WESTERN REGION by nw17D34

VIEWS: 4 PAGES: 22

									WESTERN REGION

                  2008
                  TRIAL HSC
               EXAMINATION




 Mathematics


  SOLUTIONS
Question 1    Trial HSC Examination- Mathematics                2008
Part Solution                                                    Marks Comment
(a)    1      
                 2                                               2     1 for answer
       2.5  1  0.84256  0.843 (3 sig fig)                         1 for
      e                                                              rounding
(b)      2x  4  2                                              2     1 for each
                                                                       part of the
      2  2 x  4  2
                                                                       inequality
        2  2x  6
         1 x  3
(c)      4                                                       2     1 for correct
                ab 3                                                 method
       2 3
         4    4   2 3                                                 1 for values
               
       2 3 2 3 2 3                                                  of a and b
               8 4 3
                
                43
      a b 3  8 4 3
           a  8 and b  4
(d)   4½ + 3 + 1½ + …                                            2
                                           1            1
      Series is arithmetic with a  4        and d  1
                                           2            2
             n
      Sn       (2a  (n  1)d )                                       1 for formula
             2
             10           1
      S10  (9  (9) 1 )
              2           2
                1
            5  22 
                2
                  1                                                    1 for answer
            112
                  2
(e)   2 z  6 zy  xz  3xy  2 z  z  3 y   x  z  3 y 
         2                                                       2     1 for partial
                                                                       factorisation
                                2 z  x  z  3 y                  or simple
                                                                       mistake
                                                                       leading to
                                                                       incorrect
                                                                       factorisation

                                                                       2 for correct
                                                                       factorisation.
Question 1    Trial HSC Examination- Mathematics   2008
Part Solution                                       Marks Comment
(f)                                                 2     1 for use of
           6 1  8  3  5
      d                                                  formula
              62   8 
                            2

                                                          1 for answer
           13
       
           100
        1.3
Question 2    Trial HSC Examination- Mathematics                     2008
Part Solution                                                         Marks Comment

          2 x3  x3   6 x2  3x4
(a)   d                                                               2     1 for each
i)    dx                                                                    part of
                                                                            derivative.
ii)   d  1             d 2 x                                       2     1 for each
          2 x  sin x    e  sin x                                     part of
      dx  e            dx
                                                                            derivative.
                                         2e2 x  cos x

      
(b)                                                e4 x               2     1 for each
i)         sec 2 x  e 4 x            dx  tan x       c                  part of
                                                    4
                                                                            integral.
ii)                      x3       
                                                   e                  3     1 for each
      
           e
                  2
               x  dx    2 ln x 
                2
                                                                            part of the
       1          x     3         1                                       integral.
        e3             1
             2 ln e   2 ln1
        3              3
          3
        e          1
       2 0
        3          3                                                        1 for
        e 1
          3
                                                                            substitution.
               2
            3
(c)   y  2cos 3x                                                     3     1 for using
           y                                                                correct
       3                                                                    integral
                                                                            including
       2                                                                    units.

       1
                                                                            1 for
                                                                            integration

                                                                x
                                                             6
                                                                            1 for
      -1                                                                    evaluation
                             


                        
                            12
      Area                      2 cos 3 x dx
                         0
                                         
                      2sin 3 x  12
                    
                      3 0     
                                  
                        2sin
                                  4  2sin 0
                                 3       3
                            2
                                 0
                        3 2
                         2
                          square units.
                        3
Question 3    Trial HSC Examination- Mathematics                2008
Part Solution                                                    Marks Comment
(a)    Midpoint of (1, 6) and (5, 0).                            1     1 for answer
i)           1 5 6  0   6 6 
       MP         ,          ,    3,3
             2        2  2 2
ii)    Show that (3,3) lies on 2 x  3 y  3  0                 1     1 for answer
       LHS  2  3  3  3  3
        693
        0  RHS
       So M lies on BD.
iii)                       60 6           3                     1     1 for answer
       Gradient AC  m1             
                           1  5 4        2
iv)    Find gradient m2 of BD 2 x  3 y  3  0                  2     1 for gradient
       2x  3 y  3  0                                                of BD
       3y  2x  3
           2
       y    x 1
           3
               2                                                       1 for
        m2                                                           showing
               3
                                                                       product of
                 3 2
       m1.m2   .  1                                                gradients =
                 2 3                                                   -1
        BD is perpendicular to AC

v)                                                               1     1 for answer
       AC      5  1   0  6 
                      2               2



