| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Verify, factorise, solve with substitution |
| Difficulty | Standard +0.3 This is a standard three-part factor theorem question requiring: (a) substitution to find a constant, (b) polynomial division and factorisation, and (c) solving a trigonometric equation using cosec. All techniques are routine for P2 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(x = -3\), equate to zero and attempt solution for \(a\) | M1 | |
| Obtain \(a = 17\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Divide by \(x + 3\) at least as far as the \(x\) term | M1 | |
| Obtain \(6x^2 - x - 1\) | A1 | |
| Conclude \((x+3)(3x+1)(2x-1)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt solution of \(\sin\theta = k\) where \(-1 \leq k \leq 1\) | M1 | |
| Obtain \(199.5\) | A1 | |
| Obtain \(340.5\) | A1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $x = -3$, equate to zero and attempt solution for $a$ | M1 | |
| Obtain $a = 17$ | A1 | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Divide by $x + 3$ at least as far as the $x$ term | M1 | |
| Obtain $6x^2 - x - 1$ | A1 | |
| Conclude $(x+3)(3x+1)(2x-1)$ | A1 | |
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt solution of $\sin\theta = k$ where $-1 \leq k \leq 1$ | M1 | |
| Obtain $199.5$ | A1 | |
| Obtain $340.5$ | A1 | |
---6 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 6 x ^ { 3 } + a x ^ { 2 } - 4 x - 3$$
where $a$ is a constant. It is given that $( x + 3 )$ is a factor of $\mathrm { p } ( x )$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$.
\item Using this value of $a$, factorise $\mathrm { p } ( x )$ completely.
\item Hence solve the equation $\mathrm { p } ( \operatorname { cosec } \theta ) = 0$ for $0 ^ { \circ } < \theta < 360 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2020 Q6 [8]}}