| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring product rule and collecting terms, followed by substituting a point. The algebra is slightly involved but follows standard procedures with no novel insight required. Slightly easier than average due to being a routine application of a well-practiced technique. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3y^2-11x^2+11xy = 20y-36x+28\) \(\Rightarrow 6y\dfrac{dy}{dx}-22x+11x\dfrac{dy}{dx}+11y = 20\dfrac{dy}{dx}-36\) | M1 M1 A1 | M1: \(y^2 \to Ay\dfrac{dy}{dx}\); M1: \(11xy \to px\dfrac{dy}{dx}+qy\); A1: All correct |
| \((6y+11x-20)\dfrac{dy}{dx} = 22x-11y-36 \Rightarrow \dfrac{dy}{dx} = ...\) | M1 | Collects terms in \(\dfrac{dy}{dx}\) (must be 3 and from appropriate terms) and makes \(\dfrac{dy}{dx}\) the subject |
| \(\dfrac{dy}{dx} = \dfrac{22x-11y-36}{6y+11x-20}\) | A1 | Correct expression or correct equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=4 \Rightarrow 3y^2-176+44y = 20y-144+28\) | M1 | Substitutes \(x=4\) into \(C\) to obtain a 3TQ in \(y\) |
| \(3y^2+24y-60 = 0 \Rightarrow y = ...\) | M1 | Solves for \(y\) |
| \(y = -10\ (,2)\) | A1 | Correct value |
| \((4,-10) \rightarrow \dfrac{dy}{dx} = \dfrac{88+110-36}{-60+44-20}\) | M1 | Substitutes \(x=4\) and their negative \(y\) into their \(\dfrac{dy}{dx}\) |
| \(\dfrac{dy}{dx} = -\dfrac{9}{2}\) | A1 | Correct value |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3y^2-11x^2+11xy = 20y-36x+28$ $\Rightarrow 6y\dfrac{dy}{dx}-22x+11x\dfrac{dy}{dx}+11y = 20\dfrac{dy}{dx}-36$ | M1 M1 A1 | M1: $y^2 \to Ay\dfrac{dy}{dx}$; M1: $11xy \to px\dfrac{dy}{dx}+qy$; A1: All correct |
| $(6y+11x-20)\dfrac{dy}{dx} = 22x-11y-36 \Rightarrow \dfrac{dy}{dx} = ...$ | M1 | Collects terms in $\dfrac{dy}{dx}$ (must be 3 and from appropriate terms) and makes $\dfrac{dy}{dx}$ the subject |
| $\dfrac{dy}{dx} = \dfrac{22x-11y-36}{6y+11x-20}$ | A1 | Correct expression or correct equivalent |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=4 \Rightarrow 3y^2-176+44y = 20y-144+28$ | M1 | Substitutes $x=4$ into $C$ to obtain a 3TQ in $y$ |
| $3y^2+24y-60 = 0 \Rightarrow y = ...$ | M1 | Solves for $y$ |
| $y = -10\ (,2)$ | A1 | Correct value |
| $(4,-10) \rightarrow \dfrac{dy}{dx} = \dfrac{88+110-36}{-60+44-20}$ | M1 | Substitutes $x=4$ and their negative $y$ into their $\dfrac{dy}{dx}$ |
| $\dfrac{dy}{dx} = -\dfrac{9}{2}$ | A1 | Correct value |
---3. The curve $C$ has equation
$$3 y ^ { 2 } - 11 x ^ { 2 } + 11 x y = 20 y - 36 x + 28$$
\begin{enumerate}[label=(\alph*)]
\item Find, in simplest form, $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
The point $P ( 4 , k )$, where $k$ is a constant, lies on $C$.\\
Given that $k < 0$
\item find the value of the gradient of $C$ at $P$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2022 Q3 [10]}}