AQA Paper 1 2024 June — Question 8 5 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeCoefficient relationship between terms
DifficultyModerate -0.8 This is a straightforward binomial expansion question requiring recall of the binomial theorem formula and basic algebraic manipulation. Part (a) is routine application of (a+b)^n expansion, and part (b) involves setting up and solving a simple linear equation. The question requires no problem-solving insight and is more mechanical than the average A-level question.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n

\begin{enumerate}[label=(\alph*)] \item Find the first three terms, in ascending powers of \(x\), in the expansion of $$(2 + kx)^5$$ where \(k\) is a positive constant. [3 marks] \item Hence, given that the coefficient of \(x\) is four times the coefficient of \(x^2\), find the value of \(k\) [2 marks]
Question 8:

AnswerMarks Guidance
8(a)Obtains the correct constant term
321.1b B1
Obtains 516kx or 108(kx)2
OE
5 k 5  4  k x  2
PI by x or
AnswerMarks Guidance
2 2 ! 21.1a M1
Obtains 3 2 + 8 0 k x + 8 0 k 2 x 2 ( + . . . )
Accept list of correct terms.
No ISW
If more terms are given it must be
obvious which are their first three
AnswerMarks Guidance
terms.1.1b A1
Subtotal3
QMarking instructions AO

AnswerMarks
8(b)Forms the equation
their A k = 4  t h e ir B k 2 OE
AnswerMarks Guidance
May recover if x is initially included.3.1a M1
1
k = o 0 r
4
1
k =
4
Since k  0
1
Deduces k = only
4
A
Or their k = their
4 B
Justification of rejection k = 0 not
AnswerMarks Guidance
required.2.2a A1F
Subtotal2
Question 8 Total5
QMarking instructions AO
Question 8:
--- 8(a) ---
8(a) | Obtains the correct constant term
32 | 1.1b | B1 | ( 2+kx)5 =32+80kx+80k2x2+...
Obtains 516kx or 108(kx)2
OE
5 k 5  4  k x  2
PI by x or
2 2 ! 2 | 1.1a | M1
Obtains 3 2 + 8 0 k x + 8 0 k 2 x 2 ( + . . . )
Accept list of correct terms.
No ISW
If more terms are given it must be
obvious which are their first three
terms. | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 8(b) ---
8(b) | Forms the equation
their A k = 4  t h e ir B k 2 OE
May recover if x is initially included. | 3.1a | M1 | 8 0 k = 4  8 0 k 2
1
k = o 0 r
4
1
k =
4
Since k  0
1
Deduces k = only
4
A
Or their k = their
4 B
Justification of rejection k = 0 not
required. | 2.2a | A1F
Subtotal | 2
Question 8 Total | 5
Q | Marking instructions | AO | Marks | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Find the first three terms, in ascending powers of $x$, in the expansion of
$$(2 + kx)^5$$
where $k$ is a positive constant.
[3 marks]

\item Hence, given that the coefficient of $x$ is four times the coefficient of $x^2$, find the value of $k$
[2 marks]
</end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2024 Q8 [5]}}