Post on 11-Feb-2016
description
1. (a) METHOD 1
f(x) = ln(1 + ex); f(0) = ln 2 A1
f ′(x) =
; f ′(0) =
A1x
x
e1
e
2
1
Note: Award A0 for f′(x) =
; f ′(0) =
xe1
1
2
1
f ″(x) =
M1A14
1)0(;
)e1(
e2)e1(e2
2
f
x
xxx
Note: Award M0A0 for f ″(x) if f ′(x) =
is usedxe1
1
ln(1 + ex) = ln 2 +
+ ... M1A12
8
1
2
1xx
METHOD 2
ln(1 + ex) = ln(1 + 1 + x +
+ ...) M1A12
2
1x
= ln 2 + ln(1 +
+ ...) A12
4
1
2
1xx
= ln 2 +
A1......4
1
2
1
2
1...
4
1
2
12
22
xxxx
= ln 2 +
+ ... A122
8
1
4
1
2
1xxx
= ln 2 +
+ ... A12
8
1
2
1xx
(b) METHOD 1
2
34
020
4ln above & terms4
2ln2lim
4ln)e1ln(2lim
x
xxx
x
x
xx
x
x
M1A1
=
M1A14
1 of powers
4
1lim
0
x
x
Note: Accept + … as evidence of recognition of cubic and higher powers.
Note: Award M1A0M1A0 for a solution, which omits the cubic and higher powers.
METHOD 2
using l’Hôpital’s Rule
M1A1xx
x xx
x
x
x 2
1)e1(e2lim
4ln)e1ln(2lim
020
M1A14
1
2
)e1(e2lim
2
0
xx
x
[10]
2. (a) use of y → y + h
(M1)x
y
d
d
x y x
y
d
dh
x
y
d
d
0 1 1 0.1 A10.1 1.1 1.22 0.122 A10.2 1.222 1.533284 0.1533284 A10.3 1.3753284 1.981528208 0.1981528208 A10.4 1.573481221 (A1)
approximate value of y = 1.57 A1
Note: Accept values in the tables correct to 3 significant figures.
(b) the approximate value is less than the actual value because it is
assumed that
remains constant throughout each intervalx
y
d
d
whereas it is actually an increasing function R1[8]
3. put y = vx so that =
M1x
vxv
x
y
d
d
d
d
substituting, M1
v +
= (v2 + 3v + 2) (A1)2
2222 23
d
d
x
xvxxv
x
vx
= v2 + 2v + 2 A1
x
vx
d
d
M1
x
x
vv
v d
22
d2
(A1)
x
x
v
v d
1)1(
d2
arctan (v + 1) = ln x + c A1
Note: Condone absence of c at this stage.
arctan(
+ 1) = ln x + c M1x
y
When x = 1, y = –1 M1c = 0 A1
+ 1 = tan ln x
x
y
y = x(tan ln x – 1) A1[11]
4. (a) I0 =
M1xxx dsineπ
0
Note: Award M1 for I0 =
xxx dsineπ
0
Attempt at integration by parts, even if inappropriate modulussigns are present. M1
=
A1 π
0
π
0
π
0
π
0 dcosesine dcosecose xxxxxx xxxx or
=
π
0
π
0
π
0 dsinesinecose xxxx xxx
A1
π
0
π
0 dsinecosesine xxxx xxxor
= M1 0
π
00
π
0
π
0 cosesine sinecose IxxIxx xxxx or
Note: Do not penalise absence of limits at this stage
I0 = e–π + 1 – I0 A1
I0 =
(1+ e–π) AG2
1
Note: If modulus signs are used around cos x, award no accuracy marksbut do not penalise modulus signs around sin x .
(b) In =
xxn
n
x dsineπ)1(
π
Attempt to use the substitution y = x – nπ M1(putting y = x – nπ, dy = dx and [nπ, (n + 1)π] → [0, π])
so In =
A1ynyny d)πsin(eπ
0
π)(
=
A1ynyyn d)πsin(eeπ
0
π
=
A1yyyn dsineeπ
0
π
= e–nπI0 AG
(c)
M1
0
0dsine
nn
x Ixx
=
(A1)00
πe In
n
the Σ term is an infinite geometric series with common ratio e–π (M1)therefore
(A1)
π0
0 e1dsine
Ixxx
=
A1
)1e(2
1e
)e1(2
e1π
π
π
π
[15]
5. (a) using a ratio test,
M1A1
1
!
)!1(
11
n
x
x
n
n
x
T
Tn
n
n
n
Note: Condone omission of modulus signs.
