2011 Mathematical Methods (CAS) Exam Assessment Report Exam 1.pdf

7/27/2019 2011 Mathematical Methods (CAS) Exam Assessment Report Exam 1.pdf

1/9

Maths Methods

20

Asses

Rep

2011

GENERIn 2011, 15

While the quof their writtwere less likSeveral quesalternatives

Students neequestions, bTime managattempted. Scompleting aquestions.

In 2011, studratios; simpliirregular polStudents sho

A = 0, B = 0

cases. In add

through by

impossible sapproach inexample, f(

Students sho

values in tri

determine fedistributionsimultaneou

When answeapproach is tequal to thework separathave the saof the questi

used is clear.

Question 10worked out twell) they w

BD = 2CD.

The necessitpenalised foIt should bepresented wi

(CAS) 1 GA 2 E

1

ment

ort

L CO94 students s

ality of respoen logic/reasly to lose ma

tions had a vaiscussed bel

d to make got they should

ment appearudents are res much as the

ents would hfication of frgons; area o

uld be awareor both could

ition to this,

( )x unless th

lutions indicuch situation

x)g(x) f

uld be able to

onometry; a

tures of circunctions; detlinear equati

ring a showo start with tight-hand sidely on the rige result. Then, the stude

. required sthe rule for thuld get 0 as

to write dxomitting this

nderstood ththout workin

xam

MENTSat Mathemati

nses was varining. Studenrks for carele

riety of solutiw.

d use of thetry to identif

d to be an isinded that t

y can of the s

ve drawn onctions, ratiostriangles; ba

that when solbe 0. Similar

hen solving

ey have consi

ates studentss is to collectx) k(x)

use the chai

tidifferentiate

lar functionrmine when

ons; formulat

hat questione left side ofe). A variatioht side to seewording of tht must provid

dents to shoderivative t

he answer. T

at the end ofthan in previ

at answers ar, the maxim

VICTORIAN C

Mathe

cal Methods (

d, it was ples who were ass errors.

on pathways,

5 minutes ofthose that h

ue for someey do not netraightforwar

knowledge aand surds; co

sic probabilit

ing equationly, students s

an equation o

dered whethe

ho are demterms, factori(x) (g(x)

rule for diffe

expressions

raphs; detero solutions, i

e and calcula

there are seva given equat

of this is toif the same pe question ise clear worki

that the derie student woe implicatio

an integralous years.

only acceptem score that

URRICULUM

atical M

CAS) exami

sing to see tstute in their

some more e

reading time.ve familiar c

tudents thisd to answer

d questions b

d skills relatllecting like t, Venn diagr

s involving faould know t

f the form (f

r ( )f x could

nstrating thase, use the nk(x)) 0 so

rentiation in

f the forma

ine a maximnfinitely mae the area bet

ral approachion and manistart with theint of simpliften a guideg so it is ob

ative was zeld use the fawas not to s

as understoo

d if they arean be award

ND ASSESSM

thods (

ation 1.

at students pse of the con

fficient than

Not only shooncepts and r

ear and Quesuestions seqfore they em

d to: algebraerms; factorims and the n

ctored expresat if ( )f x

) ( )x g x f

e equal to ze

they understll factor law(x) 0 and/o

ombination

1

x l; recogni

l domain; utiy solutions oween two cur

es students caulate it to obleft side andication can bto the approaious where d

o when BD =t thatBD = 2art with the d

by the majo

result of cord is the one

NT AUTHORI

AS) 1: G

id attention tventions and

thers. Questi

uld they readutines.

tion 10 was eentially. The

bark on the l

c manipulatiation of quadull factor law

sions ifA ( ) 0x then

( ) ( )x k x the

ro. The expli

nd what is hnd eliminaterg(x) k(x)

ith the prod

se a simple c

lise the propeunique solut

ves; and ans

n take, depenain the rightimplify it asreached t

h that can becisions are

2CD. This iCD in their derivative equ

ity of studen

rect working.ark for the a

Y 2012

A 2: Exa

the clarity anotation of m

ons 5, 6, 8 an

to make sens

ther rushed oy should thinnger, more i

n; use of trigratics; perime.

