| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Product with reciprocal term binomial |
| Difficulty | Standard +0.3 This requires expanding (3 - x/2)^6 using the binomial theorem, identifying terms that yield x^5 when multiplied by either 5 or 8x^2, then combining coefficients. It's a standard binomial expansion with an extra multiplication step, slightly above average due to the product and fractional coefficient, but still routine for AS-level. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts term in \(x^3\) or term in \(x^5\) of \(\left(3-\frac{1}{2}x\right)^6\): look for \(^6C_3 3^3\left(-\frac{1}{2}x\right)^3\) or \(^6C_5 3^1\left(-\frac{1}{2}x\right)^5\) | M1 | 3.1a — condone missing brackets and sign slips |
| Correct term: \(-\frac{135}{2}x^3\) or \(-\frac{9}{16}x^5\) | A1 | 1.1b — \(^6C_3\) or \(^6C_5\) must be processed |
| Attempts one required term in \(x^5\) of \(\left(5+8x^2\right)\left(3-\frac{1}{2}x\right)^6\): either \(5\times{^6C_5}3^1\left(-\frac{1}{2}x\right)^5\) or \(8x^2\times{^6C_3}3^3\left(-\frac{1}{2}x\right)^3\) | M1 | 1.1b — \(x^5\) may be missing as only coefficient required |
| Attempts sum of \(5\times{^6C_5}3^1\left(-\frac{1}{2}x\right)^5\) and \(8x^2\times{^6C_3}3^3\left(-\frac{1}{2}x\right)^3\) | dM1 | 2.1 — dependent on previous M; condone missing brackets and signs |
| Coefficient of \(x^5 = -\frac{45}{16}-540 = -\frac{8685}{16}\) | A1 | 1.1b — must be exact; rounded decimals e.g. \(-542.81\) will not score |
## Question 14:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts term in $x^3$ or term in $x^5$ of $\left(3-\frac{1}{2}x\right)^6$: look for $^6C_3 3^3\left(-\frac{1}{2}x\right)^3$ or $^6C_5 3^1\left(-\frac{1}{2}x\right)^5$ | M1 | 3.1a — condone missing brackets and sign slips |
| Correct term: $-\frac{135}{2}x^3$ **or** $-\frac{9}{16}x^5$ | A1 | 1.1b — $^6C_3$ or $^6C_5$ must be processed |
| Attempts one required term in $x^5$ of $\left(5+8x^2\right)\left(3-\frac{1}{2}x\right)^6$: either $5\times{^6C_5}3^1\left(-\frac{1}{2}x\right)^5$ **or** $8x^2\times{^6C_3}3^3\left(-\frac{1}{2}x\right)^3$ | M1 | 1.1b — $x^5$ may be missing as only coefficient required |
| Attempts sum of $5\times{^6C_5}3^1\left(-\frac{1}{2}x\right)^5$ and $8x^2\times{^6C_3}3^3\left(-\frac{1}{2}x\right)^3$ | dM1 | 2.1 — dependent on previous M; condone missing brackets and signs |
| Coefficient of $x^5 = -\frac{45}{16}-540 = -\frac{8685}{16}$ | A1 | 1.1b — must be exact; rounded decimals e.g. $-542.81$ will not score |
**(5 marks)**
---\begin{enumerate}
\item Find, in simplest form, the coefficient of $x ^ { 5 }$ in the expansion of
\end{enumerate}
$$\left( 5 + 8 x ^ { 2 } \right) \left( 3 - \frac { 1 } { 2 } x \right) ^ { 6 }$$
\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q14 [5]}}