| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Complex partial fractions with multiple techniques |
| Difficulty | Challenging +1.3 This is a multi-step Further Maths integration question requiring improper fraction division, factorization of a cubic denominator, partial fractions decomposition (including a quadratic factor), and integration involving both logarithmic and inverse tangent forms. While it requires multiple techniques and careful algebraic manipulation, each individual step follows standard Further Pure procedures without requiring novel insight. The structured parts guide students through the solution systematically. |
| Spec | 4.05c Partial fractions: extended to quadratic denominators4.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | DR |
| Answer | Marks | Guidance |
|---|---|---|
| So A = 1, B = 5 and C = –5 | B1 | |
| [1] | 3.1a | Attempt to divide out improper |
| Answer | Marks |
|---|---|
| substitution of values for x) | Allow embedded answers |
| (b) | DR |
| Answer | Marks |
|---|---|
| x+1 x2 +4 | B1 |
| Answer | Marks |
|---|---|
| [5] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | Correct factorisation of cubic |
| Answer | Marks |
|---|---|
| correctly. | Could be from improper fraction |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (c) | DR |
| Answer | Marks |
|---|---|
| 9 8 | *M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | Split term with x2 + 4 in |
Question 7:
7 | (a) | DR
x3 +x2 +9x−1 x3 +x2 +4x+4+5x−5
≡
x3+x2 +4x+4 x3 +x2 +4x+4
5x−5
=1+
x3+x2 +4x+4
So A = 1, B = 5 and C = –5 | B1
[1] | 3.1a | Attempt to divide out improper
fraction. Could be by symbolic
division or other valid method
(eg comparing coefficients or
substitution of values for x) | Allow embedded answers
(b) | DR
x3+x2 +4x+4=(x+1)( x2 +4 )
5x−5 D Ex+F
= +
x3 +x2 +4x+4 x+1 x2 +4
D ( x2 +4 ) +(x+1)(Ex+F)=5x−5
x=−1⇒5D=−10⇒ D=−2
x=0⇒−2−F =−5⇒ F =3
x2 :D+E =0⇒ E =2
2 2x+3
1− +
x+1 x2 +4 | B1
M1
A1
A1
A1
[5] | 3.1a
1.2
1.1
1.1
1.1 | Correct factorisation of cubic
seen in working
Correct form for partial fractions
equated to their remainder
rational fraction from (a).
Follow through their division
and factorisation.
Or equivalent to find D
correctly. | Could be from improper fraction
Allow ft A1 for second and third
coefficients found.
2 2x 3
Or 1− + +
x+1 x2 +4 x2 +4
7 | (c) | DR
2 x3+x2+9x−1 2 2 2x 3
∫ dx=∫ 1− + + dx
0 x3+x2+4x+4 0 x+1 x2+4 x2+4
2
= x−2ln(x+1)+ln ( x2 +4 ) + 3 tan−1 x
2 2
0
3π
2−2ln3+ln8+ −ln4
8
2 3
2+ln + π
9 8 | *M1
dep*M1
M1
A1
[4] | 3.1a
1.1
1.1
1.1 | Split term with x2 + 4 in
denominator and ax + b in
numerator
Correctly integrate their
expression (ignore limits)
Correctly substitute limits to
produce exact values and
evaluate their tan-1 term
a = 2, b = 2 , c = 3
9 87 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $A , B$ and $C$ for which $\frac { x ^ { 3 } + x ^ { 2 } + 9 x - 1 } { x ^ { 3 } + x ^ { 2 } + 4 x + 4 } \equiv A + \frac { B x + C } { x ^ { 3 } + x ^ { 2 } + 4 x + 4 }$.
\item Hence express $\frac { x ^ { 3 } + x ^ { 2 } + 9 x - 1 } { x ^ { 3 } + x ^ { 2 } + 4 x + 4 }$ using partial fractions.
\item Using your answer to part (b), determine $\int _ { 0 } ^ { 2 } \frac { x ^ { 3 } + x ^ { 2 } + 9 x - 1 } { x ^ { 3 } + x ^ { 2 } + 4 x + 4 } \mathrm {~d} x$ expressing your answer in the form $a + \ln b + c \pi$ where $a$ is an integer, and $b$ and $c$ are both rational.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2021 Q7 [10]}}