| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| 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 AS-level differentiation question requiring basic power rule application (including negative powers) and solving a quadratic inequality. While it involves multiple steps, each is routine: differentiate, set derivative > 0, solve. The algebra is simple and the question type is standard textbook fare, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^n \rightarrow x^{n-1}\) | M1 | 1.1b – Score for \(x^2 \rightarrow x\) or \(x^{-1} \rightarrow x^{-2}\); allow unprocessed indices |
| Sight of either \(6x\) or \(-\frac{24}{x^2}\) (may be unsimplified) | A1 | 1.1b – Condone additional term e.g. \(+2\); indices must now be processed |
| \(\frac{dy}{dx} = 6x - \frac{24}{x^2}\) | A1 | 1.1b – Accept \(\frac{dy}{dx} = 6x^1 - 24x^{-2}\); do not need to see \(\frac{dy}{dx}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts \(6x - \frac{24}{x^2} > 0 \Rightarrow x >\) | M1 | 1.1b – Sets allowable \(\frac{dy}{dx}...0\); \(\frac{dy}{dx}\) must be in form \(Ax + Bx^{-2}\), \(A,B \neq 0\); intermediate equation of form \(x^p...q\); may be implied by \(x =\) awrt 1.59 |
| \(x > \sqrt[3]{4}\) or \(x \geqslant \sqrt[3]{4}\) | A1 | 2.5 – Accept \(x > 4^{\frac{1}{3}}\) or \(x \geqslant 2^{\frac{2}{3}}\) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^n \rightarrow x^{n-1}$ | M1 | 1.1b – Score for $x^2 \rightarrow x$ or $x^{-1} \rightarrow x^{-2}$; allow unprocessed indices |
| Sight of either $6x$ or $-\frac{24}{x^2}$ (may be unsimplified) | A1 | 1.1b – Condone additional term e.g. $+2$; indices must now be processed |
| $\frac{dy}{dx} = 6x - \frac{24}{x^2}$ | A1 | 1.1b – Accept $\frac{dy}{dx} = 6x^1 - 24x^{-2}$; do not need to see $\frac{dy}{dx}$ |
**(3 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts $6x - \frac{24}{x^2} > 0 \Rightarrow x >$ | M1 | 1.1b – Sets allowable $\frac{dy}{dx}...0$; $\frac{dy}{dx}$ must be in form $Ax + Bx^{-2}$, $A,B \neq 0$; intermediate equation of form $x^p...q$; may be implied by $x =$ awrt 1.59 |
| $x > \sqrt[3]{4}$ or $x \geqslant \sqrt[3]{4}$ | A1 | 2.5 – Accept $x > 4^{\frac{1}{3}}$ or $x \geqslant 2^{\frac{2}{3}}$ |
**(2 marks)**
---\begin{enumerate}
\item A curve has equation
\end{enumerate}
$$y = 3 x ^ { 2 } + \frac { 24 } { x } + 2 \quad x > 0$$
(a) Find, in simplest form, $\frac { \mathrm { d } y } { \mathrm {~d} x }$\\
(b) Hence find the exact range of values of $x$ for which the curve is increasing.
\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q5 [5]}}