| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | March |
| Marks | 7 |
| 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 standard application of the product rule and chain rule, followed by algebraic rearrangement. Part (b) requires setting the numerator to zero and showing no real solutions exist, which is routine problem-solving for this topic. Slightly easier than average due to the clear structure and standard techniques. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(4y\frac{dy}{dx}\) as the derivative of \(2y^2\) | B1 | SC If \(\frac{dy}{dx}\) introduced instead of \(\frac{d}{dx}\) then allow B1 for both, followed by correct method M1 Max 2 |
| State or imply \(3y + 3x\frac{dy}{dx}\) as the derivative of \(3xy\) | B1 | Allow extra \(\frac{dy}{dx}\) = correct expression to collect all marks if correct |
| Complete the differentiation, all 4 terms, isolate 2 \(\frac{dy}{dx}\) terms on LHS or bracket \(\frac{dy}{dx}\) terms and solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain \(\frac{dy}{dx} = \frac{2x - 3y - 1}{4y + 3x}\) | A1 | Answer given – need to have seen \(4y\frac{dy}{dx} + 3x\frac{dy}{dx} = 2x - 3y - 1\) or \((4y+3x)\frac{dy}{dx} - 2x + 3y = -1\). Need to see \(= 2x\) or \(= 0\) consistently throughout otherwise M1 A0. No recovery allowed. When all terms are included then must be an equation |
| Allow all marks if using \(dx\) and \(dy\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate numerator to zero, obtaining \(2x = 3y + 1\) or \(3y = 2x - 1\) and form equation in \(x\) only or \(y\) only from \(2y^2 + 3xy + x = x^2\) | M1* | e.g. \(\frac{2}{9}(2x-1)^2 + x(2x-1) + x = x^2\) or \(2y^2 + \frac{3}{2}(1+3y)y + \frac{1}{2}(1+3y) = \frac{1}{4}(1+3y)^2\). Allow errors |
| Obtain \(\frac{2}{9}(2x-1)^2 = -x^2\) or a 3 term quadratic in one unknown and try to solve. If errors in quadratic formulation allow solution, applying usual rules for solution of quadratic equation, and allow M1 | DM1 | e.g. \(17x^2 - 8x + 2 = 0\) \((b^2 - 4ac = -72)\) or \(17y^2 + 6y + 1 = 0\) \((b^2 - 4ac = -32)\). \(x = 4/17 \pm (3\sqrt{2}/17)\)i, \(y = -3/17 \pm (2\sqrt{2}/17)\)i |
| Conclude that the equation has no [real] roots | A1 | Given Answer. CWO |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $4y\frac{dy}{dx}$ as the derivative of $2y^2$ | B1 | **SC** If $\frac{dy}{dx}$ introduced instead of $\frac{d}{dx}$ then allow B1 for both, followed by correct method M1 Max 2 |
| State or imply $3y + 3x\frac{dy}{dx}$ as the derivative of $3xy$ | B1 | Allow extra $\frac{dy}{dx}$ = correct expression to collect all marks if correct |
| Complete the differentiation, all 4 terms, isolate 2 $\frac{dy}{dx}$ terms on LHS or bracket $\frac{dy}{dx}$ terms and solve for $\frac{dy}{dx}$ | M1 | |
| Obtain $\frac{dy}{dx} = \frac{2x - 3y - 1}{4y + 3x}$ | A1 | Answer given – need to have seen $4y\frac{dy}{dx} + 3x\frac{dy}{dx} = 2x - 3y - 1$ or $(4y+3x)\frac{dy}{dx} - 2x + 3y = -1$. Need to see $= 2x$ or $= 0$ consistently throughout otherwise M1 A0. No recovery allowed. When all terms are included then must be an equation |
| | | Allow all marks if using $dx$ and $dy$ |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate numerator to zero, obtaining $2x = 3y + 1$ or $3y = 2x - 1$ and form equation in $x$ only or $y$ only from $2y^2 + 3xy + x = x^2$ | M1* | e.g. $\frac{2}{9}(2x-1)^2 + x(2x-1) + x = x^2$ or $2y^2 + \frac{3}{2}(1+3y)y + \frac{1}{2}(1+3y) = \frac{1}{4}(1+3y)^2$. Allow errors |
| Obtain $\frac{2}{9}(2x-1)^2 = -x^2$ or a 3 term quadratic in one unknown and try to solve. If errors in quadratic formulation allow solution, applying usual rules for solution of quadratic equation, and allow M1 | DM1 | e.g. $17x^2 - 8x + 2 = 0$ $(b^2 - 4ac = -72)$ or $17y^2 + 6y + 1 = 0$ $(b^2 - 4ac = -32)$. $x = 4/17 \pm (3\sqrt{2}/17)$i, $y = -3/17 \pm (2\sqrt{2}/17)$i |
| Conclude that the equation has no [real] roots | A1 | Given Answer. CWO |6 The equation of a curve is $2 y ^ { 2 } + 3 x y + x = x ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 2 \mathrm { x } - 3 \mathrm { y } - 1 } { 4 \mathrm { y } + 3 \mathrm { x } }$.
\item Hence show that the curve does not have a tangent that is parallel to the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q6 [7]}}