→ 0 as n → ∞ for all values of x R1the series is therefore convergent for x A1
(b) (i) ex – 1 = x +
+ ... M1322
22
xx
< x +
+ ... (for x > 0) A1222
22
xx
=
(for x < 2) A1
21
xx
=
(for 0 < x < 2) AGx
x
2
2
(ii) ex < 1 +
A1x
x
x
x
2
2
2
2
ex <
A1x
x
x1
2
2
replacing x by
(and noting that the result is true for n >
andn
1
2
1
therefore +) M1
e <
AGn
n
n
12
12
(c) (i) 1 – e–x = x
+ ... A162
32 xx
for 0 < x < 2 , the series is alternating with decreasing termsso that the sum is greater than the sum of an even number of terms R1therefore
1 – e–x > x
AG2
2x
(ii) e–x < 1 – x + 2
2x
ex >
M1
21
12x
x
e >
A1x
xx
1
222
2
replacing x by
(and noting that the result is true for n >
andn
1
2
1
therefore +)
e >
AG
n
nn
n
122
22
2
(d) from (b) and (c), e < 2.718282… and e > 2.718281… A1we conclude that e = 2.71828 correct to 5 decimal places A1
[16]
6. (a) (i)
, n ≠ –1 M1
bnbn
n
xxx
1
1
1 1d
=
A11
1
1
1
nn
b n
= ln b when n = –1 A1 bb
n xxx 11
lnd
if n + 1 > 0,
does not exist since bn+1 increases
1
1
1lim
1
nn
b n
b
without limit R1
if n + 1 < 0,
exists since bn+1 → 0 as b → ∞ R1
1
1
1lim
1
nn
b n
b
if n = –1,
does not exist since ln b increases without limit R1 bb
lnlim
(so integral exists when n < –1)
(ii)
, (n < –1) A11
1d
1 n
xxb
n
(b) (cos x – sin x)
+ (cos x + sin x)y = cos x + sin xx
y
d
d
M1
xx
xxy
xx
xx
x
y
sincos
sincos
sincos
sincos
d
d
IF =
M1A1A1xx
xxx
xx
xx
sincos
1ee )sinln(cos
dsincos
sincos
(M1)x
xx
xx
xx
yd
)sin(cos
sincos
sincos 2
=
+ k A1xx sincos
1
Note: Award the above A1 even if k is missing.
y = 1 + k(cos x – sin x)
x =
, y = –1,2
π
–1 = 1 + k(–1) M1k = 2y = 1 + 2(cos x – sin x) A1
Note: It is acceptable to solve the equation using separation of variables.[15]
7. (a) EITHER
x
xxcot
1lim
0
=
M1A1
xx
xxx tan
tanlim
0
=
, using l’Hopital A1
xxx
xx tansec
1seclim
2
2
0
=
A1A1
xxxx
xxx tansec2sec2
tansec2lim
22
2
0
= 0 A1
OR
x
xxcot
1lim
0
=
M1A1
xx
xxxx sin
cossinlim
0
=
, using l’Hopital A1
xxx
xxx cossin
sinlim
0
=
A1A1
xxx
xxxx sincos2
cossinlim
0
= 0 A1
(b) un =
A1n
xn
n
3
)2(
M1A1)1(3
)2(
3
)2(
)1(3
)2(1
1
1
n
nx
n
x
n
x
u
u
n
n
n
n
n
n
M1A1
3
)2(
)1(3
)2(lim
x
n
nxn
–5 < x < 1 M1A1
1
3
)2(x
if x = 1, series is 1 +
+ ... which diverges A13
1
2
1
if x = –5, series is – 1 +
which converges A1n
n)1(...
3
1
2
1
hence interval is –5 ≤ x < 1 A1
(c) (i) f(x) = ln(1 + sinx), f(0) = 0 A1
f′(x) =
, f′(0) = 1, A1x
x
sin1
cos
f″(x) =
, f″(0) = –1 A1xx
x
x
xxx
sin1
1
)sin1(
)sin1(
)sin1(
cos)sin1(sin22
2
f′′′(x) =
, f′′′(0) = 1, A12)sin1(
cos
x
x
ln(1 + sin x) ≈ x
A1...62
32
xx
(ii) –sin x = sin (–x) M1
so, ln (1 – sin x) ≈ –x
A1...62
32
xx
(iii) ln (1 + sin x) – ln (1 – sin x)
=
M1A13
2sin1
sin1ln
3xx
x
x
let x =
M1A1A13
6
π
6
π23ln
2
11
2
11
ln then,6
π
3
=
AG
216
π1
3
π 2
[28]
8.
= ex + 2y2 (A1)x
y
d
d
x y dy/dx δy
0 1 3 0.3 M1A1
0.1 1.3 4.485170918 0.4485170918 A1
0.2 1.7485170918 7.336026799 0.7336026799 A1
0.3 2.482119772 13.67169593 1.367169593 A1
0.4 3.849289365 A1
required approximation = 3.85 A1[8]
9. (a)
M1A1 xx
xxxx xxx d)e(d
dsinsinedcose
00
0
since e–x → 0 as x → ∞ and sin x is bounded e–x sin x → 0 as x → ∞ R1(or alternative convincing argument)
e–x sin x = 0 when x = 0 R1
the second term =
A1xxx dsine0
so
AGxxxx xx dsinedcose00
(b) continuing the process
M1A1 x
xxxxx xxx d)e(
d
dcoscosedcose
00
0
the value of the first term is 1 A1
the second term =
A1xxx dcose0
so
= 1 A1xxx dcose20
the common value of the integrals is
A12
1
[11]
10. put y = vx so that
(M1)x
vxv
x
y
d
d
d
d
the equation becomes v +
A14d
d 2 vvx
vx
A1
x
x
v
v d
4
d2
= ln x + C A1A1
2arctan
2
1 v
substituting (x, v) = (1, 2)
C =
M1A18
π
the solution is
A1
4
πln2
2arctan
x
x
y
y = 2x tan
A1
4
πln2 x
[9]
11. (a) using or obtaining (1 + x)n = 1 + nx +
x2 + ... (M1)2
)1( nn
(A1)...
