| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find vertical tangent points |
| Difficulty | Standard +0.8 This question requires implicit differentiation with exponential functions (moderately challenging), then finding where dy/dx is undefined by solving a system involving the original curve equation and the denominator condition. The algebraic manipulation to find exact coordinates from the simultaneous equations elevates this above routine implicit differentiation questions. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State correct derivative of \(ye^{2x}\) with respect to \(x\) | B1 | \(2ye^{2x} + e^{2x}\frac{dy}{dx}\) |
| State correct derivative of \(y^2e^x\) with respect to \(x\) | B1 | \(2ye^x\frac{dy}{dx} + y^2e^x\) |
| Equate attempted derivative of the LHS to zero and solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain \(\frac{dy}{dx} = \frac{2ye^x - y^2}{2y - e^x}\) | A1 | Obtain the given answer correctly. Condone multiplication by \(\frac{-1}{-1}\) and cancelling of \(e^x\) without comment |
| Alternative method: Rearrange as \(y = \frac{2}{e^{2x}-ye^x} \Rightarrow \frac{d}{dx}(e^{2x}-ye^x) = 2e^{2x} - ye^x - e^x\frac{dy}{dx}\) | B1 | Other rearrangements possible e.g. \(y = 2e^{-2x}+y^2e^{-x}\), \(\frac{d}{dx}(y^2e^{-x}) = 2ye^{-x}\frac{dy}{dx} - y^2e^{-x}\) |
| \(\frac{dy}{dx} = -\frac{2}{(e^{2x}-ye^x)^2} \times \left(2e^{2x} - ye^x - e^x\frac{dy}{dx}\right)\) | B1 | \(\Rightarrow \frac{dy}{dx} = -4e^{-x} + 2ye^{-x}\frac{dy}{dx} - y^2e^{-x}\) |
| Solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain \(\frac{dy}{dx} = \frac{2ye^x - y^2}{2y - e^x}\) | A1 | Obtain the given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equate denominator to zero and substitute for \(y\) or for \(e^x\) in the equation of the curve | *M1 | |
| Obtain equation of the form \(ae^{3x} = b\) or \(cy^3 = d\) | DM1 | \((e^{3x}=8,\quad y^3=1)\) SOI |
| Obtain \(x = \ln 2\) | A1 | Accept \(\frac{1}{3}\ln 8\) ISW |
| Obtain \(y = 1\) | A1 |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State correct derivative of $ye^{2x}$ with respect to $x$ | B1 | $2ye^{2x} + e^{2x}\frac{dy}{dx}$ |
| State correct derivative of $y^2e^x$ with respect to $x$ | B1 | $2ye^x\frac{dy}{dx} + y^2e^x$ |
| Equate attempted derivative of the LHS to zero and solve for $\frac{dy}{dx}$ | M1 | |
| Obtain $\frac{dy}{dx} = \frac{2ye^x - y^2}{2y - e^x}$ | A1 | Obtain the **given answer** correctly. Condone multiplication by $\frac{-1}{-1}$ and cancelling of $e^x$ without comment |
| **Alternative method:** Rearrange as $y = \frac{2}{e^{2x}-ye^x} \Rightarrow \frac{d}{dx}(e^{2x}-ye^x) = 2e^{2x} - ye^x - e^x\frac{dy}{dx}$ | B1 | Other rearrangements possible e.g. $y = 2e^{-2x}+y^2e^{-x}$, $\frac{d}{dx}(y^2e^{-x}) = 2ye^{-x}\frac{dy}{dx} - y^2e^{-x}$ |
| $\frac{dy}{dx} = -\frac{2}{(e^{2x}-ye^x)^2} \times \left(2e^{2x} - ye^x - e^x\frac{dy}{dx}\right)$ | B1 | $\Rightarrow \frac{dy}{dx} = -4e^{-x} + 2ye^{-x}\frac{dy}{dx} - y^2e^{-x}$ |
| Solve for $\frac{dy}{dx}$ | M1 | |
| Obtain $\frac{dy}{dx} = \frac{2ye^x - y^2}{2y - e^x}$ | A1 | Obtain the **given answer** correctly |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate denominator to zero and substitute for $y$ or for $e^x$ in the equation of the curve | *M1 | |
| Obtain equation of the form $ae^{3x} = b$ or $cy^3 = d$ | DM1 | $(e^{3x}=8,\quad y^3=1)$ SOI |
| Obtain $x = \ln 2$ | A1 | Accept $\frac{1}{3}\ln 8$ ISW |
| Obtain $y = 1$ | A1 | |
---9 The equation of a curve is $y \mathrm { e } ^ { 2 x } - y ^ { 2 } \mathrm { e } ^ { x } = 2$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y \mathrm { e } ^ { x } - y ^ { 2 } } { 2 y - \mathrm { e } ^ { x } }$.
\item Find the exact coordinates of the point on the curve where the tangent is parallel to the $y$-axis.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q9 [8]}}