| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vector Product and Surfaces |
| Type | Finding stationary points on surfaces |
| Difficulty | Challenging +1.2 This is a standard Further Maths partial differentiation question requiring finding stationary points and completing the square to determine when z > 0. The differentiation is routine, and while completing the square in two variables requires care, it's a well-practiced technique at this level with clear methodology. |
| Spec | 8.05a 3D surfaces: z = f(x,y) and implicit form, partial derivatives8.05e Stationary points: where partial derivatives are zero |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | a | ππ§ ππ§ |
| Answer | Marks |
|---|---|
| ππ₯ ππ¦ | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Both first partial derivatives attempted |
| Answer | Marks |
|---|---|
| b | Considering z as a function of 2x + y |
| Answer | Marks |
|---|---|
| 4 | M1 |
| Answer | Marks |
|---|---|
| B1 | 3.1a |
| Answer | Marks |
|---|---|
| 2.4 | a and b non-zero constants |
| Answer | Marks |
|---|---|
| 8π₯ + 4π¦ + 6 = 0 and/or 4x + 2y + 3 = 0 | M1 |
| Answer | Marks |
|---|---|
| 4 2 2 | M1 |
| Substituting either into z | M1 |
| Correct substitution | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | A1 | seen |
| For S always lying above the x-y planeπ§ > 0 | B1 | Must justify that a minimum value of z occurs when |
| Answer | Marks |
|---|---|
| 4 | B1 |
Question 9:
9 | a | ππ§ ππ§
= 8x + 4y + 6 and = 4x + 2y + 3
ππ₯ ππ¦ | M1
A1
[2] | 1.1
1.1 | Both first partial derivatives attempted
Both correct
b | Considering z as a function of 2x + y
z = (2x + y)2 + a(2x + y) + b
a=3 and b=k
For S always lying above the x-y planeπ§ > 0
π‘ = 2π₯ +π¦ so π§ = π‘2 +3π‘+π (> 0) soi
Discriminant of π‘2 +3π‘+π <0
Discriminant=9-4k
9
π > so least integer k is 3
4 | M1
M1
A1
B1
M1
A1
B1 | 3.1a
3.2a
2.1
1.1
3.1a
1.1
2.4 | a and b non-zero constants
2
3
Or completing the square (π‘+ ) +π (This implies
2
the first two M1) and stating π > 0
2
3 9
If completing the square, (π‘+ ) β +π
2 4
This implies the first A1
SC1 k=3 with no evidence
Alternative method
8π₯ + 4π¦ + 6 = 0 and/or 4x + 2y + 3 = 0 | M1
3 1 3
π₯ = β β π¦ or y = β2π₯ β
4 2 2 | M1
Substituting either into z | M1
Correct substitution | A1
9
β +π
4 | A1 | seen
For S always lying above the x-y planeπ§ > 0 | B1 | Must justify that a minimum value of z occurs when
4x + 2y + 3 = 0 (eg through a sketch)
9
ο π > so least integer k is 3
4 | B1
[7]
PMT
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Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.9 For all real values of $x$ and $y$ the surface $S$ has equation $z = 4 x ^ { 2 } + 4 x y + y ^ { 2 } + 6 x + 3 y + k$, where $k$ is a constant and an integer.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \partial z } { \partial x }$ and $\frac { \partial z } { \partial y }$.
\item Determine the smallest value of the integer $k$ for which the whole of $S$ lies above the $x - y$ plane.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2022 Q9 [9]}}