2007 maths methods cas 3 & 4 exam 2 solutions · ©the heffernan group 2007 maths methods (cas)...

_____________________________________________________________________ ©THE HEFFERNAN GROUP 2007 Maths Methods (CAS) 3 & 4 Trial Exam 2 solutions

MATHS METHODS (CAS) 3 & 4 TRIAL EXAMINATION 2 SOLUTIONS 2007 Section 1 – Answers _____________________________________________________________________

1. D 9. D 17. B 2. B 10. E 18. D 3. C 11. E 19. C 4. C 12. B 20. E 5. B 13. D 21. D 6. B 14. A 22. E 7. E 15. D 8. C 16. E

Section 1 – Multiple-choice solutions Question 1

For ( )34

2−

=x

xf , the maximal domain is given by

The answer is D. Question 2

( )

14

2cos3

12

2cos3

+⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛+=

+⎟⎠

⎞⎜⎝

⎛+=

π

π

x

xxf

The period is and the amplitude is 3.

The answer is B.

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Phone 9836 5021Fax 9836 5025

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2

Question 3

( )( )

( )

68,

67,

62,

6

38,

37,

32,

32

232sin

420322sin4

200322sin4

ππππ

ππππ

π

π

=

=

=

≤≤=

≤≤=−

x

x

x

xx

xx

The sum is .

The answer is C. Question 4 Draw a quick tree diagram. There are 8 possible outcomes with equal probability, 1 of those is to get 3 heads, 3 of those are to get 2 heads, 3 of those are to get 1 head, and 1 of those is to get 0 heads. The answer is C.

S

T

A

C

0.5

0.5

0.5

0.5

0.5

0.5

0.5

0.5

0.50.5

0.50.5

0.5

0.5

H

H

H

H H

H

H

H

TT

T

T

H

H

T

T

HT

T

H

H

T

H

T

TH

T

H

Outcomes

H

H

HT

T

T

T

T

T

T


3

Question 5

222231232

=++

=−−

=++

zyxzyxzyx

Option A does not give the right hand side of the equation. Option C has one incorrect element in the 33× matrix. Option D has the two column matrices in the wrong positions. Option E actually solves the system, which is not what the question has asked. So option E is incorrect. Only option B shows an equivalent matrix equation. The answer is B. Question 6 A translation of 2 units to the left changes the rule xey −=1 to become .

A reflection in the y-axis changes the rule to The answer is B. Question 7

))(( xfg exists if gf dr ⊆ .

Now )[ ∞= ,0gd .

So we require that )[ ∞⊆ ,0fr .

For option A, Rrf = .

For option B, Rrf = .

For option C, Rrf = .

For option D, )[ ∞−= ,1fr .

For option E, [ ).,0 ∞=fr The answer is E.


4

Question 8 This is the topic of “functionalities” where you need to consider the type of functions that follow the rules given. Consider )2(log)( xxf e= so )2(log)( xyxyf e= This then follows the rule for log addition where )(log)2(log)2(log yxxy eee += Only option C shows this correct addition. The answer is C. Question 9 Method 1

( )

4114

123123or

2332

123 so123

=

−=−

+=+−

+=−−

=

=

+=−

+=−

p

ppppp

p

ppppp

The answer is D. Method 2 Sketch 1 and )23(abs 21 +=−= xyxy .

Points of intersection occur at 23or

41 so,,

23at and

41

==== ppxx .

The answer is D. Question 10 A graph of the product of functions can help here to eliminate the incorrect answers. The graph of the function )()()( xgxfxh ×= gives xxxxh elog)2()( 2 ×−= . The graph of the function )(xh exists for the domain that is common to both graphs of )(xf and )(xg ; that is, the domain { }0\R . This means that option A has the incorrect domain and also option B cannot be correct. The graph is not continuous at 0=x . By graphing )(xhy = , we see that there is a local maximum at approximately )61.0,30.0( so option D is incorrect and there is no stationary point at 1=x so option C is incorrect. Also xxxxh elog)1(22)(' −+−= . This derivative is offered in option E. The answer is E.