2
3
2
1
2
)(
2
1)(1)1(
2222
12
xxx
= 1 +
A1...8
3
2
1 42 xx
(b) integrating, and changing sign
arccos x = – x –
M1A1...40
3
6
1 53 Cxx
put x = 0,
= C M1
2
π
AG
53
40
3
6
1
2
πarccos xxxx
(c) EITHER
using arccos x2
M1A11062
30
3
6
1
2
πxxx
M1A1
6
6
06
22
0
powershigher 6lim
arccos2
π
limx
x
x
xx
xx
=
A16
1
OR
using l’Hôpital’s Rule M1
limit =
M15
4
0 6
221
1
limx
xxx
x
=
A14
4
0 3
11
1
limxx
x
=
M13
3
2
34
0 12
4
)1(
1
2
1
limx
x
xx
=
A16
1
(d)
M1xxxxxx d40
3
6
1
2
πdarccos
2.0
0
2
5
2
3
2
12.0
0
=
(A1)
2.0
0
2
7
2
5
2
3
140
3
15
1
3
2
2
π
xxxx
=
(A1)2
7
2
5
2
3
2.0140
32.0
15
12.0
3
22.0
2
π
= 0.25326 (to 5 decimal places) A1
Note: Accept integration of the series approximation using a GDC.
using a GDC, the actual value is 0.25325 A1so the approximation is not correct to 5 decimal places R1
[17]
12. (a) (i) consider
M1
n
n
n
n
n
n
nx
xn
T
T
2
2
)1(1
1
1
=
A1n
xn
2
)1(
as n → ∞ A1
2
x
the radius of convergence satisfies
= 1, i.e. R = 2 A1
2
R
(ii) the series converges for – 2 < x < 2, we need to consider x = ±2 (R1)when x = 2, the series is 1 + 2 + 3 + ... A1this is divergent for any one of several reasons e.g. findingan expression for Sn or a comparison test with the harmonic
series or noting that
≠ 0 etc. R1nn
u
lim
when x = –2, the series is –1 + 2 – 3 + 4... A1this is divergent for any one of several reasonse.g. partial sums are–1, 1, –2, 2, –3, 3... or noting that
≠ 0 etc. R1n
nu
lim
the interval of convergence is – 2 < x < 2 A1
(b) (i) this alternating series is convergent because the moduli ofsuccessive terms are monotonic decreasing R1
and the nth term tends to zero as n → ∞ R1
(ii) consider the partial sums0.333, 0.111, 0.269, 0.148, 0.246 M1A1
since the sum to infinity lies between any pair of successivepartial sums, it follows that the sum to infinity lies between0.148 and 0.246 so that it is less than 0.25 R1
Note: Accept a solution which looks only at 0.333, 0.269, 0.246 andstates that these are successive upper bounds.
[15]
13. integrating factor = M1 xxd2tane
= e2 ln sec x A1
= sec2x A1
it follows that
ysec2 x = ∫sin x sec2x dx M1 = ∫sec x tan x dx (A1) = sec x + C A1substituting, 0 = 2 + C so C = –2 M1A1the solution is
y = cos x – 2 cos2 x A1
EITHER
using a GDCmaximum value of y is 0.125 A2
OR
y′ = –sin x + 4 sin x cos x = 0 M1
(or sin x = 0, which leads to a minimum)
4
1cos x
A1
8
1 y
[11]
14. METHOD 1
f(0) =
hence using l’Hôpital’s Rule, (M1)0
0
g(x) = 1 – cos(x6), h(x) = x12;
A1A16
6
11
65
2
)sin(
12
)sin(6
)(
)(
x
x
x
xx
xh
xg
EITHER
, using l’Hôpital’s Rule again, (M1)0
0
)(
)(
xh
xg
A1A1
2
)cos(
12
)cos(6
)(
)( 6
5
65 x
x
xx
xh
xg
, hence the limit is
A1
2
1
)(
)(
xh
xg
2
1
OR
So
A16
6
012
6
0 2
sinlim
cos1lim
x
x
x
xxx
=
A16
6
0 2
sinlim
2
1
x
xx
=
= 1 A1(R1)6
6
0 2
sinlim since
2
1
x
xx
METHOD 2
substituting x6 for x in the expansion cos x = 1 –
... (M1)242
42 xx
M1A1
12
2412
12
6...
