2008-2009 F7 Alm Mock Exam Marking Scheme

2008-20098 57 ALM 2nd TermExaminatron Markins Scheme

l. ( a)

,,,r 2000- N )

dl{

- = Alyl

dt

2000 )

\

2000dN .'

f^

|

Jr o o oN( 2 0 0 0 _ ( t

,u

|

I l. r r ooo\ N

2000_N)

Jo

N) \

IM

tt . =l)"dr .t

VtN=lAdt Jo

IA

[ln] w l -ln | 2ooo- N lflo,,= t"t N t000 , 2000- 1/

-

rrr-

-

2000- 1000 --

l-

l tt

IA

= etl

'o*;uL= 2ooo -.N 2000

1V = ---------=

IA

l+e "'

1.(b)

Forr>0 e-^'>o l+ e-1'>1

1M

-1 ^'-.r 1+ e 2ooo < 2ooo

1+ e-"'

< 2000 ..'1V .'. Thenumberof daisiesplantedin thegardenwill neverexceed2000for r > 0. 2"

IA

x- e' d \, dt dy

d v (d r\

dx

d t\&)

,a v dt

t 4= ! (, 4\ q= " , (,,+ -,,4 \= ,,,(1 -4 \ dY'

d t\

dt ) d\

dt'

\

d,)

l dt'

dr )

dv . ,d ' y ) x --.;-i r-.^ * 4_1= 1dxdx

( n 'rY' ' l ' ' ! - u ' l - r r 'u -!'1* q ,= G' Y dr' /

,t

,

\

dt .)

\

dt

!l]--o4Y+ 4 y - s 2 ' dt'

ctt

IM IA

Characteristicequation: IM

m 2 -4 m + 4 :0 m:2 ."yr :

(repeated)

cl€ 2 t + c2 te 2 t

IA

Letyo: Af e2'. 4,, =2trrrrr, +2Ate2l dt t2

a v. ;

= 4 A !' e ' ' + 8 A Ie ' ' + 2 A e 2'

IM

2008-20098 57 ALM 2nd TermExaminationMarkins Scheme

.'.4At2e2'+ 8Ate2'+ 2Ae2'- 4Q.Afe'' 2A =l

IA

.t ,-

n--

z

| , ., IA

z 1. . v = c , e' ' + c , t e"' + . t ' e' '

'2

I = e '.,( ' l c .+ c .t+ -t' I ,) '2) \ ( | = x 2 lc , + c , l n x + -(l n x )' z" \ I ' 2 \' ) 3. ( a)

IA

Let X V be the breakdownvoltageof a randomly chosendiode. x - N (4 0 , 1 .5 2 ) P (3 9 < X< 4 2 )

= p,39-+o .2.42-40\ ' 1.5 1.5 = P(-0.6667
IM

IA

Let the value be c V. P (X> c ): 0 .1 5 { r ' - 4 O\ " PIZ>l=0.15 1.5 \ )

r - 4O

IM

1.5 c=40+1.5(1.0365) = 41.5548 .'. only l5% of all diodeshavevoltages 41.5548V. exceeding 3.(c)

IA

P(X> 42) = P (Z > '

4 -) -4 0-)

1 .5 = P(Z > 1 .3 3 3 3 ) IA

= 0 .0 9 1 2 P(at leastone hasa voltageexceeding42)

:r-[ -P (x>44]4 :l-(1 -0.09n)4 :0.3179 4.(a)

IA IA

P(Ru)

5f4) 3f3)

8[t.1.8ltj

IM

29

IA

n 4 (b)

P (G) :1 -2 :9 72

IA

72

-2-

2008-20098 57 ALM 2nd TermExaminationMarkins Scheme

P(G,iG.r-

P(G'aGr) p(G,)

IM

3r6)

r(r, 43 72 l8 43 5. ( a)

IA

360= o.z5 480

P ,:

IA

The95V:oconfidenceinterval forp

_{ , - lr . I

0.75(l- 0.75) ,0.75+1.96 480

=lo75

\ = (0.7l 13,0.7887) 5 (b)

1M IA

H s :p : 0 .7 H ;p * 0 .7 Under ffe and using normal approximation,

U' - ,r,tro.z.0'7(0'3)l

IM

480

o.z s -o'5-o.z

480 =23407>1.96 0.7(0.3)

IA IM

6.(a)

f (x) = xlnx /'(x) :l + ln x

f" (*)=|

-+ .ft"(*)= ... p ,(x ) = f 6 \+ /' (l )(* lt

1 1 * ,f" (l ),r-1i , 2l

IM

= (x - t)*f,*-t)' -l t x - t )' 2' 6' =-!e, -6x2+3x+2 ) 6'

6 (b) [' o ' (* )= ]

w h i c h i s d e c re asi ng on [0.5. I .5].