5

Question 11 Method 1

( )

( )

( )

325

1At 22

41

41121

4121

rule)(chain

4121

2 where

2

2

2

21

21

221

21

2

2

=

=+

+=

+⋅⋅=

+=

⋅=

+==

+==

+=

+=

−

−

dxdyx

xx

x

xu

xu

dxdu

dudy

dxdy

xdxduu

dudy

xxuu

xx

xxy

The answer is E. Question 12 Let v equal the volume of the cylinder.

π

π

π

π

π

5.2 ,10When

125

rule)(Chain So

.10when find torequired are We

1 Also

25

255 since

2

==

×=

⋅=

=

=

=

=

=

=

dtdvt

t

dtdh

dhdv

dtdv

tdtdv

tdtdhdhdv

hvr

hrv

The answer is B.

Method 2

Sketch 21

2 )2( xxy += .

Calculate 1 when =xdxdy .

325

4433757.1

=

=dxdy

The answer is E.


6

Question 13 ( )xf is positive for 22 >∪−< xx . ( )xf ' is the gradient of ( )xf and is positive for .

So ( )xf and ( )xf ' are both positive for . The answer is D. Question 14 Let the antiderivative function of f be called g. So, ( ) ( )xfxg =' . For ( ) ( ) 0 ,, >−∈ xfaax , so the graph of g will have a positive gradient. Note that for there to be a stationary point (which may be a local min/max or point of inflection) ( ) 0=xf . For ( ) ( ) 0 ,, ≠−∈ xfaax . The answer is A.

x-2

-1

2

y

)(xfy =

x

y

)(' xfy =


7

Question 15 Method 1 – by hand

( )( )( )

( )( )

( )

a

a

aa

ax

axaxax

axaxxaaaxx

xa

ax

ax

ax

ax

1

121

21lim

2lim

2lim

23lim 22

=

−

−=

−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−=

⎟⎠

⎞⎜⎝

⎛

+−

−

→

→

→

→

Options B, C and E are options that show the common mistake where a is substituted for x

without first simplifying the denominator in the fraction 22 23 aaxx

xa+−

− . These options give

zero in the denominator so options B, C and E are incorrect. Only option D shows the correct answer. The answer is D. Method 2- using a CAS calculator The CAS calculator can calculate this limit in one step. The answer is D. Question 16

( ) ( )( ) ( )

( ) ( ) cxgxxh

dxxgxdxdxxh

xgxxh

+−=

−=

−=

∫ ∫ ∫3

'32'

'32'

2

Now ( ) ( ) 32 and 42 == gh

( ) ( ) 93 So

9944

,2 when So

2 +−=

=

+−=

=

xgxxh

cc

x

The answer is E.


8

Question 17 Sketch the graph.

( ) ( )

( ) curve sine theofsymmetry theof because sin2

sinsin Area

0

0

2

∫

∫ ∫

=

−+−−=

π

π π

π

dxx

dxxdxx

The answer is B. Question 18 This is a question on average value of a function. The average value of a function from

bxax == to is given by ∫−

b

a

dxxfab

)(1

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−

−=

24

012

cos2212024

1 dttπ

Evaluating this using the CAS calculator gives the answer as an exact value of 212 . Option E gives the incorrect answer commonly made with the mistake of forgetting to divide by ab − . Option A gives the incorrect answer confusing average value with average rate of change. Only option D shows this correct answer. The answer is D.

x

1

O

-1

y

π2


9

Question 19 The shaded area is given by normalcdf(39, 100, 42,2) = 0.93319… = 93% to the nearest percent (Alternatively, normalcdf(-1.5,100,0,1) = 0.93319…) The answer is C. Question 20 This represents a binomial distribution because there is a fixed probability of an event happening i.e. 0.78 and there is a fixed number of “trials” i.e. 4 and we want a particular number of events required i.e. 3 or 4. Method 1 ( ) ( )( ) ( ) ( ) ( )

