| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Sigma Notation Manipulation |
| Difficulty | Challenging +1.2 This is an Euler's theorem for homogeneous functions question requiring partial differentiation and algebraic verification. While it involves Further Maths content (partial derivatives, homogeneous functions), the execution is mechanical: compute partial derivatives, substitute into given expressions, and verify the identity. The pattern recognition (homogeneous function of degree 9) and algebraic manipulation are straightforward with no novel insight required beyond applying standard techniques. |
| Spec | 8.05d Partial differentiation: first and second order, mixed derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | f = 4xy7 +15x4y4 −40x7y and f = 14x2y6 +12x5y3 −5x8 |
| Answer | Marks |
|---|---|
| = 18x2y7 +27x5y4 −45x8y = 9 f | M1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Clear attempt at partial differentiation |
| Answer | Marks |
|---|---|
| (b) | f = 4y7 +60x3y4 −280x6y and f = 28xy6 +60x4y3 −40x7 |
| Answer | Marks |
|---|---|
| x | B1 B1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| a | b | c |
| a | b | d |
| b | c | a |
| c | a | b |
| d | d | c |
Question 2:
2 | (a) | f = 4xy7 +15x4y4 −40x7y and f = 14x2y6 +12x5y3 −5x8
x y
Then xf + yf = 4x2y7 +15x5y4 −40x8y + 14x2y7 +12x5y4 −5x8y
x y
= 18x2y7 +27x5y4 −45x8y = 9 f | M1
A1 A1
M1
A1
[5] | 1.1
1.1 1.1
2.1
1.1 | Clear attempt at partial differentiation
n = 9 must be clearly stated
(b) | f = 4y7 +60x3y4 −280x6y and f = 28xy6 +60x4y3 −40x7
xx xy
Then xf + yf = 4xy7 +60x4y4 −280x7y + 28xy7 +60x4y4 −40x7y
xx xy
= 32xy7 +120x4y4 −320x7y = 8f
x | B1 B1
M1
A1 | 1.1 1.1
2.1
1.1
Alternative method
Differentiate. (a)’s result w.r.t. x : xf + f + yf = nf M1 A1
xx x yx x
f = f (by the Mixed Derivative Theorem) B1
yx xy
⇒ xf + yf = (n – 1)f A1
xx xy x
[4]
a | b | c | d
a | b | d | a | c
b | c | a | d | b
c | a | b | c | d
d | d | c | b | a
Column 1
Column 3
Any 3rd R or C2 For $x , y \in \mathbb { R }$, the function f is given by $\mathrm { f } ( x , y ) = 2 x ^ { 2 } \mathrm { y } ^ { 7 } + 3 x ^ { 5 } y ^ { 4 } - 5 x ^ { 8 } y$.
\begin{enumerate}[label=(\alph*)]
\item Prove that $\mathrm { xf } _ { \mathrm { x } } + \mathrm { yf } _ { \mathrm { y } } = \mathrm { nf }$, where $n$ is a positive integer to be determined.
\item Show that $\mathrm { xf } _ { \mathrm { xx } } + \mathrm { yf } _ { \mathrm { xy } } = ( \mathrm { n } - 1 ) \mathrm { f } _ { \mathrm { x } }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2020 Q2 [9]}}