IA IM

= Z =tU f,o,(0.5) 0.5' "f(4)(1.5):',=tu 1.5' 27 .'.1.f'o)(r)l< l6 if lx- ll<0.5

IA

2008-2009857 ALM 2nd Term ExaminationMarkine Scheme

6.(c)

Errorsljjit,', - p,(iltu]

=li)ln*>o,r,>1*

IM

=I-ff<.-tl*

wheref e (0.5,1.5)

'3Ji;o

-t)odx

IM IA

=1Io-r)'l;: :0.0083 7.(a)

IA

[' 3kt'dt* J['r Lft + s\dt= t

Jo

IM

2'

=1 +sr1l fu')',.f;t|,2 . k (e

r _)l=l

/r+.:l -+15---5 2\22)

8ft=l

*=!

IA

8

7.(b)

E(r)=Jol'"lr +stpt at* Jtl,'+Q' 8 16.

IM

=fl,,l'*ai1,, *1,,.l' L32 lo r6L3 2 ), =L* L(g* 45-l -I)

32 16\232) _ 181 ,96

IA

+st'pt E(r\=Jf)lr a,*t:+t' rr 1 6 o8

1M

tft,_ t' + _ t'5 . -Il'

Ir.l ' | + -t -t-t-

140 lo 1614

3 L

r t r(et =-+-l -+4)---40 16\4 4

s )| 3)

_ r2 l 30

IA

var(r)=o(r\-In(D]'

IM

= r2r-[(rer)' 3 0 %.J IA

=0.4785

-4-

2008-20098 57 ALM 2nd TermExaminatron MarkrnsScheme

7.(c)

F o r0 < t< 1 , nr i

F (tl = | -r' d- l r0 8

Ir ,lI '

= | - /'

L 8 -J N l.

IA

- - t-

8 F o rl < l < 3 , nr 1

I

F(/)=l_0*5),#+:(tr) r t l6 8

IM

tll , -l' I = _l_t r _) 1 t + _ t6L2 l, 8

r [ r , + )-I ___)| - l r l+ _

: - l- t -

16L2 l" 5 32 16

2 '7

)

IA

32 r<0

0 l,

0
-a

.'. F(t) = 8

1, 32

5 t6

IA

7 l
I 7 (d)

8

t>3

The requiredprobability

= c; .c: cl [,c1ry]'[r - Fe)]'zlF@- F(l)l

2M

= 3 o f l l ' z f r - r 7 ) ' z f r 7- r ) 32)\32

\8i \ = 0.0418

8)

IA

8.(aXi) .f (x) -- In(l + cosx) -s i n x J tx 1 = . l + c o s -r - (l + c o sx ) c o sx + s i nx(- si nx) .f"(x) = (1+ cosx)2 -c o s x * c o s ' x -s i n t (l + cosx)2 -c o s x -l (l + cosx)' _

-l I + cosx

-11, .

J"\x)=

- sin x

(-l + cosr)-

IA

2008-20098 57 ALM 2nd TermExaminationMarkineScheme fto'(*) =

- (1+ cosx)'?cosx + sinx[2(1+ cosxX- sinx)] (1+ cosx)a -c o s r-c o s ' x

-2 s i n 2 x

(1+ cosx)l c o s ' x -c o s x -2 (l + cosx)3 (c o s x + l )(c o s x -2 ) (l + cosx)3

IA

c o s x -2

(1.."*t' 8.(a)(ii) /(5 ), _ (1+ c o s x )2 (-s i n x )- (cosx-2)(2)(l + cosx)(-si n:r) t \^-.\ l

_ - s i nx (l + c o s x - 2 c osx + 4) (l + c o s x )3

- -sinx(5-cosx).0 (l+cosx)'

v * .-l-' [ o . za)l

IA

T-1

.'..f'o'(t) is strictlydecreasing - or I o.i l.

L 2)

,

- t 4, . ^. l v t- - - - -

, , 0, ( r \ -

c os 0- 2

1M

|

(l + cos0)' 4 (-\ cosla l-2 \ 2/

"tt-t----L \2) \-/

-

|

/-\l'

L

\2rl

l

l t*.o tl 1 l l

.' . ma x l fto )(x ' tl :2

IA

r€ 0.| 2)

r

r5

IE _"l nl l'r Error: -" ! f'or(E\ 180n"

l^rl where4elo.-| L 2)

IM

5 lt

l.rar,

,l - -______ _( maxl/,-,(x)l 180(2')n " ,.[6rlr'

2

5

:

" < - 106 -2880na

IM

5

1T'

n-^ 2---------------2880(10') n'2t06256.835... n > 18.05...

