| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Trigonometric substitution equations |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining routine factor theorem application (finding a constant), polynomial factorization, and a trigonometric substitution that directly follows from the factorization. All steps are standard techniques 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 graphs1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = -2\) and equate to zero | M1 | |
| Solve equation to confirm \(a = -4\) | A1 | [2] |
| (ii) (a) Find quadratic factor by division, inspection, identity, ... | M1 | |
| Obtain \(6x^2 - x - 2\) | A1 | |
| Conclude \((x + 2)(3x - 2)(2x + 1)\) | A1 | [3] |
| (b) State or imply at least \(\sec \theta = -2\) and attempt solution | M1 | |
| Obtain \(120°\) and no others in range | A1 | [2] |
(i) Substitute $x = -2$ and equate to zero | M1 |
Solve equation to confirm $a = -4$ | A1 | [2]
(ii) (a) Find quadratic factor by division, inspection, identity, ... | M1 |
Obtain $6x^2 - x - 2$ | A1 |
Conclude $(x + 2)(3x - 2)(2x + 1)$ | A1 | [3]
(b) State or imply at least $\sec \theta = -2$ and attempt solution | M1 |
Obtain $120°$ and no others in range | A1 | [2]
---4 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 6 x ^ { 3 } + 11 x ^ { 2 } + a x + a$$
where $a$ is a constant. It is given that $( x + 2 )$ is a factor of $\mathrm { p } ( x )$.\\
(i) Use the factor theorem to show that $a = - 4$.\\
(ii) When $a = - 4$,
\begin{enumerate}[label=(\alph*)]
\item factorise $\mathrm { p } ( x )$ completely,
\item solve the equation $6 \sec ^ { 3 } \theta + 11 \sec ^ { 2 } \theta + a \sec \theta + a = 0$ for $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2015 Q4 [7]}}