= 0then eithey need to t

y should not

it eliminatio

ppening. A ror accept sol.

ct rule; find a

mposite qua

rties of probaions occur foer a show t

ding on the qside (but notmuch as possus showing tused. Whateade and so th

plies that aftrivative andl to 0 and pr

s. Fewer stud

When a corrnswer. The m

ination

nd legibilityathematics

9 have

e of the

r notabout

volved

onometricter of

erst both

ivide

of

commendedtions; for

nd use exact

ratic form;

bilitysystems of

at question.

estion. Onessume it

ble, thenat they bother the nature

e process

er having(if all wasve that

ents were

ct answer isarks awarded

1


2/9

Maths Methods

20

Asses

Rep

are an indicaseemingly cIn general, s( 8)( 2x x

the quadraticBCD had sid

SPECIFQuestion 1a

Marks

%

d 4 x dx

This was a dtranscribed t

3

2 .

Question 1b

Marks

%

g(x) 2xsin(

g

6

2

6

sin

3

3

2

Answer: g

The product

students wit

cos

3

.

Question 2a

Marks

%1

3loge ( 3x 4

(The constan

Common err

the general a

(CAS) 1 GA 2 E

1

ment

ort

tion of workirrect answer

udents strug0 but also

graph. Queses in the ratio

C INFO

.

0

40

((4 x)

1

2 )

dx

1

2

fferentiationeir final ans

.

0

14

2x) 2x2cos(2

sin

3

2

and cos

3

3

6

2

36

rule was gene

the correct d

.

0

57

) +c or1

3lo

t c is not nec

ors included

nti-derivative

xam

g involved.from incorrec

led with Quethat this mea

ion 10d. relieof 1:2: 5 (

RMATI

1 Ave

60 0

(4 x)

1

2 1

question invoer incorrectl

1

42 4

x)

2

36 cos(

3)

1

2

rally used we

erivative did

1 Ave

43 0

e( x 4

3) +c

ssary. It can

issing the1

3

, ( )f x dx , r

tudents attet working an

tions 4b. ands 8 0 or x

d on the study using Pyth

N

rage

.6

1

2 4 x

lving the use, omitting th

Avera

4 1.3

ll by students

not get the se

rage

.4

be any real n

and omitting

quires an arb

Published: 8

pting Questiso have a ze

10d. Questio2 0x . T

nt using thegoras theor

of the chain re negative si

ge

. However, t

ond mark be

mber, includi

the modulus

itrary consta

March 2012

n 2b. and Quro score for t

n 4b. requireis could be re

nowledge from).

ule. Some stun. Some had

e determinati

cause they lef

ng zero.)

(absolute val

t to be includ

estion 9 coulese question

them not onladily observe

m 10c. to de

dents differeifficulty wit

on of exact v

t their answe

e) signs. Stu

ed.

end up with.

y to understad by looking

uce that the r

tiated correctthe index, s

lues was poo

s containing

dents should

the

d thatat a sketch of

ight triangle

ly butowing it as

r. Many

sin

3

and

e aware that

2


3/9

Maths Methods

20

Asses

Rep

Question 2b

Marks

%

2

2

4 15 2 =

2 15

let 2

15

16

2 16 sin

4

x x

x

x

x

a

a a

a a

x

This questioSeveral stud

Question 3a

Marks

%

: [1,7]

4

range

period

Common inc

Question 3b

Marks

%

sin 2 +3

(2 + ) = ..311

= , a4 12

x

x

x

Most studenin radians. Tof poor alge

Question 4a

Marks

%

( ( )) (

(

so 2 a

g x

c

The majorityfactorised in

(CAS) 1 GA 2 E

1

ment

ort

.

0

61

6

16 0

16 0

1 0

ce 2 1 0

x

x

was poorlynts resorted

.

0

20

orrect respon

.

0

26

.

