| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Trigonometric substitution equations |
| Difficulty | Moderate -0.3 This is a standard multi-part question on the Factor and Remainder Theorem requiring routine algebraic manipulation. Part (i) involves setting up two simultaneous equations using p(-3)=0 and p(-2)=18, which is textbook application. Part (ii)(a) requires showing one real root via discriminant or sign analysis, and part (ii)(b) involves a straightforward substitution with sec y. While it has multiple parts and requires careful algebra, all techniques are standard A-level procedures with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs |
The polynomial $p(x)$ is defined by
$$p(x) = ax^3 + 3x^2 + bx + 12,$$
where $a$ and $b$ are constants. It is given that $(x + 3)$ is a factor of $p(x)$. It is also given that the remainder is 18 when $p(x)$ is divided by $(x + 2)$.
\begin{enumerate}[label=(\roman*)]
\item Find the values of $a$ and $b$. [5]
\item When $a$ and $b$ have these values,
\begin{enumerate}[label=(\alph*)]
\item show that the equation $p(x) = 0$ has exactly one real root, [4]
\item solve the equation $p(\sec y) = 0$ for $-180° < y < 180°$. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2016 Q7 [12]}}