2011 Mathematical Methods (CAS) Exam 2

7/27/2019 2011 Mathematical Methods (CAS) Exam 2

http://slidepdf.com/reader/full/2011-mathematical-methods-cas-exam-2 1/28

MATHEMATICAL METHODS (CAS)

Written examination 2

Wednesday 9 November 2011

Reading time: 11.45 am to 12.00 noon (15 minutes)

Writing time: 12.00 noon to 2.00 pm (2 hours)

QUESTION AND ANSWER BOOK

Structure of book

Section Number of

questions

Number of questions

to be answered

Number of

marks

12

224

224

2258

Total 80

6WXGHQWVDUHSHUPLWWHGWREULQJLQWRWKHH[DPLQDWLRQURRPSHQVSHQFLOVKLJKOLJKWHUVHUDVHUVVKDUSHQHUVUXOHUVDSURWUDFWRUVHWVTXDUHVDLGVIRUFXUYHVNHWFKLQJRQHERXQGUHIHUHQFHRQHDSSURYHG&$6FDOFXODWRUPHPRU\'2(6127QHHGWREHFOHDUHGDQGLIGHVLUHGRQHVFLHQWL¿FFDOFXODWRU)RUDSSURYHGFRPSXWHUEDVHG&$6WKHLUIXOOIXQFWLRQDOLW\PD\EHXVHG

6WXGHQWVDUH127SHUPLWWHGWREULQJLQWRWKHH[DPLQDWLRQURRPEODQNVKHHWVRISDSHUDQGRUZKLWHRXWOLTXLGWDSH

Materials supplied 4XHVWLRQDQGDQVZHUERRNRISDJHVZLWKDGHWDFKDEOHVKHHWRIPLVFHOODQHRXVIRUPXODVLQWKH

FHQWUHIROG $QVZHUVKHHWIRUPXOWLSOHFKRLFHTXHVWLRQV

Instructions

'HWDFKWKHIRUPXODVKHHWIURPWKHFHQWUHRIWKLVERRNGXULQJUHDGLQJWLPH :ULWH\RXUstudent numberLQWKHVSDFHSURYLGHGDERYHRQWKLVSDJH &KHFNWKDW\RXUnameDQGstudent numberDVSULQWHGRQ\RXUDQVZHUVKHHWIRUPXOWLSOHFKRLFH

TXHVWLRQVDUHFRUUHFWandVLJQ\RXUQDPHLQWKHVSDFHSURYLGHGWRYHULI\WKLV

$OOZULWWHQUHVSRQVHVPXVWEHLQ(QJOLVK

At the end of the examination 3ODFHWKHDQVZHUVKHHWIRUPXOWLSOHFKRLFHTXHVWLRQVLQVLGHWKHIURQWFRYHURIWKLVERRN

Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic

devices into the examination room.

9,&725,$1&855,&8/80$1'$66(660(17$87+25,7<

SUPERVISOR TO ATTACH PROCESSING LABEL HERE

Figures

Words

STUDENT NUMBER Letter

9LFWRULDQ&HUWL¿FDWHRI(GXFDWLRQ

2011



0$7+0(7+&$6(;$0

SECTION 1±FRQWLQXHG

Question 1

7KHPLGSRLQWRIWKHOLQHVHJPHQWMRLQLQJ±WRd LV

A.d

2

5

2, −

⎛

⎝ ⎜

⎞

⎠⎟

B.

