Question 1 - A-Level Maths

Edexcel C4 2005 June — Question 1 5 marks

Exam Board	Edexcel
Module	C4 (Core Mathematics 4)
Year	2005
Session	June
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Generalised Binomial Theorem
Type	Factoring out constants first
Difficulty	Moderate -0.3 This is a straightforward application of the binomial expansion for fractional powers. Students must factor out the constant (√4 = 2) and expand (1 - 9x/4)^(1/2), then apply the standard binomial formula. While it requires careful algebraic manipulation and simplification, it's a routine C4 question with no conceptual surprises—slightly easier than average due to its mechanical nature.
Spec	1.04c Extend binomial expansion: rational n, \|x\|<1 1.04d Binomial expansion validity: convergence conditions

Use the binomial theorem to expand

$$\sqrt { } ( 4 - 9 x ) , \quad | x | < \frac { 4 } { 9 }$$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, simplifying each term.

Show mark scheme Show mark scheme source

Answer	Marks
$(4-9x)^{\frac{1}{2}} = 2\left(1-\frac{9x}{4}\right)^{\frac{1}{2}}$	B1
$= 2\left(1+\frac{1}{1}\left(-\frac{9x}{4}\right) + \frac{\frac{1}{2}(-\frac{1}{2})}{1.2}\left(-\frac{9x}{4}\right)^2 + \frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{1.2.3}\left(-\frac{9x}{4}\right)^3 + \ldots\right)$	M1
$= 2\left(1-\frac{9}{8}x - \frac{81}{128}x^2 - \frac{729}{1024}x^3 + \ldots\right)$	A1, A1, A1
$= 2 - \frac{9}{4}x - \frac{81}{64}x^2 - \frac{729}{512}x^3 + \ldots$	A1, A1, A1
Note: The M1 is gained for $\frac{\frac{1}{2}(-\frac{1}{2})}{1.2}(\ldots)^2$ or $\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{1.2.3}(\ldots)^3$	M1
Special Case: If the candidate reaches $= 2\left(1-\frac{9}{8}x - \frac{81}{128}x^2 - \frac{729}{1024}x^3 + \ldots\right)$ and goes no further, allow A1 A0 A0

Total: [5]

$(4-9x)^{\frac{1}{2}} = 2\left(1-\frac{9x}{4}\right)^{\frac{1}{2}}$ | B1 |

$= 2\left(1+\frac{1}{1}\left(-\frac{9x}{4}\right) + \frac{\frac{1}{2}(-\frac{1}{2})}{1.2}\left(-\frac{9x}{4}\right)^2 + \frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{1.2.3}\left(-\frac{9x}{4}\right)^3 + \ldots\right)$ | M1 |

$= 2\left(1-\frac{9}{8}x - \frac{81}{128}x^2 - \frac{729}{1024}x^3 + \ldots\right)$ | A1, A1, A1 |

$= 2 - \frac{9}{4}x - \frac{81}{64}x^2 - \frac{729}{512}x^3 + \ldots$ | A1, A1, A1 |

Note: The M1 is gained for $\frac{\frac{1}{2}(-\frac{1}{2})}{1.2}(\ldots)^2$ or $\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{1.2.3}(\ldots)^3$ | M1 |

**Special Case:** If the candidate reaches $= 2\left(1-\frac{9}{8}x - \frac{81}{128}x^2 - \frac{729}{1024}x^3 + \ldots\right)$ and goes no further, allow A1 A0 A0 | |

**Total: [5]**

---

Show LaTeX source

\begin{enumerate}
  \item Use the binomial theorem to expand
\end{enumerate}

$$\sqrt { } ( 4 - 9 x ) , \quad | x | < \frac { 4 } { 9 }$$

in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, simplifying each term.\\

\hfill \mbox{\textit{Edexcel C4 2005 Q1 [5]}}

This paper (8 questions)

View full paper

Q1 5 Q2 7 Q3 8 Q4 7 Q5 10 Q6 12 Q7 13 Q8 13

Answer	Marks
\((4-9x)^{\frac{1}{2}} = 2\left(1-\frac{9x}{4}\right)^{\frac{1}{2}}\)	B1
\(= 2\left(1+\frac{1}{1}\left(-\frac{9x}{4}\right) + \frac{\frac{1}{2}(-\frac{1}{2})}{1.2}\left(-\frac{9x}{4}\right)^2 + \frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{1.2.3}\left(-\frac{9x}{4}\right)^3 + \ldots\right)\)	M1
\(= 2\left(1-\frac{9}{8}x - \frac{81}{128}x^2 - \frac{729}{1024}x^3 + \ldots\right)\)	A1, A1, A1
\(= 2 - \frac{9}{4}x - \frac{81}{64}x^2 - \frac{729}{512}x^3 + \ldots\)	A1, A1, A1
Note: The M1 is gained for \(\frac{\frac{1}{2}(-\frac{1}{2})}{1.2}(\ldots)^2\) or \(\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{1.2.3}(\ldots)^3\)	M1
Special Case: If the candidate reaches \(= 2\left(1-\frac{9}{8}x - \frac{81}{128}x^2 - \frac{729}{1024}x^3 + \ldots\right)\) and goes no further, allow A1 A0 A0