78780220780220780

goals 4Prgoals 3Pr04

4413

34

⋅=

⋅⋅+⋅⋅=

+

CC

The answer is E. Method 2 binompdf(4,0.78,3) + binompdf(4,0.78,4) = 0.7878 The answer is E. Question 21 This is the topic of Probability Density Functions (PDF), where you need to consider functions in the domain given that have an area under the curve equal to 1. The mean is given by

∫=e

dxxxf1

)(µ

ee

dxexxe

e

e

41

43

)(log1

1

−=

×= ∫

This means options C or D could be correct. The mode is the x-value that gives the maximum point on the graph of )(xfy = .

( )

xexf

exexexf e

11)(' so

elsewhere0

1,log1)(

×=

⎪⎩

⎪⎨

⎧≤≤

=

This gives no solution for a stationary point in the domain. )(xf is an increasing function so the maximum point is at the end of the domain at ex = .

So the mode is e. The answer is D.

36 38 39 4240 44 46 4810 2 3-3 -2 -1.5 -1Z

X


10

Question 22 The graph of ( )xfy = could be or Both of these graphs produce the graph of ( )xfy = .

The equation of the first could be ( ) ( )52 2 −−= xxy . This is option E.

The equation of the second could be ( ) ( )52 2 −−−= xxy . This is not offered. The answer is E.

x

y

52

x

y

52


11

SECTION 2 Question 1 a. The x-intercept occurs when ( ) 02log6 2 =xx e . Solve this equation for x.

So 21

=x

The x-intercept occurs at as required

(1 mark)

b. The function ( ) ( )xxxf e 2log6 2= is only defined for values of x such that ; that is, for because ( )xe 2log is only defined for these values.

(1 mark)

c. The turning point is located at ( )30,30 ⋅−⋅ where each coordinate is expressed correct to one decimal place.

(1 mark) d. From part c. [ )∞⋅−= ,30fr where 3.0− is expressed correct to one decimal place.

(1 mark)

e. i. The function f does not have an inverse function because it is not a function.

(1 mark)

ii. The function g must be a function since exists. The maximal domain of g is therefore [ )∞⋅ ...,30320 where the value of a is the x-coordinate of the turning point of the curve. The answer is where a is correct to one decimal place.

(1 mark) f. i. The two transformations are:

• a dilation by a factor of k units from the y axis (1 mark) • a reflection in the x-axis (1 mark)

ii. We need to solve 0=⎟⎠

⎞⎜⎝

⎛−

kxf for x. (1 mark)

That is, we need to solve xkx

kx

e for 02log62

=⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛− .

So 2kx = . (1 mark)


12

g. Differentiate ( ) ( )xxxf e 2log6 2= . ( ) ( ) xxxxf e 62log12' +=

Since the gradient of the tangent is 1, solve ( ) 162log12 =+ xxx e ...37805.0=x (1 mark)

From the earlier graph, check that this x-coordinate is feasible. (For example, for 3.00 << x the gradient would be negative and therefore couldn’t be 1.)

Now ( ) ...2397.0...37085.0 −=f So 2.0 and 4.0 −== ba where a and b are both expressed correct to one decimal place. (1 mark)

h.

i. ( ) ( )

places) decimal 2 o(correct t units square 3811250750250 area eApproximat

⋅=

×⋅+⋅×⋅= ff

(1 mark)

ii. Since the rectangles extend above the function f the approximation will be greater than the exact area.