1= , is b

2 6

5 13. , ,

6 6 6

re the only so

s could deterhe majority ora, poor arit

.

0

8

25) 9

2)( 8)

nd 8 or

x

d

of studentsorrectly or d

xam

1

6

one. Many so guessing a

1

29 5

ses were a pe

1

43 3

se angle

..

lutions withi

mine the basf students whmetic, choosi

1

17 7

8 and 2c d

ained full mad not expand

3

28

udents started checking t

Avera

1 1.3

iod of 4 or

Avera

1 1.1

domain

angle. Withdid get the

ng a value ou

Avera

5 1.7

rks for this q(x + 5)2 corre

Published: 8

Averag

1

by writingatx = 4. Atte

ge

4

and a ran

ge

some preferrase angle di

tside the requ

ge

estion. Howectly.

March 2012

2x 4x mpts involvin

e of [1,7] o

ng to work inot get both

ired domain o

ver, many di

2a rather thag logarithms

r [7,1].

degrees, thecorrect answr not looking

not state a v

n 2 4xa rarely helped.

final answerrs. This wasfor multiple

lue forc and

2xa .

was requiredeither a resulolutions.

d, and others

3


4/9

Maths Methods

20

Asses

Rep

Question 4b

Marks

%

( , 8]

maximal do

x

This questiointersectionproperties ofsquare root fQuestion 5a

Marks

%

Pr(X 3.5)

Method 1.usi

3

2

Method 2.

(3 )

0.625

By

x dx

This questioprobability (values and vThe followi

3.5

2| 3 |x dx

(CAS) 1 GA 2 E

1

ment

ort

.

0

84when (

[ 2, ) or

ain x

was very po ofx 8 wicomposite f

unction work.

.

0

71Pr(X 3) Pr(

g triangles : (

3.5

3

:

( 3)

integration

x d

was very poiven by the alues greater

g incorrect d

3.52

2

32

xx

1.5

y

0.2

0.4

0.6

0.8

1

1.2

xam

1

6 1)( ) 0

\ ( 8, 2)

c x d

orly done. Cothx 2); ornctions tende.

1

11 13 X 3.5)

.511) (0.5

3.5

32

2

| 3 |

32

x dx

xx

orly done. Ferea under thehan 1.

termination

0.375

2

Avera

0 0.3

mmon incorrx 8 (as thed to get conf

Avera

8 0.5

0.5 0.5) 0.

3.52

3

32

xx

students ungraph) had to

f the area wa

.5 3

Published: 8

ge

ct responsesunion ofx

sed. Students

ge

5 0.125 0.6

derstood whabe between

s common.

3.5

March 2012

included [3, -8 withx needed to lo

5

t the graph ofand 1. Man

4 4.5

3], (the doma2). Those wk for a doma

the functionstudents ga

x

in off(x));xho attemptedin that would

looked like oe answers wi

2 (as theto use themake the

that theh negative

4


5/9

Maths Methods

20

Asses

Rep

Question 5b

Marks

%

Pr( 2.5 |

Pr( 2.5

Pr(

0.375 3

0.625 5

X

X

The majority

Pr( 2.5)X

denominator

Question 6a

Marks

%

The followi

Method 1:

1 1

2 2

,3

4,

7

from the grad

4

from the inter

4

km c

m ck

k or

k c

Answer: kMethod 2:

Comparison o

k

4

3

k 7

k

Usingk

4

k

k2 7k 12

Using3

k 7

k2 10k 21

Using k4

k1

k 4 satisfie

Answer: k

(CAS) 1 GA 2 E

1

ment

ort

.

0

543.5)

3.5)

3.5)

0.6

X

of students

Pr( 3.5X

.

.

0

43

g are three di

sing gradie

1 2

3

3

1

7

ient ( 7)

3

cepts 3

1so t

3

k

k

k k

k

k

c

4

sing ratios

f coefficients a

3

3

7

k 4 ork

k 3

1

3 k 4

3 k 4

s all 3 ratios a

4

xam

1

32 1

r( 2.5)

r( 3.5)

X

X

ere aware th

, which is no

1

22 1

fferent metho

t,m, andy-i

1 2

12 so

0 and

ere are infin

c c

ite

s common rati

3

rk 6

d the lines will

Avera

4 0.6

t Pr( 2.5X

t correct. Mo

3

0 25

ds that can be

tercept,c.

