Edexcel C34 Specimen — Question 4 13 marks

Exam BoardEdexcel
ModuleC34 (Core Mathematics 3 & 4)
SessionSpecimen
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem
TypeTwo unknowns from two coefficient conditions
DifficultyStandard +0.3 This is a standard two-part binomial expansion question requiring students to first expand (2-3x)^{-2}, then use coefficient conditions to find two unknowns. While it involves multiple steps and algebraic manipulation, the techniques are routine for C3/C4 students: apply the binomial theorem formula, multiply by a linear numerator, and solve simultaneous equations from coefficients. The problem-solving is straightforward with no novel insights required.
Spec1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions

4. (a) Use the binomial theorem to expand $$( 2 - 3 x ) ^ { - 2 } , \quad | x | < \frac { 2 } { 3 }$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\). Give each coefficient as a simplified fraction. $$\mathrm { f } ( x ) = \frac { a + b x } { ( 2 - 3 x ) ^ { 2 } } , \quad | x | < \frac { 2 } { 3 } , \quad \text { where } a \text { and } b \text { are constants. }$$ In the binomial expansion of \(\mathrm { f } ( x )\), in ascending powers of \(x\), the coefficient of \(x\) is 0 and the coefficient of \(x ^ { 2 }\) is \(\frac { 9 } { 16 }\) Find
(b) the value of \(a\) and the value of \(b\),
(c) the coefficient of \(x ^ { 3 }\), giving your answer as a simplified fraction.

Question 4:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((2-3x)^{-2} = 2^{-2}\left(1-\frac{3x}{2}\right)^{-2}\)B1
\(\left(1-\frac{3x}{2}\right)^{-2} = 1+(-2)\left(-\frac{3x}{2}\right)+\frac{(-2)(-3)}{2\times1}\left(-\frac{3x}{2}\right)^2+\frac{(-2)(-3)(-4)}{3\times2\times1}\left(-\frac{3x}{2}\right)^3+\ldots\)M1 A1
\(= 1+3x+\frac{27}{4}x^2+\frac{27}{2}x^3+\ldots\)
\((2-3x)^{-2} = \frac{1}{4}+\frac{3}{4}x+\frac{27}{16}x^2+\frac{27}{8}x^3+\ldots\)M1 A1
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(x) = (a+bx)\left(\frac{1}{4}+\frac{3}{4}x+\frac{27}{16}x^2+\frac{27}{8}x^3+\ldots\right)\)
Coefficient of \(x\): \(\frac{3a}{4}+\frac{b}{4}=0 \quad (3a+b=0)\)M1
Coefficient of \(x^2\): \(\frac{27a}{16}+\frac{3b}{4}=\frac{9}{16} \quad (9a+4b=3)\)M1
Either correctA1
Leading to \(a=-1,\ b=3\)dM1 A1
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Coefficient of \(x^3\): \(\frac{27a}{8}+\frac{27b}{16}=\frac{27}{8}\times-1+\frac{27}{16}\times3\)M1 A1ft
\(=\frac{27}{16}\left(=1\frac{11}{16}\right)\)A1 cao
# Question 4:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2-3x)^{-2} = 2^{-2}\left(1-\frac{3x}{2}\right)^{-2}$ | B1 | |
| $\left(1-\frac{3x}{2}\right)^{-2} = 1+(-2)\left(-\frac{3x}{2}\right)+\frac{(-2)(-3)}{2\times1}\left(-\frac{3x}{2}\right)^2+\frac{(-2)(-3)(-4)}{3\times2\times1}\left(-\frac{3x}{2}\right)^3+\ldots$ | M1 A1 | |
| $= 1+3x+\frac{27}{4}x^2+\frac{27}{2}x^3+\ldots$ | | |
| $(2-3x)^{-2} = \frac{1}{4}+\frac{3}{4}x+\frac{27}{16}x^2+\frac{27}{8}x^3+\ldots$ | M1 A1 | |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = (a+bx)\left(\frac{1}{4}+\frac{3}{4}x+\frac{27}{16}x^2+\frac{27}{8}x^3+\ldots\right)$ | | |
| Coefficient of $x$: $\frac{3a}{4}+\frac{b}{4}=0 \quad (3a+b=0)$ | M1 | |
| Coefficient of $x^2$: $\frac{27a}{16}+\frac{3b}{4}=\frac{9}{16} \quad (9a+4b=3)$ | M1 | |
| Either correct | A1 | |
| Leading to $a=-1,\ b=3$ | dM1 A1 | |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Coefficient of $x^3$: $\frac{27a}{8}+\frac{27b}{16}=\frac{27}{8}\times-1+\frac{27}{16}\times3$ | M1 A1ft | |
| $=\frac{27}{16}\left(=1\frac{11}{16}\right)$ | A1 | cao |

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4. (a) Use the binomial theorem to expand

$$( 2 - 3 x ) ^ { - 2 } , \quad | x | < \frac { 2 } { 3 }$$

in ascending powers of $x$, up to and including the term in $x ^ { 3 }$. Give each coefficient as a simplified fraction.

$$\mathrm { f } ( x ) = \frac { a + b x } { ( 2 - 3 x ) ^ { 2 } } , \quad | x | < \frac { 2 } { 3 } , \quad \text { where } a \text { and } b \text { are constants. }$$

In the binomial expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, the coefficient of $x$ is 0 and the coefficient of $x ^ { 2 }$ is $\frac { 9 } { 16 }$

Find\\
(b) the value of $a$ and the value of $b$,\\
(c) the coefficient of $x ^ { 3 }$, giving your answer as a simplified fraction.\\

\hfill \mbox{\textit{Edexcel C34  Q4 [13]}}