| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary points coordinates |
| Difficulty | Moderate -0.8 This is a straightforward C2 differentiation question requiring standard techniques: differentiate (including the power rule for x^{-2}), set dy/dx = 0, solve the resulting equation, and substitute back to find y. It's more routine than average A-level questions since it follows a well-practiced algorithm with no conceptual challenges or multi-step problem-solving. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(\frac{dy}{dx} = 2 - 16x^{-3}\) | M1 A1 | 1st M1: At least one term differentiated (not integrated) correctly, so \(2x \to 2\), or \(\frac{8}{x^2} \to -16x^{-3}\), or \(3 \to 0\). A1: This answer or equivalent e.g. \(2 - \frac{16}{x^3}\) |
| \(2 - 16x^{-3} = 0\) so \(x^{-3} = \ldots\) or \(x^3 = \ldots\), or \(2 - 16x^{-3} = 0\) so \(x = 2\) | M1 | 2nd M1: Sets \(\frac{dy}{dx}\) to 0, and solves to give \(x^3 = \text{value}\) or \(x^{-3} = \text{value}\) (or states \(x=2\) with no working following correctly stated \(2-16x^{-3}=0\)) |
| \(x = 2\) only (after correct derivative) | A1 | A1: \(x = 2\) cso (if \(x = -2\) is included this is A0 here) |
| \(y = 2 \times \text{"2"} + 3 + \frac{8}{\text{"2"}^2}\) | M1 | 3rd M1: Attempts to substitute their positive \(x\) (found from attempt to differentiate) into \(y = 2x + 3 + \frac{8}{x^2}\), \(x > 0\). Or may be implied by \(y = 9\) or correct follow through from their positive \(x\) |
| \(= 9\) | A1 | A1: 9 cao (Does not need to be written as coordinates) (ignore the extra \((-2, 1)\) here) |
| (6) | Total: 6 |
## Question 1:
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $\frac{dy}{dx} = 2 - 16x^{-3}$ | M1 A1 | 1st M1: At least one term **differentiated (not integrated)** correctly, so $2x \to 2$, or $\frac{8}{x^2} \to -16x^{-3}$, or $3 \to 0$. A1: This answer or equivalent e.g. $2 - \frac{16}{x^3}$ |
| $2 - 16x^{-3} = 0$ so $x^{-3} = \ldots$ or $x^3 = \ldots$, or $2 - 16x^{-3} = 0$ so $x = 2$ | M1 | 2nd M1: Sets $\frac{dy}{dx}$ to 0, and solves to give $x^3 = \text{value}$ or $x^{-3} = \text{value}$ (or states $x=2$ with no working following correctly stated $2-16x^{-3}=0$) |
| $x = 2$ only (after correct derivative) | A1 | A1: $x = 2$ cso (if $x = -2$ is included this is A0 here) |
| $y = 2 \times \text{"2"} + 3 + \frac{8}{\text{"2"}^2}$ | M1 | 3rd M1: Attempts to substitute **their positive** $x$ (found from attempt to differentiate) into $y = 2x + 3 + \frac{8}{x^2}$, $x > 0$. Or may be implied by $y = 9$ or correct follow through from their positive $x$ |
| $= 9$ | A1 | A1: 9 cao (Does not need to be written as coordinates) (ignore the extra $(-2, 1)$ here) |
| | **(6)** | **Total: 6** |\begin{enumerate}
\item Using calculus, find the coordinates of the stationary point on the curve with equation
\end{enumerate}
$$y = 2 x + 3 + \frac { 8 } { x ^ { 2 } } , \quad x > 0$$
\hfill \mbox{\textit{Edexcel C2 2013 Q1 [6]}}