IM

n=20 .'.20subintervals shouldbe used. 8 (b)

Lete(x)=l;r'+x -21=[*'+x-2 I I l-1x'+x-2) Jrt"'

+ x -21 dt =

I'

IA

if 1<x<4 ifo<x
- Q' + x - 2)clx* + x -2)dx J,o{r'

- b-

tA

2008-20098 57 ALM 2nd TermExamrnation MarkrnsScheme

In orderto find theexactvalueof

J,t

*' * x - 2l dxby Simpson's rule,thenumbers

of subintervals for thetwo intervalsmustbe even. So 8 subintervals shouldbe used. 4- U . h = - = u .5

^ _

IM

IA

8

Jnlr'+x-21rk =

h.

+ g(1.s)+ g(2.s)+g(3.5)]+2[g(l)+ g(2)+ g(3)]+s(4)] ;3 {s(ol+ a[g(O.s)

_11 3

IA

g(O.s):1.2s c (l): 0 g(l.s): l.7s c Q ): 4 g(2.5)-6.7s s (3 ): 1 0 g(3.s):13.7s { ,-\ _ 6)-r :2x +t_ :.ltor' Let f(x): (:r* t -Jtox' -s "

c(0):2 c(4): l8

e.(a)

Vx e

IM

IA

[.3 ,4 ],

(x)= 2 - :(tox' - o) I lzox) -f' 2', l0x

_a

-.)

-a

IA

ffi troor'

1 /1 o " =

l 0 (1 O -r' ? -6 + 6 ) l U x - -6

fi.-4l}x" -6

V

<2 ^ lto <0 .'./(x)is strictlydecreasing on [.3, 4].

IM

+ r - JOir it' - 6 = 0 . 2 e 8>50 f(t.3)=2(1.3) .f(4\=2(4)+t-Jl0(4f-6 =_ 3 . 4 0 e<70 .'. Ehas exactly one real root lying in the interval t r .3,41.

e.(b)

g ' (x ) =

IA

z-)00; -elIpo*1 ln.t r- - - - - - - -' -"^ -. lllx ' - 6

IA

\, I ' ox2o) ' tzoxl] 2

a )1 ,( r g " (x ) = _ ro [(1 0 ,,_ ]_ L'

-r o(rox' - ofi[ro'' - 6- I ox,]

oo(r o"'- ofi

IA

2008-20098 57 ALM 2nd TermExanrination Markins Scheme

>0

V x e [.3,4] .'.9'(x)is strictlyincreasingon [.3, 4].

c'(r.3)=3-+

IM

-_-0.s376

Jl0(1.3)'- 6

IA

s'(4)=3--#gL =-0.2233 {lo(4)' - 6

' ls'(x)l<0.9376<1 e.(c)(i)

IA

f (r.6 )= 2 (1 .6 )+ 1 -J l 0 (1 .6 )' z-6 = -0:272< 0 IM

.' .T h ero o t i s i n [1 .3 , 1 .6 ]. Vx e [.3 , 1 .6 ],

^ trooa

P ( xl= J- .1 - -

Vl0x'-6 f60 -=?" llO- r X'"' lox2-6

< 3-.,/ib <0 .' .g (t) i s s tri c tl yd e c re a s i nogn [1.3, 1.6].

I N,{

,' .c (1 .6 )< g (x ) < g (1 .3 ) 1 .3 7 2 8< g (x )< 1 .5 9 8 5 1 .3< g (x ) < 1 .6

1M

." .g (x ) e [1 .3 , 1 .6 ] .'.x6: 1.6can be used.

n 1.6000 1.3728

9 l0

1.4564 1.4691

1.5343

11

1.4147 1.5006

l2

1.4598 t.4666

IJ

|.461',7

t.4316

I4 t5

t.46s2 1.4626

16

|.4645

1.4831 t.4498

IM

1.4739 IA

.the root is l 46.(cor.to 2 d.p.)

e.(cXii)

10. ( a)

c (3 .5 )= 3 (3 .s )+ 1- J l 0 (3 .s )' ?* 6 = 0.7065

IM

.' g(3.5) e [1.3, 4] and this doesnot guaranteethe convergenceof the iterative formula.

