Edexcel FP3 2014 June — Question 9 8 marks

Exam BoardEdexcel
ModuleFP3 (Further Pure Mathematics 3)
Year2014
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeRational function powers
DifficultyChallenging +1.8 This is a Further Maths FP3 reduction formula question requiring integration by parts with careful algebraic manipulation to derive the recurrence relation, followed by application to find a specific integral. The derivation involves non-trivial substitution and manipulation of terms, and finding I_2 requires working backwards through I_1 = arctan(x). While systematic, it demands strong technical facility beyond standard A-level and represents a challenging multi-step problem typical of harder Further Maths content.
Spec1.08i Integration by parts4.08a Maclaurin series: find series for function
$$I_n = \int (x^2 + 1)^{-n} dx, \quad n > 0$$
  1. Show that, for \(n > 0\) $$I_{n+1} = \frac{x(x^2 + 1)^{-n}}{2n} + \frac{2n - 1}{2n}I_n$$ [5]
  2. Find \(I_2\) [3]
Question 9:
AnswerMarks Guidance
9(a)(x2 1)ndx x(x2 1)n xn(x2 1)n12xdx M1A1
M1: Integration by parts in the correct direction
A1: Correct expression
(If the parts formula is not quoted and the expression is wrong, score M0A0)
 x(x2 1)n 2nx2(x2 1)n1dx
AnswerMarks
 x(x2 1)n 2n(x2 1)n (x2 1)n1dxUse of x2  x2 11 or equivalent.
Dependent on the previous
AnswerMarks
method mark.dM1
I  x(x2 1)n 2nI 2nI
AnswerMarks
n n n1Correctly replaces
(x2 1)ndxand(x2 1)n1dx by
I and I .
n n+1
Dependent on both previous
AnswerMarks
method marks.ddM1
x(x2 1)n 2n1
I   I
AnswerMarks
n1 2n 2n nCorrect completion to the printed
answer with no errors.A1cso
(5)
AnswerMarks
(b)x(x2 1)1 1
I   I
AnswerMarks
2 2 2 1Correct application of the given
reduction formula using n = 1 onlyM1
dx
I  =arctanxC
AnswerMarks
1 x2 1I karctanx(must be x and not just
1
AnswerMarks
for arctan)M1
x 1
I   arctanxC
AnswerMarks Guidance
2 2(x2 1) 2Cao (constant not needed) A1
(3)

Total 8

PMT PMT
Extra Notes
5 10 8
1 
2. (a) M1  0 0 1
 
5
 
0 5 4
 
1 0 2
 
MTM 0 41 4
 
 
2 4 5
 
3. (b) Parts once then exponentials
1
1 1  ex ex 1 ex ex
I 1e2xsinhx  1e2xcoshxdx 1e2x  1e2x dx
2  0 0 2   2 2   0 2 2
0
M1 integrates by parts and writes coshx as exponentials
A1 Correct expression
  1e2x ex ex 1   1 e2x  1 ex  1 1  1e3x ex 1  1  1e3e1 1  1e0 e0
  2 2    12 4   2 3  0 2 3 2 3
0 0
M1 epxdxqepxat least once and correct use of the limits 0 and 1
e3 e 1
    A1
 6 2 3
Any exact equivalent (allow e1) but all like terms collected but isw following a correct answer.
PMT PMT
Pearson Education Limited. Registered company number 872828
with its registered office at Edinburgh Gate, Harlow, Essex CM20 2JE
Question 9:
--- 9(a) ---
9(a) | (x2 1)ndx x(x2 1)n xn(x2 1)n12xdx | M1A1
M1: Integration by parts in the correct direction
A1: Correct expression
(If the parts formula is not quoted and the expression is wrong, score M0A0)
 x(x2 1)n 2nx2(x2 1)n1dx
 x(x2 1)n 2n(x2 1)n (x2 1)n1dx | Use of x2  x2 11 or equivalent.
Dependent on the previous
method mark. | dM1
I  x(x2 1)n 2nI 2nI
n n n1 | Correctly replaces
(x2 1)ndxand(x2 1)n1dx by
I and I .
n n+1
Dependent on both previous
method marks. | ddM1
x(x2 1)n 2n1
I   I
n1 2n 2n n | Correct completion to the printed
answer with no errors. | A1cso
(5)
(b) | x(x2 1)1 1
I   I
2 2 2 1 | Correct application of the given
reduction formula using n = 1 only | M1
dx
I  =arctanxC
1 x2 1 | I karctanx(must be x and not just
1
for arctan) | M1
x 1
I   arctanxC
2 2(x2 1) 2 | Cao (constant not needed) | A1
(3)
Total 8
PMT PMT
Extra Notes
5 10 8
1 
2. (a) M1  0 0 1
 
5
 
0 5 4
 
1 0 2
 
MTM 0 41 4
 
 
2 4 5
 
3. (b) Parts once then exponentials
1
1 1  ex ex 1 ex ex
I 1e2xsinhx  1e2xcoshxdx 1e2x  1e2x dx
2  0 0 2   2 2   0 2 2
0
M1 integrates by parts and writes coshx as exponentials
A1 Correct expression
  1e2x ex ex 1   1 e2x  1 ex  1 1  1e3x ex 1  1  1e3e1 1  1e0 e0
  2 2    12 4   2 3  0 2 3 2 3
0 0
M1 epxdxqepxat least once and correct use of the limits 0 and 1
e3 e 1
    A1
 6 2 3
Any exact equivalent (allow e1) but all like terms collected but isw following a correct answer.
PMT PMT
Pearson Education Limited. Registered company number 872828
with its registered office at Edinburgh Gate, Harlow, Essex CM20 2JE
$$I_n = \int (x^2 + 1)^{-n} dx, \quad n > 0$$

\begin{enumerate}[label=(\alph*)]
\item Show that, for $n > 0$
$$I_{n+1} = \frac{x(x^2 + 1)^{-n}}{2n} + \frac{2n - 1}{2n}I_n$$
[5]
\item Find $I_2$ [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP3 2014 Q9 [8]}}
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