CAIE P1 2018 November — Question 1 3 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionNovember
Marks3
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TopicBinomial Theorem (positive integer n)
TypeBinomial with negative or fractional powers of x
DifficultyModerate -0.3 This is a straightforward binomial expansion requiring identification of the correct term where powers of x combine to give x^(-3). It's slightly easier than average because it's a single-step problem with clear methodology (finding r where 7-r - r = -3), though students must be careful with the algebra of negative powers.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n
1 Find the coefficient of \(\frac { 1 } { x ^ { 3 } }\) in the expansion of \(\left( x - \frac { 2 } { x } \right) ^ { 7 }\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(7C5 \cdot x^2(-2/x)^5\)B1 Can appear in an expansion. Allow 7C2
\(21 \times -32\)B1 Identified. Allow \((21x^2) \times (-32x^{-5})\). Implied by correct answer
\(-672\)B1 Allow \(\frac{-672}{x^3}\). If 0/3 scored, 672 scores SCB1
3 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $7C5 \cdot x^2(-2/x)^5$ | **B1** | Can appear in an expansion. Allow 7C2 |
| $21 \times -32$ | **B1** | Identified. Allow $(21x^2) \times (-32x^{-5})$. Implied by correct answer |
| $-672$ | **B1** | Allow $\frac{-672}{x^3}$. If 0/3 scored, 672 scores SCB1 |
| | **3** | |
1 Find the coefficient of $\frac { 1 } { x ^ { 3 } }$ in the expansion of $\left( x - \frac { 2 } { x } \right) ^ { 7 }$.\\

\hfill \mbox{\textit{CAIE P1 2018 Q1 [3]}}
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