| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Algebraic function with square root |
| Difficulty | Challenging +1.8 This is a substantial Further Maths reduction formula question requiring multiple techniques: differentiation of composite functions, integration by parts to establish a recurrence relation, and arc length for parametric curves. While the individual steps are standard for Further Maths students, the multi-part structure, algebraic manipulation needed for the reduction formula, and the connection between parts (especially recognizing the arc length simplifies to Iβ) require solid technical facility and some mathematical maturity. It's harder than a typical A-level question but represents expected Further Maths material rather than requiring exceptional insight. |
| Spec | 8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | a | 3 t 1 6 + t 2 |
| [1] | 1.1 | |
| b | I = n ο² t n β 1 . t 1 6 + t 2 d t = |
| Answer | Marks |
|---|---|
| n n β 2 | M1 |
| Answer | Marks |
|---|---|
| [5] | 3.1a |
| Answer | Marks |
|---|---|
| 2.2a | Use of integration by parts, with appropriate splitting, |
| Answer | Marks |
|---|---|
| c | β’ β’ |
| Answer | Marks |
|---|---|
| = t 3 t 2 + 1 6 d t = I 3 | M1* |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Attempt at differentiation using the Product rule |
Question 7:
7 | a | 3 t 1 6 + t 2 | B1
[1] | 1.1
b | I = n ο² t n β 1 . t 1 6 + t 2 d t =
π π‘π β 1(16+π‘2)3/2 βπβ«(πβ1)π‘π β 2(16+π‘2)3/2 dπ‘
a=b=1/3
3I = n t n β 1 (1 6 + t 2 ) 3 / 2 β ( n β 1 ) ο² t n β 2 (. 1 6 + t 2 ) 1 6 + t 2 d t
= t n β 1 (1 6 + t 2 ) 3 / 2 β ( n β 1 ) (1 6 I + I )
n β 2 n
ο (n + 2) I = (125Γ3π β 1) β 16(n β 1) I
n n β 2 | M1
A1
M1*
M1dep
A1
[5] | 3.1a
1.1
3.1a
2.1
2.2a | Use of integration by parts, with appropriate splitting,
so that (a)βs result can be used
1st stage all correct
Splitting (16+π‘2)3/2 in the integral appropriately
Integral(s) correctly expressed in terms of Iβs
AG cao (showing correct substitution of limits at some
point)
c | β’ β’
π₯ = βπ‘4sinπ‘+4π‘3cosπ‘ or π¦ = Β±π‘4cosπ‘+4π‘3sinπ‘
ο¦ο§ο¨ β’x οΆο·οΈ 2 ο¦ο§ο¨ β’y οΆο·οΈ 2 (t )
+ = 8 s in 2 t β 8 t 7 s in t c o s t + 1 6 t 6 c o s 2 t
+ (π‘8cos2π‘+8π‘7sinπ‘cosπ‘+
16π‘6sin2π‘)
L = ο² t8 +16t6 dt
3ο²0
= t 3 t 2 + 1 6 d t = I 3 | M1*
A1
M1dep
A1
[4] | 1.1
1.1
1.1
1.1 | Attempt at differentiation using the Product rule
ο¦β’οΆ 2 ο¦β’οΆ 2
Use of arc-length formula with their ο§xο· +ο§yο·
ο¨ οΈ ο¨ οΈ
13
AG www, including limits (note that I = 94 )
3
157
\begin{enumerate}[label=(\alph*)]
\item Differentiate $\left( 16 + t ^ { 2 } \right) ^ { \frac { 3 } { 2 } }$ with respect to $t$.
Let $I _ { n } = \int _ { 0 } ^ { 3 } t ^ { n } \sqrt { 16 + t ^ { 2 } } d t$ for integers $n \geqslant 1$.
\item Show that, for $n \geqslant 3 , \left. ( n + 2 ) \right| _ { n } = 125 \times 3 ^ { n - 1 } - \left. 16 ( n - 1 ) \right| _ { n - 2 }$.
\item The curve $C$ is defined parametrically by $\mathrm { x } = \mathrm { t } ^ { 4 } \cos \mathrm { t }$, $\mathrm { y } = \mathrm { t } ^ { 4 } \sin \mathrm { t }$, for $0 \leqslant t \leqslant 3$. The length of $C$ is denoted by $L$.
Show that $\mathrm { L } = \mathrm { I } _ { 3 }$. (You are not required to evaluate this integral.)
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2022 Q7 [10]}}