| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Increasing/decreasing intervals |
| Difficulty | Moderate -0.3 This is a straightforward application of differentiation requiring finding dy/dx, sketching a quadratic, and using the second derivative to determine where the gradient decreases. While it involves multiple steps and 'show detailed reasoning', the techniques are standard AS-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02h Express solutions: using 'and', 'or', set and interval notation1.02m Graphs of functions: difference between plotting and sketching1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 12x^2-12x-9\) | M1 (AO 1.1a) | DR: Attempt to differentiate seen |
| When \(\frac{dy}{dx}=0\): \(12x^2-12x-9=0\); \(3(2x+1)(2x-3)=0\) so \(x=-0.5\), \(1.5\) | M1 (dep), A1 (AO 1.1a, 1.1a) | Attempt to solve their \(\frac{dy}{dx}=0\); both values seen — may be indicated on graph |
| Sketch: correct shape through \((0,-9)\) with minimum at \(x=1.5\) and maximum at \(x=-0.5\) | B1 [4] (AO 1.1) | SC: For cubic graph of the function drawn with M0M0A0, allow SC1 for correct shape with minimum when \(x=1.5\) and maximum when \(x=-0.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Min point of gradient function when \(\frac{d^2y}{dx^2}=24x-12=0\) so \(x=\frac{1}{2}\) | M1 (AO 3.1a) | DR: Attempt to find the vertex (including completing the square or symmetry argument) |
| Gradient is decreasing for \(\left\{x: x<\frac{1}{2}\right\}\) | A1 [2] (AO 2.5) | Inequality correctly formed and expressed as a set; allow either \(<\) or \(\leq\) |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 12x^2-12x-9$ | M1 (AO 1.1a) | DR: Attempt to differentiate seen |
| When $\frac{dy}{dx}=0$: $12x^2-12x-9=0$; $3(2x+1)(2x-3)=0$ so $x=-0.5$, $1.5$ | M1 (dep), A1 (AO 1.1a, 1.1a) | Attempt to solve their $\frac{dy}{dx}=0$; both values seen — may be indicated on graph |
| Sketch: correct shape through $(0,-9)$ with minimum at $x=1.5$ and maximum at $x=-0.5$ | B1 [4] (AO 1.1) | SC: For cubic graph of the function drawn with M0M0A0, allow SC1 for correct shape with minimum when $x=1.5$ and maximum when $x=-0.5$ |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Min point of gradient function when $\frac{d^2y}{dx^2}=24x-12=0$ so $x=\frac{1}{2}$ | M1 (AO 3.1a) | DR: Attempt to find the vertex (including completing the square or symmetry argument) |
| Gradient is decreasing for $\left\{x: x<\frac{1}{2}\right\}$ | A1 [2] (AO 2.5) | Inequality correctly formed and expressed as a set; allow either $<$ or $\leq$ |7 In this question you must show detailed reasoning.\\
A curve has equation $y = 4 x ^ { 3 } - 6 x ^ { 2 } - 9 x + 4$.
\begin{enumerate}[label=(\alph*)]
\item Sketch the gradient function for this curve, clearly indicating the points where the gradient is zero.
\item Find the set of values of $x$ for which the gradient function is decreasing. Give your answer using set notation.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2020 Q7 [6]}}