C.d −⎛

⎝ ⎜

⎞

⎠⎟

5

20,

D. 05

2,

−⎛

⎝ ⎜

⎞

⎠⎟

d

E.5

20

+⎛

⎝ ⎜

⎞

⎠⎟

d ,

Question 2

7KHJUDGLHQWRIDOLQHperpendicularWRWKHOLQHZKLFKSDVVHVWKURXJK±DQG±LV

A. 1

2

B. –2

C. 1

2

D. 4

E. 2

Question 3

,I x + aLVDIDFWRURI x3 – 13 x2 – axZKHUHa 5?^`WKHQWKHYDOXHRIaLV

A. – 4

B. –3

C. – 1

D. 1

E. 2

SECTION 1

Instructions for Section 1

$QVZHUallTXHVWLRQVLQSHQFLORQWKHDQVZHUVKHHWSURYLGHGIRUPXOWLSOHFKRLFHTXHVWLRQV

&KRRVHWKHUHVSRQVHWKDWLVcorrect IRUWKHTXHVWLRQ

$FRUUHFWDQVZHUVFRUHVDQLQFRUUHFWDQVZHUVFRUHV

0DUNVZLOOnotEHGHGXFWHGIRULQFRUUHFWDQVZHUV

1RPDUNVZLOOEHJLYHQLIPRUHWKDQRQHDQVZHULVFRPSOHWHGIRUDQ\TXHVWLRQ



0$7+0(7+&$6(;$0

SECTION 1±FRQWLQXHGTURN OVER

Question 4

7KHGHULYDWLYHRIORJe f xZLWKUHVSHFWWR xLV

A. c f x

f x

( )

( )

B. 2c f x

f x

( )

( )

C. c f x

f x

( )

( )2

D. ORJe f c x

E. ORJe f c x

Question 5

7KHLQYHUVHIXQFWLRQRI g >fo R g x 2 4 x LV

A. g –1>fo R g –1 x x2 4

2

B. g –1>fo R g –1 x x±2

C. g –1>fo R g –1 x x

24

D. g –1>fo R g –1 x x2 4

2

E. g –1 R o R g –1 x x2

42

Question 6

)RUWKHFRQWLQXRXVUDQGRPYDULDEOH X ZLWKSUREDELOLW\GHQVLW\IXQFWLRQ

f x x x e

elsewhere

e( )

log ( )=

≤ ≤⎧⎨⎩

1

0

WKHH[SHFWHGYDOXHRI X E X LVFORVHVWWR

A.

B.

C. 1

D.

E.



0$7+0(7+&$6(;$0


Question 7

%\FRQVLGHULQJWKHSRLQWRQWKHJUDSKRI f x 3 x DQGXVLQJWKHOLQHDUDSSUR[LPDWLRQPHWKRG

f h| f hf cDQHVWLPDWHIRU 38 5. LV

A.

B.

C. D.

E.

Question 8

&RQVLGHUWKHIXQFWLRQ f R o R f x x x±DQGWKHIXQFWLRQ

g 3

25,

⎡

⎣⎢ ⎞

⎠⎟ o R g x x

,IWKHIXQFWLRQh f + g WKHQWKHGRPDLQRIWKHLQYHUVHIXQFWLRQRIhLV

A. >

B. −⎡

⎣⎢⎤

⎦⎥3

410,

C. −⎛

⎝ ⎜

⎤

⎦⎥3

4

15

4,

D. 3

413,

⎡

⎣⎢ ⎞

⎠⎟

E. 3

2 13,⎡

⎣⎢ ⎞

⎠⎟



0$7+0(7+&$6(;$0


Question 9

7KHJUDSKRIWKHIXQFWLRQ y f xLVVKRZQEHORZ

4

–5 –3 O

y

x

:KLFKRIWKHIROORZLQJFRXOGEHWKHJUDSKRIWKHGHULYDWLYHIXQFWLRQ y f c x"

–3 –5 O

4

A.

–3 O

B.

–3 O

D.

–3 –5 O

4

E.

–3 –5 O

C.

x x x

x x

y y

y y

y



0$7+0(7+&$6(;$0


Question 10

0DULDLVDYHU\HQWKXVLDVWLF\RXQJWHQQLVSOD\HUZKRSOD\VDPDWFKHYHU\GD\,IVKHZLQVRQRQHSDUWLFXODUGD\WKHSUREDELOLW\WKDWVKHZLQVWKHQH[WGD\LV,IVKHORVHVRQHGD\WKHSUREDELOLW\WKDWVKHORVHVWKHQH[WGD\LV

7KHORQJWHUPSUREDELOLW\WKDW0DULDZLQVDPDWFKLV

A.

