| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2017 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.8 This question requires implicit differentiation with product rule (moderate complexity), then solving dy/dx = 0 which leads to a cubic relationship y³ = x²y. The algebraic manipulation to find stationary points and verifying only two exist requires careful reasoning beyond routine differentiation exercises, placing it moderately above average difficulty. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| State or imply \(3x^2y + x^3\frac{dy}{dx}\) as derivative of \(x^3y\) | B1 | |
| State or imply \(9xy^2\frac{dy}{dx} + 3y^3\) as derivative of \(3xy^3\) | B1 | |
| Equate derivative of the LHS to zero and solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain the given answer | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| Equate numerator to zero and use \(x = -y\) to obtain an equation in \(x\) or in \(y\) | M1 | |
| Obtain answer \(x = a\) and \(y = -a\) | A1 | |
| Obtain answer \(x = -a\) and \(y = a\) | A1 | |
| Consider and reject \(y = 0\) and \(x = y\) as possibilities | B1 |
## Question 6(i):
| Answer | Mark | Notes |
|--------|------|-------|
| State or imply $3x^2y + x^3\frac{dy}{dx}$ as derivative of $x^3y$ | B1 | |
| State or imply $9xy^2\frac{dy}{dx} + 3y^3$ as derivative of $3xy^3$ | B1 | |
| Equate derivative of the LHS to zero and solve for $\frac{dy}{dx}$ | M1 | |
| Obtain the given answer | A1 | AG |
## Question 6(ii):
| Answer | Mark | Notes |
|--------|------|-------|
| Equate numerator to zero and use $x = -y$ to obtain an equation in $x$ or in $y$ | M1 | |
| Obtain answer $x = a$ and $y = -a$ | A1 | |
| Obtain answer $x = -a$ and $y = a$ | A1 | |
| Consider and reject $y = 0$ and $x = y$ as possibilities | B1 | |6 The equation of a curve is $x ^ { 3 } y - 3 x y ^ { 3 } = 2 a ^ { 4 }$, where $a$ is a non-zero constant.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 x ^ { 2 } y - 3 y ^ { 3 } } { 9 x y ^ { 2 } - x ^ { 3 } }$.\\
(ii) Hence show that there are only two points on the curve at which the tangent is parallel to the $x$-axis and find the coordinates of these points.\\
\hfill \mbox{\textit{CAIE P3 2017 Q6 [8]}}