| 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 | Show dy/dx equals given expression |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring product rule and chain rule application, followed by solving a cubic equation that factors nicely. While it involves multiple steps, the techniques are standard for P3 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| 5(i) | dy |
| Answer | Marks |
|---|---|
| dx | B1 |
| Answer | Marks |
|---|---|
| dx | B1 |
| Answer | Marks |
|---|---|
| dx | M1 |
| Obtain the given answer | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | Equate numerator to zero | *M1 |
| Obtain y = −2x, or equivalent | A1 | |
| Obtain an equation in x or y | DM1 | |
| Obtain final answer x = −1, y = 2 and x = 1, y = −2 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | dy
State or imply y3 +3xy2 as derivative of xy3
dx | B1
dy
State or imply 4y3 as derivative of y4
dx | B1
dy
Equate derivative of the LHS to zero and solve for
dx | M1
Obtain the given answer | A1
4
--- 5(ii) ---
5(ii) | Equate numerator to zero | *M1
Obtain y = −2x, or equivalent | A1
Obtain an equation in x or y | DM1
Obtain final answer x = −1, y = 2 and x = 1, y = −2 | A1
4
Question | Answer | MarksThe equation of a curve is $2x^4 + xy^3 + y^4 = 10$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{dy}{dx} = -\frac{8x^3 + y^3}{3xy^2 + 4y^3}$. [4]
\item Hence show that there are two points on the curve at which the tangent is parallel to the $x$-axis and find the coordinates of these points. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2017 Q5 [8]}}