1so there ar

4

ly many soluti

s (all ratios wo

be identical.

Published: 8

ge

3.5)X w

e correct exp

Averag

1.2

used to iden

no solutions i

ns in this case

uld require the

March 2012

as Pr( 2.5X

ressions for t

ify and justif

n this case

same value of

) . However,

e numerator

k= -4.

k to have infini

few express

were evident

te solutions).

ed it as

than for the

5


6/9

Maths Methods

20

Asses

Rep

Method 3:

determinant

determinant

k2 7k12

k -4 or

when k 3 t

3x 3y

when k 4 t

4x 3y

Answer: k

Many studeseen. Which

Question 6b

Marks

%For unique s

2 7 12k k

\{ 4,k R

Most studen

was using thQuestion 7a

Marks

%

3p

This questi

also a numQuestion 7a

Marks

%

23 (1 )p

The majorityQuestion 7b

Marks

%

p3 3p2 3

4p3 3p2

p 0 andp

The cancelli

(CAS) 1 GA 2 E

1

ment

ort

sing determ

k(k 7) 12

0 for infinitel

(k 4)(k 3)

3

he two equatio

0 and 4x 4

he two equatio

1 and 4x

4

ts who tried tver method

.

0

67lutions0 3}

s were aware

determinant

i.

0

29

n was most

er of respon

ii.

0

59

of incorrect

.

0

243

2(4p 3) 0

3

4

g out ofp w

xam

inant

many solutio

0

ns become

y 1 so there a

ns become

3y 1 so there

o solve simulas chosen, st

1 Ave

33 0

of the two va

. Many stude

1 Ave

71 0

y well done

ses where st

1 Ave

41 0

esponses neg

1

58 1

s rarely supp

s or no solutio

re no solutions

are infinitely

aneous equatudents had to

rage

.4

lues ofkthat

ts chose a va

rage

.7

, although p

udents incor

rage

.4

lected the 3 (

Avera

8 1

orted. Many s

Published: 8

s

in this case

any solutions i

ions rarely sujustify their

led to either

lue ofkthat

p p 3p

rectly assig

or the numbe

ge

tudents incor

March 2012

n this case

cceeded. Othelection ofk.

o or infinite

as not select

was a com

ed a numeri

r of possible

ectly assume

r variations

olutions. The

d as the ans

on incorrec

cal value to

rrangements

dp could not

f these meth

most succes

er to part a.

response. T

.

).

be zero.

ds were

ful approach

here were

6


7/9

Maths Methods

20

Asses

Rep

Question 8a

Marks

%

Pr( A B)=Pr

Alternatively

Pr(A B) Pr(

Pr(AB)

B

B'

From the tablePr( ')A B

A simple Vevarying mea

who used thi

Question 8b

Marks

%

Pr( ) = 0B

The most co

Question 9

Marks

%

0

0

42

Point of inter

Area= (

or Area =

4

8 as

m

ax

xax

a m

The majorityof the pointcurves by redefinite inteincorrectly c

point of inte

(CAS) 1 GA 2 E

1

ment

ort

.

0

34

(A B) Pr(A)

:

A) Pr(B) Pr(

Pr(A) Pr(B)

A

1

10

3

5

1 1 3

4 10 20

nn diagramures of succe

s but wrote P

.

