| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard technique to find dy/dx, then solving dy/dx=0 for stationary points. The algebra is routine (factoring 3x²+6x=0 gives x=0,-2, then substituting back), making it slightly easier than average but still requiring multiple steps and careful execution. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(2y\frac{dy}{dx}\) as the derivative of \(y^2\) | B1 | Allow for \(3x^2dx + 2ydy\) or \(F_x = 3x^2+6x\) and \(F_y = 2y+3\). |
| Equate derivative of LHS to zero and solve for \(\frac{dy}{dx}\) | M1 | \(3x^2+2y\frac{dy}{dx}+6x+3\frac{dy}{dx}=0\) or \(3x^2dx+2ydy+6xdx+3dy=0\) or \(\frac{dy}{dx}=-\frac{F_x}{F_y}\) need evidence from B1 mark or formula must be seen. Allow errors. |
| Obtain the given answer | A1 | AG \(\frac{dy}{dx}=-\frac{3x^2+6x}{2y+3}\) not \(\frac{-3x^2-6x}{2y+3}\). Must factorise with \(\frac{dy}{dx}\) e.g. \(3x^2+6x+\frac{dy}{dx}(2y+3)=0\) or \(3x^2dx+6xdx+dy(2y+3)=0\). |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equate numerator to zero and solve for \(x\) | *M1 | Allow for just one \(x\) value. |
| Obtain \(x=0\) and \(x=-2\) only | A1 | |
| Substitute their \(x\), [\(x=0\) or \(x=-2\)] in curve equation to obtain quadratic equation in \(y\) equal to 0 | DM1 | \(y^2+3y-4=0\) or \(y^2+3y=0\). |
| Obtain \(y=1\) and \(y=-4\) [when \(x=0\)] | A1 | |
| Obtain \(y=0\) and \(y=-3\) [when \(x=-2\)] | A1 | ISW If forget \(x=0\) then max 3/5. |
| Total | 5 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $2y\frac{dy}{dx}$ as the derivative of $y^2$ | B1 | Allow for $3x^2dx + 2ydy$ or $F_x = 3x^2+6x$ and $F_y = 2y+3$. |
| Equate derivative of LHS to zero and solve for $\frac{dy}{dx}$ | M1 | $3x^2+2y\frac{dy}{dx}+6x+3\frac{dy}{dx}=0$ or $3x^2dx+2ydy+6xdx+3dy=0$ or $\frac{dy}{dx}=-\frac{F_x}{F_y}$ need evidence from B1 mark or formula must be seen. Allow errors. |
| Obtain the given answer | A1 | AG $\frac{dy}{dx}=-\frac{3x^2+6x}{2y+3}$ not $\frac{-3x^2-6x}{2y+3}$. Must factorise with $\frac{dy}{dx}$ e.g. $3x^2+6x+\frac{dy}{dx}(2y+3)=0$ or $3x^2dx+6xdx+dy(2y+3)=0$. |
| **Total** | **3** | |
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## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate numerator to zero and solve for $x$ | *M1 | Allow for just one $x$ value. |
| Obtain $x=0$ and $x=-2$ only | A1 | |
| Substitute their $x$, [$x=0$ or $x=-2$] in curve equation to obtain quadratic equation in $y$ equal to 0 | DM1 | $y^2+3y-4=0$ or $y^2+3y=0$. |
| Obtain $y=1$ and $y=-4$ [when $x=0$] | A1 | |
| Obtain $y=0$ and $y=-3$ [when $x=-2$] | A1 | ISW If forget $x=0$ then max 3/5. |
| **Total** | **5** | |
---7 The equation of a curve is $x ^ { 3 } + y ^ { 2 } + 3 x ^ { 2 } + 3 y = 4$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 3 x ^ { 2 } + 6 x } { 2 y + 3 }$.
\item Hence find the coordinates of the points on the curve at which the tangent is parallel to the $x$-axis. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q7 [8]}}