24211
cos1
x
xx
x
x
=
A1A1...242
1 12
x
M1A1
2
1cos1lim
12
6
0
x
xx
Note: Accept solutions using Maclaurin expansions.[7]
15. (a) ex = 1 + x +
+ ...!4!3!2
432 xxx
putting x =
(M1)2
2x
A2
48821
!32!2221e
642
3
6
2
422
2
xxxxxxx
(b)
M1(A1)
xx
uuuu
uu0
3
7
2
53
0
2
!327!22523de
2
=
A1!327!22523 3
7
2
53
xxxx
336406
753 xxxx
(c) putting x = 1 in part (b) gives
≈ 0.85535... (M1)(A1)xx
de1
0
2
2
≈ 0.341 A1x
x
deπ2
1 1
0
2
2
[9]
16. writing the differential equation in standard form gives
= e–x M1y
x
x
x
y
1d
d
dx = x + ln(x – 1) M1A1
1
11d
1 xx
x
x
hence integrating factor is ex+ln(x–1) = (x – 1)ex M1A1
hence, (x – 1)
+ xex y = x – 1 (A1)x
yx
d
de
= (x – 1) (A1)
x
yx x
d
]e)1[(d
(x – 1)ex y =
dx A1 )1(x
A1cx
xyx x
2e)1(
2
substituting (0, 1), c = –1 (M1)A1
(A1)
2
22e)1(
2
xxyx x
hence, y =
(or equivalent) A1xx
xx
e)1(2
222
[13]
17. (a) applying the alternating series test as
+ M1
nnn
ln
1,2
A1
nnnnn
ln
1
)1ln()1(
1,
A10
ln
1lim
nnn
hence, by the alternating series test, the series converges R1
(b) as
is a continuous decreasing function, apply the integral testxx ln
1
to determine if it converges absolutely (M1)
M1A1x
xxx
xx
b
bd
ln
1limd
ln
122
let u = ln x then du =
(M1)A1xx
d1
= ln u (A1)u
ud
1
hence,
which does not exist M1A1A1b
b
b
bxx
xx 22
)][ln(lnlimdln
1lim
hence, the series does not converge absolutely (A1)the series converges conditionally A1
[15]
18. (a)
M1A1A1x
x
xx
xxx 21
seclim
tanlim
2
020
A11
1
1tanlim
20
xx
xx
(b)
M1A1A1
2
πcos
2
πln422
lim
2
πsin1
ln21lim
1
22
1 xxxxx
xxxx
xx
M1A1A1
2
πsin
4
π
ln44lim
21 x
xx
A122
22
1 π
16
4
π
4
2
πsin1
ln21lim
x
xxxx
[11]
19. (a) from
= y tan x + cos x, f′(0) = 1 A1x
y
d
d
now
M1A1A1A1xxx
yxy
x
ysintan
d
dsec
d
d 2
2
2
Note: Award A1 for each term on RHS.
A12
π)0( f
A1
4
π
2
π 2xxy
(b) recognition of integrating factor (M1)
integrating factor is xxdtan
e
= eln cosx (A1)= cos x (A1)
M1 xxxy dcoscos 2
A1 xxxy d)2cos1(
2
1cos
A1k
xxxy
4
2sin
2cos
when x = π, y = 0
M1A12
π k
(A1)
2
π
4
2sin
2cos
xxxy
A1
2
π
4
2sin
2sec
xxxy
[17]
20. (a) comparing with the series
A1
1
1
n n
using the limit comparison test (M1)
M1A11sin
lim1
1sin
lim0
x
x
n
nxn
since
A1diverges 1
sindiverges,1
11
nn nn
(b) using integral test (M1)let u = ln x (M1)
xx
u 1
d
d
A1
xu
uu
xxx ln
11d
1d
)(ln
122
a
a xx
xx 22 2 ln
1limd
)(ln
1
=
(M1)(A1)
2ln
1
ln
1lim
aa
as a → ∞,
→ 0 (A1)aln
1
A1
2ln
1d
)(ln
12 2
xxx
hence the series is convergent AG[12]
21. (a) Using an increment of 0.25 in the x-values A1
n xn yn f(xn, yn) hf(xn, yn) yn+1 = yn + hf(xn, yn)
0 1 –1 1 0.25 –0.75 (M1)A1
1 1.25 –0.75 0.68 0.17 –0.58 A1
2 1.5 –0.58 0.574756 0.143689 –0.4363... A1
3 1.75 –0.436311 0.531080 0.132770 –0.3035... A1
Note: The A1 marks are awarded for final column.
y(2) ≈ –0.304 A1
(b) (i) let y = vx M1
(A1)
x
vxv
x
y
d
d
d
d
(M1)
2
222
2d
d
x
xxv
x
vxv
(A1)
2
21
d
d 2vv
x
vx
A1
2
)1(
d
d 2v
x
vx
M1x
xv
vd
1d
)1(
22
2(1 – v)–1 = ln x + c A1A1
= ln x + c
x
y
1
2
when x = 1, y = –1 c = 1 M1A1
= ln x + 1
yx
x
2
M1A1
x
xxx
x
xxy
ln1
ln
ln1
2
(ii) when x = 2, y = –0.362
A1
2ln1
42accept
[20]
22. (a)
M1)23)(13(
3
239
32
nnnn
=
A1A1)23(
1
13
1
nn
nth partial sum =
23
1
13
1...
14
1
11
1
11
1
8
1
8
1
5
1
5
1
2
1
nn
A1
=
A123
1
2
1
n
A1
12 2
1
23
1
2
1
239
3lim
nn nnn
(b) (i)
= 1 + x + x2 + x3 + x4 + ... =
A1
0r
rxx1
1
(ii) (a) replacing x by –x2 gives (M1)
= 1 + (–x2) + (–x2)2 + (–x2)3 + (–x2)4 + ... A1
)(1
12x
= 1 – x2 + x4 – x6 + x8 – ... (A1)
21
1
x
=
A1 N2
0
2)1(r
rr x
(b) arctan x =
M1A1cxxx
xx
x
...7531
d 753
2
x = 0 c = 0 A1
arctan x =
A112
)1(12
0
r
x r
r
r
(c) by taking x =
M13
1
arctan
A1
0
12
123
1)1(
6
π
3
1
r
rr
r
(c)
= 1! + 2! + 3! + 4! + 5! + ... M1
100
1
!n
n
= 1 + 2 + 6 + 24 + 120 + ...≡ 1 + 2 + 6 + 24 + 0 + 0 + 0 + ... (mod 15) M1A1≡ 33(mod 15) A1≡ 3 (mod 15) AG
[21]
23. (a) ex = 1 + x +
...!4!3!2
432
xxx
+ ... M1A1
!4
2
!3
2
!2
2
21e
423222
22
2
xxx
xx
A1
38448821
π2
1e
π2
1 86422
2
xxxxx
(b) (i)
M1tttttx
d3844882
1π2
1 86
0
42
=
A1
3456336406π2
1 9753 xxxxx
P(Z ≤ x) = 0.5 +
R1A1
...