(1 mark)

i. The required area is given by ( )∫⋅

1

50

2 2log6 dxxx e . (1 mark)

Simplifying this gives 127)2(log2 −e square units. (1 mark)

Total 17 marks

x

4

0.75 10.5

y

)(xfy =

(1 mark)


13

Question 2 a. At midnight on Thursday, 0=t , so

( ) ( )30

300sin300=

+=T

The temperature is C°30 . (1 mark) b. The temperature of this food is in the “hot safe zone” if its temperature is greater than

°60 Since ( ) 306

sin30 +⎟⎠

⎞⎜⎝

⎛=

ttT π , the maximum temperature that this food reaches

is 6030130 =+× since the maximum value for 6

sin ⎟⎠

⎞⎜⎝

⎛ tπ is one. So the temperature

is never greater than °60 and so the temperature of the food is never in the “hot safe zone”. (1 mark)

c. Solve .for 306

sin3060 tt+⎟⎠

⎞⎜⎝

⎛=

π

So 15or 3 == tt . So the temperature of the food is C°60 at 3am and 3pm on Friday.

(1 mark) d. The food is safe between C00 and C05 . This occurs for 12.1088.7 << t and

12.2288.19 << t . (1 mark)

The food is therefore safe between 7.53am and 10.07am on Friday and again between 7.53pm and 10.07pm on Friday. (1 mark)

e. i. The rate at which the temperature of this food is changing is given by the

gradient of the function ( )tT . From the graph of ( )tTy = we see that the gradient is negative for ( ) ( )21,159,3 ∪∈t . So on Friday between 3am and 9am and then again between 3pm and 9pm, the rate is negative. (1 mark)

ii. The rate at which the temperature of this food is changing is given by the gradient of the function ( )tT ; that is by the derivative ( )tT ' .

Now ( ) 306

sin30 +⎟⎠

⎞⎜⎝

⎛=

ttT π

so ( ) ⎟⎠

⎞⎜⎝

⎛=

6cos5' ttT π

π by hand or by CAS

Solve 56

cos5 =⎟⎠

⎞⎜⎝

⎛ tππ for t for 240 ≤≤ t (1 mark)

...6186.21...,3813.14...,6186.9...,3813.2=t The rate at which the temperature of the food is changing equals five degrees per hour on Friday at 2.23am, at 9.37am,at 2.23pm and at 9.37pm. (1 mark)


14

f.

plate stored at safe temperature

plate stored at unsafe temperature

x$ $5 $10

)Pr( xX = 5.0 5.0

(2 marks) g. ( ) ( ) 5.75.0105.05 =×+×=XE . Expected price is $7.50. (1 mark) h. ( ) ( ) 1415.721)(212 =−=−=− XEXE (1 mark) i.

( ) ( )( )( )

( )( )2525.64

5.75.624

)()(4

ar212ar

2

22

2

=

=

−=

−=

=−

XEXE

XVXV

(1 mark)

Total 14 marks

(1 mark)


15

Question 3

a. ⎢⎣

⎡=

4.06.0

T ⎥⎦

⎤

2.08.0

(Alternatively ⎢⎣

⎡=

8.02.0

T ⎥⎦

⎤

6.04.0

) (1 mark)

b. ⎥⎦

⎤⎢⎣

⎡=01

0S (Alternatively ⎥⎦

⎤⎢⎣

⎡=10

0S if the alternative answer was given in

part a.) (1 mark) c. There are 2 transitions to 2007

⎢⎣

⎡=

4.06.0

2S 2

2.08.0⎥⎦

⎤⎥⎦

⎤⎢⎣

⎡

01

= ⎥⎦

⎤⎢⎣

⎡

32.068.0

(1 mark)

The probability of Sue holidaying in the mountains in 2007 is 0.32. (1 mark)

d. ⎢⎣

⎡=∞ 4.0

6.0S

∞

⎥⎦

⎤

2.08.0

⎥⎦

⎤⎢⎣

⎡

01

= ⎥⎦

⎤⎢⎣

⎡

33.067.0

(1 mark)

Over the long term Sue prefers to go to the beach for her July holidays since she holidays there 67% of the time over the long term.

(1 mark)

e. Since the function )(tf is a probability density function, then

1)3(3

0

2 =−∫ dtta .

Solving this using a CAS calculator gives 91

=a . Have shown. (1 mark)

f. Let m represent the median. The median is the “middle” value in the distribution.

