Question 5 - A-Level Maths

Edexcel C12 2019 January — Question 5 7 marks

Exam Board	Edexcel
Module	C12 (Core Mathematics 1 & 2)
Year	2019
Session	January
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Binomial Theorem (positive integer n)
Type	Numerical approximation using expansion
Difficulty	Moderate -0.8 This is a straightforward application of the binomial theorem requiring routine expansion followed by substitution. Part (a) involves direct use of the binomial formula with simple arithmetic, and part (b) requires recognizing that 0.9 = 1 - 0.1 = 1 - 1/10 × 2, making x = 0.2. Both parts are mechanical with no problem-solving insight needed, making this easier than average but not trivial due to the two-part structure and fraction manipulation.
Spec	1.04a Binomial expansion: (a+b)^n for positive integer n

(a) Use the binomial theorem to find the first 4 terms, in ascending powers of $x$, of the expansion of

$$\left( 1 - \frac { x } { 2 } \right) ^ { 8 }$$ Give each term in its simplest form.
(b) Use the answer to part (a) to find an approximate value to $0.9 ^ { 8 }$ Write your answer in the form $\frac { a } { b }$ where $a$ and $b$ are integers.

Show mark scheme Show mark scheme source

Question 5:

Part (a):

Answer	Marks	Guidance
$\left(1 - \frac{x}{2}\right)^8 = 1 - 4x + \ldots$	B1	May be awarded for unsimplified expression e.g. $1^8 + 8\left(-\frac{x}{2}\right) + \ldots$
$\left(1-\frac{x}{2}\right)^8 = 1 + 8\left(-\frac{x}{2}\right) + \frac{8\times7}{2!}\left(-\frac{x}{2}\right)^2 + \frac{8\times7\times6}{3!}\left(-\frac{x}{2}\right)^3$	M1 A1	M1: correct binomial coefficient paired with correct power of $x$ in third OR fourth term. Condone missing brackets. Accept coefficients in form $\ldots + {^8C_2}\left(\pm\frac{x}{\ldots}\right)^2 + {^8C_3}\left(\pm\frac{x}{\ldots}\right)^3$. A1: correct unsimplified third OR fourth term; coefficients cannot be left as ${^8C_2}$.
$= 1 - 4x + 7x^2 - 7x^3$	A1 (4 marks)	Fully simplified. Accept as list $1, -4x, 7x^2, -7x^3$.

Part (b):

Answer	Marks	Guidance
States or uses $x = \frac{1}{5}$ or $0.2$	B1
$(0.9)^8 \approx 1 - 4\times\frac{1}{5} + 7\times\frac{1}{25} - 7\times\frac{1}{125}$	M1	Substitutes $x = \frac{1}{5}$ or $0.2$ into their four-term expansion. Condone substitution of $x = -0.2$ or $0.1$ for this mark.
$(0.9)^8 \approx \frac{53}{125}$	A1 (3 marks)	oe. Accept $\frac{424}{1000}$ or $\frac{265}{625}$. Note: if part (a) fully correct, correct answer awards all three marks.

## Question 5:

### Part (a):
$\left(1 - \frac{x}{2}\right)^8 = 1 - 4x + \ldots$ | B1 | May be awarded for unsimplified expression e.g. $1^8 + 8\left(-\frac{x}{2}\right) + \ldots$

$\left(1-\frac{x}{2}\right)^8 = 1 + 8\left(-\frac{x}{2}\right) + \frac{8\times7}{2!}\left(-\frac{x}{2}\right)^2 + \frac{8\times7\times6}{3!}\left(-\frac{x}{2}\right)^3$ | M1 A1 | M1: correct binomial coefficient paired with correct power of $x$ in third OR fourth term. Condone missing brackets. Accept coefficients in form $\ldots + {^8C_2}\left(\pm\frac{x}{\ldots}\right)^2 + {^8C_3}\left(\pm\frac{x}{\ldots}\right)^3$. A1: correct unsimplified third OR fourth term; coefficients cannot be left as ${^8C_2}$.

$= 1 - 4x + 7x^2 - 7x^3$ | A1 (4 marks) | Fully simplified. Accept as list $1, -4x, 7x^2, -7x^3$.

### Part (b):
States or uses $x = \frac{1}{5}$ or $0.2$ | B1 |

$(0.9)^8 \approx 1 - 4\times\frac{1}{5} + 7\times\frac{1}{25} - 7\times\frac{1}{125}$ | M1 | Substitutes $x = \frac{1}{5}$ or $0.2$ into their four-term expansion. Condone substitution of $x = -0.2$ or $0.1$ for this mark.

$(0.9)^8 \approx \frac{53}{125}$ | A1 (3 marks) | oe. Accept $\frac{424}{1000}$ or $\frac{265}{625}$. Note: if part (a) fully correct, correct answer awards all three marks.

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Show LaTeX source

\begin{enumerate}
  \item (a) Use the binomial theorem to find the first 4 terms, in ascending powers of $x$, of the expansion of
\end{enumerate}

$$\left( 1 - \frac { x } { 2 } \right) ^ { 8 }$$

Give each term in its simplest form.\\
(b) Use the answer to part (a) to find an approximate value to $0.9 ^ { 8 }$ Write your answer in the form $\frac { a } { b }$ where $a$ and $b$ are integers.\\

\hfill \mbox{\textit{Edexcel C12 2019 Q5 [7]}}

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Q1 3 Q2 4 Q3 4 Q4 6 Q5 7 Q6 7 Q7 5 Q8 5 Q9 8 Q10 11 Q11 8 Q12 9 Q13 10 Q14 11 Q15 11 Q16 16

Answer	Marks	Guidance
\(\left(1 - \frac{x}{2}\right)^8 = 1 - 4x + \ldots\)	B1	May be awarded for unsimplified expression e.g. \(1^8 + 8\left(-\frac{x}{2}\right) + \ldots\)
\(\left(1-\frac{x}{2}\right)^8 = 1 + 8\left(-\frac{x}{2}\right) + \frac{8\times7}{2!}\left(-\frac{x}{2}\right)^2 + \frac{8\times7\times6}{3!}\left(-\frac{x}{2}\right)^3\)	M1 A1	M1: correct binomial coefficient paired with correct power of \(x\) in third OR fourth term. Condone missing brackets. Accept coefficients in form \(\ldots + {^8C_2}\left(\pm\frac{x}{\ldots}\right)^2 + {^8C_3}\left(\pm\frac{x}{\ldots}\right)^3\). A1: correct unsimplified third OR fourth term; coefficients cannot be left as \({^8C_2}\).
\(= 1 - 4x + 7x^2 - 7x^3\)	A1 (4 marks)	Fully simplified. Accept as list \(1, -4x, 7x^2, -7x^3\).

Answer	Marks	Guidance
States or uses \(x = \frac{1}{5}\) or \(0.2\)	B1
\((0.9)^8 \approx 1 - 4\times\frac{1}{5} + 7\times\frac{1}{25} - 7\times\frac{1}{125}\)	M1	Substitutes \(x = \frac{1}{5}\) or \(0.2\) into their four-term expansion. Condone substitution of \(x = -0.2\) or \(0.1\) for this mark.
\((0.9)^8 \approx \frac{53}{125}\)	A1 (3 marks)	oe. Accept \(\frac{424}{1000}\) or \(\frac{265}{625}\). Note: if part (a) fully correct, correct answer awards all three marks.