Question 1 - A-Level Maths

Edexcel C4 2009 June — Question 1 6 marks

Exam Board	Edexcel
Module	C4 (Core Mathematics 4)
Year	2009
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Generalised Binomial Theorem
Type	Factoring out constants first
Difficulty	Standard +0.3 This is a straightforward application of the binomial expansion requiring factoring out the constant 4 first to get (1+x/4)^(-1/2), then applying the standard formula. It's slightly above average difficulty due to the fractional negative power and need to simplify coefficients, but follows a well-practiced procedure with no novel insight required.
Spec	1.04c Extend binomial expansion: rational n, \|x\|<1

1. $$f ( x ) = \frac { 1 } { \sqrt { ( 4 + x ) } } , \quad | x | < 4$$ Find the binomial expansion of $\mathrm { f } ( x )$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$. Give each coefficient as a simplified fraction.
(6)

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Question 1:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$f(x) = \frac{1}{\sqrt{4+x}} = (4+x)^{-\frac{1}{2}}$	M1
$= (4)^{-\frac{1}{2}}(1+\ldots)^{...}$	B1	$\frac{1}{2}(1+\ldots)^{...}$ or $\frac{1}{2\sqrt{1+\ldots}}$
$= \ldots\left(1+\left(-\frac{1}{2}\right)\left(\frac{x}{4}\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(\frac{x}{4}\right)^2+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}\left(\frac{x}{4}\right)^3+\ldots\right)$	M1 A1ft	ft their $\left(\frac{x}{4}\right)$
$= \frac{1}{2}-\frac{1}{16}x+\frac{3}{256}x^2-\frac{5}{2048}x^3+\ldots$	A1, A1	(6)

# Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = \frac{1}{\sqrt{4+x}} = (4+x)^{-\frac{1}{2}}$ | M1 | |
| $= (4)^{-\frac{1}{2}}(1+\ldots)^{...}$ | B1 | $\frac{1}{2}(1+\ldots)^{...}$ or $\frac{1}{2\sqrt{1+\ldots}}$ |
| $= \ldots\left(1+\left(-\frac{1}{2}\right)\left(\frac{x}{4}\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(\frac{x}{4}\right)^2+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}\left(\frac{x}{4}\right)^3+\ldots\right)$ | M1 A1ft | ft their $\left(\frac{x}{4}\right)$ |
| $= \frac{1}{2}-\frac{1}{16}x+\frac{3}{256}x^2-\frac{5}{2048}x^3+\ldots$ | A1, A1 | (6) |

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1.

$$f ( x ) = \frac { 1 } { \sqrt { ( 4 + x ) } } , \quad | x | < 4$$

Find the binomial expansion of $\mathrm { f } ( x )$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$. Give each coefficient as a simplified fraction.\\
(6)\\

\hfill \mbox{\textit{Edexcel C4 2009 Q1 [6]}}

This paper (8 questions)

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Q1 6 Q2 8 Q3 10 Q4 9 Q5 10 Q6 8 Q7 14 Q8 10

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(f(x) = \frac{1}{\sqrt{4+x}} = (4+x)^{-\frac{1}{2}}\)	M1
\(= (4)^{-\frac{1}{2}}(1+\ldots)^{...}\)	B1	\(\frac{1}{2}(1+\ldots)^{...}\) or \(\frac{1}{2\sqrt{1+\ldots}}\)
\(= \ldots\left(1+\left(-\frac{1}{2}\right)\left(\frac{x}{4}\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(\frac{x}{4}\right)^2+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}\left(\frac{x}{4}\right)^3+\ldots\right)\)	M1 A1ft	ft their \(\left(\frac{x}{4}\right)\)
\(= \frac{1}{2}-\frac{1}{16}x+\frac{3}{256}x^2-\frac{5}{2048}x^3+\ldots\)	A1, A1	(6)