OCR C1 — Question 8 11 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind normal line equation at given point
DifficultyStandard +0.3 This is a straightforward C1 differentiation question requiring finding a tangent equation at a given point, then using the parallel condition to find where a normal has the same gradient. All steps are routine: differentiate, substitute, use perpendicular gradient relationship, and solve a simple quadratic. Slightly above average due to the two-part structure and the need to connect tangent/normal concepts, but no novel insight required.
Spec1.07b Gradient as rate of change: dy/dx notation1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives

\includegraphics{figure_8} The diagram shows the curve with equation \(y = 2 + 3x - x^2\) and the straight lines \(l\) and \(m\). The line \(l\) is the tangent to the curve at the point \(A\) where the curve crosses the \(y\)-axis.
  1. Find an equation for \(l\). [5]
The line \(m\) is the normal to the curve at the point \(B\). Given that \(l\) and \(m\) are parallel,
  1. find the coordinates of \(B\). [6]

AnswerMarks Guidance
(i) \(A(0, 2)\)B1
\(\frac{dy}{dx} = 3 - 2x\)M1 A1
grad \(= 3\)M1
\(\therefore y = 3x + 2\)A1
(ii) grad of \(m = 3\)
grad of curve at \(B = \frac{-1}{3} = -\frac{1}{3}\)M1 A1
at \(B: 3 - 2x = -\frac{1}{3}\)
\(x = \frac{5}{3}\)M1 A1
\(y = 2 + 3(\frac{5}{3}) - (\frac{5}{3})^2 = 4\frac{2}{9} \therefore B(1\frac{2}{3}, 4\frac{2}{9})\)M1 A1 (11)
**(i)** $A(0, 2)$ | B1 |
$\frac{dy}{dx} = 3 - 2x$ | M1 A1 |
grad $= 3$ | M1 |
$\therefore y = 3x + 2$ | A1 |

**(ii)** grad of $m = 3$ | |
grad of curve at $B = \frac{-1}{3} = -\frac{1}{3}$ | M1 A1 |
at $B: 3 - 2x = -\frac{1}{3}$ | |
$x = \frac{5}{3}$ | M1 A1 |
$y = 2 + 3(\frac{5}{3}) - (\frac{5}{3})^2 = 4\frac{2}{9} \therefore B(1\frac{2}{3}, 4\frac{2}{9})$ | M1 A1 | (11) |
\includegraphics{figure_8}

The diagram shows the curve with equation $y = 2 + 3x - x^2$ and the straight lines $l$ and $m$.

The line $l$ is the tangent to the curve at the point $A$ where the curve crosses the $y$-axis.
\begin{enumerate}[label=(\roman*)]
\item Find an equation for $l$. [5]
\end{enumerate}

The line $m$ is the normal to the curve at the point $B$.

Given that $l$ and $m$ are parallel,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item find the coordinates of $B$. [6]
\end{enumerate}

\hfill \mbox{\textit{OCR C1  Q8 [11]}}