| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| 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 combining routine binomial expansion with basic integration. Part (a) requires applying the binomial theorem to expand (1-1/x²)³, which is mechanical with n=3. Part (b) involves integrating term-by-term using standard power rules, then evaluating a definite integral by substitution. All steps are standard procedures with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation with negative powers. |
| 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 |
| REFEREN | REFERENCE |
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| \includegraphics[max width=\textwidth, alt={}]{f9a7a4dd-f7fd-4135-8872-2c1270d46a14-5_40_1567_2637_272} | |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(1-\frac{1}{x^2}\right)^3 = 1 - \frac{3}{x^2} + \frac{3}{x^4} - \frac{1}{x^6}\) | M1 | Expanding correctly |
| \(p = -3,\quad q = 3\) | A1 | Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\left(1-\frac{1}{x^2}\right)^3 dx = x + \frac{3}{x} - \frac{3}{3x^3} + \frac{1}{5x^5} + c\) | M1 A1 A1 A1 | M1 for attempt to integrate; each term correct |
| \(= x + \frac{3}{x} - \frac{1}{x^3} + \frac{1}{5x^5} + c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute limits \(x=1\) and \(x=\frac{1}{2}\) | M1 | |
| \(= \left(1+3-1+\frac{1}{5}\right) - \left(\frac{1}{2}+6-8+\frac{32}{5}\right) = \frac{16}{5} - \frac{45}{10} = -\frac{13}{10}\) | A1 | cao |
# Question 4:
## Part (a)
| $\left(1-\frac{1}{x^2}\right)^3 = 1 - \frac{3}{x^2} + \frac{3}{x^4} - \frac{1}{x^6}$ | M1 | Expanding correctly |
|---|---|---|
| $p = -3,\quad q = 3$ | A1 | Both correct |
## Part (b)(i)
| $\int\left(1-\frac{1}{x^2}\right)^3 dx = x + \frac{3}{x} - \frac{3}{3x^3} + \frac{1}{5x^5} + c$ | M1 A1 A1 A1 | M1 for attempt to integrate; each term correct |
|---|---|---|
| $= x + \frac{3}{x} - \frac{1}{x^3} + \frac{1}{5x^5} + c$ | | |
## Part (b)(ii)
| Substitute limits $x=1$ and $x=\frac{1}{2}$ | M1 | |
|---|---|---|
| $= \left(1+3-1+\frac{1}{5}\right) - \left(\frac{1}{2}+6-8+\frac{32}{5}\right) = \frac{16}{5} - \frac{45}{10} = -\frac{13}{10}$ | A1 | cao |
---4
\begin{enumerate}[label=(\alph*)]
\item The expression $\left( 1 - \frac { 1 } { x ^ { 2 } } \right) ^ { 3 }$ can be written in the form
$$1 + \frac { p } { x ^ { 2 } } + \frac { q } { x ^ { 4 } } - \frac { 1 } { x ^ { 6 } }$$
Find the values of the integers $p$ and $q$.
\item \begin{enumerate}[label=(\roman*)]
\item Hence find $\int \left( 1 - \frac { 1 } { x ^ { 2 } } \right) ^ { 3 } \mathrm {~d} x$.
\item Hence find the value of $\int _ { \frac { 1 } { 2 } } ^ { 1 } \left( 1 - \frac { 1 } { x ^ { 2 } } \right) ^ { 3 } \mathrm {~d} x$.
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\hfill \mbox{\textit{AQA C2 2010 Q4 [8]}}