| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Product or quotient of expansions |
| Difficulty | Standard +0.3 This question requires applying the binomial expansion to a quotient (expanding numerator and denominator separately, then multiplying), but follows a standard template with clear guidance. Part (a) is routine expansion work, part (b) tests understanding of validity conditions (|x|<1), and part (c) is straightforward substitution and arithmetic. While it involves multiple steps, each is procedural 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sqrt{\frac{1+4x}{1-x}} = (1+4x)^{0.5}\times(1-x)^{-0.5}\) | B1 | Key step setting up for binomial expansions; may be implied by later work |
| \((1+4x)^{0.5} = 1 + 0.5\times(4x) + \frac{0.5\times(-0.5)}{2}\times(4x)^2\) | M1 | Three or more terms; allow missing bracket on \((4x)^2\); sign slips condoned; method implied by \(1+2x\pm 0.5x^2\) |
| \((1-x)^{-0.5} = 1+(-0.5)(-x)+\frac{(-0.5)\times(-1.5)}{2}(-x)^2\) | M1 | Three or more terms; allow missing bracket on \((-x)^2\); method implied by \(1\pm 0.5x\pm 0.375x^2\) |
| \((1+4x)^{0.5}=1+2x-2x^2\) and \((1-x)^{-0.5}=1+0.5x+0.375x^2\) | A1 | Both correct and simplified |
| \((1+4x)^{0.5}\times(1-x)^{-0.5} = 1+\frac{5}{2}x-\frac{5}{8}x^2\ldots\) | dM1 A1* | dM1: multiply two expansions to reach quadratic (key step adding six terms); A1*: correct, no errors or omissions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Valid \( | x | <\frac{1}{4}\); should not use \(x=\frac{1}{2}\) as \(\frac{1}{2}>\frac{1}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x=\frac{1}{11}\) into \(\sqrt{\frac{1+4x}{1-x}}\approx 1+\frac{5}{2}x-\frac{5}{8}x^2\) | M1 | Substitutes into both sides; as LHS is \(\frac{\sqrt{6}}{2}\) may multiply by 2 first |
| \(\sqrt{\frac{3}{2}} = \frac{1183}{968}\) | A1 | Correct equation/statement |
| \(\sqrt{6} = \frac{1183}{484}\) or \(\frac{2904}{1183}\) | A1 | \(\sqrt{6}=2\times\frac{1183}{968}=\frac{1183}{484}\) implies all 3 marks |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{\frac{1+4x}{1-x}} = (1+4x)^{0.5}\times(1-x)^{-0.5}$ | B1 | Key step setting up for binomial expansions; may be implied by later work |
| $(1+4x)^{0.5} = 1 + 0.5\times(4x) + \frac{0.5\times(-0.5)}{2}\times(4x)^2$ | M1 | Three or more terms; allow missing bracket on $(4x)^2$; sign slips condoned; method implied by $1+2x\pm 0.5x^2$ |
| $(1-x)^{-0.5} = 1+(-0.5)(-x)+\frac{(-0.5)\times(-1.5)}{2}(-x)^2$ | M1 | Three or more terms; allow missing bracket on $(-x)^2$; method implied by $1\pm 0.5x\pm 0.375x^2$ |
| $(1+4x)^{0.5}=1+2x-2x^2$ **and** $(1-x)^{-0.5}=1+0.5x+0.375x^2$ | A1 | Both correct and simplified |
| $(1+4x)^{0.5}\times(1-x)^{-0.5} = 1+\frac{5}{2}x-\frac{5}{8}x^2\ldots$ | dM1 A1* | dM1: multiply two expansions to reach quadratic (key step adding six terms); A1*: correct, no errors or omissions |
**Total: (6)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Valid $|x|<\frac{1}{4}$; should not use $x=\frac{1}{2}$ as $\frac{1}{2}>\frac{1}{4}$ | B1 | States expansion not valid when $|x|>\frac{1}{4}$; may be implied by $\frac{1}{2}>\frac{1}{4}$ or stating valid only when $|x|<\frac{1}{4}$ |
**Total: (1)**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x=\frac{1}{11}$ into $\sqrt{\frac{1+4x}{1-x}}\approx 1+\frac{5}{2}x-\frac{5}{8}x^2$ | M1 | Substitutes into both sides; as LHS is $\frac{\sqrt{6}}{2}$ may multiply by 2 first |
| $\sqrt{\frac{3}{2}} = \frac{1183}{968}$ | A1 | Correct equation/statement |
| $\sqrt{6} = \frac{1183}{484}$ or $\frac{2904}{1183}$ | A1 | $\sqrt{6}=2\times\frac{1183}{968}=\frac{1183}{484}$ implies all 3 marks |
**Total: (3)**
**Overall: (10 marks)**\begin{enumerate}
\item (a) Use binomial expansions to show that $\sqrt { \frac { 1 + 4 x } { 1 - x } } \approx 1 + \frac { 5 } { 2 } x - \frac { 5 } { 8 } x ^ { 2 }$
\end{enumerate}
A student substitutes $x = \frac { 1 } { 2 }$ into both sides of the approximation shown in part (a) in an attempt to find an approximation to $\sqrt { 6 }$\\
(b) Give a reason why the student should not use $x = \frac { 1 } { 2 }$\\
(c) Substitute $x = \frac { 1 } { 11 }$ into
$$\sqrt { \frac { 1 + 4 x } { 1 - x } } = 1 + \frac { 5 } { 2 } x - \frac { 5 } { 8 } x ^ { 2 }$$
to obtain an approximation to $\sqrt { 6 }$. Give your answer as a fraction in its simplest form.
\hfill \mbox{\textit{Edexcel Paper 1 2018 Q11 [10]}}