IA

4 =r . s r r orr..5 f l - l - / : L ) dt \20,/ &x3y +-:+ l) dt l0 40

\30,

(l)

-8-

IA

2008-20098 57 ALM 2nd TermExaminationMarkins Scheme

From (2),

Subs.(3).(a) into(1) /i

.,\

/1

rc({4 * !!z) =-(9* 1 -:o) +1L+t5 sdr)

[dr' 12

t

dt"

dt

s

\dt

) 40

rc2 *zL = -ctY* 1 + 3 o+fI+15 dt

5

40

rc!2+3dY +t v=45 dt'

dt

8'

80" ! +242+v=360 dt' dt

r0.(b) C h a ra c te ri s tiEcq u a ti o n :

IM

8 0 m 2" 2 4 m-l :0 l1 204

o r --

cr e 2 a + cr e 4

IA

Let yo: A d.1,v.

d ' v.|,-A .

dt

dt"

""A:360 '.yn: 360 i. Y= c P

20+ c r e a + 360

IA

dv c , : c , :

d t2 0 4 W h e nl : 0 , y :c t+ c z + 3 6 0 :2 0 0 c r * c z : -1 6 0 dy

0

(i )

2 0 0 ,.^

ct

L = _ _ _ r _ itr = _ _ _ _

IM

cz

d t1 0 5 2 0 4 c r + 5 c z :2 0 0

(i i )

1M

:.c 1 : -2 5 0 a n dc 2 :9 0 r!

:.!=-250e 20+90e { +360 dv25:45:tv

L=-e

Lv

_ =_ _ o

dr'

8-

4

--e

dt22 d'y 5 -;

+_ o

IA

45 -+ q

8"

-9-

2008-20098 57 ALM 2nd TermExaminarion MarkineScheme

)-,

uy

^ S e t - -u . dt

IM

* 1 =- e4 5 -:I

25

22 ;9 e'" =I

+ fe), e' " = t - |

IA

\5/

lfltlS,

d'v

s /s \.

d !'

8 \s ,

--=-

i- l

...W h e n e ro :i

+s -i/ s ti >U .

+-i

IM

815/

| .y i s a m i n i mum. \)/-

The minimum amountof salt

rq l I

:-2501-l \5i

/q \ :

+e0l-l l5i

+360

rs t j .

=l; | (-450+e0)+360 \)/ | .r I | /5 \r l =3601 1-l: I I

| \ei I

LI

=t87.3320g

IA /q \

a n dth e ti m e :5 In l - l mi n . \ 5 ,/ IL(a)(i)

IA

Let therandomvariableX mg denotetheactualtar yield of a cigarette. X- N(p,0.82) i.e.X-N(17.5,0.8'z) P(X> 18.5) ^( = f lZ>

18.5- 17.5\

\0.8) = P(Z >r.2s) =0.5-03944 : 0.1056

IM

IA

- l0 -

2008-20098 57 ALM 2nd TermExamrnatron MarkinsScheme I L(a)(ri)

Let I denotethe number of cigarettescontainingtar yields higher than 18.5mg in a pack of50 cigarettes. )' - Bi n (5 0 ,0 . 1 0 5 6 ) Using normal approximation,

v- N(50(0. l0s6).50(0.l0s6xt - 0.l0s6)) r- N(5.28,4.122432) P(v> l0)

^ (-

=

rt

t

>

_

IM IM

ro .s- 5 .2 8 ) |

14.722432 \ ) == P(Z >2.4021) = 0.5- 0.4918 = 0.0082 -l

IA

rr.(bxi) Hs:p: 17.5 IA

H; 1t> 17.5 I I (b)(ii)

r=I1=rs.r

IA

l0

Under He, o R2 x - Nfl7.5."'" )' l0

1M

18.1- 17.5

o -t

'=

IM

ffi = 2 .3 7 1 7> 2 .3 2 6 7

IA

So t1ois rejectedat lo/olevel and there is evidencethat the actualtar yields were higherthan that claimed. r l.(b xiii) Supposethat Ho is rejectedat 1%olevel when X > c.

P \x > cl x -Nt I 2.s.9{1 -' 10' = o . o , ?_11

<

P(Z ='='-) = 0.01 ' >' u.6

IM

ffi r-17

5

0.8

m

c = 18.0886

IA

P(TypeII error) : P(accept Ho j il is true) = Pd < | 8.0886x .- Nt r a.o,0,9'D ' t0"

IM

l8.o) = P(Z < l8'08q6_0.8

IM

ffi = P(Z < 0.3503) = 0.5+ 0.1369 =0.6369

IA