applied math paper 2 marking by n.f. yang

8/6/2019 Applied Math Paper 2 Marking by N.F. Yang

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AL Applied MathematicsPaper II - Marking Scheme

byN.F. Yang


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Appl ied Mathemat ics (So lu t ions)

1. (a)( 2)( 3) ( 1)( 3) ( 1)( 2) 3

( ) (sin ) (sin ) (sin )(1 2)(1 3) 4 (2 1)(2 3) 2 (3 1)(3 2) 4

x x x x x xp x

= + +

[1M]

22( 1) (4 2 2) (2 2

2x x= + + 3) [1M]

(b)(3)1

| ( ) ( ) | | ( 1)( 2)( 3) | | ( ) |3!

p x f x x x x f = where (1,3)

3

3

1| ( 1)( 2)( 3) | | cos |

6 4 x x x

4

= [1M]

31

| ( 1)( 2)( 3) | ( )6 6 x x x

2

4 2 [1M]3

2.8 1 2| (2.8) sin( ) | | (2.8 1)(2.8 2)(2.8 3) | ( )( )

4 6 64 2p

[1M]

0.017<

2. (a) 12

2dx

dt t= ,

2

1

2 3

4d x

dt t= .

22 1 1

12

4 10 63

d x dxt t x

dt dt t t t

0 + + = + = [1M]

Hence 1( )x t is a solution of (*).

(b)u

xt

=Q

2

1dx u du

dt t t dt = +

2 2

2 3 2

2 2 1d x u du d u

dt t t dt t dt

= + 2

22

3 2 2 2

2 2 1 1( ) 5 ( )

u du d u u du ut t

t t dt t dt t t dt t 3( ) 0 + + + + = [2M]

2

2

2 52 5

u du d u u du ut

t dt dt t dt t + + + =

30

2

23

d u dut

dt dt 0+ = [1M]

Putdu

vdt

= ,

3dv

t vdt

0 + = [1M]

N.F. Yan g P.1


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4/12

(c) For4

t

> , 0y = andx satisfies the differential equation u u=& .

/ 4

u t

N

dudt

u = ,

giving/ 4

500 2t

u e +

= for 4t

> .

v

u 500 2

t

e

/ 4

500

[1M+1M]

4. (a) [1M]1

2 3

0(1 ) 1nkx x dx =

1

3 3

0

1(1 ) (1 ) 1

3

nk x d x =

13 1

0

(1 )1

3 1

nk x

n

+ = +

[1M]3( 1k n= + )

(b) (i) When k= 6,

n = 1

13 5

0( ) 6( ) E X x x= dx [1M]

14 7

0

6 6

4 7

x x =

914=

[1M]

2 2( ) ( ) [ ( )]Var X E X E X =

1

4 7

0

96( ) ( )

14 x x dx=

2 [1M]

15 8

0

6 6 8

5 8 196

x x =

1

9

245= [1M]

N.F. Yan g P.3


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(c) 2 5( ) 6( ) f x x x= 4'( ) 6(2 5 ) f x x x=

Put '( )f x 0= 3 0.4x = or 0 [1M]3

"( ) 6(2 20 ) f x x=

3"( 0.4) 0f < and [1M]"(0) 0f >

( )f x attains a maximum at 3 0.4x =

So the mode is 3 0.4

5. (a) (3500 [1M]1.96 150,3500 1.96 150) +

[1M](3206,3794)=

(b) LetXbe the daily production of factory,2 2~ (6000,150 250 )X N +

90% confidence interval for the daily production of factory is

2 2 2(6000 1.645 150 250 ,6000 1.645 150 250 ) + + + 2 [1M]

(5520.4042 [1M],6479.5958)=

(c) ( 6500)P X >

2

6500 6000(

150 250P Z

2)

= >

+[1M]

( 1.715)P Z= >

5

0.5 [0.4564 (0.4573 0.4564)]10

= +

[1M]0.0432=

6. LetXand Ybe the scores of the students in F.6 and F.7 respectively,

(a) ( 68) ( 68)P X P Y > >

68 10 68 56

( ) (10 6

P Z P Z

= > > )

)

[1M]

( 0.8) ( 2P Z P Z = > >

(0.5 0.2881)(0.5 0.4772)=

[1M]0.0048=

N.F. Yan g P.4


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(b) 2 2~ (4,10 6 ) X Y N +

(0 6)P X Y< [1M]

( )f x is a strictly increasing function.

