Math Paper 3 Practice MS

47
1. (a) METHOD 1 f(x) = ln(1 + e x ); f(0) = ln 2 A1 f (x) = ; f (0) = A1 x x e 1 e 2 1 Note: Award A0 for f(x) = ; f (0) = x e 1 1 2 1 f (x) = M1A1 4 1 ) 0 ( ; ) e 1 ( e 2 ) e 1 ( e 2 2 f x x x x Note: Award M0A0 for f (x) if f (x) = is used x e 1 1 ln(1 + e x ) = ln 2 + + ... M1A1 2 8 1 2 1 x x METHOD 2 ln(1 + e x ) = ln(1 + 1 + x + + ...) M1A1 2 2 1 x = ln 2 + ln(1 + + ...) A1 2 4 1 2 1 x x = ln 2 + A1 ... ... 4 1 2 1 2 1 ... 4 1 2 1 2 2 2 x x x x = ln 2 + + ... A1 2 2 8 1 4 1 2 1 x x x = ln 2 + + ... A1 2 8 1 2 1 x x (b) METHOD 1 2 3 4 0 2 0 4 ln above & terms 4 2 ln 2 lim 4 ln ) e 1 ln( 2 lim x x x x x x x x x x M1A1 = M1A1 4 1 of powers 4 1 lim 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.

description

IB Math HL Paper 3 Series and Differential Equations practice Markscheme

Transcript of Math Paper 3 Practice MS

Page 1: Math Paper 3 Practice MS

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.

Page 2: Math Paper 3 Practice MS

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]

Page 3: Math Paper 3 Practice MS

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]

Page 4: Math Paper 3 Practice MS

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

Page 5: Math Paper 3 Practice MS

(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

Page 6: Math Paper 3 Practice MS

(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]

Page 7: Math Paper 3 Practice MS

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]

Page 8: Math Paper 3 Practice MS

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

Page 9: Math Paper 3 Practice MS

(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]

Page 10: Math Paper 3 Practice MS

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]

Page 11: Math Paper 3 Practice MS

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

Page 12: Math Paper 3 Practice MS

(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

Page 13: Math Paper 3 Practice MS

(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]

Page 14: Math Paper 3 Practice MS

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]

Page 15: Math Paper 3 Practice MS

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]

Page 16: Math Paper 3 Practice MS

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

Page 17: Math Paper 3 Practice MS

(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]

Page 18: Math Paper 3 Practice MS

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

Page 19: Math Paper 3 Practice MS

(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

Page 20: Math Paper 3 Practice MS

(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

Page 21: Math Paper 3 Practice MS

(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]

Page 22: Math Paper 3 Practice MS

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(

Page 23: Math Paper 3 Practice MS

(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]

Page 24: Math Paper 3 Practice MS

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

Page 25: Math Paper 3 Practice MS

(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

Page 26: Math Paper 3 Practice MS

(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

Page 27: Math Paper 3 Practice MS

(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.

Page 28: Math Paper 3 Practice MS

(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

Page 29: Math Paper 3 Practice MS

(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]

Page 30: Math Paper 3 Practice MS

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

Page 31: Math Paper 3 Practice MS

(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]

Page 32: Math Paper 3 Practice MS

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

Page 33: Math Paper 3 Practice MS

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]

Page 34: Math Paper 3 Practice MS

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]

Page 35: Math Paper 3 Practice MS

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

Page 36: Math Paper 3 Practice MS

(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]

Page 37: Math Paper 3 Practice MS

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]

Page 38: Math Paper 3 Practice MS

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

Page 39: Math Paper 3 Practice MS

(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

Page 40: Math Paper 3 Practice MS

(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]

Page 41: Math Paper 3 Practice MS

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]

Page 42: Math Paper 3 Practice MS

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

Page 43: Math Paper 3 Practice MS

(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]

Page 44: Math Paper 3 Practice MS

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]

Page 45: Math Paper 3 Practice MS

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

Page 46: Math Paper 3 Practice MS

(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

Page 47: Math Paper 3 Practice MS

(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]