4

5

B. 2

5

C. 2

3

D. 1

3

E. 3

4

Question 11

7KHDYHUDJHYDOXHRIWKHIXQFWLRQZLWKUXOH f xORJe xRYHUWKHLQWHUYDO>@LV

A. ORJe

B. 1

36log ( )e

C. loge

3125

43

⎛

⎝ ⎜

⎞

⎠⎟ −

D. 1

3

3125

4 3loge

⎛

⎝ ⎜

⎞

⎠⎟ −

E. 5 5 2 2 3

3

log ( ) log ( )e e

Question 12

7KHFRQWLQXRXVUDQGRPYDULDEOH X KDVDQRUPDOGLVWULEXWLRQZLWKPHDQDQGVWDQGDUGGHYLDWLRQ)RUDJLYHQQXPEHUa3U X > a

&RUUHFWWRWZRGHFLPDOSODFHVaLVHTXDOWR

A.

B. C.

D.

E.



0$7+0(7+&$6(;$0


Question 13

,QDQRUFKDUGRIDSSOHWUHHVLWLVIRXQGWKDWWUHHVKDYHDKHLJKWJUHDWHUWKDQPHWUHV7KHKHLJKWVDUHGLVWULEXWHGQRUPDOO\ZLWKDPHDQ DQGVWDQGDUGGHYLDWLRQPHWUHV

7KHYDOXHRI LVFORVHVWWR

A.

B.

C. D.

E.

Question 14

y

xO 1

1

e 2

y = e 2 x

7R¿QGWKHDUHDRIWKHVKDGHGUHJLRQLQWKHGLDJUDPVKRZQIRXUGLIIHUHQWVWXGHQWVSURSRVHGWKHIROORZLQJFDOFXODWLRQV

i.

1

2

0

xe dx³

ii. e e dx x2 2

0

1

³

iii.

2

2

1

e y

e dy³

iv.

2

1

log ( )

2

ee x

dx³

:KLFKRIWKHIROORZLQJLVFRUUHFW"

A. ii. RQO\

B. ii. DQGiii.RQO\

C. i.ii.iii.DQGiv.

D. ii.DQGiv.RQO\

E. i.DQGiv.RQO\



0$7+0(7+&$6(;$0


Question 15

3

–1

S

623S

y

xO

7KHJUDSKVKRZQFRXOGKDYHHTXDWLRQ

A. y x= +⎛

⎝ ⎜

⎞

⎠⎟ +2

61cos

π

B. y x= −⎛ ⎝ ⎜ ⎞

⎠⎟ +2 4

61cos

π

C. y x= −⎛

⎝ ⎜

⎞

⎠⎟ −4 2

121sin

π

D. y x= +⎛

⎝ ⎜

⎞

⎠⎟ −3 2

61cos

π

E. y x= +⎛

⎝ ⎜

⎞

⎠⎟ −2 4

2

31sin

π

Question 16

:KLFKRIWKHIROORZLQJLVnotWUXHDERXWWKHIXQFWLRQ

f R o R f x | x2±_±"

A. 7KHJUDSKRI f LVFRQWLQXRXVHYHU\ZKHUH

B. f ±±

C. f x±IRUDOOYDOXHVRI x.

D. f c x 2 xIRUDOO x !

E. f c x 2 xIRUDOO x ±

Question 17

7KHQRUPDOWRWKHFXUYHZLWKHTXDWLRQ y x x= +32 DWWKHSRLQWLVSDUDOOHOWRWKHVWUDLJKWOLQHZLWKHTXDWLRQ

A. 4 x y

B. 4 y + x

C. yx

= +4

1

D. x – 4 y –5

E. 4 y + 4 x 20



0$7+0(7+&$6(;$0


Question 18

7KHHTXDWLRQ x3± x2 + 15 x + wKDVRQO\RQHVROXWLRQIRU xZKHQ

A. ±w < 25

B. w±

C. w

D. w±RUw > 25

E. w > 1

Question 19

$SDUWRIWKHJUDSKRI f R o R f x x2LVVKRZQEHORZ=RH¿QGVWKHDSSUR[LPDWHDUHDRIWKHVKDGHGUHJLRQ E\GUDZLQJUHFWDQJOHVDVVKRZQLQWKHVHFRQGGLDJUDP

y

y

x

x1 2 3 4 5 6

1

O

O 2 3 4 5 6

=RH¶VDSSUR[LPDWLRQLV pPRUHWKDQWKHH[DFWYDOXHRIWKHDUHD

7KHYDOXHRI p LVFORVHVWWR

A. 10

B. 15C. 20

D. 25

E. 30



2011 MATHMETH(CAS) EXAM 2 10

SECTION 1 – continued

Question 20

A part of the graph of g : R o R, g ( x) = x2 – 4 is shown below.

