| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Substitute expression for variable |
| Difficulty | Standard +0.3 This is a standard C4 binomial expansion question with straightforward substitution. Part (a)(i) requires routine application of the formula with n=-1/3, part (a)(ii) involves simple substitution of 3x/4 for x, and part (b) requires recognizing that the cube root expression can be rewritten using the previous result. All steps are mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| \((1+x)^{\frac{1}{3}}=1\pm\frac{1}{3}x+kx^2\) | M1 | \(1\pm\frac{1}{3}x+kx^2\) |
| \(=1-\frac{1}{3}x+\frac{2}{9}x^2\) | A1 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(1+\frac{3}{4}x\right)^{-\frac{1}{3}}=1-\frac{1}{3}\times\frac{3}{4}x+\frac{2}{9}\left(\frac{3}{4}x\right)^2\) | M1 | \(x\) replaced by \(\frac{3}{4}x\); or start binomial again; condone missing brackets |
| \(=1-\frac{1}{4}x+\frac{1}{8}x^2\) | A1 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt[3]{\frac{256}{4+3x}}=k\left(1+\frac{3}{4}x\right)^{-\frac{1}{3}}\) | M1 | \(k\neq1\) |
| \(=4\left(1-\frac{1}{4}x+\frac{1}{8}x^2\right)\) | A1F | F on (a)(ii); \(k=4\), accept \(\sqrt[3]{64}\) or \(64^{\frac{1}{3}}\) |
| \(=4-x+\frac{1}{2}x^2\quad\) or \(\quad a=4\quad b=-1\quad c=\frac{1}{2}\) | A1 (3 marks) | CSO fully simplified; Be convinced |
## Question 3:
### Part (a)(i)
| $(1+x)^{\frac{1}{3}}=1\pm\frac{1}{3}x+kx^2$ | M1 | $1\pm\frac{1}{3}x+kx^2$ |
| $=1-\frac{1}{3}x+\frac{2}{9}x^2$ | A1 (2 marks) | |
### Part (a)(ii)
| $\left(1+\frac{3}{4}x\right)^{-\frac{1}{3}}=1-\frac{1}{3}\times\frac{3}{4}x+\frac{2}{9}\left(\frac{3}{4}x\right)^2$ | M1 | $x$ replaced by $\frac{3}{4}x$; or start binomial again; condone missing brackets |
| $=1-\frac{1}{4}x+\frac{1}{8}x^2$ | A1 (2 marks) | |
### Part (b)
| $\sqrt[3]{\frac{256}{4+3x}}=k\left(1+\frac{3}{4}x\right)^{-\frac{1}{3}}$ | M1 | $k\neq1$ |
| $=4\left(1-\frac{1}{4}x+\frac{1}{8}x^2\right)$ | A1F | F on (a)(ii); $k=4$, accept $\sqrt[3]{64}$ or $64^{\frac{1}{3}}$ |
| $=4-x+\frac{1}{2}x^2\quad$ or $\quad a=4\quad b=-1\quad c=\frac{1}{2}$ | A1 (3 marks) | CSO fully simplified; Be convinced |
**Total: 7 marks**
---3
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the binomial expansion of $( 1 + x ) ^ { - \frac { 1 } { 3 } }$ up to and including the term in $x ^ { 2 }$.\\
(2 marks)
\item Hence find the binomial expansion of $\left( 1 + \frac { 3 } { 4 } x \right) ^ { - \frac { 1 } { 3 } }$ up to and including the term in $x ^ { 2 }$.
\end{enumerate}\item Hence show that $\sqrt [ 3 ] { \frac { 256 } { 4 + 3 x } } \approx a + b x + c x ^ { 2 }$ for small values of $x$, stating the values of the constants $a , b$ and $c$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2010 Q3 [7]}}