(1) 1 (1) 0.3069 0f g=

( )f x 0 = has exactly one root in [1, 2]

i.e. Ehas exactly one root in [1, 2]

(b) (i) 2( ) 1 sing x x= +

Since 20 sin 1x , 1 ( )g x 2

]

[1M]

If ,0 [1,2]x 1 0( ) [1,2]x g x=

Inductively, 1( ) [1,2r r x g x = r . [1M]

N.F. Yan g P.6


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(ii) 1 0 0 01

| | | '( ) | ( ) | |2

e e g e= [1M]

2

2 1 1 1 0

1 1| | | '( ) | ( ) | | ( ) |

2 2e e g e e= | [1M]

Inductively,

2

1 2

1 1 1| | ( ) | | ( ) | | ( ) | |

2 2 2

r

r r re e e L 0e [1M]

(iii) 01

lim | | | | lim( ) 02

n

nn n

e e

= [1M]

lim | | 0nn

e

= i.e. lim nn

x s

= . [1M]

(c)

N.F. Yan g P.7

[2M]

r rx ( )rg x | ( ) |r rg x x

0 1.3 1.388684405 0.0886844051 1.388684405 1.402569166 0.0138847612 1.402569166 1.404266922 0.0016977563 1.404266922 1.404465509 0.0001985884 1.404465509 1.40448861 2.31004 5105 1.40448861 1.404491295 2.68537 6106 1.404491295 1.404491607

3.121467

10

6 1.404491x = [1M]

9. (a) Let rt x x=

[ 1 2 3( ) ( ) (0) ( )h

h]f t dt h c f h c f c f h

+ + (1) [1M]

Since (1) is exact for ,( ) 1f t = ( ) 2 f t t = and 4( ) 5 f t t = ,

(2) [1M]1 2 32 (h

hdt h h c c c

= = + + )

)

4 )

(3) [1M]1 32 0 ( 2 2h

htdt h c h c h

= = +

(4) [1M]4 5 41 35 2 (5 5h

ht dt h h c h c h

= = +

By solving, 1 31

5c c= = , 2

8

5c = . [1M]

[ ]1

11 1( ) ( ) 8 ( ) ( )5

r

r

x

r rx

h

f x dx f x f x f x

+

+ + +

r


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(b) Let , the given formula is reduced tort x x=

N.F. Yan g P.8

] [ ] [21 2 1 20

( ) (0) ( ) '(0) '( )h

f t dt h a f a f h h b f b f h + + + (5) [1M]

Putting 2 3( ) 1,2 ,3 ,4 f t t t = t

)

22 ]

3

2 ]

4

2 ]

(6)1 20 (

h

dt h h a a= = + (7) [1M]2 22 1

02 [2 ] [2

h

tdt h h a h h b b= = + +

(8) [1M]4 5 4 220

5 [5 ] [20h

t dt h h a h h b h= = +

(9) [1M]5 6 5 220

6 [6 ] [30h

t dt h h a h h b h= = +

By solving,1

2

3a = ,

2

1

3a = ,

1

1

5b = ,

2

1

30b = . [1M]

1 2

1 1

2 1 1 1( ) ( ) ( ) '( ) '( )

3 3 5 30

r

r

x

r r r r x

f x dx h f x f x h f x f x+

+ +

+ +

(c) 20

2 1 1 1( ) (0) ( ) '(0) '( )