3456336406π2
1 9753 xxxxx
(ii) P(–0.5 ≤ Z ≤ 0.5) =
M1
...
3456
5.0
336
5.0
40
5.0
6
5.05.0
π2
2 9753
= 0.38292 = 0.383 A1[9]
24. (a) the general term is
A1n
xn
n
3
)2(
(b)
M1A1A1
n
n
n
n
nn
n
n x
n
n
x
a
a
)2(
3
)1(3
)2(limlim
1
11
=
A1
)1(3
)2(lim
n
nxn
=
A1R111
lim since 3
)2(
n
nxn
the series is convergent if
< 1 R13
)2( x
then –3 < x + 2 < 3 –5 < x < 1 A1
if x = –5, series is 1 – 1 +
which converges M1A1...)1(
...3
1
2
1
n
n
if x = 1, series is 1 + 1 +
+ ... which diverges M1A1n
1...
3
1
2
1
the interval of convergence is –5 ≤ x < 1 A1[14]
25. (u + 3v3)
= 2vu
v
d
d
M1A1
2
3
22
)3(
d
d 23 v
v
u
v
vu
v
u
A1
2
3
2d
d 2v
v
u
v
u
IF is M1vv
vln
2
1d
2
1
ee
= A12
1
v
M1v
v
v
ud
2
3 2
3
=
A1cv 2
5
5
3
u =
A1vcv 3
5
3
[8]
26. put y = vx so that
M1A1x
vxv
x
y
d
d
d
d
the equation becomes v + x
= v + v2 (A1)x
v
d
d
leading to x
= v2 A1x
v
d
d
separating variables,
M1A1 2
dd
v
v
x
x
hence lnx = – v–1 + C A1A1
substituting for v, ln x =
+ C M1y
x
Note: Do not penalise absence of C at the above stages.
substituting the boundary conditions,
0 = –
+ C M12
1
C =
A12
1
the solution is ln x =
(A1)2
1
y
x
leading to y =
(or equivalent form) A1x
x
ln21
2
Note: Candidates are not required to note that x ≠ e
[13]
27. (a) ex – 1 = x +
+ ... A162
32 xx
+ ... M1A1
6
62
2
62
621e
332232
321e
xxx
xxx
xxx
x
= 1 + x +
M1A1...62262
33232
xxxxx
= 1 + x + x2 +
+ ... AG3
6
5x
(b) EITHER
f′(x) = 1 + 2x +
+ ... A12
5 2x
M1A1
...2
52
...6
5
1)(
1)(2
32
xx
xxx
xf
xf
=
A1...
2
52
...1
x
x
as x → 0 A1
2
1
OR
using l’Hopital’s rule, M1
M1A1
1e
1elim
11e
1elim
)1(e
)1(e
0)1(e
)1(e
0
xxx x
x
x
x
=
A1)1e(e
elim
)1(e
)1(e
0
xx
x
x x
x
=
A12
1
[10]
28. (a) un =
M1
2
11
2
n
→ 2 as n → ∞ A1L = 2
(b) for │2 – un│< ε then 2 – un < ε M1
and so we require 2 –
< ε M11
22
2
n
n
< ε A1
1
22
n
(M1)A11
2
n
we have shown that, given ε > 0, there exists an integer N ≥
12
such that 2 – un < ε for n > N, which establishes the limit R2
Note: Do not penalise N =
.12
[9]
29. (a) use the comparison test with
, M1 2
1
n
(which we know is convergent as it is a p-series with p = 2)
M1A1
2
1
)3(
1
nnn
hence the given series is convergent AG
(b) (i) let
M1A1
)3(
)3(
3)3(
1
nn
BnnA
n
B
n
A
nn
n = 0 gives A =
A13
1
n = –3 gives B =
A13
1
33
1
3
1
)3(
1So
nnnn
(ii)
(M1)(A1)
11 33
1
3
1
)3(
1
nn nnnn
=
M113
1
43
1
A123
1
53
1
3
3
1
63
1
4
3
1
73
1
+ ...
all terms cancel except
the required sum to infinity M1A118
11
9
1
6
1
3
1
Note: Award M1 for attempting to find an expression for Sn in the form
terms of order
.9
1
6
1
3
1
n
1
[13]
30. (a) the nth term is
un =
M1A1nxn
n
)13...(852
)12...(531
(using the ratio test to test for absolute convergence)
M1A1x
n
n
u
u
n
n
)23(
)12(1
A1
3
2lim 1 x
u
u
n
n
n
let R denote the radius of convergence
then
= 1 so R =
M1A13
2R
2
3
Note: Do not penalise the absence of absolute value signs.
