| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Expansion with algebraic manipulation then integrate |
| Difficulty | Standard +0.3 This is a straightforward C2 question combining basic binomial expansion with standard integration techniques. Part (a) requires routine expansion of (2+x²)³, part (b)(i) involves dividing through by x⁴ and integrating term-by-term (including negative powers), and part (b)(ii) is simple substitution of limits. All steps are standard procedures with no problem-solving insight required, making it slightly easier than average. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2+x^2)^3 = [(2)^3] + 3(2)^2(x^2) + 3(2)(x^2)^2 [+(x^2)^3]\) | M1 | For either (1),3,3,(1) OE unsimplified or \(\binom{3}{1}2^2x^2 + \binom{3}{2}2(x^2)^2\) OE. PI |
| \(p = 3(2)^2 = 12\) | A1 | AG. Be convinced. Condone left as \(12x^2\) |
| \(q = 6\) | B1 | Accept left as \(6x^4\) |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{(2+x^2)^3}{x^4}dx = \int x^{-4}(8+12x^2+qx^4+x^6)dx\) or \(\int\left(\frac{8}{x^4}+\frac{12}{x^2}+q+x^2\right)dx\) | M1 | Uses (a) and either an indication that \(\frac{1}{x^n} = x^{-n}\) in a product PI or cancelling to get at least 3 correct ft terms |
| \(\int(8x^{-4}+12x^{-2}+q+x^2)dx\) | A1F | Ft on c's non-zero \(q\). PI by next line in solution |
| \(= \frac{8x^{-3}}{-3}+\frac{12x^{-1}}{-1}+qx+\frac{x^3}{3}\ \{+c\}\) | M1 | Correct integration of either \(8x^{-4}\) or \(12x^{-2}\); accept unsimplified |
| A1 | Correct integration of both \(8x^{-4}\) and \(12x^{-2}\); accept unsimplified coefficients | |
| \(= \ldots\ldots + 6x + \frac{x^3}{3} + c\) | B1F | For "\(6x + \frac{x^3}{3} + c\)" simplified. The only ft is "6" replaced by c's value for \(q\) where \(q\) is a non-zero integer |
| \(\left(= -\frac{8}{3}x^{-3} - 12x^{-1} + 6x + \frac{x^3}{3} + c\right)\) | ||
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_1^2 \frac{(2+x^2)^3}{x^4}dx = \left\{-\frac{8}{3}(2)^{-3}-12(2^{-1})+6(2)+\frac{2^3}{3}\right\} - \left\{-\frac{8}{3}(1)^{-3}-12(1)^{-1}+6(1)+\frac{1^3}{3}\right\}\) | M1 | Dealing correctly with limits; \(F(2)-F(1)\) (must have attempted integration to get F ie c's F is not just the integrand) |
| \(= \left(-\frac{1}{3}-6+12+\frac{8}{3}\right) - \left(-\frac{8}{3}-12+6+\frac{1}{3}\right)\) | ||
| \(= 16\frac{2}{3}\) | A1 | OE exact answer e.g. 50/3. NMS scores 0 |
| Total: 2 marks |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2+x^2)^3 = [(2)^3] + 3(2)^2(x^2) + 3(2)(x^2)^2 [+(x^2)^3]$ | M1 | For either (1),3,3,(1) OE unsimplified or $\binom{3}{1}2^2x^2 + \binom{3}{2}2(x^2)^2$ OE. PI |
| $p = 3(2)^2 = 12$ | A1 | AG. Be convinced. Condone left as $12x^2$ |
| $q = 6$ | B1 | Accept left as $6x^4$ |
| **Total: 3 marks** | | |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{(2+x^2)^3}{x^4}dx = \int x^{-4}(8+12x^2+qx^4+x^6)dx$ or $\int\left(\frac{8}{x^4}+\frac{12}{x^2}+q+x^2\right)dx$ | M1 | Uses (a) and either an indication that $\frac{1}{x^n} = x^{-n}$ in a product PI or cancelling to get at least 3 correct ft terms |
| $\int(8x^{-4}+12x^{-2}+q+x^2)dx$ | A1F | Ft on c's non-zero $q$. PI by next line in solution |
| $= \frac{8x^{-3}}{-3}+\frac{12x^{-1}}{-1}+qx+\frac{x^3}{3}\ \{+c\}$ | M1 | Correct integration of either $8x^{-4}$ or $12x^{-2}$; accept unsimplified |
| | A1 | Correct integration of both $8x^{-4}$ and $12x^{-2}$; accept unsimplified coefficients |
| $= \ldots\ldots + 6x + \frac{x^3}{3} + c$ | B1F | For "$6x + \frac{x^3}{3} + c$" simplified. The only ft is "6" replaced by c's value for $q$ where $q$ is a non-zero integer |
| $\left(= -\frac{8}{3}x^{-3} - 12x^{-1} + 6x + \frac{x^3}{3} + c\right)$ | | |
| **Total: 5 marks** | | |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_1^2 \frac{(2+x^2)^3}{x^4}dx = \left\{-\frac{8}{3}(2)^{-3}-12(2^{-1})+6(2)+\frac{2^3}{3}\right\} - \left\{-\frac{8}{3}(1)^{-3}-12(1)^{-1}+6(1)+\frac{1^3}{3}\right\}$ | M1 | Dealing correctly with limits; $F(2)-F(1)$ (must have attempted integration to get F ie c's F is not just the integrand) |
| $= \left(-\frac{1}{3}-6+12+\frac{8}{3}\right) - \left(-\frac{8}{3}-12+6+\frac{1}{3}\right)$ | | |
| $= 16\frac{2}{3}$ | A1 | OE exact answer e.g. 50/3. NMS scores 0 |
| **Total: 2 marks** | | |
---3
\begin{enumerate}[label=(\alph*)]
\item The expression $\left( 2 + x ^ { 2 } \right) ^ { 3 }$ can be written in the form
$$8 + p x ^ { 2 } + q x ^ { 4 } + x ^ { 6 }$$
Show that $p = 12$ and find the value of the integer $q$.
\item \begin{enumerate}[label=(\roman*)]
\item Hence find $\int \frac { \left( 2 + x ^ { 2 } \right) ^ { 3 } } { x ^ { 4 } } \mathrm {~d} x$.\\
(5 marks)
\item Hence find the exact value of $\int _ { 1 } ^ { 2 } \frac { \left( 2 + x ^ { 2 } \right) ^ { 3 } } { x ^ { 4 } } \mathrm {~d} x$.\\
(2 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q3 [10]}}