        16  36
        52
        2 13
(vi    The lines AC and BD would form the diagonals of the       1     Need to
       quadrilateral ABCD.                                             mention
       BD is the perpendicular bisector of AC from ii and iv           perpendicular
       above..                                                         (or meet at
       The diagonals of a kite meet at right angles and one            right angles)
       diagonal bisects the other, so ABCD meets the criteria          and midpoint
       for a kite.                                                     (or bisect) for
                                                                       point for
                                                                       mark.
Question 3    Trial HSC Examination- Mathematics                2008
Part Solution                                                    Marks Comment
(b)    2 hours 51 minutes 2.85                                   1     1 for answer
i)                               0.95
            3 hours           3
       2 hours 42 minutes 27 seconds 2.7075
                                                0.95
             2 hours 51 minutes           2.85
        forms a geometric series with r  0.95

ii)    a  3, n  5 and r  0.95                                 1     1 for answer
       u5  ar n 1
        3  0.954
        2.4435
        2 hours 26 min 37 sec
iii)        a 1  r n                                          1     1 for answer
       sn 
              1  r 
           3 1  0.955 
       
            1  0.95
        13.573
        13 hours 34 min 23 sec
iv)    un  ar n 1                                              2     1 for
                                                                       equation to
       2 hours 6 min  3  0.95n 1                                    be solved
       3  0.95n 1  2.1
       0.95n 1  2.1  3  0.7
                                                                       1 for answer
       ln  0.95n 1   ln  0.7                                     either using
        n  1 ln  0.95   ln  0.7                                logs as
                                                                       shown or by
                  ln  0.7                                            guess and
       n 1 
                 ln  0.95                                            check.
       n  1  6.953
       n  7.953
       Would need to continue for 8 weeks to better the time.
Question 4    Trial HSC Examination- Mathematics               2008
Part Solution                                                   Marks Comment
(a)     cosec2 x  cot 2 x  cos 2 x   1  cos 2 x              2     2 marks for
                                                                     required
                  cos 2 x                cos 2 x                      result.
                                   sin 2 x                            1 mark for
                                                                     partial work
                                   cos 2 x
                                                                      toward
                                 sin x
                                                                     required
                                 cos x                                result.
                                tan x
(b)                         4 2                                 1     1 mark for
i)    P(Blue on first die) =                                         answer.
                            6 3
                            1
      P(Blue on 2nd die) =
                            6
                            2 1 1
      P(2 Blue showing) =  
                            3 6 9

(b)   P(Different)= P(BR)+ P(RB)         OR =1-P(same)          2     2 for correct
ii)                  2 5 1 1             =1-(P(RR)+PBB))              answer.
                     
                     3 6 3 6                  5 1 2 1
                                          1                    1 mark if
                     5 1                      6 3 3 6
                                                                    correct
                     9 18                      7                      strategy
                    11                    1
                                             18                      used, but
                    18                     11                         mistake
                                                                     made in the
                                           18
                                                                      process.

(c)                                                             1     1 mark if
i)                                                                    diagram
                                                                      shows angles
                                       142 m                          and distances
                                                                      correctly.
            16o                                      20o
      A                              K                     B
(c)            142                       142                    2     1 mark for
ii)   tan16                   tan 20                              answer to
               AK                        KB
                 142                       142                        each part of
          AK                       KB                               the required
               tan16                    tan 20                      distance.
              495.2                    390.1

      Distance AB = 495.2 + 390.1 = 885 m (nearest m)
Question 4    Trial HSC Examination- Mathematics          2008
Part Solution                                              Marks Comment
(d)       N                                                2     2 for
i)        P                                                      complete
                                                                 answer .
        360 km 30o                       Q
                      80o                                        1 if started
                      R         480 km                           using the cos
                                                                 rule correctly
                                                                 but simple
      PQ 2  3602  4802  2  360  480 cos110                 mistake
                                                                 made.
      PQ 2  478202
      PQ  692 km (nearest km)
(d)   First find QPR                                      2     2 for
ii)   sin QPR sin110                                           complete
                                                                answer .
          480         692          Can also be found
                    480  sin110  using cos rule using
      sin QPR                    the 3 sides..                 1 if started
                         692                                     using the sin
      sin QPR  0.652                                           or cos rule
      QPR  41                                                 correctly but
                                                                 simple
      NPR  150
                                                                 mistake
      Bearing (NPQ )  150  41                               made.
       109
Question 5            Trial HSC Examination- Mathematics                    2008
Part Solution                                                                Mark   Comment
                                                                             s
(a)   EAD  60            (equilateral  )                                  3       1 for
i)                                                                                  equilateral
      DAC  45 (isosceles right  )                                                ∆
      EAB  EAD  DAC  CAB                                                    1 for
                                                                                    isosceles ∆
       60  45  51
                                                                                    1 for
       156                                                                         answer
ii)   ABC  180  156                   (cointerior  on || lines AE and BC) 1     1 for
                                                                                    answer
               24

