| Exam Board | Edexcel |
|---|---|
| Module | PURE |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Standard +0.3 This is a straightforward multi-step calculus problem requiring students to use the relationship between gradient and normal, then integrate. Part (a) involves finding the gradient at x=3 from the normal's gradient (-1/24 means tangent gradient is 24), substituting into f'(3) to solve for k. Part (b) is routine integration with constant determination using the point on the normal. All techniques are standard with clear signposting, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
\begin{enumerate}
\item A curve $C$ has equation $y = \mathrm { f } ( x )$.
\end{enumerate}
The point $P$ with $x$ coordinate 3 lies on $C$
\section*{Given}
\begin{itemize}
\item $\mathrm { f } ^ { \prime } ( x ) = 4 x ^ { 2 } + k x + 3$ where $k$ is a constant
\item the normal to $C$ at $P$ has equation $y = - \frac { 1 } { 24 } x + 5$\\
(a) show that $k = - 5$\\
(b) Hence find $\mathrm { f } ( x )$.
\end{itemize}
\hfill \mbox{\textit{Edexcel PURE 2024 Q8}}