| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a straightforward AS-level differentiation question requiring standard techniques: finding a tangent equation using dy/dx, determining intercepts, calculating triangle area, and using second derivative test for stationary points. All steps are routine with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 2x - 2x^{-3}\) | M1 | Attempt to differentiate |
| At \(\left(2, \frac{17}{4}\right)\) gradient \(= 2\times2 - 2\times2^{-3} = \frac{15}{4}\) | A1 | Correct value in any form |
| Equation of tangent: \(y - \frac{17}{4} = \frac{15}{4}(x-2)\), giving \(y = \frac{15}{4}x - \frac{13}{4}\) | M1 | Using their gradient to find the equation of the tangent |
| Crosses \(x\)-axis when \(y=0\): \(x = \frac{13}{15}\) | A1 | Allow 0.867 or better |
| Crosses \(y\)-axis when \(x=0\): \(y = -\frac{13}{4}\) | A1 | Any form |
| Area of triangle \(= \frac{1}{2} \times \frac{13}{15} \times \frac{13}{4} = \frac{169}{120}\) [below the axis] | A1 [6] | FT their values but must be exact. Accept positive or negative value. Note: Area 1.41 gets 5/6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At stationary point \(\frac{dy}{dx} = 2x - 2x^{-3} = 0\) | M1 | Equating their derivative to zero and attempting to solve |
| \(x^4 = 1\) giving \(x = \pm 1\) | A1 | Finding both roots and no others |
| So there are only two stationary points. \(\frac{d^2y}{dx^2} = 2 + 6x^{-4}\) | M1 | Differentiating their \(\frac{dy}{dx}\). Allow for evaluating gradient at appropriate points either side |
| When \(x=1\): \(\frac{d^2y}{dx^2} = 2 + 6\times1^4 [=8] > 0\) | M1 | Evaluating at (at least one of) their stationary point(s). Arguing point is minimum from their gradients [incl sketch] |
| So minimum point. When \(x=-1\): \(\frac{d^2y}{dx^2} = 2 + 6\times(-1)^4 [=8] > 0\), so also minimum point | A1 | Argues that both points are minimum from correct working. Allow arguing that second derivative is positive everywhere or a symmetry argument |
| The two stationary points are both minimum points so there is no maximum point | E1 [6] | Deduces that there are no maximum points |
## Question 12:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 2x - 2x^{-3}$ | M1 | Attempt to differentiate |
| At $\left(2, \frac{17}{4}\right)$ gradient $= 2\times2 - 2\times2^{-3} = \frac{15}{4}$ | A1 | Correct value in any form |
| Equation of tangent: $y - \frac{17}{4} = \frac{15}{4}(x-2)$, giving $y = \frac{15}{4}x - \frac{13}{4}$ | M1 | Using their gradient to find the equation of the tangent |
| Crosses $x$-axis when $y=0$: $x = \frac{13}{15}$ | A1 | Allow 0.867 or better |
| Crosses $y$-axis when $x=0$: $y = -\frac{13}{4}$ | A1 | Any form |
| Area of triangle $= \frac{1}{2} \times \frac{13}{15} \times \frac{13}{4} = \frac{169}{120}$ [below the axis] | A1 [6] | FT their values but must be exact. Accept positive or negative value. Note: Area 1.41 gets 5/6 marks |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At stationary point $\frac{dy}{dx} = 2x - 2x^{-3} = 0$ | M1 | Equating their derivative to zero and attempting to solve |
| $x^4 = 1$ giving $x = \pm 1$ | A1 | Finding both roots and no others |
| So there are only two stationary points. $\frac{d^2y}{dx^2} = 2 + 6x^{-4}$ | M1 | Differentiating their $\frac{dy}{dx}$. Allow for evaluating gradient at appropriate points either side |
| When $x=1$: $\frac{d^2y}{dx^2} = 2 + 6\times1^4 [=8] > 0$ | M1 | Evaluating at (at least one of) their stationary point(s). Arguing point is minimum from their gradients [incl sketch] |
| So minimum point. When $x=-1$: $\frac{d^2y}{dx^2} = 2 + 6\times(-1)^4 [=8] > 0$, so also minimum point | A1 | Argues that both points are minimum from correct working. Allow arguing that second derivative is positive everywhere or a symmetry argument |
| The two stationary points are both minimum points so there is no maximum point | E1 [6] | Deduces that there are no maximum points |12 In this question you must show detailed reasoning.
Fig. 12 shows part of the graph of $y = x ^ { 2 } + \frac { 1 } { x ^ { 2 } }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a1b6c827-7d74-4527-9b60-58872e3d5ef7-7_574_574_402_233}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
The tangent to the curve $\mathrm { y } = \mathrm { x } ^ { 2 } + \frac { 1 } { \mathrm { x } ^ { 2 } }$ at the point $\left( 2 , \frac { 17 } { 4 } \right)$ meets the $x$-axis at A and meets the $y$-axis at B . O is the origin.
\begin{enumerate}[label=(\alph*)]
\item Find the exact area of the triangle OAB .
\item Use calculus to prove that the complete curve has two minimum points and no maximum point.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2020 Q12 [12]}}