| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Show gradient expression then find coordinates |
| Difficulty | Moderate -0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dt รท dx/dt), followed by two simple substitutions to find gradients at specific points. The algebra is routine with standard quotient rule application, making it slightly easier than average for A-level. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(\frac{dx}{dt} = \frac{2}{2t+3}\) | B1 | |
| Use quotient rule, or equivalent, to find \(\frac{dy}{dt}\) | M1 | |
| Obtain \(\frac{dy}{dt} = \frac{2(2t+3) - 2(2t-3)}{(2t+3)^2}\) | A1 | OE |
| Divide to confirm \(\frac{dy}{dx} = \frac{6}{2t+3}\) | A1 | AG |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to find value of \(t\) corresponding to \(x = 0\) | M1 | |
| Obtain \(t = -1\) and hence gradient is \(6\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to find value of \(t\) corresponding to \(y = 0\) | M1 | |
| Obtain \(t = \frac{3}{2}\) and hence gradient is \(1\) | A1 | |
| 2 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $\frac{dx}{dt} = \frac{2}{2t+3}$ | B1 | |
| Use quotient rule, or equivalent, to find $\frac{dy}{dt}$ | M1 | |
| Obtain $\frac{dy}{dt} = \frac{2(2t+3) - 2(2t-3)}{(2t+3)^2}$ | A1 | OE |
| Divide to confirm $\frac{dy}{dx} = \frac{6}{2t+3}$ | A1 | AG |
| | **4** | |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to find value of $t$ corresponding to $x = 0$ | M1 | |
| Obtain $t = -1$ and hence gradient is $6$ | A1 | |
| | **2** | |
---
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to find value of $t$ corresponding to $y = 0$ | M1 | |
| Obtain $t = \frac{3}{2}$ and hence gradient is $1$ | A1 | |
| | **2** | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{83d0697c-b133-47da-a745-dfdafa7dbf10-08_663_433_260_854}
The diagram shows the curve with parametric equations
$$x = \ln ( 2 t + 3 ) , \quad y = \frac { 2 t - 3 } { 2 t + 3 } .$$
The curve crosses the $y$-axis at the point $A$ and the $x$-axis at the point $B$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 } { 2 t + 3 }$.
\item Find the gradient of the curve at $A$.
\item Find the gradient of the curve at $B$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2021 Q5 [8]}}