CAIE P1 2015 November — Question 1 3 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2015
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeCoefficient zero by design (proof)
DifficultyStandard +0.3 This is a straightforward binomial expansion problem requiring students to expand (a+x)^5, multiply by the linear factor, collect x^2 terms, and show they sum to zero. While it involves multiple steps and algebraic manipulation, it's a standard textbook exercise with no novel insight required—slightly easier than average due to its mechanical nature and the fact that students are told what to prove.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n

1 In the expansion of \(\left( 1 - \frac { 2 x } { a } \right) ( a + x ) ^ { 5 }\), where \(a\) is a non-zero constant, show that the coefficient of \(x ^ { 2 }\) is zero.

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((a+x)^5 = a^5 + {}^5C_1a^4x + {}^5C_2a^3x^2 + \ldots\)M1 Ignore subsequent terms
\(\left(-\frac{2}{a} \times (\text{their } 5a^4) + (\text{their } 10a^3)\right)(x^2)\)M1
\(0\)A1 [3] AG
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(a+x)^5 = a^5 + {}^5C_1a^4x + {}^5C_2a^3x^2 + \ldots$ | M1 | Ignore subsequent terms |
| $\left(-\frac{2}{a} \times (\text{their } 5a^4) + (\text{their } 10a^3)\right)(x^2)$ | M1 | |
| $0$ | A1 [3] | AG |

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1 In the expansion of $\left( 1 - \frac { 2 x } { a } \right) ( a + x ) ^ { 5 }$, where $a$ is a non-zero constant, show that the coefficient of $x ^ { 2 }$ is zero.

\hfill \mbox{\textit{CAIE P1 2015 Q1 [3]}}