| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Second derivative test justification |
| Difficulty | Moderate -0.3 This is a straightforward stationary points question requiring routine differentiation of powers of x, solving a simple equation, and applying the second derivative test. While it involves multiple steps and simplification of surds, all techniques are standard Core 1/2 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07e Second derivative: as rate of change of gradient1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 6x^{0.5} - 24x^{-1.5}\) | M1 A1 A1 | M1 for correct power on any of the three terms including \(\sqrt{8} \to 0\). A1 for two of three terms correctly differentiated. A1 cao. Accept \(6\sqrt{x} - \frac{24}{x\sqrt{x}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = 3x^{-0.5} + 36x^{-2.5}\) | M1 A1 | M1 for reducing any fractional power by one. A1 cao. Accept \(\frac{3}{\sqrt{x}} - \frac{36}{x^2\sqrt{x}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 0 \Rightarrow 6x^{0.5} - 24x^{-1.5} = 0\) | M1 | Sets (or implies) \(\frac{dy}{dx} = 0\) |
| \(x^2 = 4 \Rightarrow x = 2\) | dM1, A1 | Dependent on previous M. Forms equation of type \(x^n = A\) following correct index work. \(x=2\) (ignore \(x=-2\)) |
| Substitutes \(x=2\) into \(y = 4x\sqrt{x} + \frac{48}{\sqrt{x}} - \sqrt{8} \Rightarrow y = 30\sqrt{2}\) | M1, A1 | All three M marks must have been scored for A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Substitutes \(x=2\) into \(\frac{d^2y}{dx^2} = 3x^{-0.5} + 36x^{-2.5}\) | M1 | For substituting \(x=2\) and finding a numerical result |
| \(\frac{d^2y}{dx^2} > 0 \Rightarrow\) minimum | A1 cso | Statement and reason required. Accept \(3\times2^{-0.5}+36\times2^{-2.5}>0\Rightarrow\) minimum. If numerical value given it must be correct: accept \(6\sqrt{2}\) or awrt 8.5 |
## Question 4:
**(a)(i)**
| $\frac{dy}{dx} = 6x^{0.5} - 24x^{-1.5}$ | M1 A1 A1 | M1 for correct power on any of the three terms including $\sqrt{8} \to 0$. A1 for two of three terms correctly differentiated. A1 cao. Accept $6\sqrt{x} - \frac{24}{x\sqrt{x}}$ |
**(a)(ii)**
| $\frac{d^2y}{dx^2} = 3x^{-0.5} + 36x^{-2.5}$ | M1 A1 | M1 for reducing any fractional power by one. A1 cao. Accept $\frac{3}{\sqrt{x}} - \frac{36}{x^2\sqrt{x}}$ |
**(b)**
| $\frac{dy}{dx} = 0 \Rightarrow 6x^{0.5} - 24x^{-1.5} = 0$ | M1 | Sets (or implies) $\frac{dy}{dx} = 0$ |
| $x^2 = 4 \Rightarrow x = 2$ | dM1, A1 | Dependent on previous M. Forms equation of type $x^n = A$ following correct index work. $x=2$ (ignore $x=-2$) |
| Substitutes $x=2$ into $y = 4x\sqrt{x} + \frac{48}{\sqrt{x}} - \sqrt{8} \Rightarrow y = 30\sqrt{2}$ | M1, A1 | All three M marks must have been scored for A1 |
**(c)**
| Substitutes $x=2$ into $\frac{d^2y}{dx^2} = 3x^{-0.5} + 36x^{-2.5}$ | M1 | For substituting $x=2$ and finding a numerical result |
| $\frac{d^2y}{dx^2} > 0 \Rightarrow$ minimum | A1 cso | Statement and reason required. Accept $3\times2^{-0.5}+36\times2^{-2.5}>0\Rightarrow$ minimum. If numerical value given it must be correct: accept $6\sqrt{2}$ or awrt 8.5 |
---4. The curve $C$ has equation $y = 4 x \sqrt { x } + \frac { 48 } { \sqrt { x } } - \sqrt { 8 } , x > 0$
\begin{enumerate}[label=(\alph*)]
\item Find, simplifying each term,
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { d } y } { \mathrm {~d} x }$
\item $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$
\end{enumerate}\item Use part (a) to find the exact coordinates of the stationary point of $C$.
\item Determine whether the stationary point of $C$ is a maximum or minimum, giving a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q4 [12]}}