| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Parametric curve crosses axis, find gradient there |
| Difficulty | Standard +0.3 This is a standard parametric differentiation question with routine iterative method application. Part (a) requires finding dy/dx using the chain rule (standard technique), part (b) involves algebraic manipulation to reach a given equation (shown result), and part (c) is straightforward iteration with a given formula. All techniques are core P2 syllabus with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Obtain \(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{6}{2t-3}\) | B1 | |
| Use product rule to find \(\frac{\mathrm{d}y}{\mathrm{d}t}\) | M1 | Allow unsimplified |
| Obtain \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(4\ln t+4)(2t-3)}{6}\) | A1 | OE |
| Attempt to find \(t\) corresponding to point \(A\) using a complete and correct method | M1 | |
| Obtain \(t=2\) and hence gradient is \(\frac{2}{3}\ln 2+\frac{2}{3}\) | A1 | Or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equate \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to 12 and attempt rearrangement to \(2t-3=\frac{k}{4\ln t+4}\) | M1 | |
| Confirm given result \(t=\frac{9}{1+\ln t}+\frac{3}{2}\) with sufficient detail | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use iteration process correctly at least once | M1 | Need to see 4.9626 |
| Obtain final answer 4.96 | A1 | Answer required to exactly 3 s.f. |
| Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval \([4.955, 4.965]\) | A1 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain $\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{6}{2t-3}$ | B1 | |
| Use product rule to find $\frac{\mathrm{d}y}{\mathrm{d}t}$ | M1 | Allow unsimplified |
| Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(4\ln t+4)(2t-3)}{6}$ | A1 | OE |
| Attempt to find $t$ corresponding to point $A$ using a complete and correct method | M1 | |
| Obtain $t=2$ and hence gradient is $\frac{2}{3}\ln 2+\frac{2}{3}$ | A1 | Or exact equivalent |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate $\frac{\mathrm{d}y}{\mathrm{d}x}$ to 12 and attempt rearrangement to $2t-3=\frac{k}{4\ln t+4}$ | M1 | |
| Confirm given result $t=\frac{9}{1+\ln t}+\frac{3}{2}$ with sufficient detail | A1 | AG |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use iteration process correctly at least once | M1 | Need to see 4.9626 |
| Obtain final answer 4.96 | A1 | Answer required to exactly 3 s.f. |
| Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval $[4.955, 4.965]$ | A1 | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{b104e2a7-06c8-4e2e-a4f9-5095ad56897a-10_803_394_269_863}
The diagram shows the curve with parametric equations
$$x = 3 \ln ( 2 t - 3 ) , \quad y = 4 t \ln t$$
The curve crosses the $y$-axis at the point $A$. At the point $B$, the gradient of the curve is 12 .
\begin{enumerate}[label=(\alph*)]
\item Find the exact gradient of the curve at $A$.
\item Show that the value of the parameter $t$ at $B$ satisfies the equation
$$t = \frac { 9 } { 1 + \ln t } + \frac { 3 } { 2 }$$
\item Use an iterative formula, based on the equation in (b), to find the value of $t$ at $B$, giving your answer correct to 3 significant figures. Use an initial value of 5 and give the result of each iteration to 5 significant figures.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q6 [10]}}