A

B

y

xaO

The area of the region marked A is the same as the area of the region marked B.

The exact value of a is

A. 0

B. 6

C. 6

D. 12

E. 2 3

Question 21For two events, P and Q, Pr( P Q) = Pr( P Q).

P and Q will be independent events exactly when

A. Pr( P ) = Pr(Q)

B. Pr( P Q ) = Pr( P Q)

C. Pr( P Q) = Pr( P ) + Pr(Q)

D. Pr( P Q ) = Pr( P Q)

E. Pr( P ) =

1

2



11 0$7+0(7+&$6(;$0

END OF SECTION 1

TURN OVER

Question 22

7KHH[SUHVVLRQ

ORJcaORJ

abORJ

bc

LVHTXDOWR

A.

1 1 1

log ( ) log ( ) log ( )c a ba b c

B. 1 1 1

log ( ) log ( ) log ( )a b cc a b

C. 1 1 1

log ( ) log ( ) log ( )a b cb c a

D. 1 1 1

log ( ) log ( ) log ( )a b ca b c

E. 1 1 1

log ( ) log ( ) log ( )c b aab ac cb




SECTION 2 – Question 1 – continued

Question 1

Two ships, the Elsa and the Violet, have collided. Fuel immediately starts leaking from the Elsa into the sea.

The captain of the Elsa estimates that at the time of the collision his ship has 6075 litres of fuel on board and

he also forecasts that it will leak into the sea at a rate of t 2

5litres per minute, where t is the number of minutes

that have elapsed since the collision.

a. At this rate how long, in minutes, will it take for all the fuel from the Elsa to leak into the sea?

3 marks

SECTION 2

Instructions for Section 2

Answer all questions in the spaces provided.

In all questions where a numerical answer is required an exact value must be given unless otherwiseVSHFL¿HG

In questions where more than one mark is available, appropriate working must be shown.

Unless otherwise indicated, the diagrams in this book are not drawn to scale.



13 0$7+0(7+&$6(;$0

SECTION 2 – Question 1±FRQWLQXHGTURN OVER

,QIDFWWKHFDSWDLQ¶VDVVHVVPHQWRIWKHVLWXDWLRQLVZURQJ7KHDPRXQWRIIXHORQERDUGDWWKHWLPHRIWKHFROOLVLRQis not OLWUHV7KHIXHODFWXDOO\OHDNVLQWRWKHVHDIRUPLQJDFLUFXODURLOVOLFN7KHDUHDRIWKLVFLUFOHLVLQFUHDVLQJDWWKHFRQVWDQWUDWHRIVTXDUHPHWUHVSHUPLQXWH

$WWLPHt PLQXWHVDIWHUWKHFROOLVLRQWKHUDGLXVRIWKHFLUFOHLVr PHWUHV

b. 6KRZWKDWZKHQWKHUDGLXVRIWKHFLUFOHLVPHWUHVWKHUDGLXVLVLQFUHDVLQJDW10

3S

PHWUHVSHUPLQXWH

PDUNV



0$7+0(7+&$6(;$0

SECTION 2 – Question 1±FRQWLQXHG

,QWKHIROORZLQJDOOPHDVXUHPHQWVRIGLVWDQFHDUHLQPHWUHV

6RPHWLPHODWHUWKHIXHOVWRSVOHDNLQJ$XWKRULWLHVZLOOXVHWZRÀRDWLQJEDUULHUVWRVXUURXQGWKHRLOVOLFN2QHRIWKHEDUULHUVEDUULHULVLQWKHVKDSHRIWKHJUDSKRIWKHIXQFWLRQ

g >@o R g xVLQπ x

80

⎛

⎝ ⎜

⎞

⎠⎟

7KLVJUDSKLVVKRZQEHORZ

y (m)

