| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find constant from gradient condition |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question with straightforward algebraic manipulation. Part (a) requires routine application of the product rule and chain rule to find dy/dx, then matching coefficients. Part (b) involves using the point condition and gradient of the normal to set up two simultaneous equations. While multi-step, all techniques are standard A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{d}{dx}(3y^2) = 6y\dfrac{dy}{dx}\) or \(\dfrac{d}{dx}(qxy) = qx\dfrac{dy}{dx} + qy\) | M1 | Selecting appropriate method. Allow \(3y^2\to\alpha y\dfrac{dy}{dx}\) or \(qxy\to\alpha x\dfrac{dy}{dx}+\beta y\) |
| \(3px^2 + qx\dfrac{dy}{dx} + qy + 6y\dfrac{dy}{dx} = 0\) | A1 | Fully correct differentiation; ignore spurious \(\dfrac{dy}{dx}=\ldots\) |
| \((qx+6y)\dfrac{dy}{dx} = -3px^2 - qy \Rightarrow \dfrac{dy}{dx} = \ldots\) | dM1 | Valid attempt to make \(\dfrac{dy}{dx}\) the subject with 2 terms in \(\dfrac{dy}{dx}\). Depends on first M1 |
| \(\dfrac{dy}{dx} = \dfrac{-3px^2 - qy}{qx + 6y}\) | A1 | Fully correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(p(-1)^3 + q(-1)(-4) + 3(-4)^2 = 26\) | M1 | Uses \(x=-1\), \(y=-4\) in equation of \(C\) to get equation in \(p\) and \(q\) |
| \(19x + 26y + 123 = 0 \Rightarrow m = -\dfrac{19}{26}\) | B1 | Deduces correct gradient of normal (may be implied) |
| \(\dfrac{-3p(-1)^2 - q(-4)}{q(-1)+6(-4)} = \dfrac{26}{19}\) or \(\dfrac{q(-1)+6(-4)}{-3p(-1)^2+q(-4)} = -\dfrac{19}{26}\) | M1 | Fully correct strategy to get equation connecting \(p\) and \(q\) using \(x=-1\), \(y=-4\) in \(\dfrac{dy}{dx}\) and gradient of normal |
| \(p - 4q = 22,\quad 57p - 102q = 624\) | dM1 | Solves simultaneously. Depends on both previous M marks |
| \(p = 2,\quad q = -5\) | A1 | Correct values |
## Question 8:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{d}{dx}(3y^2) = 6y\dfrac{dy}{dx}$ or $\dfrac{d}{dx}(qxy) = qx\dfrac{dy}{dx} + qy$ | M1 | Selecting appropriate method. Allow $3y^2\to\alpha y\dfrac{dy}{dx}$ or $qxy\to\alpha x\dfrac{dy}{dx}+\beta y$ |
| $3px^2 + qx\dfrac{dy}{dx} + qy + 6y\dfrac{dy}{dx} = 0$ | A1 | Fully correct differentiation; ignore spurious $\dfrac{dy}{dx}=\ldots$ |
| $(qx+6y)\dfrac{dy}{dx} = -3px^2 - qy \Rightarrow \dfrac{dy}{dx} = \ldots$ | dM1 | Valid attempt to make $\dfrac{dy}{dx}$ the subject with 2 terms in $\dfrac{dy}{dx}$. **Depends on first M1** |
| $\dfrac{dy}{dx} = \dfrac{-3px^2 - qy}{qx + 6y}$ | A1 | Fully correct expression |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $p(-1)^3 + q(-1)(-4) + 3(-4)^2 = 26$ | M1 | Uses $x=-1$, $y=-4$ in equation of $C$ to get equation in $p$ and $q$ |
| $19x + 26y + 123 = 0 \Rightarrow m = -\dfrac{19}{26}$ | B1 | Deduces correct gradient of normal (may be implied) |
| $\dfrac{-3p(-1)^2 - q(-4)}{q(-1)+6(-4)} = \dfrac{26}{19}$ or $\dfrac{q(-1)+6(-4)}{-3p(-1)^2+q(-4)} = -\dfrac{19}{26}$ | M1 | Fully correct strategy to get equation connecting $p$ and $q$ using $x=-1$, $y=-4$ in $\dfrac{dy}{dx}$ and gradient of normal |
| $p - 4q = 22,\quad 57p - 102q = 624$ | dM1 | Solves simultaneously. **Depends on both previous M marks** |
| $p = 2,\quad q = -5$ | A1 | Correct values |\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$p x ^ { 3 } + q x y + 3 y ^ { 2 } = 26$$
where $p$ and $q$ are constants.\\
(a) Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { a p x ^ { 2 } + b q y } { q x + c y }$$
where $a$, $b$ and $c$ are integers to be found.
Given that
\begin{itemize}
\item the point $P ( - 1 , - 4 )$ lies on $C$
\item the normal to $C$ at $P$ has equation $19 x + 26 y + 123 = 0$\\
(b) find the value of $p$ and the value of $q$.
\end{itemize}
\hfill \mbox{\textit{Edexcel Paper 2 2021 Q8 [9]}}