0

56

Pr( )A B

mon incorre

0

27

3

23

2

2

0

ection ( , )

( ))

(2 )

(2

16 4 sin

a

a

m am

x ax dx

ax x d

a

of students cf intersectiolating all arerals or the arncelling ax i

section in ter

xam

1

29 3

3

4

3

5

3

20

A B)

Pr(A B) 3

5

A'

Pr(A B)

ould have assss, to utilise t

r( A B) whe

1 Ave

44 0

1Pr( ) =

4

B

ct response

1

21 23

22

64

64

) 644

e both and

am m

aa

a m

ould set up thto achievings back to thea of a triangl

n the integral

s ofx with

Avera

8 1.1

1

4

3

4

1

10

1

4

1.0

isted many sthe relation Pr

rearranging

rage

.5

as due to stu

3

7 8

, 0

are greater tha

m m

e correct intethe final ans

x-axis. Insteae and two defand then quic

3 ,x x ax

Published: 8

ge

udents in ans(AB) Pr(A)

it. Arithmetic

ents interpre

4

162 2

n zero

a

ral equationers. Many st

d of a singleinite integralkly getting th

2 2x a a

March 2012

ering this q Pr(B) Pr(A

was a weakn

ing mutuall

Average

1.7

for the area,udents spent tefinite integr. Poor use ofe answerm =

2

2

xand sub

estion. MostB) . There

ess for some

exclusive a

ut only a fewime determinal to determibrackets led t4. Some stu

tituted this d

students triedere a number

tudents.

independen

realised theing the areae the area, tho a number oents incorrec

rectly into th

, withof students

t.

ignificanceetween twoy had threestudentsly left the

integral.

7


8/9

Maths Methods

20

Asses

Rep

Question 10

Marks

%

cos( )BD a

This questiostudents triefew students

Question 10

Marks

%

2 2

4 2

L a

L a

The secondquestion.

Question 10

Marks

%

dL

d 2asin(

WhenBD =

substitute in

dL

d 2a

Most studendifferentiatiacos(). Thbut making

Question 10

Marks

%

Using trigon

tan() 1

2

BD acos()

CD asin()

L 4 a 2B

L 19 3 5

(CAS) 1 GA 2 E

1

ment

ort

a.

0

34and siCD a

required a sto use Pytha

had only one

b.

0

42

4

BD CD A

D CD

orizontal len

c.0

58

) acos()

CD acos(

(1)

in() 2asin(

s who hadL ig with respece show thato mention of

d.

0

95

metry:

sin() 1

5,c

3 5 2

5

3 5 1

5

D CD 4 3

xam

1

4 6( )

raightforwargoras theoreof the correc

1 Ave

58 0

2 cos( )

E

a

th of wire (A

1

14 2

(1)

) 2asin()

) 0 as requir

n terms oft to a, attemppart of the qeitherBD or

1 Ave

5 0

s() 2

5

6

5 2 6 3

Avera

2 1.3

application, reciprocalanswers. So

rage

.6

sin( )a

EorBD) was

Avera

8 0.7

d

ould differenting to differeestion was raCD.

rage

.1

Published: 8

ge

f trigonometor inverse fu

e students d

missing fro

ge

tiate it correcntiate with reely done wel

March 2012

y for a right-ctions. Other

id not answer

some respo

tly. Some incspect to a andl, with many

angled triangls mixed variathis question

ses, despite p

rrect respon, or differestudents maki

2

e. However,bles as well a.

articular men

es were a cotiating asin(ng a direct su

omes units. Very

tion in the

sequence of) as

bstitution,

8


9/9

Maths Methods

20

Asses

Rep

Alternativel

MaximumL

a2 (2CD)2

Ifa 3 5

L 4 a 2B

L 19 3 5

Few studentconnection tcalculate the

(CAS) 1 GA 2 E

1

ment

ort

:

henBD = 2C

(CD)2 5(CD

a2 45 CD

DCD 4 3

obtained theat the conditlengths of all

xam

D,

)2

3 BD 6

5 12 3 19

correct answon stated inthe sides nec

3 5

r. It was notuestion 10c.

essary to dete

Published: 8

a simple one-provided enormine the ma

March 2012

step problemugh informatiximumL.

and most stuon to use Pyt

ents did notagoras theo

ake theem and

9

Documents

2011 Mathematical Methods (CAS) Exam Assessment Report Exam 1.pdf