(b) using the compound angle formula or a graphical methodthe series can be written in the form (M1)
where un = (–1)n sin
A2
1nnu
n
1
since
– i.e. an angle in the first quadrant, R12
π1
nit is an alternating series R1un → 0 as n → ∞ R1and │un+1│ < │un│ R1it follows that the series is convergent R1
[15]
31. Consider
M1A1
101
101 10
10
1
n
n
u
u n
nn
n
A110
11
10
1
n
as n A110
1
R1110
1
So by the Ratio Test the series is convergent. R1[6]
32. (a)
M1A1xxxx
x
e
1lim
elim
= 0 AG
(b) Using integration by parts M1
A1A1
axax
ax xxxx
00
0deede
A1 axaa 0ee
= 1 aea ea A1
(c) Since ea and aea are both convergent (to zero), the integral isconvergent. R1Its value is 1. A1
[9]
33. (a) Rewrite the equation in the form
M1A1
1
2
d
d2
2
x
xy
xx
y
Integrating factor = M1 x
xd
2
e
= A1x2lne
=
A12
1
x
Note: Accept
as applied to the original equation.3
1
x
(b) Multiplying the equation,
(M1)
1
12
d
d1232
x
yxx
y
x
(M1)(A1)
1
1
d
d22
xx
y
x
M1
1
d22 x
x
x
y
= arctan x + C A1
Substitute x = 1, y = 1. M1
A1
41
41
CC
A1
41arctan2
xxy
[13]
34. (a) The area under the curve is sandwiched between the sum of theareas of the lower rectangles and the upper rectangles. M2Therefore
A1
3 3333333...
5
11
4
11
3
11
d...
6
11
5
11
4
11
x
x
which leads to the printed result.
(b) We note first that
M1A1
18
1
2
1d
33 23
xx
x
Consider first
M1A1
1333333
...6
1
5
1
4
1
3
1
2
11
1
n n
M1A118
1
27
1
8
11
=
A1 boundupper an iswhich 22.1216
263
M1A1
133333
...5
1
4
1
3
1
2
11
1
n n
M1A118
1
8
11
=
(which is a lower bound) A1 18.1216
255
72
85
[15]
35. (a) Constant term = 0 A1
(b)
A1x
xf
1
1)(
A1
21
1)(
xxf
f ′′′ (x) =
A1 31
2
x
f (0) = 1; f (0) = 1; f (0) = 2 A1
Note: Allow FT on their derivatives.
M1A1...
!3
2
!2
1
!1
10)(
32
xxx
xf
=
AG22
32 xxx
(c)
(A1)2
12
1
1
x
x
ln 2
M124
1
8
1
2
1
=
A1 667.03
2
(d) Lagrange error =
(M1) 11
2
1
)!1(
)(
nn
n
cf
=
A1
4
4 2
1
24
1
1
6
c
A216
1
24
1
2
11
64
giving an upper bound of 0.25. A1
(e) Actual error = ln 2
A10265.03
2
The upper bound calculated is much larger that the actualerror therefore cannot be considered a good estimate. R1
[17]
36. (a) Using l’Hopital’s rule,
M1A1
x
xx
xxx π2cosπ2
1
limπ2sin
lnlim
11
=
A1π2
1
(b)
M1A1A1
...
!4!211
...!3!2
11
limcos1
e1lim
42
642
00
2
xx
xxx
x x
x
x
Note: Award M1 for evidence of using the two series.
=
A1
...
!4!2
...!3!2
lim42
642
0 xx
xxx
x
EITHER
=
M1A1
...
!4!2
1
...!3!2
1
lim2
42
0 x
xx
x
=
A12
2
11
OR
=
M1A1
...
!4
4
!2
2
...!3
6
!2
42
lim3
53
0 xx
xxx
x
=
...
!4
41
...!3
6
!2
42
lim2
42
0 x
xx
x
=
A121
2
[10]
37.
M1A1 122
212
122122
1Let
xx
xBxA
x
B
x
A
xx
A1
3
12 Ax
A1 N3
3
2
2
1 Bx
M1
h
xxx
I0
d2
1
12
2
3
1
=
A1 hxx 02ln12ln3
1
=
A1
2
1ln
2
12lnlim
3
1
h
hh
=
A1
2
1ln2ln
3
1
=
A12ln3
2
Note: If the logarithms are not combined in the third from last linethe last three A1 marks cannot be awarded.
[9]
38. (a) (i)
yxxx
y 212
d
d
xi yi yi y
1 2 0 0
1.1 2 0.4620 0.0462
1.2 2.0462 0.9451 0.0945
1.3 2.1407
M1
A2
Notes: Award A2 for complete table.
Award A1 for a reasonable attempt.
f (1.3) = 2.14 (accept 2.141) A1
(ii) Decrease the step size A1
(b)
yxxx
y 212
d
d
M1 2122
d
dxxxy
x
y
Integrating factor is M1A12
eed2 xxx
So,
A1 xxxxy xxx de2e2e 2222
=
M1A1 xxx xxx de2ee222 2
kx xxx 222
eee 2
= A1kx x 2
e2
y = 2
e2 xkx
x = 1, y = 2 2 = 1 + ke1 M1
k = e
y = x2 + A12 1e x
[14]
39. (a) f (x) = ln cos x
M1A1x
x
xxf tan
cos
sin)(
f (x) = sec2 x M1
f ′′′(x) = 2 sec x sec x tan x A1
f iv (x) = 2 sec2 x (sec2 x) 2 tan x (2 sec2 x tan x)
= 2 sec4 x 4 sec2 x tan2 x A1
...0
!40
!30
!200
432
ivfx
fx
fx
fxfxf
f (0) = 0, M1
f (0) = 0,
f (0) = 1,
f ′′ (0) = 0,
f iv (0) = 2, A1
Notes: Award the A1 if all the substitutions are correct.
