| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Two equations from coefficients |
| Difficulty | Standard +0.3 This is a straightforward binomial theorem question requiring students to find specific coefficients using the binomial formula, then solve a resulting equation. While it involves two expansions and solving a quartic that factors nicely, the techniques are standard and the algebraic manipulation is routine for A-level students who have practiced binomial theorem problems. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^4\) term is \([10\times](2x^2)^3\left(\frac{k^2}{x}\right)^2\) | M1 | For selecting the term in \(x^4\) |
| \(80k^4x^4 \Rightarrow a = 80k^4\) | A1 | For correct value of \(a\). Allow \(80k^4x^4\) |
| \([x^2\) term is \([6\times](2kx)^2 \times 1 = 24k^2x^2 \Rightarrow] b = 24k^2\) | B1 | For correct value of \(b\). Allow \(24k^2x^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(80k^4 + 24k^2 - 216[=0]\ \ [\Rightarrow 10k^4 + 3k^2 - 27 = 0]\) | M1 | Forming a 3-term equation in \(k\) (all terms on one side) with their \(a\) and \(b\) and no \(x\)'s |
| \((2k^2-3)(5k^2+9)\ [=0]\ [\Rightarrow k^2 = \frac{3}{2}\) or \(-\frac{9}{5}]\) | M1 | Attempt to solve 3-term quartic (or quadratic in another variable) by factorisation, formula or completing the square |
| \([k] = \pm\sqrt{\frac{3}{2}}\) | A1 | OE e.g. \(\pm\frac{\sqrt{6}}{2}\), \(\pm\sqrt{1.5}\), AWRT \(\pm1.22\). Omission of \(\pm\) A0. Additional answers A0. If M1 M0, SC B1 can be awarded for correct final answer, max 2/3 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^4$ term is $[10\times](2x^2)^3\left(\frac{k^2}{x}\right)^2$ | M1 | For selecting the term in $x^4$ |
| $80k^4x^4 \Rightarrow a = 80k^4$ | A1 | For correct value of $a$. Allow $80k^4x^4$ |
| $[x^2$ term is $[6\times](2kx)^2 \times 1 = 24k^2x^2 \Rightarrow] b = 24k^2$ | B1 | For correct value of $b$. Allow $24k^2x^2$ |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $80k^4 + 24k^2 - 216[=0]\ \ [\Rightarrow 10k^4 + 3k^2 - 27 = 0]$ | M1 | Forming a 3-term equation in $k$ (all terms on one side) with their $a$ and $b$ and no $x$'s |
| $(2k^2-3)(5k^2+9)\ [=0]\ [\Rightarrow k^2 = \frac{3}{2}$ or $-\frac{9}{5}]$ | M1 | Attempt to solve 3-term quartic (or quadratic in another variable) by factorisation, formula or completing the square |
| $[k] = \pm\sqrt{\frac{3}{2}}$ | A1 | OE e.g. $\pm\frac{\sqrt{6}}{2}$, $\pm\sqrt{1.5}$, AWRT $\pm1.22$. Omission of $\pm$ A0. Additional answers A0. If M1 M0, **SC B1** can be awarded for correct final answer, max 2/3 |3 The coefficient of $x ^ { 4 }$ in the expansion of $\left( 2 x ^ { 2 } + \frac { k ^ { 2 } } { x } \right) ^ { 5 }$ is $a$. The coefficient of $x ^ { 2 }$ in the expansion of $( 2 k x - 1 ) ^ { 4 }$ is $b$.
\begin{enumerate}[label=(\alph*)]
\item Find $a$ and $b$ in terms of the constant $k$.
\item Given that $a + b = 216$, find the possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q3 [6]}}