| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard differentiation techniques: finding a tangent equation, finding a normal equation, and solving simultaneous linear equations. All steps are routine procedures with no problem-solving insight required, making it easier than average, though the multi-part structure prevents it from being trivial. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx}=3x^2-10x+6\) | M1 | AO 1.1 — At least two terms correct |
| When \(x=0\), \(\frac{dy}{dx}=6\) | M1 | AO 1.1 — FT their \(\frac{dy}{dx}\) |
| Tangent goes through origin so equation is \(y=6x\) cao | A1 | AO 1.1 — Need reasoning e.g. \(y-0=6(x-0)\) or use of \(y=mx+c\); \(y=6x\) implies previous M mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x=1\), \(\frac{dy}{dx}=-1\) | M1 | AO 1.1 — FT their \(\frac{dy}{dx}\) from (a) |
| Gradient of normal is \(1\) | M1 | AO 1.1 — FT negative reciprocal of their \(\frac{dy}{dx}\) |
| \([(y-2)=(x-1)]\) so \(y=x+1\) | A1 | AO 1.1 — oe but constant terms should be collected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6x=x+1\) | M1 | AO 1.1 — Eliminate a variable. FT their equations from (a) and (b) for M1 only |
| \(x=\frac{1}{5},\ y=\frac{6}{5}\) oe | A1 | AO 1.1 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx}=3x^2-10x+6$ | M1 | AO 1.1 — At least two terms correct |
| When $x=0$, $\frac{dy}{dx}=6$ | M1 | AO 1.1 — FT their $\frac{dy}{dx}$ |
| Tangent goes through origin so equation is $y=6x$ cao | A1 | AO 1.1 — Need reasoning e.g. $y-0=6(x-0)$ or use of $y=mx+c$; $y=6x$ implies previous M mark |
**Total: [3]**
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x=1$, $\frac{dy}{dx}=-1$ | M1 | AO 1.1 — FT their $\frac{dy}{dx}$ from (a) |
| Gradient of normal is $1$ | M1 | AO 1.1 — FT negative reciprocal of their $\frac{dy}{dx}$ |
| $[(y-2)=(x-1)]$ so $y=x+1$ | A1 | AO 1.1 — oe but constant terms should be collected |
**Total: [3]**
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6x=x+1$ | M1 | AO 1.1 — Eliminate a variable. FT their equations from (a) and (b) for M1 only |
| $x=\frac{1}{5},\ y=\frac{6}{5}$ oe | A1 | AO 1.1 |
**Total: [2]**
---5 In this question you must show detailed reasoning.\\
This question is about the curve $y = x ^ { 3 } - 5 x ^ { 2 } + 6 x$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent, $T$, to the curve at the point ( 0,0 ).
\item Find the equation of the normal, $N$, to the curve at the point ( 1,2 ).
\item Find the coordinates of the point of intersection of $T$ and $N$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2023 Q5 [8]}}