2010 Maths Methods CAS Units 3 & 4 Exam 1 solutions

_____________________________________________________________________ © THE HEFFERNAN GROUP 2010 Maths Methods (CAS) 3 & 4 Trial Exam 1 solutions

MATHS METHODS (CAS) 3 & 4 TRIAL EXAMINATION 1 SOLUTIONS 2010 Question 1 a.

( ) line) (optional 323

233

2)1(3

23

3332

323

−+=

+−=

×+−=

xxe

xeeex

xexedxdy

x

xxx

xx

(1 mark) – correct answer

b. Method 1 – short way

0)tan()()tan()cos()sin()(

))(cos(log)(

=

−=

−=

−=

=

ππ'fxxxx'f

xxf e

(1 mark) Method 2 – long way

0)tan()(So)tan(

)sin()cos(

1

)sin(1

rule)(chain

)sin(1)cos( where)(log

))(cos(logLet

=

−=

−=

−×=

−×=

⋅=

−==

==

=

ππ'fx

xx

xu

dxdu

dudy

dxdy

xdxdu

ududy

xuuyxy

e

e

(1 mark)

(1 mark)

(1 mark)

(1 mark) –using product rule

©THE HEFFERNAN GROUP 2010 Maths Methods (CAS) 3 & 4 Trial Exam 1 solutions

2

Question 2 a. ( )∫ + dxex x2

cex

dxex

x

x

++=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+= ∫

223

221

21

32

(1 mark) b. )3sin()( xx'f =

cx

dxxxf

+−=

= ∫)3cos(

31

)3sin()(

(1 mark)

Given

So

(1 mark) Question 3

Show )(12)( 2uvfv

fuf =⎟⎠

⎞⎜⎝

⎛−

required. as )(

)(log

1log

1log)(log

1log)(log

1log2)(log

12)(

2

2

2

2

2

RHSuvf

uvv

u

vu

vu

vu

vfufLHS

e

e

ee

ee

ee

=

=

=

⎟⎠

⎞⎜⎝

⎛÷=

⎟⎠

⎞⎜⎝

⎛−=

⎟⎠

⎞⎜⎝

⎛−=

⎟⎠

⎞⎜⎝

⎛−=

⎟⎠

⎞⎜⎝

⎛−=

(1 mark)

(1 mark)


3

Question 4 a. Draw a diagram.

b.

BLACK 1 2 3 4 5 6 1 x . . . . . 2 . x . . . . 3 . . x . . . 4 . . . x . . 5 . . . . x . 6 . . . . . x

R E D

(1 mark)

(1 mark)

BLACK 1 2 3 4 5 6 1 . x x x x x 2 . . x x x x 3 . . . x x x 4 . . . . x x 5 . . . . . x 6 . . . . . .

R E D


4

c. This represents a binomial distribution with 4=n . Since odd numbers occur on both die on 9 occasions (from the diagram),

41

369==p . (1 mark)

)0Pr(1)1Pr( =−=≥ XX

256175

256811

141 and 1Note:

431

43

411

0

04

4

40

04

=

−=

=⎟⎠

⎞⎜⎝

⎛=⎟

⎠

⎞⎜⎝

⎛−=

⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−=

C

C

(1 mark) Question 5

Let Swap x and y for inverse

(1 mark)

Rearrange

Do a quick sketch of .

Rrd

rRd

gg

gg

=∞−=

∞−==

−− 11 and ),2( So,

),2(,

1)2(log)(,),2(: So 11 −+=→∞− −− xxgRg e Note that to define a function you must give the rule (equation) and the domain.

(1 mark) – correct domain (1 mark) – correct rule

x

y

2−=y

)(xgy =

BLACK 1 2 3 4 5 6 1 x . x . x . 2 . . . . . . 3 x . x . x . 4 . . . . . . 5 x . x . x . 6 . . . . . .

R E D


5

Question 6

For no solutions or infinite solutions the determinant of the matrix ⎥⎦

⎤⎢⎣

⎡

−121m

m equals zero.

That is,

(1 mark)

(1 mark)

They are the same equation hence there are an infinite number of solutions. If ,

(3) and (5) describe parallel lines with different y-intercepts so there are no points of intersection and hence no solutions. So for there is no solution.

(1 mark)

Question 7

X 2 3 4 5 6

0.2 0.4 0.1 0.2 0.1

a. Median = 3

(1 mark)

b. )63Pr( <≥ XX

(1 mark)

(1 mark)


6

Question 8 a.

