| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem and Partial Fractions |
| Type | Partial fractions with validity range |
| Difficulty | Standard +0.3 This is a structured multi-part question testing standard binomial expansion techniques and partial fractions. Parts (a) and (b) are routine applications of the binomial theorem for negative indices. Part (b) involves algebraic manipulation and showing terms form a GP. Part (c) combines previous results in a scaffolded way. The validity range requires understanding convergence conditions. While it has multiple parts (13 marks total), each step follows standard A-level procedures with clear signposting, making it slightly easier than average overall. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks |
|---|---|
| 9(a)(i) | Obtains |
| Answer | Marks | Guidance |
|---|---|---|
| with at least two terms correct | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(a)(ii) | ( 2−3x)−1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 4 8 | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 4 8 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone one sign error | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2−3x 2 4 8 | 2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 2.2a | B1 |
| Subtotal | 5 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(b) | Uses a valid method to find P or |
| Answer | Marks | Guidance |
|---|---|---|
| comparison of coefficients | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains P = 8 | 1.1b | A1 |
| Obtains Q = – 4 | 1.1b | A1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(c)(i) | Multiplies their P by their |
| Answer | Marks | Guidance |
|---|---|---|
| Condone a sign error | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(c)(ii) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 2.2a | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question 9 Total | 13 | |
| Q | Marking instructions | AO |
Question 9:
--- 9(a)(i) ---
9(a)(i) | Obtains
(−1 )(−2 )( )2
3x
1+(−1 )( )+
3x
2!
OE
with at least two terms correct | 1.1a | M1 | (−1 )(−2 )( 3x)2
( 1+3x)−1≈1+(−1 )( 3x)+
2!
=1−3x+9x2
1−3x+9x2
Obtains | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 9(a)(ii) ---
9(a)(ii) | ( 2−3x)−1
Writes fraction as
1 3 9
PI by + x+ x2
2 4 8 | 1.1b | B1 | 1 =( 2−3x )−1
2−3x
−1
3x
=2 −1 1+ −
2
≈ 1 1+ (−1 ) − 3x + (−1 )(−2 ) − 3x 2
2 2 2! 2
1 1 3 9
≈ + x+ x2
2−3x 2 4 8
1 3 9
, x and x2 form a geometric
2 4 8
3
sequence with common ratio x
2
Factorises to obtain the form
2 −1( 1− Ax)−1
1 3 9
PI by + x+ x2
2 4 8 | 1.1a | M1
−1
3x
Expands 1− to obtain
2
(−1 )(−2 ) 2
3x 3x
1+(−1)± + ±
2 2! 2
OE
Condone one sign error | 1.1a | M1
Completes a correct argument
1 1 3 9
to show ≈ + x+ x2
2−3x 2 4 8 | 2.1 | R1
States that the common ratio is
3
x
2 | 2.2a | B1
Subtotal | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | Uses a valid method to find P or
Q
1 2
Substitution of x=− or x=
3 3
Or
Rearranging and substitution or
comparison of coefficients | 1.1a | M1 | 36x P Q
= +
( 1+3x)( 2−3x) ( 2−3x) ( 1+3x)
36x=P( 1+3x)+Q( 2−3x)
1
Let x=−
3
⇒−12=3Q
Q=−4
2
x=
3
⇒24=3P
P=8
Obtains P = 8 | 1.1b | A1
Obtains Q = – 4 | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 9(c)(i) ---
9(c)(i) | Multiplies their P by their
expansion in (a)(ii) and
multiplies their Q by their
expansion in (a)(i)
Condone a sign error
Or
P
Multiplies their by their
3
expansion in (a)(ii) and
Q
multiplies their by their
3
expansion in (a)(i)
Condone a sign error
Or
Writes the product of 12x or 36x
with their three-term expansion
in (a)(i) and their three-term
expansion in (a)(ii)
Condone a sign error | 3.1a | M1 | 8 4
−
( 2−3x ) ( 1+3x )
≈8 2 1 + 3 4 x+ 8 9 x2 −4 1−3x+9x2
=18x−27x2
12x
∴ ≈6x−9x2
( 1+3x )( 2−3x )
6x−9x2
Obtains | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 9(c)(ii) ---
9(c)(ii) | 1
Deduces x <
3
ACF | 2.2a | R1 | 1
x <
3
Subtotal | 1
Question 9 Total | 13
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the binomial expansion of $(1 + 3x)^{-1}$ up to and including the term in $x^2$
[2 marks]
\item Show that the first three terms in the binomial expansion of
$$\frac{1}{2 - 3x}$$
form a geometric sequence and state the common ratio.
[5 marks]
\end{enumerate}
\item It is given that
$$\frac{36x}{(1 + 3x)(2 - 3x)} = \frac{P}{(2 - 3x)} + \frac{Q}{(1 + 3x)}$$
where $P$ and $Q$ are integers.
Find the value of $P$ and the value of $Q$
[3 marks]
\item \begin{enumerate}[label=(\roman*)]
\item Using your answers to parts (a) and (b), find the binomial expansion of
$$\frac{12x}{(1 + 3x)(2 - 3x)}$$
up to and including the term in $x^2$
[2 marks]
\item Find the range of values of $x$ for which the binomial expansion of
$$\frac{12x}{(1 + 3x)(2 - 3x)}$$
is valid.
[1 mark]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2024 Q9 [13]}}