So, ( ) 5.0391

0

2 =−∫m

dtt .

Solving this for m gives 0.6188… So the median time to the nearest minute is 37 minutes (to the nearest minute). (1 mark) g. The probability that Sue does less than two hours of yard duty in a week is given by

( )

2726

)3(912Pr

2

0

2

=

−=< ∫ dxxX

(1 mark)


16

h. The probability that Sue does more than one hour of yard duty in a week given that

she does less than two hours in the week is given by

)2Pr()21Pr()2\1Pr(

<

<<=<>

XXXX (1 mark)

∫

∫= 2

0

2

1

)(

)(

dxxf

dxxf

2726

277÷=

267

= (1 mark)

i. Again, since )(xg is a probability density function

∫ =−a

dxxa0

2 1)3( (1 mark)

Solving this for a gives 316.0or ,354.0 −=a but 0>a so 354.0=a correct to 3 decimal places. (1 mark)

Total 13 marks


17

Question 4 a. i. )1(log715)( ++= ttV e

( )

( ) 01log since 151log715)0(

==

+=

e

eV

The initial volume occupied is 15 3cm . (1 mark) ii. )16(log715)15( eV += 3cm (1 mark) iii. )1(log715)( ++= ttV e

17)('+

=t

tV (1 mark)

The graph of )(' tVy = for 0≥t , has a maximum value at 0=t , so the volume is increasing at its greatest rate at 0=t . (1 mark)

b. Method 1 – using a CAS calculator Solve )1ln(71517 ++= t .

172

−= et (1 mark) Method 2 – by hand

1

1

)1ln(72

)1ln(72)1ln(71517

72

72

−=

=+

+=

+=

++=

et

et

t

tt

(1 mark) c. Method 1 – using a CAS calculator Solve )1ln(71540 ++= t .

1725

−= et Have shown (1 mark)

Method 2 – by hand

1

1

)1ln(725

)1ln(725)1ln(71540

725

725

−=

=+

+=

+=

++=

et

et

t

tt

Have shown (1 mark)


18

d. From part c., when 1 so1, ,40 725

1725

−=−== emetV because the function )(tVm is continuous. (1 mark)

Similarly, when 40104020 =+− t

50=t

so 502 =m (1 mark) e. The average volume is given by

( )( ) ...723.331log71501

17

25

1

0725 =++

−+−∫+−

dtte

e

e (1 mark)

The average volume is 3cm34 to the nearest 3cm (1 mark) f. 0)( =tVm

520104020

=

=+−

tt

(1 mark) g. Method 1

The function ( )tVk is an increasing function and therefore we need to be sure that the initial value; that is when 0=t , of the function is greater than 20.

)2(log15)( kttV ek ++= , 0≥t . We require that ( ) 20>tVk (1 mark) that is,

( ) 202log15 >++ kte When 0=t

( )( )

2

2

52log202log15

5

5

ek

ek

kk

e

e

>

>

>

>+

(1 mark)

So the range of values of k are given by ⎟⎟⎠

⎞⎜⎜⎝

⎛∞∈ ,

2

5ek . (1 mark)


19

Method 2

)2(log15)( kttV ek ++= , 0≥t . We require that ( ) 20>tVk First solve ( ) 202log15 =++ kte for k using a CAS calculator

2

5 tek −= (1 mark)

The graph of ( )tVy k= is shown and is an increasing function.

When ,0=t 2

5ek =

( ) .2

,20 and 0For 5ektVt k === If

2

5ek < then )2(log15)( kttV ek ++= <20

(1 mark)

So, for ( )2

,205ektVk >> .

So the range of values of k are given by ⎟⎟⎠

⎞⎜⎜⎝

⎛∞∈ ,

2

5ek . (1 mark)

Total 14 marks

t

(is indicated by the solid line)

-2k

y

tVy k=

2007 maths methods cas 3 & 4 exam 2 solutions · ©the heffernan group 2007 maths methods (cas)...

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