3 3 5 30

h

f t dt h f f h h f f h

+ +

If 6( ) f t t = , the error is2 7

6 6 5

0

( ) (6 )

3 30 105

h h ht dt h h + =

h[1M]

For any ( )f t ,

(6 )6

5

( )( ) ( )

6!

f f t p t t

= + , [0, ]h [1M]

where 5 ( )p t is a polynomial in twith degree 5.

the error for ( )f t is

2

0

(6) 7 7

(6)

2 1 1 1( ) (0) ( ) '(0) '( )

3 3 5 30

( ) ( )6! 105 75600

h

f t dt h f f h h f f h

f h h f

+

= =

[2M]

Putting ,rt x x=

1

72 (

1 1

2 1 1 1( ) ( ) ( ) '( ) '( ) (

3 3 5 30 75600

r

r

x

r r r r x

hf x dx h f x f x h f x f x f 6) )

+

+ +

= + + +

[1M]

for 1r rx x + .7

(6) ( )75600

hf for 1r rx x +


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10. (a) Without lost of generality, we assume thatXis the winner, Yis the loser andZis the

pausing player in the nth match so that

nx = P(Xis the final victory)

= P(XwinsZin the (n + 1)th game

+ P(XlossesZin the (n + 1)th game and the loserXwill be the final victory) [1M]

1

3 1

4 4ny += +

ny = P(Yis the final victory)

= P(XlossesZin the (n + 1)th game and the pausing player Yin the (n + 1)th game will

be the final victory) [1M]

1

1

4nz +=

= P(Zis the final victory)nz

= P(ZwinsXin the (n + 1)th game and the winnerZwill be the final victory) [1M]

1

1

4nx +=

(b) (i) Let X, Y and Z be the events thatA,B and Cwins a game respectively.

Assume thatA is the winner of the first game so that

1x 3 63 1 3 1 3( ) ( ) ( ) ( )

4 4 4 4 4= + + +L [1M]

3

3

41

1 ( )4

=

16

21= [1M]

(ii) 13 1

4 4n nx y += +

2

3 1 1(

4 4 4nz += + ) [1M]

3

3 1 1( )

4 16 4nx += +

3

3 1

4 64nx += + [1M]

N.F. Yan g P.9


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N.F. Yan g P.10

2(iii) Since 1(4 )n

n x k k = + , we have

3

1 2 1

3 1(4 ) [ (4 ) ]

4 64

nk k k

++ == + + 2n

k i.e. 216

21k = [1M]

1

16

(4 ) 21

n

nx k = +

Since 116

21x = 1

16 164

21 21k = + i.e. 1 0k = [1M]

16

21nx = for . [1M]1n

Hence 11 1 16

4 4 21 21n nz x += = =

4[1M]

and1

1 1 4

4 4 21 21n ny z

+

= = =1

[1M]

(c) P(Xwill be the final victory)

= P(Xwins the 1st game and will be the final victory)

+ P(Xloses the 1st game and will be the final victory)

= 1 13 1 3 16 1 1 4

4 4 4 21 4 21 84x y+ = + =

9[1M]

P(B will be the final victory) = 1 11 3 1 16 3 1 1

4 4 4 21 4 21 84

x y+ = + =9

[1M]

P(Cwill be the final victory) =16 4

84 21= [1M]

11. (a) LetX(in minutes) be the late time,2~ ( , )X N

( 0) 0.1003P X < =

0( ) 0.1003P Z

< =

1.280

= (1) [1M]

( 4) 0.1793P X =

4

( ) 0.1793P Z

=

40.918

= (2) [1M]

By solving (1) and (2), 2.33 = and 1.82 = [1M]


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(b) (i)15.3

1.27512 12

xx = = = [1M]

0

1

: 2.3

: 2.3

H

H

3

3

=

applied math paper 2 marking by n.f. yang

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