| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2017 |
| Session | Specimen |
| Marks | 11 |
| Topic | Parametric differentiation |
| Type | Gradient condition leads to trig equation |
| Difficulty | Standard +0.8 This is a multi-part parametric differentiation question requiring finding dy/dx, forming tangent equations, and solving a trigonometric equation. Part (a) is routine, but parts (b) and (c) require algebraic manipulation to derive the given equation and then solve a non-standard trigonometric equation (likely requiring substitution or auxiliary angle methods), making this moderately challenging for Pure 1 level. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
The parametric equations of a curve are given by $x = 2\cos\theta$ and $y = 3\sin\theta$ for $0 \leq \theta < 2\pi$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac{dy}{dx}$ in terms of $\theta$. [2]
\end{enumerate}
The tangents to the curve at the points P and Q pass through the point (2, 6).
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the values of $\theta$ at the points P and Q satisfy the equation $2\sin\theta + \cos\theta = 1$. [4]
\item Find the values of $\theta$ at the points P and Q. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2017 Q12 [11]}}