| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Gradient condition leads to trig equation |
| Difficulty | Standard +0.3 This is a standard parametric differentiation question requiring dy/dx = (dy/dt)/(dx/dt), setting equal to 2, then using R-formula to solve a trigonometric equation. All techniques are routine A-level methods with clear signposting, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(\dfrac{dx}{dt} = 2 - 2\cos 2t\) | B1 | |
| Obtain \(\dfrac{dy}{dt} = 5 - 2\sin 2t\) | B1 | |
| Equate attempt at \(\dfrac{dy}{dx}\) to 2 and rearrange | M1 | |
| Confirm equation \(2\sin 2t - 4\cos 2t = 1\) | A1 | Answer given; necessary detail needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(R = \sqrt{20}\) or \(4.47\) | B1 | |
| Use appropriate trigonometry to find \(\alpha\) | M1 | |
| Obtain \(\alpha = 1.107\) with no errors seen | A1 | |
| Carry out correct method to find value of \(t\) | M1 | |
| Obtain \(t = 0.666\) | A1 | |
| Substitute value of \(t\) between 0 and \(\frac{1}{2}\pi\) into expressions for \(x\) and \(y\) | M1 | |
| Obtain \(x = 0.361\), \(y = 3.57\) | A1 |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $\dfrac{dx}{dt} = 2 - 2\cos 2t$ | B1 | |
| Obtain $\dfrac{dy}{dt} = 5 - 2\sin 2t$ | B1 | |
| Equate attempt at $\dfrac{dy}{dx}$ to 2 and rearrange | M1 | |
| Confirm equation $2\sin 2t - 4\cos 2t = 1$ | A1 | Answer given; necessary detail needed |
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $R = \sqrt{20}$ or $4.47$ | B1 | |
| Use appropriate trigonometry to find $\alpha$ | M1 | |
| Obtain $\alpha = 1.107$ with no errors seen | A1 | |
| Carry out correct method to find value of $t$ | M1 | |
| Obtain $t = 0.666$ | A1 | |
| Substitute value of $t$ between 0 and $\frac{1}{2}\pi$ into expressions for $x$ and $y$ | M1 | |
| Obtain $x = 0.361$, $y = 3.57$ | A1 | |7 The parametric equations of a curve are
$$x = 2 t - \sin 2 t , \quad y = 5 t + \cos 2 t$$
for $0 \leqslant t \leqslant \frac { 1 } { 2 } \pi$. At the point $P$ on the curve, the gradient of the curve is 2 .\\
(i) Show that the value of the parameter at $P$ satisfies the equation $2 \sin 2 t - 4 \cos 2 t = 1$.\\
(ii) By first expressing $2 \sin 2 t - 4 \cos 2 t$ in the form $R \sin ( 2 t - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { 1 } { 2 } \pi$, find the coordinates of $P$. Give each coordinate correct to 3 significant figures.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE P2 2019 Q7 [11]}}