| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| 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 basic polynomial differentiation, finding a gradient from a linear equation, and substituting to find an unknown constant. All steps are routine with no problem-solving insight needed, making it slightly easier than average but not trivial due to the multi-part structure and algebraic manipulation required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx}=3kx^2-2x+1\) | M1A1 | \(x^n\to x^{n-1}\); \(+c\) scores A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of line is \(\frac{7}{2}\) | B1 | Must confirm using \(m=\frac{7}{2}\) |
| When \(x=-\frac{1}{2}\): \(3k\times(\frac{1}{4})-2\times(-\frac{1}{2})+1=\frac{7}{2}\) | M1, M1 | Sub \(x=-\frac{1}{2}\) into \(\frac{dy}{dx}\); form equation in \(k\) |
| \(\frac{3k}{4}=\frac{3}{2}\Rightarrow k=2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x=-\frac{1}{2}\Rightarrow y=k\times(-\frac{1}{8})-(\frac{1}{4})-\frac{1}{2}-5=-6\) | M1, A1 | Sub their \(k\) and \(x=-\frac{1}{2}\) into \(y\) |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx}=3kx^2-2x+1$ | M1A1 | $x^n\to x^{n-1}$; $+c$ scores A0 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of line is $\frac{7}{2}$ | B1 | Must confirm using $m=\frac{7}{2}$ |
| When $x=-\frac{1}{2}$: $3k\times(\frac{1}{4})-2\times(-\frac{1}{2})+1=\frac{7}{2}$ | M1, M1 | Sub $x=-\frac{1}{2}$ into $\frac{dy}{dx}$; form equation in $k$ |
| $\frac{3k}{4}=\frac{3}{2}\Rightarrow k=2$ | A1 | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=-\frac{1}{2}\Rightarrow y=k\times(-\frac{1}{8})-(\frac{1}{4})-\frac{1}{2}-5=-6$ | M1, A1 | Sub their $k$ and $x=-\frac{1}{2}$ into $y$ |
---The curve $C$ has equation $y = k x ^ { 3 } - x ^ { 2 } + x - 5$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
The point $A$ with $x$-coordinate $- \frac { 1 } { 2 }$ lies on $C$. The tangent to $C$ at $A$ is parallel to the line with equation $2 y - 7 x + 1 = 0$.
Find
\item the value of $k$,
\item the value of the $y$-coordinate of $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2008 Q9 [8]}}