OR

(1 mark) – correct left branch (1 mark) – correct middle branch

(1 mark) – correct right branch (curved or straight)

b. }2,1{\' −= Rd f (1 mark)

x

y

-1 2

)(xfy =

)(' xfy =

x

y

-1 2)(xfy =

)(' xfy =


7

Question 9 Method 1 – intuitively

If we were given a restricted domain like then [ ]π4,02 ∈x so

811,

89,

83,

8

411,

49,

43,

42

ππππ

ππππ

=

=

x

x

(1 mark) However, the domain is not restricted; we are looking for the general solution where each of the solutions found above is repeated for every clockwise and anticlockwise rotation by . We can express these values as

integers) ofset (the 8

3or 8

Znnxnx ∈+=+= ππ

ππ

(1 mark) (1 mark) Method 2 – graphically

Sketch the graphs of 2

1 and )2sin( == yxy .

We know that in the first quadrant 21)2sin( =x

4

2 π=x

8π

=x

By symmetry and using the graph, the points of intersection occur at

,...811,

89,

83,

8,

85,

87...

,...84

5,84

5,84

,84

,84

3,84

3...

ππππππ

ππππππππππππ

−−=

+−+−+−−−=x

We can express these values as

integers) ofset (the 8

3or 8

Znnxnx ∈+=+= ππ

ππ

(1 mark) (1 mark) (1 mark) for graph

S

T

A

C

x

y

-1

1

4π

2π

43π

45π

23π

47π π2

)2sin( xy =

21

=y

4π

−2π

−43π

−


8

Method 3 – using a formula For

Znanxanx ∈−+=+= −− ),(sin)12( or )(sin2 11 ππ

For

82)12(,

8

4)12(2

422

21sin)12(2or

21sin22 11

ππππ

ππ

ππ

ππ

−+

=∈+=

−+=+=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+= −−

nxZnnx

nxnx

nxnx

Znn

n

n

n

∈+=

+=

−+=

−+=

,838388488)12(4

ππ

ππ

πππ

ππ

(1 mark) (1 mark) (1 mark) for showing

Method 4 – using a formula For , Znanx n ∈−+= − ),(sin)1( 1π

For

Znnx

nx

nx

n

n

n

∈−+=

−+=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+= −

,8

)1(2

4)1(2

21sin)1(2 1

ππ

ππ

π

(1 mark) – correct first term (1 mark) – correct second term

(1 mark) for showing

(Note that for this formula

83

82,1when,

85

82,1when

8,0when,

87

8,2when

ππππππ

ππππ

=−==−=−−=−=

==−=+−=−=

xnxn

xnxn

and so on.)


9

Question 10 a.

(1 mark) – correct endpoint (1 mark) – correctly labelled asymptote

(1 mark) – correct shape

b.

),3(or 3> So3

26 sign) inequality the 4210

reverse tohavet don' weso02 so ,0 that (note )2(210

22

10

∞∈

<

<

+<

≥+≥+<

<+

ttttt

tttt

(1 mark)

c. distance travelled dtt∫ ⎟

⎠

⎞⎜⎝

⎛+

+=1

02

101 (1 mark)

[ ]

metres 23log101

)2(log10)3(log101))}2(log100())3(log101{(

)2(log10 1

0

⎟⎠

⎞⎜⎝

⎛+=

−+=

+−+=

++=

e

ee

ee

e tt

(1 mark)

t

23

3 4 5 6 71 2 8 9 10 11 12 13 14 15 16 17

4

56

)(tv

2101)(+

+=t

tv

1=v

)6,0(


10

Question 11 a.

12

12)(

)(221

heightbase21

2

2

+=

+×=

×=

××=

××=

aaa

a

afa

afa

A

(1 mark)

b. (quotient rule)

(1 mark) correct use of quotient rule

For min/max

(1 mark)

x

y

-a aO

12)( 2 +

=x

xf

))(,( afa))(,( afa −−


11

025

625

82)1(

22,2At

02

)1(22,0At

22

2

22

2

<

−=

−=

+

−==

>

=

+

−==

aa

dadAa

aa

dadAa

So at there is a local maximum. (1 mark)

From part a., area unit. square 122

122 ==+

=aa

So the maximum area is 1 square unit. (1 mark)

0=a 1=a 2=a

2010 Maths Methods CAS Units 3 & 4 Exam 1 solutions

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