| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Multiply by polynomial |
| Difficulty | Standard +0.3 This is a standard two-part binomial expansion question requiring routine application of the generalized binomial theorem with n=-1/2, followed by polynomial multiplication. Part (a) involves factoring out constants and applying the formula mechanically, while part (b) simply multiplies the result by (3+x). The techniques are well-practiced in C4, requiring no novel insight—slightly easier than average due to its predictable structure. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{1}{\sqrt{9-10x}}\right) = (9-10x)^{-\frac{1}{2}}\) | B1 | Writes down \((9-10x)^{-\frac{1}{2}}\) or uses power of \(-\frac{1}{2}\) |
| \(= (9)^{-\frac{1}{2}}\left(1-\frac{10x}{9}\right)^{-\frac{1}{2}} = \frac{1}{3}\left(1-\frac{10x}{9}\right)^{-\frac{1}{2}}\) | B1 | \((9)^{-\frac{1}{2}}\) or \(\frac{1}{3}\) outside brackets |
| \(= \left\{\frac{1}{3}\right\}\left[1+\left(-\frac{1}{2}\right)(kx)+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(kx)^2+...\right]\) | M1 | At least two correct terms. Expands \((...+kx)^{-\frac{1}{2}}\) to give any 2 terms out of 3 simplified or unsimplified, where \(k \neq 1\) |
| \(= \frac{1}{3}+\frac{5}{27}x+\frac{25}{162}x^2+...\) | A1; A1 | A1: \(\frac{1}{3}+\frac{5}{27}x\) (simplified fractions); A1: Accept only \(\frac{25}{162}x^2\) |
| Total | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3+x}{\sqrt{9-10x}} = (3+x)(9-10x)^{-\frac{1}{2}}\) | M1 | Writes down \((3+x)\)(their part (a) answer, at least 2 of the 3 terms). Can be implied by later work |
| \(= (3+x)\left(\frac{1}{3}+\frac{5}{27}x+\left\{\frac{25}{162}x^2\right\}+...\right)\) | M1 | Multiplies out to give exactly one constant term, exactly 2 terms in \(x\) and exactly 2 terms in \(x^2\). Ignore terms in \(x^3\). Can be implied. |
| \(= 1+\frac{8}{9}x+\frac{35}{54}x^2+...\) | A1 | |
| Total | [3] | |
| Question Total | [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = (9-10x)^{-\frac{1}{2}}\) | B1 | |
| \(f''(x) = 75(9-10x)^{-\frac{5}{2}}\) | B1 oe | Correct \(f''(x)\) |
| \(f'(x) = (-\frac{1}{2})(9-10x)^{-\frac{3}{2}}(-10)\) | M1 | \(\pm a(9-10x)^{-\frac{3}{2}}\); \(a \neq \pm 1\) |
| \(f(0) = \frac{1}{3}\), \(f'(0) = \frac{5}{27}\), \(f''(0) = \frac{75}{243} = \frac{25}{81}\) | ||
| \(f(x) = \frac{1}{3} + \frac{5}{27}x + \frac{25}{162}x^2 + \ldots\) | A1; A1 |
## Question 1:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{1}{\sqrt{9-10x}}\right) = (9-10x)^{-\frac{1}{2}}$ | B1 | Writes down $(9-10x)^{-\frac{1}{2}}$ or uses power of $-\frac{1}{2}$ |
| $= (9)^{-\frac{1}{2}}\left(1-\frac{10x}{9}\right)^{-\frac{1}{2}} = \frac{1}{3}\left(1-\frac{10x}{9}\right)^{-\frac{1}{2}}$ | B1 | $(9)^{-\frac{1}{2}}$ or $\frac{1}{3}$ outside brackets |
| $= \left\{\frac{1}{3}\right\}\left[1+\left(-\frac{1}{2}\right)(kx)+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(kx)^2+...\right]$ | M1 | At least two correct terms. Expands $(...+kx)^{-\frac{1}{2}}$ to give any 2 terms out of 3 simplified or unsimplified, where $k \neq 1$ |
| $= \frac{1}{3}+\frac{5}{27}x+\frac{25}{162}x^2+...$ | A1; A1 | A1: $\frac{1}{3}+\frac{5}{27}x$ **(simplified fractions)**; A1: Accept only $\frac{25}{162}x^2$ |
| **Total** | **[5]** | |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3+x}{\sqrt{9-10x}} = (3+x)(9-10x)^{-\frac{1}{2}}$ | M1 | Writes down $(3+x)$(their part (a) answer, at least 2 of the 3 terms). Can be implied by later work |
| $= (3+x)\left(\frac{1}{3}+\frac{5}{27}x+\left\{\frac{25}{162}x^2\right\}+...\right)$ | M1 | Multiplies out to give exactly one constant term, exactly 2 terms in $x$ and exactly 2 terms in $x^2$. Ignore terms in $x^3$. Can be implied. |
| $= 1+\frac{8}{9}x+\frac{35}{54}x^2+...$ | A1 | |
| **Total** | **[3]** | |
| **Question Total** | **[8]** | |
---
**Special Cases (Part a):**
- If candidate would otherwise score A0A0, allow Special Case 1st A1 for either:
- SC: $\frac{1}{3}\left[1+\frac{5}{9}x;...\right]$ or $SC: \lambda\left[1+\frac{5}{9}x+\frac{25}{54}x^2+...\right]$ or $SC:\left[\lambda+\frac{5\lambda}{9}x+\frac{25\lambda}{54}x^2+...\right]$
- Special case M1: Award for correct simplified or unsimplified $1+n(kx)+\frac{n(n-1)}{2!}(kx)^2$ with $n \neq -\frac{1}{2}$, $n \neq$ **positive integer** and consistent $(kx)$, where $k \neq 1$
**Note (Part a ctd):** You cannot recover correct work for part (a) in part (b).
## Question 1:
### Part (a) – Maclaurin Expansion Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (9-10x)^{-\frac{1}{2}}$ | B1 | |
| $f''(x) = 75(9-10x)^{-\frac{5}{2}}$ | B1 oe | Correct $f''(x)$ |
| $f'(x) = (-\frac{1}{2})(9-10x)^{-\frac{3}{2}}(-10)$ | M1 | $\pm a(9-10x)^{-\frac{3}{2}}$; $a \neq \pm 1$ |
| $f(0) = \frac{1}{3}$, $f'(0) = \frac{5}{27}$, $f''(0) = \frac{75}{243} = \frac{25}{81}$ | | |
| $f(x) = \frac{1}{3} + \frac{5}{27}x + \frac{25}{162}x^2 + \ldots$ | A1; A1 | |
---\begin{enumerate}
\item (a) Find the binomial expansion of
\end{enumerate}
$$\frac { 1 } { \sqrt { } ( 9 - 10 x ) } , \quad | x | < \frac { 9 } { 10 }$$
in ascending powers of $x$ up to and including the term in $x ^ { 2 }$.\\
Give each coefficient as a simplified fraction.\\
(b) Hence, or otherwise, find the expansion of
$$\frac { 3 + x } { \sqrt { } ( 9 - 10 x ) } , \quad | x | < \frac { 9 } { 10 }$$
in ascending powers of $x$, up to and including the term in $x ^ { 2 }$.\\
Give each coefficient as a simplified fraction.\\
\hfill \mbox{\textit{Edexcel C4 2014 Q1 [8]}}