C

O x (m)

7KHVHFRQGEDUULHUHQFORVLQJWKHRLOVOLFNEDUULHULVLQWKHVKDSHRIWKHJUDSKRIWKHIXQFWLRQhJLYHQE\

h>@o R h x −⎛

⎝

⎜⎞

⎠

⎟6040

sinπ x

c. 2QWKHD[HVDERYHVNHWFKWKHJUDSKRI y h x/DEHODOOD[HVLQWHUFHSWVDQGWXUQLQJSRLQWVZLWKWKHLUFRRUGLQDWHV

PDUNV



15 2011 MATHMETH(CAS) EXAM 2

SECTION 2 – continued TURN OVER

d. 6RPHWLPHODWHUWKHRLOVOLFNFRPSOHWHO\¿OOVWKHDUHDHQFORVHGE\WKHWZREDUULHUV

i. Find sinπ x

dx40

0

40⎛

⎝ ⎜

⎞

⎠⎟∫

ii. )LQGWRWKHQHDUHVWVTXDUHPHWUHWKHWRWDODUHDRIWKHRLOVOLFNDWWKDWWLPH

1 + 2 = 3 marks

Total 11 marks



0$7+0(7+&$6(;$0


Question 2

,QDFKRFRODWHIDFWRU\WKHPDWHULDOIRUPDNLQJHDFKFKRFRODWHLVVHQWWRRQHRIWZRPDFKLQHVPDFKLQH ARUPDFKLQH B

7KHWLPH X VHFRQGVWDNHQWRSURGXFHDFKRFRODWHE\PDFKLQH A,LVQRUPDOO\GLVWULEXWHGZLWKPHDQDQGVWDQGDUGGHYLDWLRQ

7KHWLPHY VHFRQGVWDNHQWRSURGXFHDFKRFRODWHE\PDFKLQH BKDVWKHIROORZLQJSUREDELOLW\GHQVLW\IXQFWLRQ

f y

y

y y

e y y

( )

. . ( )

=

<

≤ ≤

>

⎧

⎨

⎪⎪

⎩

⎪⎪ − −

0 0

160 4

0 25 40 5 4

a. )LQGFRUUHFWWRIRXUGHFLPDOSODFHV

i. 3U X

ii. 3UY

PDUNV

b. )LQGWKHPHDQRIY FRUUHFWWRWKUHHGHFLPDOSODFHV

PDUNV



0$7+0(7+&$6(;$0


c. i. )LQGWKHPHGLDQRIY

ii. )LQGWKHYDOXHRIaFRUUHFWWRWZRGHFLPDOSODFHVVXFKWKDW3UY a

PDUNV

d. ,WFDQEHVKRZQWKDWPr( )Y ≤ =39

32 $UDQGRPVDPSOHRIFKRFRODWHV produced by machine BLVFKRVHQ)LQGWKHSUREDELOLW\FRUUHFWWRIRXUGHFLPDOSODFHVWKDWH[DFWO\RIWKHVHFKRFRODWHVWRRNRUOHVVVHFRQGVWRSURGXFH

PDUNV




SECTION 2 – continued

All of the chocolates produced by machine A and machine B are stored in a large bin. There is an equal number of chocolates from each machine in the bin.

It is found that if a chocolate, produced by either machine, takes longer than 3 seconds to produce then it canHDVLO\EHLGHQWL¿HGE\LWVGDUNHUFRORXU

e. A chocolate is selected at random from the bin. It is found to have taken longer than 3 seconds to produce.Find, correct to four decimal places, the probability that it was produced by machine A.

