| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Classifying stationary points on surfaces |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring partial differentiation, solving simultaneous equations, finding stationary points, linear approximation using the tangent plane, and finding normals to surfaces. While these are standard FP3 techniques, the multi-part structure, algebraic manipulation required (especially factoring in part i), and the conceptual understanding needed for the tangent plane approximation in part iii make this significantly harder than average A-level questions but not exceptionally difficult for Further Maths. |
| Spec | 8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05e Stationary points: where partial derivatives are zero8.05f Nature of stationary points: classify using Hessian matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{\partial z}{\partial x} = 6x^2+6x+12y\) | B1 | |
| \(\dfrac{\partial z}{\partial y} = 6y^2+6y+12x\) | B1 | |
| If \(\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial y}\): \(6x^2+6x+12y = 6y^2+6y+12x\) | ||
| \(x^2-y^2-x+y=0\) | ||
| \((x-y)(x+y-1)=0\) | M1 | Identifying factor \((x-y)\) |
| \(y=x\) or \(y=1-x\) | E1E1 [5] | SC if M0: B1 for verifying \(y=x\); B1 for verifying \(y=1-x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial y} = 0\) | M1 | Can be implied |
| If \(y=x\) then \(6x^2+6x+12x=0\) | M1 | Obtaining quadratic in \(x\) (or \(y\)) |
| \(x=0,\; -3\) | M1 | Obtaining a non-zero value of \(x\) |
| Stationary points \((0,\;0,\;0)\) and \((-3,\;-3,\;54)\) | B1A1 | Condone \((0,0)\) for B1 |
| If \(y=1-x\) then \(6x^2+6x+12(1-x)=0\) | M1 | Obtaining quadratic with no real roots |
| \(x^2-x+2=0\) | ||
| Which has no real roots \((D=-7<0)\) | A1 [7] | Correctly shown; Just stating 'No real roots' M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At P: \(\dfrac{\partial z}{\partial x} = \dfrac{21}{2}\), \(\dfrac{\partial z}{\partial y} = \dfrac{21}{2}\) | M1, A1 | Substituting into \(\dfrac{\partial z}{\partial x}\) or \(\dfrac{\partial z}{\partial y}\); Correct value or substitution seen |
| \(\delta z \approx \dfrac{\partial z}{\partial x}\delta x + \dfrac{\partial z}{\partial y}\delta y\) | M1 | |
| \(w \approx \dfrac{21}{2}h + \dfrac{21}{2}h\) | A1 ft | |
| \(h \approx \dfrac{w}{21}\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial y} = 24\) | M1 | Allow sign error; \(24\lambda\) is M0 unless \(\lambda=\pm1\) appears later |
| If \(y=x\) then \(6x^2+6x+12x=24\); \(x=1,\;-4\) | M1 | Obtaining quadratic in \(x\) (or \(y\)) |
| Points \((1,\;1,\;22)\) and \((-4,\;-4,\;32)\) | A1A1 | If neither correct, give A1 for \(x=1,\,-4\) |
| If \(y=1-x\) then \(6x^2+6x+12(1-x)=24\) | M1 | Obtaining quadratic in \(x\) (or \(y\)); Or third linear factor of quartic |
| \(x=2,\;-1\) | ||
| Points \((2,\;-1,\;5)\) and \((-1,\;2,\;5)\) | A1A1 [7] | If neither correct, give A1 for \(x=2,\,-1\) |
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{\partial z}{\partial x} = 6x^2+6x+12y$ | B1 | |
| $\dfrac{\partial z}{\partial y} = 6y^2+6y+12x$ | B1 | |
| If $\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial y}$: $6x^2+6x+12y = 6y^2+6y+12x$ | | |
| $x^2-y^2-x+y=0$ | | |
| $(x-y)(x+y-1)=0$ | M1 | Identifying factor $(x-y)$ |
| $y=x$ or $y=1-x$ | E1E1 [5] | SC if M0: B1 for verifying $y=x$; B1 for verifying $y=1-x$ |
---
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial y} = 0$ | M1 | Can be implied |
| If $y=x$ then $6x^2+6x+12x=0$ | M1 | Obtaining quadratic in $x$ (or $y$) |
| $x=0,\; -3$ | M1 | Obtaining a non-zero value of $x$ |
| Stationary points $(0,\;0,\;0)$ and $(-3,\;-3,\;54)$ | B1A1 | Condone $(0,0)$ for B1 |
| If $y=1-x$ then $6x^2+6x+12(1-x)=0$ | M1 | Obtaining quadratic with no real roots |
| $x^2-x+2=0$ | | |
| Which has no real roots $(D=-7<0)$ | A1 [7] | Correctly shown; Just stating 'No real roots' M1A0 |
---
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| At P: $\dfrac{\partial z}{\partial x} = \dfrac{21}{2}$, $\dfrac{\partial z}{\partial y} = \dfrac{21}{2}$ | M1, A1 | Substituting into $\dfrac{\partial z}{\partial x}$ or $\dfrac{\partial z}{\partial y}$; Correct value or substitution seen |
| $\delta z \approx \dfrac{\partial z}{\partial x}\delta x + \dfrac{\partial z}{\partial y}\delta y$ | M1 | |
| $w \approx \dfrac{21}{2}h + \dfrac{21}{2}h$ | A1 ft | |
| $h \approx \dfrac{w}{21}$ | A1 [5] | |
---
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial y} = 24$ | M1 | Allow sign error; $24\lambda$ is M0 unless $\lambda=\pm1$ appears later |
| If $y=x$ then $6x^2+6x+12x=24$; $x=1,\;-4$ | M1 | Obtaining quadratic in $x$ (or $y$) |
| Points $(1,\;1,\;22)$ and $(-4,\;-4,\;32)$ | A1A1 | If neither correct, give A1 for $x=1,\,-4$ |
| If $y=1-x$ then $6x^2+6x+12(1-x)=24$ | M1 | Obtaining quadratic in $x$ (or $y$); Or third linear factor of quartic |
| $x=2,\;-1$ | | |
| Points $(2,\;-1,\;5)$ and $(-1,\;2,\;5)$ | A1A1 [7] | If neither correct, give A1 for $x=2,\,-1$ |
---2 A surface has equation $z = 2 \left( x ^ { 3 } + y ^ { 3 } \right) + 3 \left( x ^ { 2 } + y ^ { 2 } \right) + 12 x y$.\\
(i) For a point on the surface at which $\frac { \partial z } { \partial x } = \frac { \partial z } { \partial y }$, show that either $y = x$ or $y = 1 - x$.\\
(ii) Show that there are exactly two stationary points on the surface, and find their coordinates.\\
(iii) The point $\mathrm { P } \left( \frac { 1 } { 2 } , \frac { 1 } { 2 } , 5 \right)$ is on the surface, and $\mathrm { Q } \left( \frac { 1 } { 2 } + h , \frac { 1 } { 2 } + h , 5 + w \right)$ is a point on the surface close to P . Find an approximate expression for $h$ in terms of $w$.\\
(iv) Find the four points on the surface at which the normal line is parallel to the vector $24 \mathbf { i } + 24 \mathbf { j } - \mathbf { k }$.
\hfill \mbox{\textit{OCR MEI FP3 2013 Q2 [24]}}