| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show no stationary points exist |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring product rule and chain rule application, followed by routine tangent equation finding and analyzing when dy/dx = 0. Part (c) requires showing the numerator cannot be zero (since e^(2x) and y are always positive on the curve), which is a simple logical step once the derivative is found. Slightly above average due to the implicit differentiation and the 'show that' proof elements, but all techniques are standard P2 material. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use product rule to differentiate \(2e^{2x}y\) | M1 | Must be in the form \(k_1 y e^{2x} + k_2 e^{2x}\dfrac{dy}{dx}\) |
| Obtain \(4e^{2x}y + 2e^{2x}\dfrac{dy}{dx}\) | A1 | |
| Differentiate \(-y^3\) to obtain \(-3y^2\dfrac{dy}{dx}\) | B1 | |
| Obtain \(\dfrac{dy}{dx} = \dfrac{4e^{2x}y}{3y^2 - 2e^{2x}}\) | A1 | AG |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(0\) and \(2\) to find gradient of tangent | M1 | |
| Attempt to find equation of tangent through \((0, 2)\) with numerical gradient | M1 | |
| Obtain \(4x - 5y + 10 = 0\) or equivalent of required form | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equate numerator of derivative to zero and state that at least one of \(e^{2x}\) and \(y\) cannot be zero | M1 | |
| Complete argument | A1 | |
| Total | 2 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use product rule to differentiate $2e^{2x}y$ | M1 | Must be in the form $k_1 y e^{2x} + k_2 e^{2x}\dfrac{dy}{dx}$ |
| Obtain $4e^{2x}y + 2e^{2x}\dfrac{dy}{dx}$ | A1 | |
| Differentiate $-y^3$ to obtain $-3y^2\dfrac{dy}{dx}$ | B1 | |
| Obtain $\dfrac{dy}{dx} = \dfrac{4e^{2x}y}{3y^2 - 2e^{2x}}$ | A1 | AG |
| **Total** | **4** | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $0$ and $2$ to find gradient of tangent | M1 | |
| Attempt to find equation of tangent through $(0, 2)$ with numerical gradient | M1 | |
| Obtain $4x - 5y + 10 = 0$ or equivalent of required form | A1 | |
| **Total** | **3** | |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate numerator of derivative to zero and state that at least one of $e^{2x}$ and $y$ cannot be zero | M1 | |
| Complete argument | A1 | |
| **Total** | **2** | |
---5 The equation of a curve is $2 \mathrm { e } ^ { 2 x } y - y ^ { 3 } + 4 = 0$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 \mathrm { e } ^ { 2 x } y } { 3 y ^ { 2 } - 2 \mathrm { e } ^ { 2 x } }$.
\item The curve passes through the point $( 0,2 )$.
Find the equation of the tangent to the curve at this point, giving your answer in the form $a x + b y + c = 0$.
\item Show that the curve has no stationary points.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2020 Q5 [9]}}