| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Binomial times quadratic coefficient |
| Difficulty | Moderate -0.3 Part (a) is straightforward application of binomial theorem with positive integer n=10, requiring routine calculation of coefficients. Part (b) requires multiplying the expansion by a quadratic and collecting x³ terms—a standard 'hence' extension that adds one extra step but remains a textbook exercise with no novel insight required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(1-\frac{1}{2}x\right)^{10} = 1 + \binom{10}{1}\left(-\frac{1}{2}x\right) + \binom{10}{2}\left(-\frac{1}{2}x\right)^2 + \binom{10}{3}\left(-\frac{1}{2}x\right)^3 \ldots\) | M1 | Attempt at binomial expansion to get third and/or fourth term. Correct binomial coefficient combined with correct power of \(x\). Ignore bracket errors and omission of or incorrect powers of \(\pm\frac{1}{2}\). Accept \(^{10}C_2\), \(^{10}C_3\), e.g. \(\binom{10}{2}\) or \(\binom{10}{3}\) or 45 or 120 from Pascal's triangle |
| \(= 1 - 5x + \frac{45}{4}x^2 - 15x^3 + \ldots\) | B1, A1, A1 | Allow terms to be "listed". Allow equivalents for \(\frac{45}{4}\) e.g. \(11\frac{1}{4}\), 11.25. Allow \(+\frac{45}{4}x^2\) to come from \(\binom{10}{2}\left(\frac{1}{2}x\right)^2\). Do not allow \(1 + -5x\) for \(1-5x\) or \(+-15x^3\) for \(-15x^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((3+5x-2x^2)\left(1-5x+\frac{45}{4}x^2-15x^3\right) = \ldots\) | M1 | Uses expansion from (a) to identify correct pairing of terms and adds the \(x^3\) terms or \(x^3\) coefficients together. Look for \(3\times(-15) + 5\times\left(\frac{45}{4}\right) + (-2)\times(-5)\) with or without the \(x^3\)'s |
| \(\frac{85}{4}\) | A1 | cao (Allow \(\frac{85}{4}x^3\)) |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(1-\frac{1}{2}x\right)^{10} = 1 + \binom{10}{1}\left(-\frac{1}{2}x\right) + \binom{10}{2}\left(-\frac{1}{2}x\right)^2 + \binom{10}{3}\left(-\frac{1}{2}x\right)^3 \ldots$ | M1 | Attempt at binomial expansion to get third and/or fourth term. Correct binomial coefficient combined with correct power of $x$. Ignore bracket errors and omission of or incorrect powers of $\pm\frac{1}{2}$. Accept $^{10}C_2$, $^{10}C_3$, e.g. $\binom{10}{2}$ or $\binom{10}{3}$ or 45 or 120 from Pascal's triangle |
| $= 1 - 5x + \frac{45}{4}x^2 - 15x^3 + \ldots$ | B1, A1, A1 | Allow terms to be "listed". Allow equivalents for $\frac{45}{4}$ e.g. $11\frac{1}{4}$, 11.25. Allow $+\frac{45}{4}x^2$ to come from $\binom{10}{2}\left(\frac{1}{2}x\right)^2$. Do not allow $1 + -5x$ for $1-5x$ or $+-15x^3$ for $-15x^3$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(3+5x-2x^2)\left(1-5x+\frac{45}{4}x^2-15x^3\right) = \ldots$ | M1 | Uses expansion from (a) to identify correct pairing of terms and adds the $x^3$ terms or $x^3$ coefficients together. Look for $3\times(-15) + 5\times\left(\frac{45}{4}\right) + (-2)\times(-5)$ with or without the $x^3$'s |
| $\frac{85}{4}$ | A1 | cao (Allow $\frac{85}{4}x^3$) |
---\begin{enumerate}
\item (a) Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of
\end{enumerate}
$$\left( 1 - \frac { 1 } { 2 } x \right) ^ { 10 }$$
giving each term in its simplest form.\\
(b) Hence find the coefficient of $x ^ { 3 }$ in the expansion of
$$\left( 3 + 5 x - 2 x ^ { 2 } \right) \left( 1 - \frac { 1 } { 2 } x \right) ^ { 10 }$$
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel C12 2018 Q5 [6]}}