| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Combined expansions then integrate |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question requiring routine application of binomial expansion and basic integration. Part (a) involves simple substitution or coefficient matching, part (b) requires recognizing that even powers cancel when subtracting expansions (a standard observation), and part (c) is direct integration of a polynomial. All steps are mechanical with no problem-solving insight required, making this easier than average. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = 16\) | B1 | Obtains 16; not incorrectly labelled |
| \(B = 600\) | B1 | Obtains 600; not incorrectly labelled |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2+5x)^4 - (2-5x)^4 = A - 160x + Bx^2 - 1000x^3 + 625x^4\) | M1 | Obtains expansion of \((2-5x)^4\); accept \(A\) and \(B\) unsubstituted; or uses valid method and obtains one of \(C = 320\) or \(D = 2000\) |
| \(= 320x + 2000x^3\) | R1F | Completes reasoned argument; accept \(A\) and \(B\) unsubstituted; must finish with \(320x + 2000x^3\); don't accept just \(C=320\) and \(D=2000\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int\!\left((2+5x)^4-(2-5x)^4\right)dx = \int\!(320x + 2000x^3)\,dx\) | M1 | Integrates one term correctly; accept \(C\) and \(D\) unsubstituted; or uses reverse of chain rule to obtain at least one term of form \(P(2\pm5x)^5\), \(P = \pm\frac{1}{5}\) or \(\pm\frac{1}{25}\) |
| \(= 160x^2 + 500x^4 + c\) | A1F | Obtains \(\frac{320}{2}x^2 + \frac{2000}{4}x^4 + c\); FT \(C\) and \(D\) unsubstituted; or \(\frac{(2+5x)^5}{5\times5}+\frac{(2-5x)^5}{5\times5}+c\); condone missing \(+c\) |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = 16$ | B1 | Obtains 16; not incorrectly labelled |
| $B = 600$ | B1 | Obtains 600; not incorrectly labelled |
---
## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2+5x)^4 - (2-5x)^4 = A - 160x + Bx^2 - 1000x^3 + 625x^4$ | M1 | Obtains expansion of $(2-5x)^4$; accept $A$ and $B$ unsubstituted; or uses valid method and obtains one of $C = 320$ or $D = 2000$ |
| $= 320x + 2000x^3$ | R1F | Completes reasoned argument; accept $A$ and $B$ unsubstituted; must finish with $320x + 2000x^3$; don't accept just $C=320$ and $D=2000$ |
---
## Question 5(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\!\left((2+5x)^4-(2-5x)^4\right)dx = \int\!(320x + 2000x^3)\,dx$ | M1 | Integrates one term correctly; accept $C$ and $D$ unsubstituted; or uses reverse of chain rule to obtain at least one term of form $P(2\pm5x)^5$, $P = \pm\frac{1}{5}$ or $\pm\frac{1}{25}$ |
| $= 160x^2 + 500x^4 + c$ | A1F | Obtains $\frac{320}{2}x^2 + \frac{2000}{4}x^4 + c$; FT $C$ and $D$ unsubstituted; or $\frac{(2+5x)^5}{5\times5}+\frac{(2-5x)^5}{5\times5}+c$; condone missing $+c$ |
---5 The binomial expansion of $( 2 + 5 x ) ^ { 4 }$ is given by
$$( 2 + 5 x ) ^ { 4 } = A + 160 x + B x ^ { 2 } + 1000 x ^ { 3 } + 625 x ^ { 4 }$$
5
\begin{enumerate}[label=(\alph*)]
\item Find the value of $A$ and the value of $B$.\\[0pt]
[2 marks]\\
L\\
5
\item Show that
$$( 2 + 5 x ) ^ { 4 } - ( 2 - 5 x ) ^ { 4 } = C x + D x ^ { 3 }$$
where $C$ and $D$ are constants to be found.\\
5
\item Hence, or otherwise, find
$$\int \left( ( 2 + 5 x ) ^ { 4 } - ( 2 - 5 x ) ^ { 4 } \right) \mathrm { d } x$$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2022 Q5 [6]}}