| Exam Board | OCR |
|---|---|
| Module | AS Pure (AS Pure Mathematics) |
| Year | 2017 |
| Session | Specimen |
| Marks | 9 |
| Topic | Factor & Remainder Theorem |
| Type | Trigonometric substitution equations |
| Difficulty | Standard +0.3 This is a straightforward two-part question combining routine factor theorem verification (substituting x=1/2) with a simple trigonometric substitution (sin θ = 1/2). Both parts require standard techniques with no novel insight—slightly easier than average due to the direct nature of the substitution and the simple solution θ = 30°, 150°. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05o Trigonometric equations: solve in given intervals |
6 In this question you must show detailed reasoning.
The cubic polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 4 x ^ { 3 } + 4 x ^ { 2 } + 7 x - 5$.
\begin{enumerate}[label=(\alph*)]
\item Show that $( 2 x - 1 )$ is a factor of $\mathrm { f } ( x )$.
\item Hence solve the equation $4 \sin ^ { 3 } \theta + 4 \sin ^ { 2 } \theta + 7 \sin \theta - 5 = 0$ for $0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{OCR AS Pure 2017 Q6 [9]}}