| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| 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 with standard techniques. Part (a) requires routine application of implicit differentiation rules, and part (b) involves setting dy/dx = 0 and solving the resulting system—both are textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(6xy+3x^2\dfrac{dy}{dx}\) as derivative of \(3x^2y\) | B1 | Allow B1 B1 for \((3x^2dx+)6xy\,dx+3x^2dy-3y^2dy[=0]\) |
| State or imply \(3y^2\dfrac{dy}{dx}\) as derivative of \(y^3\) | B1 | |
| Equate attempted derivative of LHS to zero and solve to obtain equation with \(\dfrac{dy}{dx}\) as subject | M1 | Allow if zero implied by subsequent working |
| Obtain \(\dfrac{dy}{dx}=\dfrac{x^2+2xy}{y^2-x^2}\) correctly | A1 | AG. Accept \(y'\) for \(\dfrac{dy}{dx}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate numerator to zero | \*M1 | Must be using the given derivative |
| Obtain \(x=-2y\), or equivalent | A1 | An equation with \(x\) or \(y\) as subject SOI |
| Use \(x^3+3x^2y-y^3=3\) to obtain an equation in \(x\) or \(y\) | DM1 | \(-8y^3+12y^3-y^3=3\) or \(x^3-\frac{3}{2}x^3+\frac{1}{8}x^3=3\) or equivalent |
| Obtain the point \((-2,1)\) and no others from solving cubic equation | A1 | Allow if each component stated separately. ISW |
| State the point \((0,-\sqrt[3]{3})\), or equivalent from correct work | B1 | Accept \((0,-1.44)\); allow if each component stated separately. ISW |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $6xy+3x^2\dfrac{dy}{dx}$ as derivative of $3x^2y$ | B1 | Allow B1 B1 for $(3x^2dx+)6xy\,dx+3x^2dy-3y^2dy[=0]$ |
| State or imply $3y^2\dfrac{dy}{dx}$ as derivative of $y^3$ | B1 | |
| Equate attempted derivative of LHS to zero and solve to obtain equation with $\dfrac{dy}{dx}$ as subject | M1 | Allow if zero implied by subsequent working |
| Obtain $\dfrac{dy}{dx}=\dfrac{x^2+2xy}{y^2-x^2}$ correctly | A1 | AG. Accept $y'$ for $\dfrac{dy}{dx}$ |
**Total: 4 marks**
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## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate numerator to zero | \*M1 | Must be using the given derivative |
| Obtain $x=-2y$, or equivalent | A1 | An equation with $x$ or $y$ as subject SOI |
| Use $x^3+3x^2y-y^3=3$ to obtain an equation in $x$ or $y$ | DM1 | $-8y^3+12y^3-y^3=3$ or $x^3-\frac{3}{2}x^3+\frac{1}{8}x^3=3$ or equivalent |
| Obtain the point $(-2,1)$ and no others from solving cubic equation | A1 | Allow if each component stated separately. ISW |
| State the point $(0,-\sqrt[3]{3})$, or equivalent from correct work | B1 | Accept $(0,-1.44)$; allow if each component stated separately. ISW |
**Total: 5 marks**
---7 The equation of a curve is $x ^ { 3 } + 3 x ^ { 2 } y - y ^ { 3 } = 3$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 2 } + 2 x y } { y ^ { 2 } - x ^ { 2 } }$.
\item Find the coordinates of the points on the curve where the tangent is parallel to the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q7 [9]}}