| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a standard C1 differentiation application requiring finding a normal equation (routine derivative and negative reciprocal) and solving a simultaneous equation. While part (b) involves some algebraic manipulation, both parts follow well-practiced procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2} - \frac{27}{x^2}\) | M1A1 | M1: \(\frac{1}{2}\) or \(-\frac{27}{x^2}\); A1: fully correct, e.g. \(\frac{1}{2}x^0 - 27x^{-2}\) |
| \(x=3 \Rightarrow \frac{dy}{dx} = \frac{1}{2} - \frac{27}{9} = \left(-\frac{5}{2}\right)\) | M1 | Substitutes \(x=3\) into their \(\frac{dy}{dx}\) to obtain a numerical gradient |
| \(m_T = -\frac{5}{2} \Rightarrow m_N = -1 \div -\frac{5}{2}\); \(y - \left(-\frac{3}{2}\right) = \frac{2}{5}(x-3)\) | M1 | Correct method for equation of normal; uses \(-\frac{1}{m_T}\) with \(\left(3, -\frac{3}{2}\right)\); \(m_T\) from calculus |
| \(10y = 4x - 27\) | A1* | cso (correct equation must be seen in (a)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}x + \frac{27}{x} - 12 = \frac{4x-27}{10}\) or equivalent in \(y\) | M1 | Equate equations to produce equation in \(x\) only or \(y\) only; allow sign slips only |
| \(x^2 - 93x + 270 = 0\) or \(20y^2 - 636y - 999 = 0\) | A1 | Correct 3-term quadratic (or any multiple); allow terms on both sides |
| \((x-90)(x-3)=0 \Rightarrow x = \ldots\) or quadratic formula applied | dM1 | Attempt to solve 3TQ leading to at least one value of \(x\) or \(y\). Dependent on first M1 |
| \(x = 90\) or \(y = 33.3\) oe | A1 | cso; \(x\) must be 90 and \(y\) equivalent to \(\frac{333}{10}\) |
| \(x = 90\) and \(y = 33.3\) oe | A1 | cso; both values required |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{2} - \frac{27}{x^2}$ | M1A1 | M1: $\frac{1}{2}$ or $-\frac{27}{x^2}$; A1: fully correct, e.g. $\frac{1}{2}x^0 - 27x^{-2}$ |
| $x=3 \Rightarrow \frac{dy}{dx} = \frac{1}{2} - \frac{27}{9} = \left(-\frac{5}{2}\right)$ | M1 | Substitutes $x=3$ into their $\frac{dy}{dx}$ to obtain a numerical gradient |
| $m_T = -\frac{5}{2} \Rightarrow m_N = -1 \div -\frac{5}{2}$; $y - \left(-\frac{3}{2}\right) = \frac{2}{5}(x-3)$ | M1 | Correct method for equation of normal; uses $-\frac{1}{m_T}$ with $\left(3, -\frac{3}{2}\right)$; $m_T$ from calculus |
| $10y = 4x - 27$ | A1* | cso (correct equation must be seen in (a)) |
---
## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}x + \frac{27}{x} - 12 = \frac{4x-27}{10}$ or equivalent in $y$ | M1 | Equate equations to produce equation in $x$ only or $y$ only; allow sign slips only |
| $x^2 - 93x + 270 = 0$ or $20y^2 - 636y - 999 = 0$ | A1 | Correct 3-term quadratic (or any multiple); allow terms on both sides |
| $(x-90)(x-3)=0 \Rightarrow x = \ldots$ or quadratic formula applied | dM1 | Attempt to solve 3TQ leading to at least one value of $x$ or $y$. **Dependent on first M1** |
| $x = 90$ **or** $y = 33.3$ oe | A1 | cso; $x$ must be 90 and $y$ equivalent to $\frac{333}{10}$ |
| $x = 90$ **and** $y = 33.3$ oe | A1 | cso; both values required |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{937246f9-2b6a-48df-b919-c6db3d6f863b-28_643_1171_260_518}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve $C$ with equation
$$y = \frac { 1 } { 2 } x + \frac { 27 } { x } - 12 , \quad x > 0$$
The point $A$ lies on $C$ and has coordinates $\left( 3 , - \frac { 3 } { 2 } \right)$.
\begin{enumerate}[label=(\alph*)]
\item Show that the equation of the normal to $C$ at $A$ can be written as $10 y = 4 x - 27$
The normal to $C$ at $A$ meets $C$ again at the point $B$, as shown in Figure 3.
\item Use algebra to find the coordinates of $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2018 Q10 [10]}}