| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find vertical tangent points |
| Difficulty | Standard +0.8 This question requires implicit differentiation using the quotient rule, then finding where dx/dy is undefined (vertical tangents). While the differentiation is standard P3 content, recognizing that vertical tangents occur when dx/dy → ∞ (denominator = 0) and solving the resulting quadratic requires solid conceptual understanding beyond routine exercises. The algebraic simplification and interpretation elevate this above average difficulty. |
| Spec | 1.02y Partial fractions: decompose rational functions1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = \frac{2y^2+6}{3y-3} \Rightarrow \frac{dx}{dy} = \frac{4y(3y-3)-3(2y^2+6)}{(3y-3)^2}\) | M1 A1 | Attempts quotient rule; look for \(\frac{Ay(3y-3)-B(2y^2+6)}{(3y-3)^2}\), \(A,B>0\) |
| \(\frac{dx}{dy} = \frac{6y^2-12y-18}{9(y-1)^2} = \frac{2y^2-4y-6}{3(y-1)^2}\) | dM1, A1 | Requires attempt to get single fraction with simplification; common factor 3 must be cancelled |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P\) and \(Q\) where \(\frac{dx}{dy}=0\) or where \(2y^2-4y-6=0\) | B1 | Indicates \(P\) and \(Q\) are where \(\frac{dx}{dy}=0\) |
| \(2y^2-4y-6=0 \Rightarrow 2(y-3)(y+1)=0 \Rightarrow y=3,-1\) | M1 | Solves their 3TQ from attempt at \(\frac{dx}{dy}=0\) |
| Subs \(y=-1\) and \(y=3\) in \(x=\frac{2y^2+6}{3y-3}\) | dM1 | Substitutes found \(y\) values back into \(x\) expression |
| \(x = -\frac{4}{3}\) and \(x=4\) | A1cso | Both values correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(x = \frac{2y^2+6}{3y-3} \Rightarrow 3xy - 3x = 2y^2 + 6 \Rightarrow 3x + 3y\frac{dx}{dy} - 3\frac{dx}{dy} = 4y\) | M1 A1 | Implicit differentiation of \(3xy - 3x = 2y^2 + 6\) |
| \(\frac{dx}{dy} = \frac{4y-3x}{3(y-1)}\) | dM1, A1 | Correct rearrangement to single fraction form |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(x = \frac{2y^2+6}{3y-3} = \frac{2y}{3} + \frac{2}{3} + \frac{8}{3(y-1)} \Rightarrow \frac{dx}{dy} = \frac{2}{3} - \frac{8}{3(y-1)^2}\) | M1 A1 | Long division or equivalent method to achieve \(Ay + B + \frac{C}{3y-3}\), then differentiate |
| \(\frac{dx}{dy} = \frac{2(y-1)^2 - 8}{3(y-1)^2} = \frac{2y^2-4y-6}{3(y-1)^2}\) oe | dM1, A1 | Single fraction, simplify numerator to 3TQ or use difference of squares |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| States \(P\) and \(Q\) are where \(\frac{dx}{dy} = 0\) or where \(4y - 3x = 0\) | B1 | Must state condition clearly |
| \(\Rightarrow \frac{4}{3}y = \frac{2y^2+6}{3y-3} \Rightarrow 4y^2 - 4y = 2y^2 + 6 \Rightarrow 2y^2 - 4y - 6 = 0\) | M1 | Substitution leading to quadratic (main scheme) |
| \(x = -\frac{4}{3}\) and \(x = 4\) only | A1cso | Must be equations not just values; must come from correct derivative; must be exact |
## Question 10:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \frac{2y^2+6}{3y-3} \Rightarrow \frac{dx}{dy} = \frac{4y(3y-3)-3(2y^2+6)}{(3y-3)^2}$ | M1 A1 | Attempts quotient rule; look for $\frac{Ay(3y-3)-B(2y^2+6)}{(3y-3)^2}$, $A,B>0$ |
| $\frac{dx}{dy} = \frac{6y^2-12y-18}{9(y-1)^2} = \frac{2y^2-4y-6}{3(y-1)^2}$ | dM1, A1 | Requires attempt to get single fraction with simplification; common factor 3 must be cancelled |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P$ and $Q$ where $\frac{dx}{dy}=0$ or where $2y^2-4y-6=0$ | B1 | Indicates $P$ and $Q$ are where $\frac{dx}{dy}=0$ |
| $2y^2-4y-6=0 \Rightarrow 2(y-3)(y+1)=0 \Rightarrow y=3,-1$ | M1 | Solves their 3TQ from attempt at $\frac{dx}{dy}=0$ |
| Subs $y=-1$ and $y=3$ in $x=\frac{2y^2+6}{3y-3}$ | dM1 | Substitutes found $y$ values back into $x$ expression |
| $x = -\frac{4}{3}$ and $x=4$ | A1cso | Both values correct |
## Question (Part a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $x = \frac{2y^2+6}{3y-3} \Rightarrow 3xy - 3x = 2y^2 + 6 \Rightarrow 3x + 3y\frac{dx}{dy} - 3\frac{dx}{dy} = 4y$ | M1 A1 | Implicit differentiation of $3xy - 3x = 2y^2 + 6$ |
| $\frac{dx}{dy} = \frac{4y-3x}{3(y-1)}$ | dM1, A1 | Correct rearrangement to single fraction form |
**(Alt a):**
| Working/Answer | Marks | Guidance |
|---|---|---|
| $x = \frac{2y^2+6}{3y-3} = \frac{2y}{3} + \frac{2}{3} + \frac{8}{3(y-1)} \Rightarrow \frac{dx}{dy} = \frac{2}{3} - \frac{8}{3(y-1)^2}$ | M1 A1 | Long division or equivalent method to achieve $Ay + B + \frac{C}{3y-3}$, then differentiate |
| $\frac{dx}{dy} = \frac{2(y-1)^2 - 8}{3(y-1)^2} = \frac{2y^2-4y-6}{3(y-1)^2}$ oe | dM1, A1 | Single fraction, simplify numerator to 3TQ or use difference of squares |
---
## Question (Part b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| States $P$ and $Q$ are where $\frac{dx}{dy} = 0$ or where $4y - 3x = 0$ | B1 | Must state condition clearly |
| $\Rightarrow \frac{4}{3}y = \frac{2y^2+6}{3y-3} \Rightarrow 4y^2 - 4y = 2y^2 + 6 \Rightarrow 2y^2 - 4y - 6 = 0$ | M1 | Substitution leading to quadratic (main scheme) |
| $x = -\frac{4}{3}$ and $x = 4$ only | A1cso | Must be equations not just values; must come from correct derivative; must be exact |10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bef290fb-fbac-4c9c-981e-5e323ac7182e-30_719_876_246_598}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the curve with equation
$$x = \frac { 2 y ^ { 2 } + 6 } { 3 y - 3 }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } x } { \mathrm {~d} y }$ giving your answer as a fully simplified fraction.
The tangents at points $P$ and $Q$ on the curve are parallel to the $y$-axis, as shown in Figure 4.
\item Use the answer to part (a) to find the equations of these two tangents.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2023 Q10 [8]}}