2013 kilbaha vce mathematical methods trial exam 2 ... year 2013 vce mathematical methods cas trial

Year 2013

VCE Mathematical

Methods CAS

Trial Examination 2

Suggested Solutions

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Mathematical Methods CAS Trial Exam 2 Solutions Section 1 Page 3

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SECTION 1

ANSWERS

1 A B C D E

2 A B C D E

3 A B C D E

4 A B C D E

5 A B C D E

6 A B C D E

7 A B C D E

8 A B C D E

9 A B C D E

10 A B C D E

11 A B C D E

12 A B C D E

13 A B C D E

14 A B C D E

15 A B C D E

16 A B C D E

17 A B C D E

18 A B C D E

19 A B C D E

20 A B C D E

21 A B C D E

22 A B C D E

Mathematical Methods CAS Trial Exam 2 Solutions Section 1 Page 4

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SECTION 1

Question 1 Answer B

 

 

2 2

2 2 30

2 4 2

f x b x

b b b f b f b

 

      

 

average rate of change

  3 230 22 2 3 2

0 2 2 2

b b bf f b f

b b b

      

       



Question 2 Answer B

   

 

 

: ,3 , 2

3

3

f a a R f x a x

f a a

f a a

   

 

 

The range is  ,3a a

Question 3 Answer B

The period 3

3

T bn b

 

   

Question 4 Answer A

      logef x g x h x using the product rule

          

 

          

 

       

   

2

2

2 2

log

2 2 2 log 22

2

Now 2 3 , 2 4 , 2 and 2 2

3 2 6 2 4 log 8

e

e

e

g x h x f x g x h x

h x

g h g hf

h

g g h e h

f e e e

   

   

    

      

x

y

0 2 3a a a

3a

2a

a

Mathematical Methods CAS Trial Exam 2 Solutions Section 1 Page 5

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Question 5 Answer A

  2 1 1

2 1 2 2 1

1 1 1

k

k k k k k



          

When 0  or 2 , 1k k   there is no unique solution.

When 1k   there is no solution.

When 2k  there is infinitely many solutions.

Question 6 Answer C

       

     

 

2

0 0

2 2

0

1 tan

cos

60 1 3 180

1 1 tan 3 4

3 3 3 1cos 3 2

tan 61 3 4 1.8019 180

f x x f x x

x h

f f

f x h f x hf x

 

  





 

   

               

               

  

   

Mathematical Methods CAS Trial Exam 2 Solutions Section 1 Page 6

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Question 7 Answer A

 

 

    

3 2

2

2

4 3 2

3 8 3

3 8 3

3 1 3

f x x x x

f x x x

x x

x x

    

    

   

   

Turning points at 1

and 3 3

x x  

To restrict the domain to make f a

one-one function we require 1

3 a  

Question 8 Answer E

 

   

2 2log 2 2

2 log 2 2 log

2

2 and 2 2 2 2 and 1

2 2

1 10

2 2

0 2

e

e e

y x

y x y x

y x y x x y y x

x x x T

y y y

   

    



            



                                

Question 9 Answer C

    1

cos sin when sin 6 6 2

T

dy b y bx b bx x m b

dx b

     

          

 

normal 2 1

4 2

Nm b b 

   

Question 10 Answer E

   2 2 2f x ax bx x ax b    , the graph

crosses the x-axis at 2

0 and b

x x a

  .

  2 2f x ax b   .

A decreasing function has a negative gradient,

so that   0 2 2 0f x ax b     . if 0 and 0ax b a b   

then b

x a

 .

x

y

-3 -2 -1 0 1 2 3 4 5

-4

-2

2

4

6

8

10

12

14

16

18

x

y

2 0

b b

a a









0

 

Mathematical Methods CAS Trial Exam 2 Solutions Section 1 Page 7

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Question 11 Answer C

T catches the train and D drives to work

0.7 0.3

0.2 0.8

T T T D

D T D D

    

    

100 0.7 0.2 0.4 0.4

0.3 0.8 0.6 0.6

T D

T

D

       

   

long-term drives is 3

0.6 5



Question 12 Answer C

2

2

3 3 3 3 given 36

2

36 36 6 3 2 3 cm/min

3 3 3 3 2 3 3x

dA dA A x x

dx dt

dx dx dA dx

dt dA dt dtx 

   

      

Question 13 Answer D

 

2 2

0

2 2

0

1 3

0 2

2 1 1 1 3

2

k

x

k

x k

e dx k

e e k k





     

 



solving for k gives, 1.904k 

Question 14 Answer E

     2 1andf x x g x x g