| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Quadratic inequality solving |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing routine calculus and quadratic techniques. Parts (a)-(b) involve standard differentiation to find and classify a stationary point, (c) requires a basic sketch with shading, and (d) asks for solving a quadratic inequality. All parts are textbook exercises requiring only direct application of learned methods with no problem-solving insight needed. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02h Express solutions: using 'and', 'or', set and interval notation1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 4x - 3 = 0\) at a stationary point | M1 | Attempt to differentiate and equate to zero. Differentiation must be used |
| \(x = 0.75\) | A1 | cao any form |
| When \(x = 0.75\), \(y = 2 \times 0.75^2 - 3 \times 0.75 - 2 = -3.125\) | A1 | cao |
| So stationary point at \((0.75, -3.125)\) | [3] | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = 4 > 0\) so minimum point | M1 | Finding the second derivative FT their derivative. Do not allow from an argument based on the coefficient of \(x^2\) |
| — | E1 [2] | Clear conclusion from consideration of the sign of the second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to sketch a parabola using their labelled minimum point | M1 | Also allow M1 for attempt to sketch parabola using the intersection with at least one axis |
| Parabola with minimum point \((0.75, -3.125)\) and one other correct point clearly shown, e.g. \((0, -2)\) | A1 | A1 parabola through 3 correct points e.g. \((-0.5, 0)\), \((2, 0)\) and \((0, -2)\) |
| \(y \geq 2x^2 - 3x - 2\) is the shaded region above the curve including the boundary | A1 [3] | Area above their curve indicated and the boundary clearly included |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2x+1)(x-2) > 0\) | M1 | Factorising the quadratic or attempting to solve \(2x^2 - 3x - 2 = 0\). Allow M1A1 for roots of quadratic equation BC |
| Boundary values \(x = -\frac{1}{2}\) and \(2\) | A1 | Correct roots of quadratic equation |
| Indicates that the required sets are less than their lower root and more than their upper root | A1 | — |
| \(\left\{x : x < -\frac{1}{2}\right\} \cup \{x : x > 2\}\) | A1 [4] | Correct set notation must be used FT their roots. Allow M1A1A1A0 if solved BC without set notation seen |
# Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 4x - 3 = 0$ at a stationary point | M1 | Attempt to differentiate and equate to zero. Differentiation must be used |
| $x = 0.75$ | A1 | cao any form |
| When $x = 0.75$, $y = 2 \times 0.75^2 - 3 \times 0.75 - 2 = -3.125$ | A1 | cao |
| So stationary point at $(0.75, -3.125)$ | [3] | — |
---
# Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2} = 4 > 0$ so minimum point | M1 | Finding the second derivative FT their derivative. Do not allow from an argument based on the coefficient of $x^2$ |
| — | E1 [2] | Clear conclusion from consideration of the sign of the second derivative |
---
# Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to sketch a parabola using their labelled minimum point | M1 | Also allow M1 for attempt to sketch parabola using the intersection with at least one axis |
| Parabola with minimum point $(0.75, -3.125)$ and one other correct point clearly shown, e.g. $(0, -2)$ | A1 | A1 parabola through 3 correct points e.g. $(-0.5, 0)$, $(2, 0)$ and $(0, -2)$ |
| $y \geq 2x^2 - 3x - 2$ is the shaded region above the curve including the boundary | A1 [3] | Area above their curve indicated and the boundary clearly included |
---
# Question 8(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2x+1)(x-2) > 0$ | M1 | Factorising the quadratic or attempting to solve $2x^2 - 3x - 2 = 0$. Allow M1A1 for roots of quadratic equation BC |
| Boundary values $x = -\frac{1}{2}$ and $2$ | A1 | Correct roots of quadratic equation |
| Indicates that the required sets are less than their lower root and more than their upper root | A1 | — |
| $\left\{x : x < -\frac{1}{2}\right\} \cup \{x : x > 2\}$ | A1 [4] | Correct set notation must be used FT their roots. Allow M1A1A1A0 if solved BC without set notation seen |
---8 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Use differentiation to find the coordinates of the stationary point on the curve with equation $y = 2 x ^ { 2 } - 3 x - 2$.
\item Use the second derivative to determine the nature of the stationary point.
\item Show by shading on a sketch the region defined by the inequality $y \geqslant 2 x ^ { 2 } - 3 x - 2$, indicating clearly whether the boundary is included or not.
\item Solve the inequality $2 x ^ { 2 } - 3 x - 2 > 0$ using set notation for your answer.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2021 Q8 [12]}}