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

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  • 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

    Kilbaha Multimedia Publishing http://kilbaha.com.au

    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

    Kilbaha Multimedia Publishing http://kilbaha.com.au

    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

    Kilbaha Multimedia Publishing http://kilbaha.com.au

    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

    Kilbaha Multimedia Publishing http://kilbaha.com.au

    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

    Kilbaha Multimedia Publishing http://kilbaha.com.au

    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

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