| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Standard +0.3 Part (a) is a standard normal line calculation requiring differentiation and negative reciprocal of gradient. Part (b) requires finding the minimum point and determining inequalities from a diagram, which is routine but involves multiple steps. Overall slightly easier than average as these are well-practiced techniques with no novel problem-solving required. |
| Spec | 1.02i Represent inequalities: graphically on coordinate plane1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{dy}{dx}=\right)8x - 10\) | B1 | Correct derivative |
| At \(\left(\frac{1}{2}, 3\right)\): \(m_T = -6 \Rightarrow m_N = \frac{1}{6}\) | M1* | Substitutes \(x = 0.5\) into their two-term derivative and using product of gradients is \(-1\) |
| \(y - 3 = \frac{1}{6}\left(x - \frac{1}{2}\right)\) | M1dep* | Using \(y-3=m\left(x-\frac{1}{2}\right)\) with \(m \neq -6\) or their tangent gradient (so must have attempted normal gradient). Or using \(y = mx + c\) |
| \(2x - 12y + 35 = 0\) | A1 | Must \(= 0\) and integer coefficients. All terms on one side |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x \geq 1.25\) | B1 | (\(y'=0 \Rightarrow x = 1.25\)) |
| \(y \geq 4x^2 - 10x + 7\) | B1 | |
| \(2x - 12y + 35 \geq 0\) | B1FT | o.e. Follow through their (a). SCB2 if all "correct" (including FT from (a)), but either all strict or a mix of strict and non-strict inequalities used |
## Question 7:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{dy}{dx}=\right)8x - 10$ | B1 | Correct derivative |
| At $\left(\frac{1}{2}, 3\right)$: $m_T = -6 \Rightarrow m_N = \frac{1}{6}$ | M1* | Substitutes $x = 0.5$ into their two-term derivative and using product of gradients is $-1$ |
| $y - 3 = \frac{1}{6}\left(x - \frac{1}{2}\right)$ | M1dep* | Using $y-3=m\left(x-\frac{1}{2}\right)$ with $m \neq -6$ or their tangent gradient (so must have attempted normal gradient). Or using $y = mx + c$ |
| $2x - 12y + 35 = 0$ | A1 | Must $= 0$ and integer coefficients. All terms on one side |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x \geq 1.25$ | B1 | ($y'=0 \Rightarrow x = 1.25$) |
| $y \geq 4x^2 - 10x + 7$ | B1 | |
| $2x - 12y + 35 \geq 0$ | B1FT | o.e. Follow through their (a). SCB2 if all "correct" (including FT from (a)), but either all strict or a mix of strict and non-strict inequalities used |7\\
\includegraphics[max width=\textwidth, alt={}, center]{8c0b68bd-2257-4994-b444-def0b3f64334-5_944_938_260_244}
The diagram shows the curve $C$ with equation $y = 4 x ^ { 2 } - 10 x + 7$ and two straight lines, $l _ { 1 }$ and $l _ { 2 }$. The line $l _ { 1 }$ is the normal to $C$ at the point $\left( \frac { 1 } { 2 } , 3 \right)$. The line $l _ { 2 }$ is the normal to $C$ at the minimum point of $C$.
\begin{enumerate}[label=(\alph*)]
\item Determine the equation of $l _ { 1 }$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers to be determined.
The shaded region shown in the diagram is bounded by $C , l _ { 1 }$ and $l _ { 2 }$.
\item Determine the inequalities that define the shaded region, including its boundaries.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q7 [7]}}