| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Substitution into binomial expansion |
| Difficulty | Moderate -0.8 Part (i) is direct application of binomial expansion formula requiring only the first three terms. Part (ii) involves substitution and collecting terms, but the algebra is straightforward with only two terms to track. This is a standard textbook exercise testing basic binomial theorem mechanics with minimal problem-solving required. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 + 6y + 15y^2\) | B1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 + 6(px - 2x^2) + 15(px - 2x^2)^2\) | M1 | SOI. Allow \(6C1 \times 1^5(px - 2x^2)\), \(6C2 \times 1^4(px-2x^2)^2\) |
| \((15p^2 - 12)(x^2) = 48(x^2)\) | A1 | 1 term from each bracket and equate to 48 |
| \(p = 2\) | A1 | SC: A1 \(p = 4\) from \(15p - 12 = 48\) |
## Question 1:
**Part (i)**
$1 + 6y + 15y^2$ | **B1** | CAO
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**Part (ii)**
$1 + 6(px - 2x^2) + 15(px - 2x^2)^2$ | **M1** | SOI. Allow $6C1 \times 1^5(px - 2x^2)$, $6C2 \times 1^4(px-2x^2)^2$
$(15p^2 - 12)(x^2) = 48(x^2)$ | **A1** | 1 term from each bracket and equate to 48
$p = 2$ | **A1** | **SC:** A1 $p = 4$ from $15p - 12 = 48$
---1 (i) Expand $( 1 + y ) ^ { 6 }$ in ascending powers of $y$ as far as the term in $y ^ { 2 }$.\\
(ii) In the expansion of $\left( 1 + \left( p x - 2 x ^ { 2 } \right) \right) ^ { 6 }$ the coefficient of $x ^ { 2 }$ is 48 . Find the value of the positive constant $p$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q1 [4]}}