| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Tangent with given gradient |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question with standard techniques. Part (a) is routine application of the chain rule to differentiate implicitly. Part (b) requires setting the derivative equal to 2 (from the given line) and solving simultaneously with the original equation—a multi-step process but using well-practiced methods with no novel insight required. Slightly above average due to the algebraic manipulation needed in part (b). |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(6y\frac{dy}{dx}\) as the derivative of \(3y^2\) | B1 | Allow \(y'\) for \(\frac{dy}{dx}\) throughout. Accept \(\frac{\partial f}{\partial x} = 6x + 4y\) |
| State or imply \(4x\frac{dy}{dx} + 4y\) as the derivative of \(4xy\) | B1 | Accept \(\frac{\partial f}{\partial y} = 4x + 6y\) |
| Equate derivative of LHS to zero and solve for \(\frac{dy}{dx}\) | M1 | Allow an extra \(\frac{dy}{dx}\) in front of their differentiated equation. Allow if \(= 0\) is implied but not seen. Allow \(\frac{dy}{dx} = -\dfrac{\partial f/\partial x}{\partial f/\partial y}\) |
| Obtain \(\frac{dy}{dx} = -\dfrac{3x+2y}{2x+3y}\) | A1 | AG – must come from correct working. The position of the negative must be clear. |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate \(\frac{dy}{dx}\) to \(-2\) and solve for \(x\) in terms of \(y\) or for \(y\) in terms of \(x\) | *M1 | Must be using the given derivative. |
| Obtain \(x = -4y\) or \(y = -\dfrac{x}{4}\) | A1 | Seen or implied by correct later work. |
| Substitute their \(x = -4y\) or their \(y = -\dfrac{x}{4}\) in curve equation | DM1 | Allow unsimplified. |
| Obtain \(y = \pm\dfrac{1}{\sqrt{7}}\) or \(x = \pm\dfrac{4}{\sqrt{7}}\) | A1 | Or exact equivalent. Or \(x = \dfrac{4}{\sqrt{7}}\) and \(y = -\dfrac{1}{\sqrt{7}}\) or exact equivalent. |
| Obtain both pairs of values | A1 | Or \(x = -\dfrac{4}{\sqrt{7}}\) and \(y = \dfrac{1}{\sqrt{7}}\) or exact equivalent. A1 A0 for incorrect final pairing. |
| 5 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $6y\frac{dy}{dx}$ as the derivative of $3y^2$ | B1 | Allow $y'$ for $\frac{dy}{dx}$ throughout. Accept $\frac{\partial f}{\partial x} = 6x + 4y$ |
| State or imply $4x\frac{dy}{dx} + 4y$ as the derivative of $4xy$ | B1 | Accept $\frac{\partial f}{\partial y} = 4x + 6y$ |
| Equate derivative of LHS to zero and solve for $\frac{dy}{dx}$ | M1 | Allow an extra $\frac{dy}{dx}$ in front of their differentiated equation. Allow if $= 0$ is implied but not seen. Allow $\frac{dy}{dx} = -\dfrac{\partial f/\partial x}{\partial f/\partial y}$ |
| Obtain $\frac{dy}{dx} = -\dfrac{3x+2y}{2x+3y}$ | A1 | AG – must come from correct working. The position of the negative must be clear. |
| | 4 | |
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## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate $\frac{dy}{dx}$ to $-2$ and solve for $x$ in terms of $y$ or for $y$ in terms of $x$ | *M1 | Must be using the given derivative. |
| Obtain $x = -4y$ or $y = -\dfrac{x}{4}$ | A1 | Seen or implied by correct later work. |
| Substitute their $x = -4y$ or their $y = -\dfrac{x}{4}$ in curve equation | DM1 | Allow unsimplified. |
| Obtain $y = \pm\dfrac{1}{\sqrt{7}}$ or $x = \pm\dfrac{4}{\sqrt{7}}$ | A1 | Or exact equivalent. Or $x = \dfrac{4}{\sqrt{7}}$ and $y = -\dfrac{1}{\sqrt{7}}$ or exact equivalent. |
| Obtain both pairs of values | A1 | Or $x = -\dfrac{4}{\sqrt{7}}$ and $y = \dfrac{1}{\sqrt{7}}$ or exact equivalent. A1 A0 for incorrect final pairing. |
| | 5 | |
---7 The equation of a curve is $3 x ^ { 2 } + 4 x y + 3 y ^ { 2 } = 5$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 3 x + 2 y } { 2 x + 3 y }$.
\item Hence find the exact coordinates of the two points on the curve at which the tangent is parallel to $y + 2 x = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q7 [9]}}