| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Second derivative test justification |
| Difficulty | Moderate -0.3 This is a standard C2 stationary points question requiring routine differentiation (including negative power), solving dy/dx=0, finding second derivative, and applying the second derivative test. All steps are algorithmic with no problem-solving insight needed, making it slightly easier than average, though the negative power and two turning points add minor complexity. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07f Convexity/concavity: points of inflection1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = -3+\frac{12}{x^4}\) or \(-3+12x^{-4}\) | M1 A1 | M1: \(x^n \to x^{n-1}\); A1: Correct derivative |
| \(\frac{dy}{dx}=0 \Rightarrow -3+\frac{12}{x^4}=0 \Rightarrow x=...\) or substitutes \(x=\sqrt{2}\) into \(y'\) | M1 | \(y'=0\) and attempt to solve for \(x\) |
| \(x^4=4\) and \(x=\sqrt{2}\), or \(\frac{dy}{dx}=-3+\frac{12}{(\sqrt{2})^4}=-3+12(\sqrt{2})^{-4}=0\) | A1 | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = -\sqrt{2}\) | B1 | Awrt \(-1.41\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = \frac{-48}{x^5}\) or \(-48x^{-5}\) | B1ft | Follow through their first derivative from part (a) |
| Answer | Marks | Guidance |
|---|---|---|
| An appreciation that either \(y''>0 \Rightarrow\) a minimum or \(y''<0 \Rightarrow\) a maximum | B1 | |
| Maximum at P as \(y''<0\) | B1 | Cso; need fully correct solution, reference to P or \(\sqrt{2}\) and negative/\(< 0\) and maximum; no incorrect/contradictory statements |
| Minimum at Q as \(y''>0\) | B1 | Cso; need fully correct solution, part (b) must be correct, reference to P or \(-\sqrt{2}\) and positive/\(> 0\) and minimum; no incorrect/contradictory statements |
## Question 8:
$y = 6-3x-\frac{4}{x^3}$
### Part (a):
| $\frac{dy}{dx} = -3+\frac{12}{x^4}$ or $-3+12x^{-4}$ | M1 A1 | M1: $x^n \to x^{n-1}$; A1: Correct derivative |
|---|---|---|
| $\frac{dy}{dx}=0 \Rightarrow -3+\frac{12}{x^4}=0 \Rightarrow x=...$ or substitutes $x=\sqrt{2}$ into $y'$ | M1 | $y'=0$ and attempt to solve for $x$ |
| $x^4=4$ and $x=\sqrt{2}$, or $\frac{dy}{dx}=-3+\frac{12}{(\sqrt{2})^4}=-3+12(\sqrt{2})^{-4}=0$ | A1 | Correct completion with no errors |
### Part (b):
| $x = -\sqrt{2}$ | B1 | Awrt $-1.41$ |
|---|---|---|
### Part (c):
| $\frac{d^2y}{dx^2} = \frac{-48}{x^5}$ or $-48x^{-5}$ | B1ft | Follow through their first derivative from part (a) |
|---|---|---|
### Part (d):
| An appreciation that either $y''>0 \Rightarrow$ a minimum or $y''<0 \Rightarrow$ a maximum | B1 | |
|---|---|---|
| Maximum at P as $y''<0$ | B1 | Cso; need fully correct solution, reference to P or $\sqrt{2}$ and negative/$< 0$ and maximum; no incorrect/contradictory statements |
| Minimum at Q as $y''>0$ | B1 | Cso; need fully correct solution, part (b) must be correct, reference to P or $-\sqrt{2}$ and positive/$> 0$ and minimum; no incorrect/contradictory statements |
---8. The curve $C$ has equation $y = 6 - 3 x - \frac { 4 } { x ^ { 3 } } , x \neq 0$
\begin{enumerate}[label=(\alph*)]
\item Use calculus to show that the curve has a turning point $P$ when $x = \sqrt { } 2$
\item Find the $x$-coordinate of the other turning point $Q$ on the curve.
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item Hence or otherwise, state with justification, the nature of each of these turning points $P$ and $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2013 Q8 [9]}}