Edexcel C1 — Question 9 11 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks11
PaperDownload PDF ↗
TopicTangents, normals and gradients
TypeFind normal line equation at given point
DifficultyModerate -0.3 This is a straightforward C1 integration and differentiation question with standard techniques: finding a normal (negative reciprocal of gradient), integrating to find the curve equation using a boundary condition, and analyzing when a derivative equals a constant. All parts are routine textbook exercises requiring no problem-solving insight, though the multi-step nature and 11 total marks place it slightly below average difficulty rather than being trivial.
Spec1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation

The gradient of the curve \(C\) is given by $$\frac{dy}{dx} = (3x - 1)^2.$$ The point \(P(1, 4)\) lies on \(C\).
  1. Find an equation of the normal to \(C\) at \(P\). [4]
  2. Find an equation for the curve \(C\) in the form \(y = f(x)\). [5]
  3. Using \(\frac{dy}{dx} = (3x - 1)^2\), show that there is no point on \(C\) at which the tangent is parallel to the line \(y = 1 - 2x\). [2]

The gradient of the curve $C$ is given by
$$\frac{dy}{dx} = (3x - 1)^2.$$

The point $P(1, 4)$ lies on $C$.

\begin{enumerate}[label=(\alph*)]
\item Find an equation of the normal to $C$ at $P$. [4]
\item Find an equation for the curve $C$ in the form $y = f(x)$. [5]
\item Using $\frac{dy}{dx} = (3x - 1)^2$, show that there is no point on $C$ at which the tangent is parallel to the line $y = 1 - 2x$. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1  Q9 [11]}}