| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Moderate -0.8 This is a straightforward C1 question requiring basic differentiation of a polynomial using the power rule, then finding a normal line equation at a given point. All steps are routine: differentiate, substitute x=1 to find gradient, use perpendicular gradient property, and rearrange to required form. No problem-solving or insight needed beyond standard technique application. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 4x^3 - 16x\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 1\): \(y = 1 - 8 + 3 = -4\) | B1 | |
| At \(x = 1\), \(\frac{dy}{dx} = 4 - 16 = -12\) | B1ft | Follow through on their derivative |
| Gradient of normal \(= -\frac{1}{m}\) \(\left(= \frac{1}{12}\right)\) | M1 | |
| \(y - (-4) = \frac{1}{12}(x-1)\) leading to \(x - 12y - 49 = 0\) | M1 A1 | (7 marks) |
## Question 3:
### Part (a)
$\frac{dy}{dx} = 4x^3 - 16x$ | M1 A1 | (2 marks)
### Part (b)
$x = 1$: $y = 1 - 8 + 3 = -4$ | B1 |
At $x = 1$, $\frac{dy}{dx} = 4 - 16 = -12$ | B1ft | Follow through on their derivative
Gradient of normal $= -\frac{1}{m}$ $\left(= \frac{1}{12}\right)$ | M1 |
$y - (-4) = \frac{1}{12}(x-1)$ leading to $x - 12y - 49 = 0$ | M1 A1 | **(7 marks)**
---3. For the curve $C$ with equation $y = x ^ { 4 } - 8 x ^ { 2 } + 3$,
\begin{enumerate}[label=(\alph*)]
\item find $\frac { \mathrm { d } y } { \mathrm {~d} x }$,
The point $A$, on the curve $C$, has $x$-coordinate 1 .
\item Find an equation for the normal to $C$ at $A$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\[0pt]
[P1 June 2003 Question 8*]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q3 [7]}}