3 marks

Total 15 marks



0$7+0(7+&$6(;$0


CONTINUES OVER PAGE



0$7+0(7+&$6(;$0


Question 3

a. &RQVLGHUWKHIXQFWLRQ f R o R f x x3 + 5 x±

i. )LQG f c x

ii. ([SODLQZK\ f c xIRUDOO x

PDUNV

b. 7KHFXELFIXQFWLRQ pLVGH¿QHGE\ p R o R p xax3 + bx2 + cx + k ZKHUHabcDQGk DUHUHDOQXPEHUV

i. ,I pKDVmVWDWLRQDU\SRLQWVZKDWSRVVLEOHYDOXHVFDQmKDYH"

ii. ,I pKDVDQLQYHUVHIXQFWLRQZKDWSRVVLEOHYDOXHVFDQmKDYH"

PDUNV

c. 7KHFXELFIXQFWLRQqLVGH¿QHGE\q R o Rq x± x3 i. :ULWHGRZQDQH[SUHVVLRQIRUq –1 x



21 0$7+0(7+&$6(;$0


ii. 'HWHUPLQHWKHFRRUGLQDWHVRIWKHSRLQWVRILQWHUVHFWLRQRIWKHJUDSKVRI yq xDQG yq –1 x

PDUNVd. 7KHFXELFIXQFWLRQ g LVGH¿QHGE\ g R o R g x x3 + 2 x2 + cx + k ZKHUHcDQGk DUHUHDOQXPEHUV

i. ,I g KDVH[DFWO\RQHVWDWLRQDU\SRLQW¿QGWKHYDOXHRIc

ii. ,I WKLV VWDWLRQDU\SRLQW RFFXUV DWD SRLQWRILQWHUVHFWLRQRI y g xDQG y g –1 x¿QGWKHYDOXHRIk

PDUNV

7RWDOPDUNV




SECTION 2 – Question 4 – continued

Question 4

Deep in the South American jungle, Tasmania Jones has been working to help the Quetzacotl tribe to getdrinking water from the very salty water of the Parabolic River. The river follows the curve with equation

y = x2 – 1, xDVVKRZQEHORZ$OOOHQJWKVDUHPHDVXUHGLQNLORPHWUHV

Tasmania has his camp site at (0, 0) and the Quetzacotl tribe’s village is at (0, 1). Tasmania builds a desalination plant, which is connected to the village by a straight pipeline.

y

x1

Parabolic River

village (0, 1)

desalination plant (m, n)

camp

(0, 0)

(–1, 0)

p i p e l i n e

a. If the desalination plant is at the point (m, n) show that the length, L kilometres, of the straight pipelinethat carries the water from the desalination plant to the village is given by

L m m= − +4 23 4.

3 marks



23 0$7+0(7+&$6(;$0


b. ,IWKHGHVDOLQDWLRQSODQWLVEXLOWDWWKHSRLQWRQWKHULYHUWKDWLVFORVHVWWRWKHYLOODJH

i. ¿QGdL

dmDQGKHQFH¿QGWKHFRRUGLQDWHVRIWKHGHVDOLQDWLRQSODQW

ii. ¿QGWKHOHQJWKLQNLORPHWUHVRIWKHSLSHOLQHIURPWKHGHVDOLQDWLRQSODQWWRWKHYLOODJH

PDUNV



0$7+0(7+&$6(;$0


7KHGHVDOLQDWLRQSODQWLVDFWXDOO\EXLOWDW7

2

3

4,

⎛

⎝ ⎜⎜

⎞

⎠⎟⎟

,IWKHGHVDOLQDWLRQSODQWVWRSVZRUNLQJ7DVPDQLDQHHGVWRJHWWRWKHSODQWLQWKHPLQLPXPWLPH

7DVPDQLDUXQVLQDVWUDLJKWOLQHIURPKLVFDPSWRDSRLQW x yRQWKHULYHUEDQNZKHUH x 7

2.+HWKHQVZLPV

XSWKHULYHUWRWKHGHVDOLQDWLRQSODQW

7DVPDQLDUXQVIURPKLVFDPSWRWKHULYHUDWNPSHUKRXU7KHWLPHWKDWKHWDNHVWRVZLPWRWKHGHVDOLQDWLRQ SODQWLVSURSRUWLRQDOWRWKHGLIIHUHQFHEHWZHHQWKH yFRRUGLQDWHVRIWKHGHVDOLQDWLRQSODQWDQGWKHSRLQWZKHUHKHHQWHUVWKHULYHU

