| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent parallel to given line |
| Difficulty | Moderate -0.3 This is a straightforward C1 differentiation question requiring verification of a point on a curve, finding a tangent equation, and using the parallel tangent condition (equal gradients). All steps are routine applications of standard techniques with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x=3\), \(y=9-36+24+3=0\) | B1 | \((9-36+27=0\) is OK\()\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx}=\frac{3}{3}x^2-2\times4\times x+8\) \((=x^2-8x+8)\) | M1 A1 | 1st M1: some correct differentiation (\(x^n\to x^{n-1}\) for one term); 1st A1: correct unsimplified (all 3 terms) |
| When \(x=3\), \(\frac{dy}{dx}=9-24+8\Rightarrow m=-7\) | M1 | Substituting \(x_p(=3)\) in their \(\frac{dy}{dx}\); clear evidence |
| Equation of tangent: \(y-0=-7(x-3)\); \(y=-7x+21\) | M1, A1 c.a.o. | Using their \(m\) to find tangent at \(p\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx}=m\) gives \(x^2-8x+8=-7\) | M1 | Forming correct equation: "their \(\frac{dy}{dx}=\) gradient of their tangent" |
| \((x^2-8x+15=0)\); \((x-5)(x-3)=0\); \(x=(3)\) or \(5\) | M1, A1 | For solving quadratic based on their \(\frac{dy}{dx}\) leading to \(x=\) |
| \(y=\frac{1}{3}(5)^3-4(5)^2+8(5)+3\) | M1 | For using their \(x\) value in \(y\) to obtain \(y\) coordinate |
| \(y=-15\frac{1}{3}\) or \(-\frac{46}{3}\) | A1 |
## Question 10:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=3$, $y=9-36+24+3=0$ | B1 | $(9-36+27=0$ is OK$)$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx}=\frac{3}{3}x^2-2\times4\times x+8$ $(=x^2-8x+8)$ | M1 A1 | 1st M1: some correct differentiation ($x^n\to x^{n-1}$ for one term); 1st A1: correct unsimplified (all 3 terms) |
| When $x=3$, $\frac{dy}{dx}=9-24+8\Rightarrow m=-7$ | M1 | Substituting $x_p(=3)$ in their $\frac{dy}{dx}$; clear evidence |
| Equation of tangent: $y-0=-7(x-3)$; $y=-7x+21$ | M1, A1 c.a.o. | Using their $m$ to find tangent at $p$ |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx}=m$ gives $x^2-8x+8=-7$ | M1 | Forming correct equation: "their $\frac{dy}{dx}=$ gradient of their tangent" |
| $(x^2-8x+15=0)$; $(x-5)(x-3)=0$; $x=(3)$ or $5$ | M1, A1 | For solving quadratic based on their $\frac{dy}{dx}$ leading to $x=$ |
| $y=\frac{1}{3}(5)^3-4(5)^2+8(5)+3$ | M1 | For using their $x$ value in $y$ to obtain $y$ coordinate |
| $y=-15\frac{1}{3}$ or $-\frac{46}{3}$ | A1 | |10. The curve $C$ has equation $y = \frac { 1 } { 3 } x ^ { 3 } - 4 x ^ { 2 } + 8 x + 3$.
The point $P$ has coordinates $( 3,0 )$.
\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.
Another point $Q$ also lies on $C$. The tangent to $C$ at $Q$ is parallel to the tangent to $C$ at $P$.
\item Find the coordinates of $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2005 Q10 [11]}}