| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| 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 requiring standard technique application. Part (a) involves routine differentiation of each term and algebraic rearrangement to isolate dy/dx. Part (b) requires setting the derivative to zero and solving a simple linear equation. While it involves multiple steps, it follows a well-practiced procedure with no conceptual challenges or novel insights required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x + 2y\frac{dy}{dx} + 10 + 2\frac{dy}{dx} - \left(4y + 4x\frac{dy}{dx}\right) = 0\) | M1 A1 M1 | First M1: differentiates implicitly to include \(\pm 4x\frac{dy}{dx}\) or \(y^2 \rightarrow 2y\frac{dy}{dx}\) or \(2y \rightarrow 2\frac{dy}{dx}\). A1: \(x^2+y^2+10x+2y \rightarrow 2x+2y\frac{dy}{dx}+10+2\frac{dy}{dx}\) and \(10\rightarrow 0\). Second M1: \(-4xy \rightarrow \pm 4y \pm 4x\frac{dy}{dx}\) |
| \(2x + 10 - 4y + (2y + 2 - 4x)\frac{dy}{dx} = 0\) | dM1 | Dependent on first M1; attempt to factorise out all terms in \(\frac{dy}{dx}\) (at least two terms) |
| \(\frac{dy}{dx} = \frac{2x+10-4y}{4x-2y-2}\) | ||
| \(\frac{dy}{dx} = \frac{x+5-2y}{2x-y-1}\) \(\left\{= \frac{-x-5+2y}{-2x+y+1}\right\}\) | A1 cso oe | Must be simplified. cso: if candidate's solution not completely correct, do not give this mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 0 \Rightarrow x + 5 - 2y = 0\) | M1 | Sets numerator of their \(\frac{dy}{dx}\) equal to zero (or denominator of their \(\frac{dx}{dy}\) equal to zero) oe. Note: if numerator involves one variable only, then *only* the first M1 mark is possible in part (b) |
| So \(x = 2y - 5\); substitutes into printed equation | M1 | Substitutes their \(x\) or \(y\) into printed equation to give equation in one variable only |
| \((2y-5)^2 + y^2 + 10(2y-5) + 2y - 4(2y-5)y = 10\) gives \(3y^2 - 22y + 35 = 0\) | A1 oe | For obtaining either \(-3y^2+22y-35\{=0\}\) or \(3y^2-22y+35\{=0\}\) |
| \((3y-7)(y-5) = 0\) | ddM1 | Dependent on previous 2 M marks. Method mark for solving a quadratic equation |
| \(y = \frac{7}{3},\ 5\) | A1 cao | \(y = \frac{7}{3}\), 5. cao (2.33 or 2.3 without reference to \(\frac{7}{3}\) or \(2\frac{1}{3}\) not allowed) |
## Question 3:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x + 2y\frac{dy}{dx} + 10 + 2\frac{dy}{dx} - \left(4y + 4x\frac{dy}{dx}\right) = 0$ | M1 A1 M1 | First M1: differentiates implicitly to include $\pm 4x\frac{dy}{dx}$ or $y^2 \rightarrow 2y\frac{dy}{dx}$ or $2y \rightarrow 2\frac{dy}{dx}$. A1: $x^2+y^2+10x+2y \rightarrow 2x+2y\frac{dy}{dx}+10+2\frac{dy}{dx}$ **and** $10\rightarrow 0$. Second M1: $-4xy \rightarrow \pm 4y \pm 4x\frac{dy}{dx}$ |
| $2x + 10 - 4y + (2y + 2 - 4x)\frac{dy}{dx} = 0$ | dM1 | Dependent on first M1; attempt to factorise out all terms in $\frac{dy}{dx}$ (at least two terms) |
| $\frac{dy}{dx} = \frac{2x+10-4y}{4x-2y-2}$ | | |
| $\frac{dy}{dx} = \frac{x+5-2y}{2x-y-1}$ $\left\{= \frac{-x-5+2y}{-2x+y+1}\right\}$ | A1 cso oe | Must be simplified. cso: if candidate's solution not completely correct, do not give this mark |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 0 \Rightarrow x + 5 - 2y = 0$ | M1 | Sets numerator of their $\frac{dy}{dx}$ equal to zero (or denominator of their $\frac{dx}{dy}$ equal to zero) oe. **Note:** if numerator involves one variable only, then *only* the first M1 mark is possible in part (b) |
| So $x = 2y - 5$; substitutes into printed equation | M1 | Substitutes their $x$ or $y$ into printed equation to give equation in one variable only |
| $(2y-5)^2 + y^2 + 10(2y-5) + 2y - 4(2y-5)y = 10$ gives $3y^2 - 22y + 35 = 0$ | A1 oe | For obtaining either $-3y^2+22y-35\{=0\}$ or $3y^2-22y+35\{=0\}$ |
| $(3y-7)(y-5) = 0$ | ddM1 | Dependent on previous 2 M marks. Method mark for solving a quadratic equation |
| $y = \frac{7}{3},\ 5$ | A1 cao | $y = \frac{7}{3}$, 5. **cao** (2.33 or 2.3 without reference to $\frac{7}{3}$ or $2\frac{1}{3}$ not allowed) |3.
$$x ^ { 2 } + y ^ { 2 } + 10 x + 2 y - 4 x y = 10$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$, fully simplifying your answer.
\item Find the values of $y$ for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2014 Q3 [10]}}