| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | March |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Iterative method for special point |
| Difficulty | Standard +0.8 This question combines implicit differentiation (standard P2 topic) with an iterative method to find a special point where the tangent is vertical. Parts (a)-(c) require careful algebraic manipulation and understanding of vertical tangents (dx/dy = 0), while part (d) is routine iteration. The conceptual leap to recognize vertical tangent conditions and derive the iterative formula elevates this above typical textbook exercises, though the individual techniques are all standard A-level. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use product rule to differentiate \(e^{2x}y\) | M1 | |
| Obtain \(2e^{2x}y + e^{2x}\frac{dy}{dx}\) | A1 | |
| Obtain \(2e^{2x}y + e^{2x}\frac{dy}{dx} - e^y\frac{dy}{dx} = 0\) and rearrange to confirm given result | A1 | AG – necessary detail needed |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Consider \(e^{2x}y = 0\) and either state \(e^{2x} \neq 0\) or substitute \(y = 0\) in equation of curve | M1 | |
| Complete argument with \(e^{2x} \neq 0\) or \(e^{2x} > 0\) and substitution to show \(y\) cannot be zero | A1 | AG – necessary detail needed |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(e^y - e^{2x} = 0\) and hence \(y = 2x\) | B1 | |
| Substitute for \(y\) in equation of curve and attempt rearrangement as far as \(e^{2x} = ...\) | M1 | |
| Use relevant logarithm properties | M1 | |
| Confirm equation \(x = \ln 10 - \frac{1}{2}\ln(2x-1)\) | A1 | AG – necessary detail needed |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use iteration process correctly at least once | M1 | |
| Obtain final answer 1.82 | A1 | Answer required to exactly 3 sf |
| Show sufficient iterations to 5 sf to justify answer or show sign change in interval \([1.815, 1.825]\) | A1 | |
| Total | 3 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use product rule to differentiate $e^{2x}y$ | M1 | |
| Obtain $2e^{2x}y + e^{2x}\frac{dy}{dx}$ | A1 | |
| Obtain $2e^{2x}y + e^{2x}\frac{dy}{dx} - e^y\frac{dy}{dx} = 0$ and rearrange to confirm given result | A1 | AG – necessary detail needed |
| **Total** | **3** | |
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Consider $e^{2x}y = 0$ and either state $e^{2x} \neq 0$ or substitute $y = 0$ in equation of curve | M1 | |
| Complete argument with $e^{2x} \neq 0$ or $e^{2x} > 0$ and substitution to show $y$ cannot be zero | A1 | AG – necessary detail needed |
| **Total** | **2** | |
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $e^y - e^{2x} = 0$ and hence $y = 2x$ | B1 | |
| Substitute for $y$ in equation of curve and attempt rearrangement as far as $e^{2x} = ...$ | M1 | |
| Use relevant logarithm properties | M1 | |
| Confirm equation $x = \ln 10 - \frac{1}{2}\ln(2x-1)$ | A1 | AG – necessary detail needed |
| **Total** | **4** | |
## Question 7(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use iteration process correctly at least once | M1 | |
| Obtain final answer 1.82 | A1 | Answer required to exactly 3 sf |
| Show sufficient iterations to 5 sf to justify answer or show sign change in interval $[1.815, 1.825]$ | A1 | |
| **Total** | **3** | |7 A curve has equation $\mathrm { e } ^ { 2 x } y - \mathrm { e } ^ { y } = 100$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 \mathrm { e } ^ { 2 x } y } { \mathrm { e } ^ { y } - \mathrm { e } ^ { 2 x } }$.
\item Show that the curve has no stationary points.\\
It is required to find the $x$-coordinate of $P$, the point on the curve at which the tangent is parallel to the $y$-axis.
\item Show that the $x$-coordinate of $P$ satisfies the equation
$$x = \ln 10 - \frac { 1 } { 2 } \ln ( 2 x - 1 )$$
\item Use an iterative formula, based on the equation in part (c), to find the $x$-coordinate of $P$ correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 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.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2022 Q7 [12]}}