| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find range where function increasing/decreasing |
| Difficulty | Moderate -0.3 This is a straightforward application of differentiation to find increasing/decreasing intervals. Students differentiate a polynomial (routine), solve a quadratic inequality (standard AS technique), and sketch a gradient function. While it requires multiple steps and 'detailed reasoning', the techniques are all standard AS-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07c Sketch gradient function: for given curve1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 24 - 6x - 3x^2\) | M1 | 1.1a: Expression for derivative seen |
| When \(x=0\), \(\frac{dy}{dx} = 24\) | A1 | 1.1: May be shown on graph or in the working |
| \(\frac{dy}{dx} = 0\): \(-3(x^2 + 2x - 8) = 0 \Rightarrow (x+4)(x-2) = 0\) | M1 | 1.1a: Method for solving their quadratic (allow any algebraic method) |
| \(x = -4,\ 2\) | A1 | 3.1a: Must be seen on graph |
| Correct graph: correct shape, maximum point to left of \(y\)-axis | B1 [5] | 1.1: Maximum point should be to the left of the \(y\)-axis but need not be exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Decreasing function when \(\frac{dy}{dx} < 0\) | M1 | 1.1a: Attempt to give the values of \(x\) for which \(\frac{dy}{dx} < 0\) from their graph; condone use of \(\leq\) for M mark |
| \(\{x: x < -4\} \cup \{x: x > 2\}\) | A1 [2] | 2.5: FT their graph if quadratic; allow "\(x < -4\) or \(x > 2\)"; must be correct use of language or set notation; do not allow \(x<-4,\ x>2\) or \(-4>x>2\) |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 24 - 6x - 3x^2$ | M1 | 1.1a: Expression for derivative seen |
| When $x=0$, $\frac{dy}{dx} = 24$ | A1 | 1.1: May be shown on graph or in the working |
| $\frac{dy}{dx} = 0$: $-3(x^2 + 2x - 8) = 0 \Rightarrow (x+4)(x-2) = 0$ | M1 | 1.1a: Method for solving their quadratic (allow any algebraic method) |
| $x = -4,\ 2$ | A1 | 3.1a: Must be seen on graph |
| Correct graph: correct shape, maximum point to left of $y$-axis | B1 [5] | 1.1: Maximum point should be to the left of the $y$-axis but need not be exact |
---
## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Decreasing function when $\frac{dy}{dx} < 0$ | M1 | 1.1a: Attempt to give the values of $x$ for which $\frac{dy}{dx} < 0$ from their graph; condone use of $\leq$ for M mark |
| $\{x: x < -4\} \cup \{x: x > 2\}$ | A1 [2] | 2.5: FT their graph if quadratic; allow "$x < -4$ or $x > 2$"; must be correct use of language or set notation; do not allow $x<-4,\ x>2$ or $-4>x>2$ |
---10 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Sketch the gradient function for the curve $y = 24 x - 3 x ^ { 2 } - x ^ { 3 }$.
\item Determine the set of values of $x$ for which $24 x - 3 x ^ { 2 } - x ^ { 3 }$ is decreasing.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2019 Q10 [7]}}