| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2017 |
| Session | December |
| Marks | 9 |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvalues of 2×2 matrix |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring multi-variable calculus (partial derivatives), analysis of sections to locate critical points, and application of the second derivative test for saddle points. While the algebraic manipulation is manageable, the conceptual demand of working with surfaces and proving the existence of a saddle point elevates this significantly above standard A-level questions. |
| Spec | 8.05d Partial differentiation: first and second order, mixed derivatives8.05e Stationary points: where partial derivatives are zero8.05f Nature of stationary points: classify using Hessian matrix |
| Answer | Marks | Guidance |
|---|---|---|
| (i) (a) either When \(y = -3\), \(z = 3x^2 + 6x + 134\); \(\frac{dz}{dx} = 6x + 6 = 0\) | B1 M1 | 3.1a 1.1 |
| or \(\frac{\partial z}{\partial x} = -12xy + 60y - 30x + 186\); When \(y = -3\) \(\frac{\partial z}{\partial x} = 6x + 6 = 0\); Stationary point at \((-1, 131)\) | B1 M1 A1 [3] | |
| (b) either When \(x = -1\), \(z = 8y^3 - 216y - 301\); \(\frac{dz}{dy} = 24y^2 - 216 = 0\) | B1 M1 | 3.1a 1.1 |
| or \(\frac{\partial z}{\partial y} = 24y^3 - 6x^2 + 60x - 150\); When \(\frac{\partial z}{\partial y} = 24y^2 - 216 = 0\); Stationary points at \((-3, 131)\) and \((3, -733)\) | B1 M1 M1 A1 A1 [4] | |
| (ii) The point \((-1, -3, 131)\) is a stationary point on both sections, so it is a stationary point on the surface. It is a minimum for section given by \(y = -3\) and a maximum for the section given by \(x = -1\) so it is a saddle point. | E1 E1 [2] | 3.2a 2.2a |
**(i)** **(a)** **either** When $y = -3$, $z = 3x^2 + 6x + 134$; $\frac{dz}{dx} = 6x + 6 = 0$ | B1 M1 | 3.1a 1.1 | Find the equation of the section; Differentiate and set equal to zero
**or** $\frac{\partial z}{\partial x} = -12xy + 60y - 30x + 186$; When $y = -3$ $\frac{\partial z}{\partial x} = 6x + 6 = 0$; Stationary point at $(-1, 131)$ | B1 M1 A1 [3] | | Find partial differential with respect to x; Substitute $y = -3$ and set equal to zero
**(b)** **either** When $x = -1$, $z = 8y^3 - 216y - 301$; $\frac{dz}{dy} = 24y^2 - 216 = 0$ | B1 M1 | 3.1a 1.1 | Find the equation of the section; Differentiate
**or** $\frac{\partial z}{\partial y} = 24y^3 - 6x^2 + 60x - 150$; When $\frac{\partial z}{\partial y} = 24y^2 - 216 = 0$; Stationary points at $(-3, 131)$ and $(3, -733)$ | B1 M1 M1 A1 A1 [4] | | Find partial differential with respect to y; Substitute $x = -1$ and set equal to zero
**(ii)** The point $(-1, -3, 131)$ is a stationary point on both sections, so it is a stationary point on the surface. It is a minimum for section given by $y = -3$ and a maximum for the section given by $x = -1$ so it is a saddle point. | E1 E1 [2] | 3.2a 2.2a8 A surface $S$ has equation $z = 8 y ^ { 3 } - 6 x ^ { 2 } y + 60 x y - 15 x ^ { 2 } + 186 x - 150 y - 100$.
\begin{enumerate}[label=(\roman*)]
\item (a) Find any stationary points of the section of $S$ given by $y = - 3$.\\
(b) Find any stationary points of the section of $S$ given by $x = - 1$.
\item Show that the surface $S$ has a least one saddle point.
\section*{OCR}
Oxford Cambridge and RSA
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2017 Q8 [9]}}