| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Direct binomial expansion then integrate |
| Difficulty | Moderate -0.8 This is a straightforward C2 question requiring routine binomial expansion of a small power (n=3) followed by term-by-term integration of simple powers. The expansion is mechanical with no problem-solving required, and the integration involves standard rules for x^n. This is easier than average A-level content, being a textbook-style exercise testing basic technique rather than insight. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\left(1 + \frac{4}{x^2}\right)^3\) | M1 (3 marks) | Any valid method as far as term(s) in \(1/x^2\) and term(s) in \(1/x^4\) |
| \(= [1] + 3(1)^2\left(\frac{4}{x^2}\right) + 3(1)\left(\frac{4}{x^2}\right)^2 + \left[\left(\frac{4}{x^2}\right)^3\right]\) | A1 | \(p = 12\) Accept \(\frac{12}{x^2}\) even within a series |
| \(= [1] + \frac{12}{x^2} + \frac{48}{x^4} + \left[\frac{64}{x^6}\right]\) | A1 (3 marks) | \(q = 48\) Accept \(\frac{48}{x^4}\) even within a series |
| (b)(i) \(\int\left(1 + \frac{4}{x^2}\right)^3 dx = \int\left(1 + \frac{p}{x^2} + \frac{q}{x^4} + \frac{64}{x^6}\right)dx\) | M1 (4 marks) | Integral of an 'expansion', at least 3 terms; PI by the next line |
| \(= x - px^{-1} - \frac{q}{3}x^{-3} - \frac{64}{5}x^{-5} (+c)\) | m1, A2F,1 | At least two powers correctly obtained; Ft on \(c's\) non-zero integer values for \(p\) and \(q\) (A1F for two terms correct; can be unsimplified); Condone missing \(c\) but check that signs have been simplified at some stage before the award of both A marks. |
| \(= x - 12x^{-1} - 16x^{-3} - \frac{64}{5}x^{-5} (+c)\) | ||
| (ii) \(\left(2 - \frac{p}{2} - \frac{q}{3(8)} - \frac{64}{5(32)}\right) - \left(1 - p - \frac{q}{3} - \frac{64}{5}\right) = 33.4\) | M1, A1 (2 marks) | F(2) – F(1), where F(x) is cand's answer or the correct answer to (b)(i).; CSO |
**(a)** $\left(1 + \frac{4}{x^2}\right)^3$ | M1 (3 marks) | Any valid method as far as term(s) in $1/x^2$ and term(s) in $1/x^4$
$= [1] + 3(1)^2\left(\frac{4}{x^2}\right) + 3(1)\left(\frac{4}{x^2}\right)^2 + \left[\left(\frac{4}{x^2}\right)^3\right]$ | A1 | $p = 12$ Accept $\frac{12}{x^2}$ even within a series
$= [1] + \frac{12}{x^2} + \frac{48}{x^4} + \left[\frac{64}{x^6}\right]$ | A1 (3 marks) | $q = 48$ Accept $\frac{48}{x^4}$ even within a series
**(b)(i)** $\int\left(1 + \frac{4}{x^2}\right)^3 dx = \int\left(1 + \frac{p}{x^2} + \frac{q}{x^4} + \frac{64}{x^6}\right)dx$ | M1 (4 marks) | Integral of an 'expansion', at least 3 terms; PI by the next line
$= x - px^{-1} - \frac{q}{3}x^{-3} - \frac{64}{5}x^{-5} (+c)$ | m1, A2F,1 | At least two powers correctly obtained; Ft on $c's$ non-zero integer values for $p$ and $q$ (A1F for two terms correct; can be unsimplified); Condone missing $c$ but check that signs have been simplified at some stage before the award of both A marks.
$= x - 12x^{-1} - 16x^{-3} - \frac{64}{5}x^{-5} (+c)$ |
**(ii)** $\left(2 - \frac{p}{2} - \frac{q}{3(8)} - \frac{64}{5(32)}\right) - \left(1 - p - \frac{q}{3} - \frac{64}{5}\right) = 33.4$ | M1, A1 (2 marks) | F(2) – F(1), where F(x) is cand's answer or the correct answer to (b)(i).; CSO
**Total for Q7: 9 marks**
---7
\begin{enumerate}[label=(\alph*)]
\item The expression $\left( 1 + \frac { 4 } { x ^ { 2 } } \right) ^ { 3 }$ can be written in the form
$$1 + \frac { p } { x ^ { 2 } } + \frac { q } { x ^ { 4 } } + \frac { 64 } { x ^ { 6 } }$$
By using the binomial expansion, or otherwise, find the values of the integers $p$ and $q$.
\item \begin{enumerate}[label=(\roman*)]
\item Hence find $\int \left( 1 + \frac { 4 } { x ^ { 2 } } \right) ^ { 3 } \mathrm {~d} x$.
\item Hence find the value of $\int _ { 1 } ^ { 2 } \left( 1 + \frac { 4 } { x ^ { 2 } } \right) ^ { 3 } \mathrm {~d} x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2008 Q7 [9]}}