| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Convert Cartesian to polar equation |
| Difficulty | Challenging +1.2 This is a structured proof by induction question with clear scaffolding. Part (a) guides students to discover the pattern through differentiation (standard chain rule applications), part (b) is a routine induction proof once the formula is conjectured, and part (c) is straightforward substitution. While it requires multiple techniques and careful algebraic manipulation, the question provides significant guidance and follows a predictable template for Further Maths induction questions. The derivatives follow a clear factorial pattern that becomes apparent quickly, making this above average but not exceptionally challenging. |
| Spec | 4.01a Mathematical induction: construct proofs4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (a) | dy a =a(1+at)−1 |
| Answer | Marks | Guidance |
|---|---|---|
| dt 1+at | M1 | 1.1 |
| derivative | or dy = a = ( a −1+t )−1 etc |
| Answer | Marks | Guidance |
|---|---|---|
| dt4 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| d t | A1 | 2.2b |
| Answer | Marks |
|---|---|
| (b) | Basis case: n = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| So true for n = 1 | *B1ft | 2.1 |
| Answer | Marks |
|---|---|
| their 1st derivative in part (a) | accept omission of working shown |
| Answer | Marks | Guidance |
|---|---|---|
| d t k + d t d t d t | M1 | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| (which is the formula with n = k + 1) | *A1 | 2.2a |
| Answer | Marks |
|---|---|
| justification must be seen | Could see substitution u = 1 + atetc |
| Answer | Marks | Guidance |
|---|---|---|
| integers n ≥ 1 | dep*A1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | d d 6 y6 = d 7 y7 = ( − 1 ) 6 a 7 ( 7 − 1 ) ! (1 + a t ) − 7 | |
| d t d t d t | M1 | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| d t 1 2 8 8 | A1 | 1.1 |
Question 9:
9 | (a) | dy a =a(1+at)−1
=
dt 1+at | M1 | 1.1 | Differentiating correctly first
derivative | or dy = a = ( a −1+t )−1 etc
dt 1+at
d 2 y2 = − a 2 ( 1 + a t ) − 2 and
d t
d 3 y3 = 2 a 3 ( 1 + a t ) − 3 and
d t
d4y
=2(−3)a4(1+at)−4
oe
dt4 | A1 | 1.1 | For all; constants need not be
evaluated and may be
unsimplified. eg Could see (–
1)2 or (–1)3
d n yn = ( − 1 ) n − 1 a n ( n − 1 ) ! ( 1 + a t ) − n
d t | A1 | 2.2b | Could be (–1)n + 1oe. | ( − 1 ) n − 1 ( n − 1 ) ! ( a − 1 + t ) − n
Omission of factorial term can
score a possible B1M1 in part (b)
[3]
(b) | Basis case: n = 1
d 1 y1 ( = − ) 1 − 1 1 a 1 (1 − 1 ) ( ! 1 + a t − ) 1
d t
= ( − 0 ) 1 a 0 ! ( 1 + a t − 1 ) = a ( 1 + a t ) − 1
B u t 1 d y1 = d y = a ( 1 + a t − ) 1
t d d t
So true for n = 1 | *B1ft | 2.1 | Convincingly showing that their
conjecture works for n = 1 for
their 1st derivative in part (a) | accept omission of working shown
in 1st line here
Assume true for n = k
dky
=(−1)k−1 ak(k−1)!(1+at)−k
ie
dtk
d k + 1 y1 = d d k yk = d ( ( − 1 ) k − 1 a k ( k − 1 ) ! (1 + a t ) − k )
d t k + d t d t d t | M1 | 3.1a | Forming the inductive
hypothesis and making it clear
that the (k + 1)th derivative is
the derivative of the kth
derivative
= ( − 1 ) k − 1 k a ( k − 1 ) ! − ( k ) a ( + 1 a ) t − − k 1
= ( − 1 ) k − 1 − ( 1 ) k a a k ( k − 1 ) ! ( 1 + a t ) − ( k + 1 )
= ( − 1 ) k + ( k 1 a k ! ) + 1 a t − + ( k 1 )
( = ( − 1 ) + − ( k 1 ) 1 a + k 1 ( ( k + 1 ) − 1 ( ) ! 1 + a t − ) ( k + 1 ) )
(which is the formula with n = k + 1) | *A1 | 2.2a | Differentiating and rewriting
into correct form. Some
intermediate working and/or
justification must be seen | Could see substitution u = 1 + atetc
So true for n = k implies true for n = k + 1.
But true for n = 1. Therefore true for all
integers n ≥ 1 | dep*A1 | 2.4 | full correct argument
[4]
(c) | d d 6 y6 = d 7 y7 = ( − 1 ) 6 a 7 ( 7 − 1 ) ! (1 + a t ) − 7
d t d t d t | M1 | 3.1a | Considering the seventh
derivative.
a = 2, t = 32 => d 7 y7 = 2 7 7 2 0 (1 + 3 ) − 7 = 7 2 0 = 4 5
d t 1 2 8 8 | A1 | 1.1 | 5.625. Ignore attempt at units.
[2]9 A function is defined by $y = f ( t )$ where $f ( t ) = \ln ( 1 + a t )$ and $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item By considering $\frac { d y } { d t } , \frac { d ^ { 2 } y } { d t ^ { 2 } } , \frac { d ^ { 3 } y } { d t ^ { 3 } }$ and $\frac { d ^ { 4 } y } { d t ^ { 4 } }$, make a conjecture for a general formula for $\frac { d ^ { n } y } { d t ^ { n } }$ in terms of $n$ and $a$ for any integer $n \geqslant 1$.
\item Use induction to prove the formula conjectured in part (a).
\item In the case where $\mathrm { f } ( t ) = \ln ( 1 + 2 t )$, find the rate at which the $6 ^ { \text {th } }$ derivative of $\mathrm { f } ( t )$ is varying when $t = \frac { 3 } { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2023 Q9 [9]}}