| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring only standard polynomial differentiation, solving a quadratic equation, and finding a tangent line equation. All parts follow routine procedures with no problem-solving insight needed, making it easier than average but not trivial due to the multi-step nature. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 3x^2 - 10x + 5\) | B1 B1 | B1 for each correct term |
| Answer | Marks |
|---|---|
| \(3x^2 - 10x + 5 = 2 \Rightarrow 3x^2 - 10x + 3 = 0\) | M1 |
| \((3x-1)(x-3) = 0 \Rightarrow x = \frac{1}{3}\) or \(x=3\) | A1 |
| \(x\)-coordinate of \(Q = \frac{1}{3}\) | A1 |
| Answer | Marks |
|---|---|
| When \(x=3\): \(y = 27 - 45 + 15 + 2 = -1\) | M1 A1 |
| Tangent: \(y-(-1) = 2(x-3) \Rightarrow y = 2x - 7\) | A1 |
| Answer | Marks |
|---|---|
| \(R\): set \(y=0\): \(x = \frac{7}{2}\), so \(R = \left(\frac{7}{2}, 0\right)\) | B1 |
| \(S\): set \(x=0\): \(y = -7\), so \(S = (0, -7)\) | B1 |
| \(RS = \sqrt{\left(\frac{7}{2}\right)^2 + 7^2} = \sqrt{\frac{49}{4}+49} = \sqrt{\frac{245}{4}} = \frac{7\sqrt{5}}{2}\) | M1 A1 |
# Question 8:
## Part (a):
| $\frac{dy}{dx} = 3x^2 - 10x + 5$ | B1 B1 | B1 for each correct term |
## Part (b):
| $3x^2 - 10x + 5 = 2 \Rightarrow 3x^2 - 10x + 3 = 0$ | M1 | |
| $(3x-1)(x-3) = 0 \Rightarrow x = \frac{1}{3}$ or $x=3$ | A1 | |
| $x$-coordinate of $Q = \frac{1}{3}$ | A1 | |
## Part (c):
| When $x=3$: $y = 27 - 45 + 15 + 2 = -1$ | M1 A1 | |
| Tangent: $y-(-1) = 2(x-3) \Rightarrow y = 2x - 7$ | A1 | |
## Part (d):
| $R$: set $y=0$: $x = \frac{7}{2}$, so $R = \left(\frac{7}{2}, 0\right)$ | B1 | |
| $S$: set $x=0$: $y = -7$, so $S = (0, -7)$ | B1 | |
| $RS = \sqrt{\left(\frac{7}{2}\right)^2 + 7^2} = \sqrt{\frac{49}{4}+49} = \sqrt{\frac{245}{4}} = \frac{7\sqrt{5}}{2}$ | M1 A1 | |
---8. A curve $C$ has equation $y = x ^ { 3 } - 5 x ^ { 2 } + 5 x + 2$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$.
The points $P$ and $Q$ lie on $C$. The gradient of $C$ at both $P$ and $Q$ is 2 . The $x$-coordinate of $P$ is 3 .
\item Find the $x$-coordinate of $Q$.
\item Find an equation for the tangent to $C$ at $P$, giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants.
This tangent intersects the coordinate axes at the points $R$ and $S$.
\item Find the length of $R S$, giving your answer as a surd.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q8 [11]}}