| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Tangent/normal with axis intercepts |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard technique to find dy/dx, then using the normal gradient to find an intercept. The algebra is slightly involved due to the 2^y term (requiring chain rule and ln 2), but the method is routine C4 material with no novel problem-solving required. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiates implicitly to include either \(\pm 4x\frac{dy}{dx}\) or \(-y^3 \to \pm\lambda y^2\frac{dy}{dx}\) or \(2^y \to \pm\mu 2^y\frac{dy}{dx}\) | 1st M1 | Ignore \(\left(\frac{dy}{dx}=\right)\); \(\lambda\), \(\mu\) are constants which can be 1 |
| Both \(4x^2 - y^3 \to 8x - 3y^2\frac{dy}{dx}\) and \(=0 \to =0\) | 1st A1 | |
| \(-4xy \to -4y - 4x\frac{dy}{dx}\) or \(4y - 4x\frac{dy}{dx}\) or \(-4y + 4x\frac{dy}{dx}\) or \(4y + 4x\frac{dy}{dx}\) | 2nd M1 | |
| \(2^y \to 2^y\ln 2\frac{dy}{dx}\) or \(2^y \to e^{y\ln 2}\ln 2\frac{dy}{dx}\) | B1 | If an extra term appears then award 1st A0 |
| Substituting \(x=-2\) and \(y=4\) into an equation involving \(\frac{dy}{dx}\) | 3rd dM1 | Dependent on first M mark. M1 gained by seeing at least one example of substituting \(x=-2\) and at least one example of \(y=4\) unless clearly applying \(x=4\) and \(y=-2\) |
| Correct final value | A1 cso | If candidate's solution not completely correct, do not give this mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiates implicitly to include either \(\pm 4y\frac{dx}{dy}\) or \(4x^2 \to \pm\lambda x\frac{dx}{dy}\) | 1st M1 | Ignore \(\left(\frac{dx}{dy}=\right)\); \(\lambda\) is a constant which can be 1 |
| Both \(4x^2 - y^3 \to 8x\frac{dx}{dy} - 3y^2\) and \(=0\to=0\) | 1st A1 | |
| \(-4xy \to -4y\frac{dx}{dy} - 4x\) or \(4y\frac{dx}{dy} - 4x\) or \(-4y\frac{dx}{dy}+4x\) or \(4y\frac{dx}{dy}+4x\) | 2nd M1 | |
| \(2^y \to 2^y\ln 2\) | B1 | |
| Substituting \(x=-2\) and \(y=4\) into equation involving \(\frac{dx}{dy}\) | 3rd dM1 | Dependent on first M mark |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates implicitly to include **either** $\pm 4x\frac{dy}{dx}$ **or** $-y^3 \to \pm\lambda y^2\frac{dy}{dx}$ **or** $2^y \to \pm\mu 2^y\frac{dy}{dx}$ | 1st M1 | Ignore $\left(\frac{dy}{dx}=\right)$; $\lambda$, $\mu$ are constants which can be 1 |
| **Both** $4x^2 - y^3 \to 8x - 3y^2\frac{dy}{dx}$ **and** $=0 \to =0$ | 1st A1 | |
| $-4xy \to -4y - 4x\frac{dy}{dx}$ or $4y - 4x\frac{dy}{dx}$ or $-4y + 4x\frac{dy}{dx}$ or $4y + 4x\frac{dy}{dx}$ | 2nd M1 | |
| $2^y \to 2^y\ln 2\frac{dy}{dx}$ **or** $2^y \to e^{y\ln 2}\ln 2\frac{dy}{dx}$ | B1 | If an extra term appears then award 1st A0 |
| Substituting $x=-2$ and $y=4$ into an equation involving $\frac{dy}{dx}$ | 3rd dM1 | Dependent on first M mark. M1 gained by seeing at least one example of substituting $x=-2$ and at least one example of $y=4$ **unless** clearly applying $x=4$ and $y=-2$ |
| Correct final value | A1 cso | If candidate's solution not completely correct, do not give this mark |
## Part (a) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates implicitly to include **either** $\pm 4y\frac{dx}{dy}$ **or** $4x^2 \to \pm\lambda x\frac{dx}{dy}$ | 1st M1 | Ignore $\left(\frac{dx}{dy}=\right)$; $\lambda$ is a constant which can be 1 |
| **Both** $4x^2 - y^3 \to 8x\frac{dx}{dy} - 3y^2$ **and** $=0\to=0$ | 1st A1 | |
| $-4xy \to -4y\frac{dx}{dy} - 4x$ or $4y\frac{dx}{dy} - 4x$ or $-4y\frac{dx}{dy}+4x$ or $4y\frac{dx}{dy}+4x$ | 2nd M1 | |
| $2^y \to 2^y\ln 2$ | B1 | |
| Substituting $x=-2$ and $y=4$ into equation involving $\frac{dx}{dy}$ | 3rd dM1 | Dependent on first M mark |
---4. The curve $C$ has equation
$$4 x ^ { 2 } - y ^ { 3 } - 4 x y + 2 ^ { y } = 0$$
The point $P$ with coordinates $( - 2,4 )$ lies on $C$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point $P$.
The normal to $C$ at $P$ meets the $y$-axis at the point $A$.
\item Find the $y$ coordinate of $A$, giving your answer in the form $p + q \ln 2$, where $p$ and $q$ are constants to be determined.\\
(3)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2017 Q4 [9]}}