Edexcel C1 — Question 9 11 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeTangent meets curve/axis — further geometry
DifficultyModerate -0.8 This is a straightforward C1 differentiation question requiring basic polynomial differentiation, solving a quadratic equation, finding a tangent equation, and calculating distance between two points. All techniques are routine with clear signposting and no problem-solving insight required. The multi-part structure guides students through standard procedures, making it easier than the average A-level question.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations

A curve \(C\) has equation \(y = x^3 - 5x^2 + 5x + 2\).
  1. Find \(\frac{dy}{dx}\) in terms of \(x\). [2]
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.
  1. Find the \(x\)-coordinate of \(Q\). [2]
  2. Find an equation for the tangent to \(C\) at \(P\), giving your answer in the form \(y = mx + c\), where \(m\) and \(c\) are constants. [3]
This tangent intersects the coordinate axes at the points \(R\) and \(S\).
  1. Find the length of \(RS\), giving your answer as a surd. [4]

Part (a)
AnswerMarks
\(\frac{dy}{dx} = 3x^2 - 10x + 5\)M1 A1 (2)
Part (b)
\(3x^2 - 10x + 5 = 2\) → \(3x^2 - 10x + 3 = 0\)
AnswerMarks
\((3x-1)(x-3) = 0\) → \(x = \frac{1}{3}\)M1 A1 (2)
Part (c)
AnswerMarks
When \(x = 3\): \(y = 27 - 45 + 15 + 2 = -1\)B1
\(y + 1 = 2(x-3)\) → \(y = 2x - 7\)M1 A1 (3)
Part (d)
AnswerMarks
\(R\): \(x = 0\) \(y = -7\); \(S\): \(y = 0\) \(x = 3.5\) (Both for M1)M1 A1 ft
\(RS = \sqrt{(72) + \left(\frac{7}{2}\right)^2} = \frac{7}{2}\sqrt{5}\) (or equivalent)M1 A1 (4)
Total: 11 marks
## Part (a)
$\frac{dy}{dx} = 3x^2 - 10x + 5$ | M1 A1 (2)

## Part (b)
$3x^2 - 10x + 5 = 2$ → $3x^2 - 10x + 3 = 0$

$(3x-1)(x-3) = 0$ → $x = \frac{1}{3}$ | M1 A1 (2)

## Part (c)
When $x = 3$: $y = 27 - 45 + 15 + 2 = -1$ | B1

$y + 1 = 2(x-3)$ → $y = 2x - 7$ | M1 A1 (3)

## Part (d)
$R$: $x = 0$ $y = -7$; $S$: $y = 0$ $x = 3.5$ (Both for M1) | M1 A1 ft

$RS = \sqrt{(72) + \left(\frac{7}{2}\right)^2} = \frac{7}{2}\sqrt{5}$ (or equivalent) | M1 A1 (4)

**Total: 11 marks**

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A curve $C$ has equation $y = x^3 - 5x^2 + 5x + 2$.

\begin{enumerate}[label=(\alph*)]
\item Find $\frac{dy}{dx}$ in terms of $x$. [2]
\end{enumerate}

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.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the $x$-coordinate of $Q$. [2]
\item Find an equation for the tangent to $C$ at $P$, giving your answer in the form $y = mx + c$, where $m$ and $c$ are constants. [3]
\end{enumerate}

This tangent intersects the coordinate axes at the points $R$ and $S$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the length of $RS$, giving your answer as a surd. [4]
\end{enumerate}

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