| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Show line is tangent, verify |
| Difficulty | Standard +0.3 This is a multi-part question involving finding x-intercepts, calculating a normal line using differentiation of a reciprocal function, and solving a simultaneous equation. While it requires multiple techniques (differentiation, gradient of normal, substitution), each step is straightforward and follows standard C1 procedures. The 'show that' in part (b) provides the answer to verify, and part (c) is routine substitution leading to a quadratic. Slightly easier than average due to the scaffolding and standard nature of all techniques. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\frac{1}{2}, 0\right)\) | B1 | Accept \(x=\frac{1}{2}\) if evidence \(y=0\) used; can be written on graph |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = x^{-2}\) | M1A1 | M1 for \(kx^{-2}\) even if '2' not differentiated to zero; A1 for \(x^{-2}\) (o.e.) only |
| At \(x=\frac{1}{2}\): \(\frac{dy}{dx} = \left(\frac{1}{2}\right)^{-2} = 4\ (=m)\) | A1 | For using \(x=0.5\) to get \(m=4\) (or \(m=1/0.25\)) |
| Gradient of normal \(= -\frac{1}{m}\ \left(= -\frac{1}{4}\right)\) | M1 | For using perpendicular gradient rule on their \(m\) |
| Equation of normal: \(y - 0 = -\frac{1}{4}\left(x - \frac{1}{2}\right)\) | M1 | For using changed gradient (based on \(y'\)) and their \(A\) to find equation of line |
| \(2x + 8y - 1 = 0\) | A1cso | Accept \(2x+8y=1\) or equivalent with \(\pm 2x\) and \(\pm 8y\); must see at least one intermediate equation after \(m=4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2 - \frac{1}{x} = -\frac{1}{4}x + \frac{1}{8}\) | M1 | For attempt to form suitable equation in one variable; do not penalise poor bracket use |
| \([2x^2 + 15x - 8 = 0]\) or \([8y^2 - 17y = 0]\) | M1 | For simplifying to 3TQ and attempting to solve; may use \(\Rightarrow x=-8\) |
| \(x = \left[\frac{1}{2}\right]\) or \(-8\) | A1 | For \(x=-8\) (ignore second value); if \(y\) found first, allow ft for \(x\) if \(x<0\) |
| \(y = \frac{17}{8}\) (or exact equivalent) | A1ft | Follow through their \(x\) value in line or curve provided answer \(>0\); dependent on both M marks |
## Question 10:
**(a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\frac{1}{2}, 0\right)$ | B1 | Accept $x=\frac{1}{2}$ if evidence $y=0$ used; can be written on graph |
**(b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = x^{-2}$ | M1A1 | M1 for $kx^{-2}$ even if '2' not differentiated to zero; A1 for $x^{-2}$ (o.e.) only |
| At $x=\frac{1}{2}$: $\frac{dy}{dx} = \left(\frac{1}{2}\right)^{-2} = 4\ (=m)$ | A1 | For using $x=0.5$ to get $m=4$ (or $m=1/0.25$) |
| Gradient of normal $= -\frac{1}{m}\ \left(= -\frac{1}{4}\right)$ | M1 | For using perpendicular gradient rule on their $m$ |
| Equation of normal: $y - 0 = -\frac{1}{4}\left(x - \frac{1}{2}\right)$ | M1 | For using changed gradient (based on $y'$) and their $A$ to find equation of line |
| $2x + 8y - 1 = 0$ | A1cso | Accept $2x+8y=1$ or equivalent with $\pm 2x$ and $\pm 8y$; must see at least one intermediate equation after $m=4$ |
**(c)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2 - \frac{1}{x} = -\frac{1}{4}x + \frac{1}{8}$ | M1 | For attempt to form suitable equation in one variable; do not penalise poor bracket use |
| $[2x^2 + 15x - 8 = 0]$ or $[8y^2 - 17y = 0]$ | M1 | For simplifying to 3TQ and attempting to solve; may use $\Rightarrow x=-8$ |
| $x = \left[\frac{1}{2}\right]$ or $-8$ | A1 | For $x=-8$ (ignore second value); if $y$ found first, allow ft for $x$ if $x<0$ |
| $y = \frac{17}{8}$ (or exact equivalent) | A1ft | Follow through their $x$ value in line or curve provided answer $>0$; dependent on both M marks |10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ff1cdb91-0286-4bc8-9e67-451500b2bf74-14_769_935_285_411}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve $C$ with equation
$$y = 2 - \frac { 1 } { x } , \quad x \neq 0$$
The curve crosses the $x$-axis at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $A$.
\item Show that the equation of the normal to $C$ at $A$ can be written as
$$2 x + 8 y - 1 = 0$$
The normal to $C$ at $A$ meets $C$ again at the point $B$, as shown in Figure 2 .
\item Find the coordinates of $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2012 Q10 [11]}}