| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Single coefficient given directly |
| Difficulty | Standard +0.3 This is a straightforward binomial expansion question requiring standard application of the binomial theorem formula. Part (a) involves writing out the coefficient of x² using nCr and simplifying—routine algebraic manipulation. Parts (b)(i) and (b)(ii) involve substituting given information to find n and a through simple arithmetic. While multi-step, each step follows directly from textbook methods with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks |
|---|---|
| 6(a) | Applies Binomial Expansion |
| Answer | Marks | Guidance |
|---|---|---|
| or a2 correct. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | 1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 6(b)(i) | Verifies that n = 6 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 6(b)(ii) | Substitutes n = 6 into formula |
| Answer | Marks | Guidance |
|---|---|---|
| PI by a = 2 or a = −2 or a2 = 4 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ACF PI by a = 2 or a = −2 | 1.1b | A1 |
| Deduces that a = –2 | 2.2a | R1 |
| Subtotal | 3 | |
| Question 6 Total | 7 | |
| Q | Marking instructions | AO |
Question 6:
--- 6(a) ---
6(a) | Applies Binomial Expansion
formula to x2 term.
At least two of nC , 3n – 2 or (ax)2
2
or a2 correct. | 1.1a | M1 | x2 term = nC 3n – 2(ax)2
2
n ( n − 1 ) 3 n
× × a2 = 4860
2 9
3na2n(n – 1) = 2 × 9 × 4860 = 87480
n ( n − 1 )
Expresses nC =
2
2
Or
3n – 2 as 3 n or 3n × 3−2
9 | 1.2 | B1
Completes reasoned argument
to obtain given expression
AG | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b)(i) ---
6(b)(i) | Verifies that n = 6 | 1.1b | B1 | Constant term is 3n
36 = 729
So n = 6
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b)(ii) ---
6(b)(ii) | Substitutes n = 6 into formula
from (a)
Or
Obtains 6C 34a2 = 4860
2
PI by a = 2 or a = −2 or a2 = 4 | 1.1a | M1 | 729a2 × 6 × 5 = 87480
a2 = 4
a = ±2
Negative x coefficient so a = –2
Obtains a2 = 4
ACF PI by a = 2 or a = −2 | 1.1b | A1
Deduces that a = –2 | 2.2a | R1
Subtotal | 3
Question 6 Total | 7
Q | Marking instructions | AO | Marks | Typical solutionIn the expansion of $(3 + ax)^n$, where $a$ and $n$ are integers, the coefficient of $x^2$ is 4860
\begin{enumerate}[label=(\alph*)]
\item Show that
$$3^n a^2 n (n - 1) = 87480$$
[3 marks]
\item The constant term in the expansion is 729
The coefficient of $x$ in the expansion is negative.
\begin{enumerate}[label=(\roman*)]
\item Verify that $n = 6$
[1 mark]
\item Find the value of $a$
[3 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2024 Q6 [7]}}