| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question with standard techniques. Part (a) requires solving a trigonometric equation using double angle formula, part (b) is routine application of dy/dx = (dy/dt)/(dx/dt), and part (c) combines these results. While it requires multiple steps and careful algebra, all techniques are standard P2 material with no novel insights needed. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | Use identity sin2t=2sintcost | B1 |
| Attempt solution of y=0 for cost | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | A1 | Or exact equivalent. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(b) | dx dy |
| Answer | Marks |
|---|---|
| dt dt | *M1 |
| Answer | Marks |
|---|---|
| dx ksec2t | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| dx | DM1 | Using correct identities. |
| Answer | Marks | Guidance |
|---|---|---|
| dx k | A1 | OE |
| Answer | Marks |
|---|---|
| 7(c) | dy |
| Answer | Marks |
|---|---|
| and cost | *M1 |
| Answer | Marks |
|---|---|
| 9 | DM1 |
| Answer | Marks |
|---|---|
| 3 | A1 |
Question 7:
--- 7(a) ---
7(a) | Use identity sin2t=2sintcost | B1
Attempt solution of y=0 for cost | M1
Obtain cost = 2
3 | A1 | Or exact equivalent.
3
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | dx dy
Differentiate to obtain at least one of and correct
dt dt | *M1
dy 6cos2t−4cost
Obtain =
dx ksec2t | A1
dy
Attempt to express in terms of cost
dx | DM1 | Using correct identities.
dy 6(2cos2t−1)cos2t−4cos3t
Obtain =
dx k | A1 | OE
4
--- 7(c) ---
7(c) | dy
Substitute value from part (a) in expression for involving k
dx
and cost | *M1
Equate to −10 and solve for k
9 | DM1
Obtain k = 4
3 | A1
3\includegraphics{figure_7}
The diagram shows the curve with parametric equations
$$x = k \tan t, \quad y = 3 \sin 2t - 4 \sin t,$$
for $0 < t < \frac{1}{2}\pi$. It is given that $k$ is a positive constant. The curve crosses the $x$-axis at the point $P$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\cos t$ at $P$, giving your answer as an exact fraction. [3]
\item Express $\frac{dy}{dx}$ in terms of $k$ and $\cos t$. [4]
\item Given that the normal to the curve at $P$ has gradient $\frac{9}{10}$, find the value of $k$, giving your answer as an exact fraction. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q7 [10]}}