Allow FT from their derivatives.
ln (cos x)
A1!4
2
!2
42 xx
=
AG122
42 xx
(b) Some consideration of the manipulation of ln 2 (M1)Attempt to find an angle (M1)
EITHER
Taking
A13
πx
ln
A1!4
3
π2
!2
3
π
2
1
42
ln 2
A1!4
81
π2
!29
π 42
ln 2
A1
108
π
2
1
9
π
972
π
18
π 2242
OR
Taking
A14
πx
ln
A1!4
4
π2
!24
π
2
1
42
A1
!4256
π2
!216
π
2ln2
1
42
ln 2
A1
192
π
2
1
8
π
1536
π
16
π 2242
[14]
40. (a) The ratio test gives
M1A1 nnn
nnn
nn
n
n xn
nx
u
u
132
311limlim
1
111
=
A1 23
1lim
n
xnn
=
A13
x
So the series converges for
< 1, A13
x
the radius of convergence is 3 A1
Note: Do not penalize lack of modulus signs.
(b) nnun 3 3 1
=
M1A1
1
113
3nn
=
A1
1...
81
5
9
1
3
11
963 nnnn
using
as the auxilliary series, M12
1
nvn
since
M1A1converges...4
1
3
1
2
1
1
1and
3
1lim
2222
n
n
n v
u
then
converges A1 nu
Note: Award M1A1A1M0M0A0A0 to candidates attemptingto use the integral test.
[13]
41. Rewrite the equation in the form
M1 2
1
21d
d
xxx
y
x
y
Integrating factor = exp
A1
21
d
xx
x
= exp
M1A1
x
xxd
2
1
1
1
= exp ln
A1
2
1
x
x
=
A12
1
x
x
Multiplying by the integrating factor,
M1 22 2
1
2d
d
2
1
x
x
x
y
x
y
x
x
=
A1 22 2
1
2
2
xx
x
Integrating,
A1A1 Cx
xyx
x
2
12ln
2
1
A1
Cx
xx
xy
2
12ln
1
2
[11]
42. (a)
M1A1 x
xxf
sin1
cos
M1
2
2
sin1
cossin1sin
x
xxxxf
=
A1 2sin1
1sin
x
x
=
AGxsin1
1
(b)
A1 2sin1
cos
x
xxf
A1 4
22
sin1
cossin12sin1sin
x
xxxxxf iv
f (0) = 0, f (0) = 1, f (0) = 1, f ′′(0) = 1, f iv (0) = 2 (A2)
Note: Award A1 for 2 errors and A0 for more than 2 errors.
ln (1 + sin x) =
M1A1...1262
432
xxx
x
(c) ln (1 sin x) = ln(1 + sin (x)) = x
M1A1...1262
432
xxx
(d) Adding, M1
ln (1 sin2 x) = ln cos2 x A1
=
A1...6
42
xx
ln cos x =
A1...122
42
xx
ln sec x =
AG...122
42
xx
(e)
M1...122
secln 2
xxx
xx
x
Limit = 0 A1[18]
43. (a) S2n = Sn +
M1nnn 2
1...
2
1
1
1
M1A1
nnnSn 2
1...
2
1
2
1
AG
2
1 nS
(b) Replacing n by 2n,
M1A1
2
124 nn SS
> Sn + 1 A1
Continuing this process,
(A1)
2
38 nn SS
In general,
M1A1
22
mSS nnm
Putting n = 2 M1
AG
222 1
mSS m
(c) Consider the (large) number N. M1
Then,
if
A1NS m 12N
mS
22
i.e. if m > 2(N S2) A1
This establishes the divergence. AG[13]
44. (a) EITHER
use the substitution y = vx
M1A11
d
d vvx
x
v
x
xv
dd
by integration
v =
= lnx + c A1x
y
OR
the equation can be rearranged as first order linear
M11
1
d
d y
xx
y
the integrating factor I is
A1
xx
xx 1
ee lnd
1
multiplying by I gives
xy
xx
11
d
d
y = ln x + c A1
x
1
THEN
the condition gives c = –1 M1A1so the solution is y = x (ln x – 1) AG
(b) (i) f ′(x) = ln x – 1 + 1 = ln x A1
f ″(x) =
A1x
1
f ′′′(x) =
A12
1
x
(ii) the Taylor series about x = 1 starts
f(x) ≈ f(1) + f′(1)(x – 1) + f″(1)
(M1)!3
)1()1(
!2
)1( 32 x
fx
= –1 +
A1A1A1!3
)1(
!2
)1( 32
xx
[12]
45. (a) (i) the integrand is non-singular on the domain if p > –1with the latter assumed, consider
M1A1x
pxxpx
pxx
RRd
111d
)(
111
=
, p ≠ 0 A1
R
px
x
p1
ln1
this evaluates to
, p ≠ 0 M1
ppR
R
p 1
1lnln
1
→
A1)1ln(1
pp
because
→ 1 as R → ∞ R1pR
R
hence the integral is convergent AG
(ii) the given series is
M1)5.0(
1)(),(
1
nnnfnf
n
the integral test and p = –0.5 in (i) establishes theconvergence of the series R1
(b) (i) as we have a series of positive terms we can apply thecomparison test, limit form
comparing with
M1
12
1
n n
M1A111
)3(
1sin
lim
2
n
nn
n
