| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| 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 with standard techniques: differentiate each term (including product rule for 2x²y and chain rule for cos(πy)), substitute given point to find gradient, then find normal equation and x-intercept. While it involves multiple steps and algebraic manipulation, all techniques are routine C4 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(4xy + 2x^2\frac{dy}{dx}\right) + 2 + 4\frac{dy}{dx} + \pi\sin(\pi y)\frac{dy}{dx} = 0\) | M1 A1 B1 | M1: differentiates implicitly to include either \(2x^2\frac{dy}{dx}\) or \(4y \to 4\frac{dy}{dx}\) or \(-\cos(\pi y) \to \pm\lambda\sin(\pi y)\frac{dy}{dx}\); B1: \(2x^2y \to 4xy + 2x^2\frac{dy}{dx}\) |
| \(\frac{dy}{dx}\left(2x^2 + 4 + \pi\sin(\pi y)\right) + 4xy + 2 = 0\) | dM1 | Attempt to factorise out all terms in \(\frac{dy}{dx}\), at least two such terms present |
| \(\frac{dy}{dx} = \frac{-4xy-2}{2x^2+4+\pi\sin(\pi y)}\) or \(\frac{4xy+2}{-2x^2-4-\pi\sin(\pi y)}\) | A1 cso | Correct answer or equivalent; final A1 cso: if not completely correct do not award |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(\left(3, \frac{1}{2}\right)\): \(m_T = \frac{dy}{dx} = \frac{-4(3)(\frac{1}{2})-2}{2(3)^2+4+\pi\sin(\frac{1}{2}\pi)} = \left\{\frac{-8}{22+\pi}\right\}\) | M1 | Substituting \(x=3\) and \(y=\frac{1}{2}\) into an equation involving \(\frac{dy}{dx}\) |
| \(m_N = \frac{22+\pi}{8}\) | M1 | Applying \(m_N = \frac{-1}{m_T}\) to find a numerical \(m_N\); can be implied by later working |
| \(y - \frac{1}{2} = \left(\frac{22+\pi}{8}\right)(x-3)\) leading to \(y = \left(\frac{22+\pi}{8}\right)x + \frac{1}{2} - \frac{66+3\pi}{8}\) | dM1 | \(y - \frac{1}{2} = m_N(x-3)\) or \(y = m_N x + c\) where \(\frac{1}{2} = (\text{their } m_N)(3) + c\); with numerical \(m_N \neq m_T\) in terms of \(\pi\); sets \(y=0\) |
| Cuts \(x\)-axis \(\Rightarrow y=0\): \(x = \frac{3\pi+62}{\pi+22}\) or \(\frac{6\pi+124}{2\pi+44}\) or \(\frac{62+3\pi}{22+\pi}\) | A1 o.e. |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(4xy + 2x^2\frac{dy}{dx}\right) + 2 + 4\frac{dy}{dx} + \pi\sin(\pi y)\frac{dy}{dx} = 0$ | M1 A1 B1 | M1: differentiates implicitly to include either $2x^2\frac{dy}{dx}$ or $4y \to 4\frac{dy}{dx}$ or $-\cos(\pi y) \to \pm\lambda\sin(\pi y)\frac{dy}{dx}$; B1: $2x^2y \to 4xy + 2x^2\frac{dy}{dx}$ |
| $\frac{dy}{dx}\left(2x^2 + 4 + \pi\sin(\pi y)\right) + 4xy + 2 = 0$ | dM1 | Attempt to factorise out all terms in $\frac{dy}{dx}$, at least two such terms present |
| $\frac{dy}{dx} = \frac{-4xy-2}{2x^2+4+\pi\sin(\pi y)}$ or $\frac{4xy+2}{-2x^2-4-\pi\sin(\pi y)}$ | A1 cso | Correct answer or equivalent; final A1 cso: if not completely correct do not award |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $\left(3, \frac{1}{2}\right)$: $m_T = \frac{dy}{dx} = \frac{-4(3)(\frac{1}{2})-2}{2(3)^2+4+\pi\sin(\frac{1}{2}\pi)} = \left\{\frac{-8}{22+\pi}\right\}$ | M1 | Substituting $x=3$ and $y=\frac{1}{2}$ into an equation involving $\frac{dy}{dx}$ |
| $m_N = \frac{22+\pi}{8}$ | M1 | Applying $m_N = \frac{-1}{m_T}$ to find a numerical $m_N$; can be implied by later working |
| $y - \frac{1}{2} = \left(\frac{22+\pi}{8}\right)(x-3)$ leading to $y = \left(\frac{22+\pi}{8}\right)x + \frac{1}{2} - \frac{66+3\pi}{8}$ | dM1 | $y - \frac{1}{2} = m_N(x-3)$ or $y = m_N x + c$ where $\frac{1}{2} = (\text{their } m_N)(3) + c$; with numerical $m_N \neq m_T$ in terms of $\pi$; sets $y=0$ |
| Cuts $x$-axis $\Rightarrow y=0$: $x = \frac{3\pi+62}{\pi+22}$ or $\frac{6\pi+124}{2\pi+44}$ or $\frac{62+3\pi}{22+\pi}$ | A1 o.e. | |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$2 x ^ { 2 } y + 2 x + 4 y - \cos ( \pi y ) = 17$$
(a) Use implicit differentiation to find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
The point $P$ with coordinates $\left( 3 , \frac { 1 } { 2 } \right)$ lies on $C$.\\
The normal to $C$ at $P$ meets the $x$-axis at the point $A$.\\
(b) Find the $x$ coordinate of $A$, giving your answer in the form $\frac { a \pi + b } { c \pi + d }$, where $a , b , c$ and $d$ are integers to be determined.
\hfill \mbox{\textit{Edexcel C4 2016 Q3 [9]}}