| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Direct quotient expansion |
| Difficulty | Moderate -0.3 This is a straightforward application of the binomial expansion formula to a quotient, requiring students to expand (1+x)^{1/2}(1-x)^{-1/2} and multiply series, then substitute a given value. While it involves the generalized binomial theorem, the question is highly structured with clear instructions and no novel problem-solving 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}}\) | B1 | Seen or implied. Also allow \(((1+x)(1-x)^{-1})^{\frac{1}{2}}\) |
| \(\left(1+\frac{1}{2}x+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}x^2+...\right)\times\left(1+\left(-\frac{1}{2}\right)(-x)+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-x)^2+...\right)\) | M1 A1 A1 | M1: Expands \((1+x)^{\frac{1}{2}}\) to give any 2 out of 3 terms, or expands \((1-x)^{-\frac{1}{2}}\) to give any 2 out of 3 terms. A1: At least one binomial expansion correct. A1: Two binomial expansions correct (ignore \(x^3\) and \(x^4\) terms) |
| \(=\left(1+\frac{1}{2}x-\frac{1}{8}x^2+...\right)\times\left(1+\frac{1}{2}x+\frac{3}{8}x^2+...\right)\) | ||
| \(=1+\frac{1}{2}x+\frac{3}{8}x^2+\frac{1}{2}x+\frac{1}{4}x^2-\frac{1}{8}x^2+...\) | M1 | Multiplies out to give 1, exactly two terms in \(x\) and exactly three terms in \(x^2\) |
| \(=1+x+\frac{1}{2}x^2\) | A1* | Answer is given in the question |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sqrt{\dfrac{1+\frac{1}{26}}{1-\frac{1}{26}}}=1+\left(\dfrac{1}{26}\right)+\dfrac{1}{2}\left(\dfrac{1}{26}\right)^2\) | M1 | Substitutes \(x=\frac{1}{26}\) into both sides |
| \(\dfrac{3\sqrt{3}}{5}=\dfrac{1405}{1352}\) | B1 | For sight of \(\sqrt{\frac{27}{25}}\) (or better) and \(\frac{1405}{1352}\) or equivalent fraction |
| \(\sqrt{3}=\dfrac{7025}{4056}\) | A1 cao | \(\frac{7025}{4056}\) or any equivalent fraction |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}}$ | B1 | Seen or implied. Also allow $((1+x)(1-x)^{-1})^{\frac{1}{2}}$ |
| $\left(1+\frac{1}{2}x+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}x^2+...\right)\times\left(1+\left(-\frac{1}{2}\right)(-x)+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-x)^2+...\right)$ | M1 A1 A1 | M1: Expands $(1+x)^{\frac{1}{2}}$ to give any 2 out of 3 terms, or expands $(1-x)^{-\frac{1}{2}}$ to give any 2 out of 3 terms. A1: At least one binomial expansion correct. A1: Two binomial expansions correct (ignore $x^3$ and $x^4$ terms) |
| $=\left(1+\frac{1}{2}x-\frac{1}{8}x^2+...\right)\times\left(1+\frac{1}{2}x+\frac{3}{8}x^2+...\right)$ | | |
| $=1+\frac{1}{2}x+\frac{3}{8}x^2+\frac{1}{2}x+\frac{1}{4}x^2-\frac{1}{8}x^2+...$ | M1 | Multiplies out to give 1, exactly two terms in $x$ and exactly three terms in $x^2$ |
| $=1+x+\frac{1}{2}x^2$ | A1* | Answer is given in the question |
**Special Case:** Award SC FINAL M1A1 for a correct $\left(1+\frac{1}{2}x-\frac{1}{8}x^2+...\right)\times\left(1+\frac{1}{2}x+\frac{3}{8}x^2+...\right)$ multiplied out with no errors leading to correct answer.
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{\dfrac{1+\frac{1}{26}}{1-\frac{1}{26}}}=1+\left(\dfrac{1}{26}\right)+\dfrac{1}{2}\left(\dfrac{1}{26}\right)^2$ | M1 | Substitutes $x=\frac{1}{26}$ into **both** sides |
| $\dfrac{3\sqrt{3}}{5}=\dfrac{1405}{1352}$ | B1 | For sight of $\sqrt{\frac{27}{25}}$ (or better) and $\frac{1405}{1352}$ or equivalent fraction |
| $\sqrt{3}=\dfrac{7025}{4056}$ | A1 cao | $\frac{7025}{4056}$ or any equivalent fraction |
---\begin{enumerate}[label=(\alph*)]
\item Use the binomial expansion to show that
$$\left. \sqrt { ( } \frac { 1 + x } { 1 - x } \right) \approx 1 + x + \frac { 1 } { 2 } x ^ { 2 } , \quad | x | < 1$$
\item Substitute $x = \frac { 1 } { 26 }$ into
$$\sqrt { \left( \frac { 1 + x } { 1 - x } \right) = 1 + x + \frac { 1 } { 2 } x ^ { 2 } }$$
to obtain an approximation to $\sqrt { } 3$\\
Give your answer in the form $\frac { a } { b }$ where $a$ and $b$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2013 Q2 [9]}}