax)^2\) with bracket correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x)=(a-x)(3+ax)^5=(a-x)(243+405ax+270a^2x^2+...)\) | M1 | Attempt to set coefficient of \(x\) in expansion of \((a-x)(3+ax)^5\) equal to 0; must see equation of form \(\pm P \pm Qa^2 = 0\) |
| \(-243+405a^2=0 \Rightarrow a^2=\frac{243}{405} \Rightarrow a=\sqrt{\frac{3}{5}}\) or equivalent | dM1A1 [3] | dM1: for \(\pm P \pm Qa^2=0 \Rightarrow a=...\); cannot score for attempt at sq rooting negative. A1: \(a=\sqrt{\frac{3}{5}}\) or exact equivalent e.g. \(a=\frac{\sqrt{15}}{5}\); may ignore reference to \(a=-\sqrt{\frac{3}{5}}\) |
# Question 11:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3+ax)^5 = 3^5 + \binom{5}{1}3^4(ax) + \binom{5}{2}3^3(ax)^2 + ...$ | M1 | Attempt at binomial expansion; correct binomial coefficient with correct power of 3 and correct power of $x$; ignore bracketing errors |
| $= 243 + 405ax + 270a^2x^2 + ...$ | B1, A1, A1 [4] | B1: simplified to 243 (just $3^5$ is B0); A1: one correct from $405ax$ and $270a^2x^2$; A1: both correct. Allow $270(ax)^2$ with bracket correct |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x)=(a-x)(3+ax)^5=(a-x)(243+405ax+270a^2x^2+...)$ | M1 | Attempt to set coefficient of $x$ in expansion of $(a-x)(3+ax)^5$ equal to 0; must see equation of form $\pm P \pm Qa^2 = 0$ |
| $-243+405a^2=0 \Rightarrow a^2=\frac{243}{405} \Rightarrow a=\sqrt{\frac{3}{5}}$ or equivalent | dM1A1 [3] | dM1: for $\pm P \pm Qa^2=0 \Rightarrow a=...$; cannot score for attempt at sq rooting negative. A1: $a=\sqrt{\frac{3}{5}}$ or exact equivalent e.g. $a=\frac{\sqrt{15}}{5}$; may ignore reference to $a=-\sqrt{\frac{3}{5}}$ |
---11. $\mathrm { f } ( x ) = ( a - x ) ( 3 + a x ) ^ { 5 }$, where $a$ is a positive constant
\begin{enumerate}[label=(\alph*)]
\item Find the first 3 terms, in ascending powers of $x$, in the binomial expansion of
$$( 3 + a x ) ^ { 5 }$$
Give each term in its simplest form.
Given that in the expansion of $\mathrm { f } ( x )$ the coefficient of $x$ is zero,
\item find the exact value of $a$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q11 [7]}}