| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Normal to curve at given point |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard techniques: finding intercepts, identifying asymptotes of a rational function, differentiating to find the normal (using negative reciprocal), and calculating distance between two points. All steps are routine C1 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(-\frac{3}{4}, 0\right)\). Accept \(x = -\frac{3}{4}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 4\) | B1 | One correct asymptote |
| \(x = 0\) or '\(y\)-axis' | B1 | Both correct asymptotes and no extra ones |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -3x^{-2}\) | M1 | \(\frac{dy}{dx} = kx^{-2}\) (allow \(\frac{dy}{dx} = kx^{-2} + 4\)) |
| At \(x = -3\), gradient of curve \(= -\frac{1}{3}\) | A1 | cao (must be a fraction with no powers, e.g. \(-3(-3)^{-2}\) scores A0 unless evaluated as \(-\frac{3}{9}\) or implied by normal gradient) |
| Gradient of normal \(= -1/m\) | dM1 | Correct perpendicular gradient rule applied to a numerical gradient that must have come from substituting \(x=-3\) into their derivative. Dependent on previous M1 |
| Normal at \(P\) is \((y-3) = 3(x+3)\) | dM1 A1 | dM1: Correct straight line method using \((-3,3)\) and a "changed" gradient. Wrong equation with no formula quoted is M0. Also dependent on first M1. A1: Any correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((-4, 0)\) and \((0, 12)\) | B1 | Both correct (may be seen on a sketch) |
| \(AB\) has length \(\sqrt{160}\) or \(AB^2 = 160\) | M1 A1cso | M1: Correct use of Pythagoras for their \(A\) and \(B\), one on \(x\)-axis and other on \(y\)-axis, obtained from their equation in (c). A1: \(\sqrt{160}\) or better e.g. \(4\sqrt{10}\) with no errors seen |
## Question 11:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(-\frac{3}{4}, 0\right)$. Accept $x = -\frac{3}{4}$ | B1 | |
**(1 mark)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 4$ | B1 | One correct asymptote |
| $x = 0$ or '$y$-axis' | B1 | Both correct asymptotes and no extra ones |
**Special case:** $x \neq 0$ **and** $y \neq 4$ scores B1B0. **(2 marks)**
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -3x^{-2}$ | M1 | $\frac{dy}{dx} = kx^{-2}$ (allow $\frac{dy}{dx} = kx^{-2} + 4$) |
| At $x = -3$, gradient of curve $= -\frac{1}{3}$ | A1 | cao (must be a fraction with no powers, e.g. $-3(-3)^{-2}$ scores A0 unless evaluated as $-\frac{3}{9}$ or implied by normal gradient) |
| Gradient of normal $= -1/m$ | dM1 | Correct perpendicular gradient rule applied to a numerical gradient that must have come from substituting $x=-3$ into their derivative. **Dependent on previous M1** |
| Normal at $P$ is $(y-3) = 3(x+3)$ | dM1 A1 | dM1: Correct straight line method using $(-3,3)$ and a "changed" gradient. Wrong equation with no formula quoted is M0. **Also dependent on first M1.** A1: Any correct equation |
**(5 marks)**
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(-4, 0)$ and $(0, 12)$ | B1 | Both correct (may be seen on a sketch) |
| $AB$ has length $\sqrt{160}$ or $AB^2 = 160$ | M1 A1cso | M1: Correct use of Pythagoras for their $A$ and $B$, one on $x$-axis and other on $y$-axis, obtained from their equation in (c). A1: $\sqrt{160}$ or better e.g. $4\sqrt{10}$ **with no errors seen** |
**(3 marks) [11 marks total]**
The images you've shared appear to be essentially blank pages — one is completely white/empty, and the other shows only publisher/contact information for Edexcel Publications (back matter of an exam paper).
There is **no mark scheme content** visible in these images. The pages shown are:
1. A blank page
2. A back cover with publisher details (Edexcel Publications contact info, Order Code UA035658 Summer 2013, and logos for Ofqual, Welsh Assembly Government, and CEA)
Could you please share the actual mark scheme pages? They would typically contain question numbers, model answers, mark allocations (M1, A1, B1, etc.), and examiner guidance notes.11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5cee336b-d9c9-4b18-ab82-52fdca1eb1e7-15_592_1394_274_283}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve $H$ with equation $y = \frac { 3 } { x } + 4 , x \neq 0$.
\begin{enumerate}[label=(\alph*)]
\item Give the coordinates of the point where $H$ crosses the $x$-axis.
\item Give the equations of the asymptotes to $H$.
\item Find an equation for the normal to $H$ at the point $P ( - 3,3 )$.
This normal crosses the $x$-axis at $A$ and the $y$-axis at $B$.
\item Find the length of the line segment $A B$. Give your answer as a surd.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2013 Q11 [11]}}