c. 6KRZWKDWWKHWRWDOWLPHWDNHQWRJHWWRWKHGHVDOLQDWLRQSODQWLVJLYHQE\

T x x k x= − + + −( )1

21

1

47 44 2 2 KRXUVZKHUHk LVDSRVLWLYHFRQVWDQWRISURSRUWLRQDOLW\

PDUNV

7KHYDOXHRIk YDULHVIURPGD\WRGD\GHSHQGLQJRQWKHZHDWKHUFRQGLWLRQV

d. ,I k 1

2 13

i. ¿QGdT

dx



25 0$7+0(7+&$6(;$0

ii. KHQFH¿QGWKHFRRUGLQDWHVRIWKHSRLQWZKHUH7DVPDQLDVKRXOGUHDFKWKHULYHULIKHLVWRJHWWRWKHGHVDOLQDWLRQSODQWLQWKHPLQLPXPWLPH

PDUNV

e. 2QRQHSDUWLFXODUGD\WKHYDOXHRIk LVVXFKWKDW7DVPDQLDVKRXOGUXQGLUHFWO\IURPKLVFDPSWRWKHSRLQWRQWKHULYHUWRJHWWRWKHGHVDOLQDWLRQSODQWLQWKHPLQLPXPWLPH)LQGWKHYDOXHRIk RQWKDWSDUWLFXODUGD\

PDUNV

f. )LQGWKHYDOXHVRIk IRUZKLFK7DVPDQLDVKRXOGUXQGLUHFWO\IURPKLVFDPSWRZDUGVWKHGHVDOLQDWLRQSODQWWRUHDFKLWLQWKHPLQLPXPWLPH

PDUNV

7RWDOPDUNV

END OF QUESTION AND ANSWER BOOK



MATHEMATICAL METHODS (CAS)

Written examinations 1 and 2

FORMULA SHEET

Directions to students

Detach this formula sheet during reading time.

This formula sheet is provided for your reference.

© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2011



MATH METH (CAS) 2

This page is blank



3 MATH METH (CAS)

Mathematical Methods (CAS)

Formulas

Mensuration

area of a trapezium:1

2 a b h+( ) volume of a pyramid:

1

3 Ah

curved surface area of a cylinder: 2S rh volume of a sphere:4

33

S r

volume of a cylinder: S r 2h area of a triangle:1

2bc Asin

volume of a cone:1

32

S r h

Calculus

d

dx x nxn n

( )=

-1

x dx

n x c nn n=

+

+ ≠ −+

∫

1

111 ,

d

dxe aeax ax( )=

e dx

ae cax ax= +∫

1

d

dx x

xelog ( )( )=1

1

xdx x ce= +∫ log

d

dxax a axsin( ) cos( )( )= sin( ) cos( )ax dx

aax c= − +∫

1

d

dxax a axcos( )( ) -= sin( )

cos( ) sin( )ax dx

aax c= +∫

1

d

dxax

a

ax

a axtan( )

( )

( ) ==

cos

sec ( )2

2

product rule:d

dxuv u

dv

dxv

du

dx( )= + quotient rule:

d

dx

u

v

vdu

dxu

dv

dx

v

⎛

⎝ ⎜

⎞

⎠⎟ =

−

2

chain rule:dy

dx

dy

du

du

dx= approximation: f x h f x h f x+( ) ≈ ( ) + ′( )

Probability

Pr( A) = 1 – Pr( Ac) Pr( A B) = Pr( A) + Pr( B) – Pr( A B)

Pr( A| B) =Pr

Pr

A B

B

∩( )

( )transition matrices: S

n = T n u S 0

mean: μ = E( X ) variance: var( X ) = V 2 = E(( X – μ)2) = E( X 2) – μ2

probability distribution mean variance

discrete Pr( X = x) = p( x) μ = ¦ x p( x) V 2 = ¦ ( x – μ)2 p( x)

continuous Pr(a < X < b) = f x dxa

b( ) μ =

−∞

∞

∫ x f x dx( ) σ μ 2 2= −

−∞

∞

∫ ( ) ( ) x f x dx

2011 Mathematical Methods (CAS) Exam 2

Documents

Transcript of 2011 Mathematical Methods (CAS) Exam 2