as sin θ ≈ θ for small θ R1
and
→ 1 R1)3(
2
nn
n
(so as the limit (of 1) is finite and non-zero, both seriesexhibit the same behavior)
converges, so this series converges R1
12
1
n n
(ii) the general term is
A1
)1(
1
nn
M1
)1)(1(
1
)1(
1
nnnn
A1
1
1
)1)(1(
1
nnn
the harmonic series diverges R1so by the comparison test so does the given series R1
[19]
46. (a) (i) f(x) = (1 + ax)(1 + bx)–1
= (1 + ax)(1 – bx + ...(–1)nbnxn + ... M1A1it follows thatcn = (–1)nbn + (–1)n–1 abn–1 M1A1
= (–b)n–1(a – b) AG
(ii) R =
A1b
1
(b) to agree up to quadratic terms requires
1 = –b + a,
= b2 – ab M1A1A12
1
from which a = –b =
A12
1
(c) ex ≈
A1x
x
5.01
5.01
putting x =
M13
1
A15
7
6
11
6
11
e 3
1
[12]
47. (a) this separable equation has general solution
∫sec2 y dy = ∫cos x dx (M1)(A1)tan y = sin x + c A1the condition gives
tan
= sin π + c c = 1 M14
π
the solution is tan y = 1 + sin x A1y = arctan (1 + sin x) AG
(b) the limit cannot exist unless a = arctan
= arctan 2 R1A1
2
πsin1
in that case the limit can be evaluated using l’Hopital’s rule (twice)limit is
M1A1
2
π2
lim
2
π2
))sin1(arctan(lim
2
π
2
πx
y
x
x
xx
where y is the solution of the differential equationthe numerator has zero limit (from the factor cos x in the differential equation) R1so required limit is
M1A12
lim
2
π
y
x
finally,y″ = –sin x cos2 y – 2 cos x cos y sin y × y′(x) M1A1
since
A15
1
2
πcos
y
y″ =
A12
πat
5
1 x
the required limit is
A110
1
[17]
48. Let f(x) =
(M1)xx
xx
sin
sin
A1A1
xxx
xxf
xx cossin
1coslim)(lim
00
=
A1A1
xxx
xx sincos2
sinlim
0
= 0 A1 N2[6]
A1the harmonic series diverges R1so by the comparison test so does the given series R1
[19]
49. For p > 1,
ispx
1
positive for x ≥ 1, and decreasing for x ≥ 1. A1A1
(M1)
L
pL
L
pL xpx
x1
11 )1(
1limd
1lim
=
A1pLp pL
1
1
)1(
1lim
1
=
A11
1
p
The convergence of this integral ensures the convergence of the seriesusing the integral test. R1AG N0
[6]
50. (a) (i) y = ln (1 + sin x)
y′ =
A1x
x
sin1
cos
y″ =
A1xsin1
1
y(3) =
A12)sin1(
cos
x
x
y(4) =
(M1)A14
22
)sin1(
cos)sin1(2)sin1(sin
x
xxxx
(ii) y(0) = 0; y′(0) = 1 A1A1y″(0) = –1; y(3)(0) = 1; y(4)(0) = –2 A1A1 A1
ln(1 + sin x) = x –
AG N0...12
1
6
1
2
1 432 xxx
(b) (i) ln(1 – sin x) = ln(1 + sin(–x)) (M1)
=
A1 N2...12
1
6
1
2
1 432 xxxx
(ii) ln(1 + sin x) + ln(1 – sin x) = ln(1 – sin2 x) (M1)
= ln cos2 x A1
So ln cos2 x = – x2 –
A1...6
1 4 x
ln cos x =
A1 N2...12
1
2
1 42 xx
(iii) Differentiating,
(M1))sin(cos
1)cos(ln
d
dx
xx
x
= –tan x A1
tan x = x +
+ ... A2 N33
3
1x
Note: No term in x4 since tan(–x) = –tan x
(c)
(M1)
...122
...3
cosln
)tan(42
42
2
xx
xx
x
x
=
A1
...122
1
...3
1
2
4
x
x
→ –2 as x → 0 A1
so
A1 N32cosln
)tan(lim
2
0
x
xx
[24]
51. (a)
241
d
d
x
xy
x
y
x y dy/dx h × dy/dx
0 1 1 0.25 A2
0.25 1.25 0.9206349206 0.2301587302 A2
0.5 1.48015873 0.8026455027 0.2006613757 A2
0.75 1.680820106 0.6332756132 0.1583189033 A2
1 1.839139009 A1
To two decimal places, when x = 1, y = 1.84. A1 N0
(b) (i) Integrating factor = (M1)
x
x
xd
4 2
e
= A1
)4ln(
2
1 2
ex
=
A124
1
x
It follows that
(M1)22 4
1
4d
d
xx
y
x
A1A1Cx
x
y
2arcsin
4 2
Putting x = 0, y = 1,
A1C2
1
Therefore, y =
A2 N0
2
1
2arcsin4 2 x
x
(ii) When x = 1, y = 1.77. A1 N1
(c)
A2
Since
is decreasing the value of y is over-estimated at each step. R1A1x
y
d
d
[24]