Question 5 - A-Level Maths

Edexcel C4 2008 June — Question 5 9 marks

Exam Board	Edexcel
Module	C4 (Core Mathematics 4)
Year	2008
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Generalised Binomial Theorem
Type	Multiply by polynomial
Difficulty	Standard +0.3 This is a straightforward application of the binomial expansion for negative/fractional powers followed by polynomial multiplication. Part (a) requires standard use of (1+x)^n formula with algebraic manipulation, while part (b) simply multiplies the result by a linear polynomial. The question is slightly above average due to the fractional power and the two-part structure, but remains a routine C4 exercise with no novel problem-solving required.
Spec	1.04c Extend binomial expansion: rational n, \|x\|<1

5. (a) Expand \(\frac { 1 } { \sqrt { } ( 4 - 3 x ) }\), where \(| x | < \frac { 4 } { 3 }\), in ascending powers of \(x\) up to and including the term in \(x ^ { 2 }\). Simplify each term.
(b) Hence, or otherwise, find the first 3 terms in the expansion of \(\frac { x + 8 } { \sqrt { } ( 4 - 3 x ) }\) as a series in ascending powers of \(x\).

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Question 5:

Part (a)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(\frac{1}{\sqrt{4-3x}} = (4-3x)^{-\frac{1}{2}} = (4)^{-\frac{1}{2}}\left(1-\frac{3x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2}\left(1-\frac{3x}{4}\right)^{-\frac{1}{2}}\)	B1	\((4)^{-\frac{1}{2}}\) or \(\frac{1}{2}\) outside brackets
\(= \frac{1}{2}\left[1 + \left(-\frac{1}{2}\right)(x) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(x)^2 + \ldots\right]\)	M1	Expands \((1+x)^{-\frac{1}{2}}\) to give \(1+(-\frac{1}{2})(x)\)
Correct simplified or unsimplified expansion with \(**\neq 1\)	A1\(\checkmark\)
\(= \frac{1}{2}\left[1 + \frac{3}{8}x + \frac{27}{128}x^2 + \ldots\right]\)	A1 isw
\(= \frac{1}{2} + \frac{3}{16}x + \frac{27}{256}x^2 + \ldots\)	A1 isw	Ignore subsequent working

Part (b)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\((x+8)\left(\frac{1}{2} + \frac{3}{16}x + \frac{27}{256}x^2 + \ldots\right)\)	M1	Writing \((x+8)\) multiplied by candidate's part (a) expansion
Multiply out brackets to find constant term, two \(x\) terms and two \(x^2\) terms	M1
\(= 4 + 2x + \frac{33}{32}x^2 + \ldots\)	A1; A1	Anything that cancels to \(4+2x\); \(\frac{33}{32}x^2\)

# Question 5:

## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{\sqrt{4-3x}} = (4-3x)^{-\frac{1}{2}} = (4)^{-\frac{1}{2}}\left(1-\frac{3x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2}\left(1-\frac{3x}{4}\right)^{-\frac{1}{2}}$ | B1 | $(4)^{-\frac{1}{2}}$ or $\frac{1}{2}$ outside brackets |
| $= \frac{1}{2}\left[1 + \left(-\frac{1}{2}\right)(**x) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(**x)^2 + \ldots\right]$ | M1 | Expands $(1+**x)^{-\frac{1}{2}}$ to give $1+(-\frac{1}{2})(**x)$ |
| Correct simplified or unsimplified expansion with $**\neq 1$ | A1$\checkmark$ | |
| $= \frac{1}{2}\left[1 + \frac{3}{8}x + \frac{27}{128}x^2 + \ldots\right]$ | A1 isw | |
| $= \frac{1}{2} + \frac{3}{16}x + \frac{27}{256}x^2 + \ldots$ | A1 isw | Ignore subsequent working |

## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x+8)\left(\frac{1}{2} + \frac{3}{16}x + \frac{27}{256}x^2 + \ldots\right)$ | M1 | Writing $(x+8)$ multiplied by candidate's part (a) expansion |
| Multiply out brackets to find constant term, two $x$ terms and two $x^2$ terms | M1 | |
| $= 4 + 2x + \frac{33}{32}x^2 + \ldots$ | A1; A1 | Anything that cancels to $4+2x$; $\frac{33}{32}x^2$ |

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5. (a) Expand $\frac { 1 } { \sqrt { } ( 4 - 3 x ) }$, where $| x | < \frac { 4 } { 3 }$, in ascending powers of $x$ up to and including the term in $x ^ { 2 }$. Simplify each term.\\
(b) Hence, or otherwise, find the first 3 terms in the expansion of $\frac { x + 8 } { \sqrt { } ( 4 - 3 x ) }$ as a series in ascending powers of $x$.\\

\hfill \mbox{\textit{Edexcel C4 2008 Q5 [9]}}

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