| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Trigonometric substitution equations |
| Difficulty | Standard +0.3 This is a structured multi-part question combining routine Factor Theorem application, polynomial division, and trigonometric substitution. Parts (a) and (b) are standard textbook exercises requiring direct application of learned techniques. Part (c) requires using the double angle formula and solving a cubic via the earlier factorization, but the question heavily scaffolds the approach. While it spans multiple topics, each step is straightforward with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
5 The polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 4 x ^ { 3 } - 11 x - 3$.
\begin{enumerate}[label=(\alph*)]
\item Use the Factor Theorem to show that ( $2 x + 3$ ) is a factor of $\mathrm { f } ( x )$.
\item Write $\mathrm { f } ( x )$ in the form $( 2 x + 3 ) \left( a x ^ { 2 } + b x + c \right)$, where $a , b$ and $c$ are integers.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation $2 \cos 2 \theta \sin \theta + 9 \sin \theta + 3 = 0$ can be written as $4 x ^ { 3 } - 11 x - 3 = 0$, where $x = \sin \theta$.
\item Hence find all solutions of the equation $2 \cos 2 \theta \sin \theta + 9 \sin \theta + 3 = 0$ in the interval $0 ^ { \circ } < \theta < 360 ^ { \circ }$, giving your solutions to the nearest degree.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2013 Q5 [11]}}