| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Solving Equations via Substitution After Division |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring polynomial division (routine), integration of a rational function after division (standard technique), and factorization followed by solving a trigonometric equation. While part (c) requires recognizing that p(x)+6 is divisible by (2x+3) and then solving cosec 2θ = -3/2, all techniques are standard A-level fare with clear scaffolding. Slightly above average due to the trigonometric substitution in part (c), but no novel insights required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.05o Trigonometric equations: solve in given intervals1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Carry out division at least as far as \(2x^2 + kx\) | M1 | |
| Obtain quotient \(2x^2 + 5x - 3\) | A1 | |
| Confirm remainder is \(-6\) | A1 | AG – necessary detail needed |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Integrate to obtain at least \(k_1 x^3\) and \(k_2\ln(2x+3)\) terms | M1 | |
| Obtain \(\frac{2}{3}x^3 + \frac{5}{2}x^2 - 3x - 3\ln(2x+3)\) | A1 | condone absence of \(...+c\) and modulus signs |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(p(x) + 6 = (2x+3)(2x^2+5x-3)\) | B1 FT | FT *their* quotient |
| Conclude \((2x+3)(2x-1)(x+3)\) | B1 | |
| State or imply \(\sin 2\theta = -\frac{2}{3}\) or \(\sin 2\theta = -\frac{1}{3}\) or both | B1 FT | FT *their* relevant factors |
| Carry out correct process to find \(\theta\) in at least one case | M1 | |
| Obtain 99.7 and 110.9 | A1 | Or greater accuracy and no others between \(0°\) and \(135°\) |
| Total | 5 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Carry out division at least as far as $2x^2 + kx$ | M1 | |
| Obtain quotient $2x^2 + 5x - 3$ | A1 | |
| Confirm remainder is $-6$ | A1 | AG – necessary detail needed |
| **Total** | **3** | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate to obtain at least $k_1 x^3$ and $k_2\ln(2x+3)$ terms | M1 | |
| Obtain $\frac{2}{3}x^3 + \frac{5}{2}x^2 - 3x - 3\ln(2x+3)$ | A1 | condone absence of $...+c$ and modulus signs |
| **Total** | **2** | |
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $p(x) + 6 = (2x+3)(2x^2+5x-3)$ | B1 FT | FT *their* quotient |
| Conclude $(2x+3)(2x-1)(x+3)$ | B1 | |
| State or imply $\sin 2\theta = -\frac{2}{3}$ or $\sin 2\theta = -\frac{1}{3}$ or both | B1 FT | FT *their* relevant factors |
| Carry out correct process to find $\theta$ in at least one case | M1 | |
| Obtain 99.7 and 110.9 | A1 | Or greater accuracy and no others between $0°$ and $135°$ |
| **Total** | **5** | |6 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 4 x ^ { 3 } + 16 x ^ { 2 } + 9 x - 15$$
\begin{enumerate}[label=(\alph*)]
\item Find the quotient when $\mathrm { p } ( x )$ is divided by $( 2 x + 3 )$, and show that the remainder is - 6 .
\item Find $\int \frac { \mathrm { p } ( x ) } { 2 x + 3 } \mathrm {~d} x$.
\item Factorise $\mathrm { p } ( x ) + 6$ completely and hence solve the equation
$$p ( \operatorname { cosec } 2 \theta ) + 6 = 0$$
for $0 ^ { \circ } < \theta < 135 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2022 Q6 [10]}}