| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Standard +0.8 This C1 question involves standard differentiation and tangent finding (parts a-b are routine), but part (c) requires solving a cubic equation where the derivative equals a specific value, then manipulating the result to match a given surd form—this multi-step problem-solving with algebraic manipulation elevates it above typical C1 exercises. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(x=2: \quad y = 8-8-2+9 = 7 \quad (*)\) | B1 (1) | Clear attempt to substitute \(x=2\) leading to 7, e.g. \(2^3 - 2\times2^2 - 2 + 9 = 7\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 4x - 1\) | M1, A1 | M1 for attempt to differentiate with at least one term fully correct; A1 for fully correct expression |
| \(x=2: \quad \frac{\mathrm{d}y}{\mathrm{d}x} = 12-8-1 (=3)\) | A1ft | For substituting \(x=2\) in their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) \((\neq y)\), accept correct expression e.g. \(3\times(2)^2 - 4\times2 - 1\) |
| \(y - 7 = 3(x-2), \qquad y = 3x+1\) | M1, A1 (5) | M1 for use of their "3" (from \(\frac{\mathrm{d}y}{\mathrm{d}x}\neq y\) and \(x=2\)) to find tangent equation; alternative: use \((2,7)\) in \(y=mx+c\) to find \(c\), award when \(c=\ldots\) seen. No attempted use of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) in (b) scores 0/5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(m = -\dfrac{1}{3}\) | B1ft | For \(-\dfrac{1}{m}\) with their \(m\) |
| \(3x^2-4x-1 = -\dfrac{1}{3}\), giving \(9x^2-12x-2=0\) or \(x^2 - \dfrac{4}{3}x - \dfrac{2}{9}=0\) | M1, A1 | M1 for forming equation from their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) \((\neq y)\) and their \(-\dfrac{1}{m}\) (must be changed from \(m\)); A1 for correct 3TQ, all terms on LHS (condone missing \(=0\)) |
| \(x = \dfrac{12+\sqrt{144+72}}{18}\), \(\left(\sqrt{216}=\sqrt{36}\sqrt{6}=6\sqrt{6}\right)\) or \((3x-2)^2=6 \rightarrow 3x = 2\pm\sqrt{6}\) | M1 | For proceeding to \(x=\ldots\) or \(3x=\ldots\) by formula or completing the square; not factorising; condone \(\pm\) |
| \(x = \dfrac{1}{3}\!\left(2+\sqrt{6}\right) \quad (*)\) | A1cso (5) | For proceeding to given answer with no incorrect working seen; can still have \(\pm\). [11] |
# Question 11:
## Part (a)
| $x=2: \quad y = 8-8-2+9 = 7 \quad (*)$ | B1 (1) | Clear attempt to substitute $x=2$ leading to 7, e.g. $2^3 - 2\times2^2 - 2 + 9 = 7$ |
## Part (b)
| $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 4x - 1$ | M1, A1 | M1 for attempt to differentiate with at least one term fully correct; A1 for fully correct expression |
| $x=2: \quad \frac{\mathrm{d}y}{\mathrm{d}x} = 12-8-1 (=3)$ | A1ft | For substituting $x=2$ in their $\frac{\mathrm{d}y}{\mathrm{d}x}$ $(\neq y)$, accept correct expression e.g. $3\times(2)^2 - 4\times2 - 1$ |
| $y - 7 = 3(x-2), \qquad y = 3x+1$ | M1, A1 (5) | M1 for use of their "3" (from $\frac{\mathrm{d}y}{\mathrm{d}x}\neq y$ and $x=2$) to find tangent equation; alternative: use $(2,7)$ in $y=mx+c$ to find $c$, award when $c=\ldots$ seen. **No attempted use of $\frac{\mathrm{d}y}{\mathrm{d}x}$ in (b) scores 0/5** |
## Part (c)
| $m = -\dfrac{1}{3}$ | B1ft | For $-\dfrac{1}{m}$ with their $m$ |
| $3x^2-4x-1 = -\dfrac{1}{3}$, giving $9x^2-12x-2=0$ or $x^2 - \dfrac{4}{3}x - \dfrac{2}{9}=0$ | M1, A1 | M1 for forming equation from their $\frac{\mathrm{d}y}{\mathrm{d}x}$ $(\neq y)$ and their $-\dfrac{1}{m}$ (must be changed from $m$); A1 for correct 3TQ, all terms on LHS (condone missing $=0$) |
| $x = \dfrac{12+\sqrt{144+72}}{18}$, $\left(\sqrt{216}=\sqrt{36}\sqrt{6}=6\sqrt{6}\right)$ or $(3x-2)^2=6 \rightarrow 3x = 2\pm\sqrt{6}$ | M1 | For proceeding to $x=\ldots$ or $3x=\ldots$ by formula or completing the square; not factorising; condone $\pm$ |
| $x = \dfrac{1}{3}\!\left(2+\sqrt{6}\right) \quad (*)$ | A1cso (5) | For proceeding to given answer with no incorrect working seen; can still have $\pm$. **[11]** |11. The curve $C$ has equation
$$y = x ^ { 3 } - 2 x ^ { 2 } - x + 9 , \quad x > 0$$
The point $P$ has coordinates (2, 7).
\begin{enumerate}[label=(\alph*)]
\item Show that $P$ lies on $C$.
\item Find the equation of the tangent to $C$ at $P$, giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants.
The point $Q$ also lies on $C$.\\
Given that the tangent to $C$ at $Q$ is perpendicular to the tangent to $C$ at $P$,
\item show that the $x$-coordinate of $Q$ is $\frac { 1 } { 3 } ( 2 + \sqrt { 6 } )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2009 Q11 [11]}}