Edexcel C4 2012 January — Question 3 9 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2012
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem
TypeSingle unknown from one coefficient condition
DifficultyStandard +0.3 This is a standard C4 binomial expansion question requiring routine application of the generalized binomial theorem with negative index, followed by algebraic manipulation to match coefficients. Part (a) is straightforward expansion, while parts (b) and (c) involve simple coefficient matching after multiplying the expansion by (2+kx). The techniques are well-practiced and the multi-step nature is typical for this topic, making it slightly easier than average.
Spec1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions
3. (a) Expand $$\frac { 1 } { ( 2 - 5 x ) ^ { 2 } } , \quad | x | < \frac { 2 } { 5 }$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 2 }\), giving each term as a simplified fraction. Given that the binomial expansion of \(\frac { 2 + k x } { ( 2 - 5 x ) ^ { 2 } } , | x | < \frac { 2 } { 5 }\), is $$\frac { 1 } { 2 } + \frac { 7 } { 4 } x + A x ^ { 2 } + \ldots$$ (b) find the value of the constant \(k\),
(c) find the value of the constant \(A\).
Question 3:
Part (a): Expand \(\frac{1}{(2-5x)^2}\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((2-5x)^{-2} = (2)^{-2}\left(1-\frac{5x}{2}\right)^{-2} = \frac{1}{4}\left(1-\frac{5x}{2}\right)^{-2}\)B1 \((2)^{-2}\) or \(\frac{1}{4}\) outside brackets
\(= \left\{\frac{1}{4}\right\}\left[1+(-2)(x)+\frac{(-2)(-3)}{2!}(x)^2+...\right]\)M1 A1ft Expands to give \(1+(-2)(x)\) or \((-2)(x)+\frac{(-2)(-3)}{2!}(x)^2\) where \(\neq 1\)
\(= \frac{1}{4}+\frac{5}{4}x+\frac{75}{16}x^2+...\)A1; A1 First A1 for \(\frac{1}{4}+\frac{5}{4}x\); second A1 for \(\frac{75}{16}x^2\)
Part (b): Find \(k\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\left\{\frac{2+kx}{(2-5x)^2}\right\} = (2+kx)\left(\frac{1}{4}+\frac{5}{4}x+\left\{\frac{75}{16}x^2+...\right\}\right)\)M1 Can be implied by later work even in part (c)
\(x\) terms: \(\frac{2(5x)}{4}+\frac{kx}{4}=\frac{7x}{4}\), giving \(10+k=7 \Rightarrow k=-3\)A1 \(k=-3\)
Part (c): Find \(A\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x^2\) terms: \(\frac{150x^2}{16}+\frac{5kx^2}{4}\)M1 Multiplies out to obtain exactly two \(x^2\) terms/coefficients, equates to 0, attempts to find \(A\)
\(A = \frac{75}{8}+\frac{5(-3)}{4} = \frac{75}{8}-\frac{15}{4} = \frac{45}{8}\)A1 \(\frac{45}{8}\) or \(5\frac{5}{8}\) or \(5.625\) 
# Question 3:

## Part (a): Expand $\frac{1}{(2-5x)^2}$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2-5x)^{-2} = (2)^{-2}\left(1-\frac{5x}{2}\right)^{-2} = \frac{1}{4}\left(1-\frac{5x}{2}\right)^{-2}$ | B1 | $(2)^{-2}$ or $\frac{1}{4}$ outside brackets |
| $= \left\{\frac{1}{4}\right\}\left[1+(-2)(**x)+\frac{(-2)(-3)}{2!}(**x)^2+...\right]$ | M1 A1ft | Expands to give $1+(-2)(**x)$ or $(-2)(**x)+\frac{(-2)(-3)}{2!}(**x)^2$ where $**\neq 1$ |
| $= \frac{1}{4}+\frac{5}{4}x+\frac{75}{16}x^2+...$ | A1; A1 | First A1 for $\frac{1}{4}+\frac{5}{4}x$; second A1 for $\frac{75}{16}x^2$ |

## Part (b): Find $k$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left\{\frac{2+kx}{(2-5x)^2}\right\} = (2+kx)\left(\frac{1}{4}+\frac{5}{4}x+\left\{\frac{75}{16}x^2+...\right\}\right)$ | M1 | Can be implied by later work even in part (c) |
| $x$ terms: $\frac{2(5x)}{4}+\frac{kx}{4}=\frac{7x}{4}$, giving $10+k=7 \Rightarrow k=-3$ | A1 | $k=-3$ |

## Part (c): Find $A$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2$ terms: $\frac{150x^2}{16}+\frac{5kx^2}{4}$ | M1 | Multiplies out to obtain exactly two $x^2$ terms/coefficients, equates to 0, attempts to find $A$ |
| $A = \frac{75}{8}+\frac{5(-3)}{4} = \frac{75}{8}-\frac{15}{4} = \frac{45}{8}$ | A1 | $\frac{45}{8}$ or $5\frac{5}{8}$ or $5.625$ |

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3. (a) Expand

$$\frac { 1 } { ( 2 - 5 x ) ^ { 2 } } , \quad | x | < \frac { 2 } { 5 }$$

in ascending powers of $x$, up to and including the term in $x ^ { 2 }$, giving each term as a simplified fraction.

Given that the binomial expansion of $\frac { 2 + k x } { ( 2 - 5 x ) ^ { 2 } } , | x | < \frac { 2 } { 5 }$, is

$$\frac { 1 } { 2 } + \frac { 7 } { 4 } x + A x ^ { 2 } + \ldots$$

(b) find the value of the constant $k$,\\
(c) find the value of the constant $A$.

\hfill \mbox{\textit{Edexcel C4 2012 Q3 [9]}}
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