| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Product with unknown constant to determine |
| Difficulty | Standard +0.8 Part (a) requires expanding (2-ax)^5 using binomial theorem, multiplying by (4+2x), collecting x^2 terms, and solving a quadratic equation—a multi-step problem requiring careful algebraic manipulation. Part (b) adds a layer of abstraction by requiring students to recognize that a unique value of a occurs when the discriminant equals zero, demanding deeper conceptual understanding beyond routine application. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Terms required for \(x^2\): \(-5 \times 2^4 \times ax + 10 \times 2^3 \times a^2x^2 \left[= -80ax + 80a^2x^2\right]\) | B1 | Can be seen as part of an expansion or in correct products |
| \(2\times (\pm\text{their coefficient of } x) + 4\times (\pm\text{their coefficient of } x^2)\) | \*M1 | |
| \(x^2\) coefficient is \(320a^2 - 160a = -15\) \(\Rightarrow 64a^2 - 32a + 3 \Rightarrow (8a-3)(8a-1)\) | DM1 | Forming a 3-term quadratic in \(a\), with all terms on the same side or correctly setting up prior to completing the square and solving using factorisation, formula or completing the square. If factorising, factors must expand to give *their* coefficient of \(a^2\) |
| \(a = \frac{1}{8}\) or \(a = \frac{3}{8}\) | A1 | OE. Special case: If DM0 for solving quadratic, SC B1 can be awarded for correct final answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(320a^2 - 160a = k \Rightarrow 320a^2 - 160a - k\ [=0]\) | M1 | Forming a 3-term quadratic in \(a\) with all terms on the same side. Allow \(\pm\) sign errors |
| Their \(b^2 - 4ac\ [=0]\), \([160^2 - 4\times320\times(-k) = 0]\) | M1 | Any use of discriminant on a 3-term quadratic |
| \(k = -20\) | A1 | |
| \(a = \frac{1}{4}\) | B1 | Condone \(a = \frac{1}{4}\) from \(k = 20\) |
| Alternative method: \(320a^2 - 160a = k\) and divide by 320 \(\left[a^2 - \frac{a}{2} = \frac{k}{320}\right]\) | M1 | Allow \(\pm\) sign errors |
| Attempt to complete the square \(\left[\left(a - \frac{1}{4}\right)^2 - \frac{1}{16} = \frac{k}{320}\right]\) | M1 | Must have \(\left(a - \frac{1}{4}\right)^2\) |
| \(a = \frac{1}{4}\) | A1 | |
| \(k = -20\) | B1 | |
| Alternative method: \(320a^2 - 160a = k\) and attempt to differentiate LHS \([640a - 160]\) | M1 | Allow \(\pm\) sign errors |
| Setting their \((640a - 160) = 0\) and attempt to solve | M1 | |
| \(a = \frac{1}{4}\) | A1 | |
| \(k = -20\) | B1 |
## Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Terms required for $x^2$: $-5 \times 2^4 \times ax + 10 \times 2^3 \times a^2x^2 \left[= -80ax + 80a^2x^2\right]$ | B1 | Can be seen as part of an expansion or in correct products |
| $2\times (\pm\text{their coefficient of } x) + 4\times (\pm\text{their coefficient of } x^2)$ | \*M1 | |
| $x^2$ coefficient is $320a^2 - 160a = -15$ $\Rightarrow 64a^2 - 32a + 3 \Rightarrow (8a-3)(8a-1)$ | DM1 | Forming a 3-term quadratic in $a$, with all terms on the same side or correctly setting up prior to completing the square and solving using factorisation, formula or completing the square. If factorising, factors must expand to give *their* coefficient of $a^2$ |
| $a = \frac{1}{8}$ or $a = \frac{3}{8}$ | A1 | OE. **Special case:** If DM0 for solving quadratic, SC B1 can be awarded for correct final answers |
---
## Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $320a^2 - 160a = k \Rightarrow 320a^2 - 160a - k\ [=0]$ | M1 | Forming a 3-term quadratic in $a$ with all terms on the same side. Allow $\pm$ sign errors |
| Their $b^2 - 4ac\ [=0]$, $[160^2 - 4\times320\times(-k) = 0]$ | M1 | Any use of discriminant on a 3-term quadratic |
| $k = -20$ | A1 | |
| $a = \frac{1}{4}$ | B1 | Condone $a = \frac{1}{4}$ from $k = 20$ |
| **Alternative method:** $320a^2 - 160a = k$ and divide by 320 $\left[a^2 - \frac{a}{2} = \frac{k}{320}\right]$ | M1 | Allow $\pm$ sign errors |
| Attempt to complete the square $\left[\left(a - \frac{1}{4}\right)^2 - \frac{1}{16} = \frac{k}{320}\right]$ | M1 | Must have $\left(a - \frac{1}{4}\right)^2$ |
| $a = \frac{1}{4}$ | A1 | |
| $k = -20$ | B1 | |
| **Alternative method:** $320a^2 - 160a = k$ and attempt to differentiate LHS $[640a - 160]$ | M1 | Allow $\pm$ sign errors |
| Setting their $(640a - 160) = 0$ and attempt to solve | M1 | |
| $a = \frac{1}{4}$ | A1 | |
| $k = -20$ | B1 | |
---8
\begin{enumerate}[label=(\alph*)]
\item It is given that in the expansion of $( 4 + 2 x ) ( 2 - a x ) ^ { 5 }$, the coefficient of $x ^ { 2 }$ is - 15 .\\
Find the possible values of $a$.
\item It is given instead that in the expansion of $( 4 + 2 x ) ( 2 - a x ) ^ { 5 }$, the coefficient of $x ^ { 2 }$ is $k$. It is also given that there is only one value of $a$ which leads to this value of $k$.
Find the values of $k$ and $a$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q8 [8]}}