(b)      4t 2  t  8                                                        2
i)    x                                                                            1 for
            4t  1
                                                                                    answer
         4 0  0  8
                  2

      x                 when t = 0
             4 0  1
       8                                                                           1 for
                                                                                    working
(b)               4t 2  t  8                                                      1 for
ii)       x                                                                        answer
                     4t  1
                   4t  18t  1   4t 2  t  8  4 
          x
                                       4t  1
                                                  2
                                                                             1
                  32t 2  12t  1  16t 2  4t  32
              
                                   4t  1
                                              2



                  16t 2  8t  31
              
                       4t  1
                                  2
Question 5                Trial HSC Examination- Mathematics            2008
Part Solution                                                            Mark   Comment
                                                                         s
                          x0                                            2
(b)    16t  8t  31
           2
iii)                        0
           4t  1
                      2



       16t 2  8t  31  0                                                      1 for
                                                                                quadratic
                                8  82  4 16  31
                          t
                                         2 16 
                              8  2048
                            
                                   32
                              8  32 2
                            
                                  32
                                1  4 2
                            
                                    4
                                      1  4 2
       So it is stationary when t 
                                          4
                                        1  4 2                                1 for
       Only us positive value so t 
                                            4                                   solution
(b)       1  4 2                                                       2      1 for
iv)    t              1.2sec v = 0, x  2.6                                   mention of
               4
                                                                                direction of
       When t  0 x  8 and v  31
                                                                                travel
       When t  2 x  2.9 and v  0.6
                                                                                before and
        Particle starts 8 units to the right of the origin, moving
                                                                                after
       toward the origin, it decelerates and stops after 1.2 sec at
                                                                                turning.
       2.6 m to right of origin, then begins to move away from
                                                                                1 for
       the origin, being 2.9 units to the right of the origin after 2
                                                                                mention of
       sec.
                                                                                position at
                                                                                at least one
                                                                                point.
(c)    3x  y  10                (1)                                    2      1 for
       x  y2                     (2)                                          substitution
                                                                                (or
       3  y  2   y  10 (3) sub (2) in (1)                                  elimination
       3 y  6  y  10                                                         )
       2y  4
                                                                                1 for
       y2                                                                      answer
       x   2  2             sub y in (2)
       x4
       solution (4, 2)
Question 6         Trial HSC Examination- Mathematics          2008
Par Solution                                                     Mark   Comment
t                                                                s
(a)   y  x6  6 x4                                              2      2 if all
(i) Crosses axis where                                                  given.
      x6  6 x4  0
                                                                        1 if only
       x4  x2  6  0                                                 one or two

                   x  6   0
                                                                        given
       x4 x  6

       Crosses axis where x  0 and x   6
(a)    y  x6  6 x4                                            4       1 for
(ii)                                                                    derivative
       y  6 x 5  24 x 3
           6 x3  x 2  4                                             1 for
                                                                        solving for
           6 x3  x  2  x  2                                      x values
       y  30 x 4  72 x 2
                                                                        1 for y
           Stationary points where
                                                                        values
        x  0, y  0, y  0
        x  2, y  32, y  192                                       1 for
        x  2, y  32, y  192                                      nature


          Stationary points  2, 32  ,  0, 0  2, 32 
       y  30 x 4  72 x 2
       At x  0 y  0 so test either side
       At x  1 y  42  concave down.
       At x  1 y  42  concave down
                                 maximum at (0, 0)
       At x  2 y  192  minimum at (2, -32).
       At x  2 y  192  minimum at (  2 , -32).
Question 6          Trial HSC Examination- Mathematics                                        2008
Par Solution                                                                                    Mark   Comment
t                                                                                               s
(a)                                                                                             2      1 for
(iii) y  30 x 4  72 x 2                                                                            Coordinate
       6 x 2  5 x 2  12 
                                                                                                       s

       6x2      5x  2 3          5x  2 3                                                        1 for
                                                                                                       testing to
      x0         y0                                                                                  determine
                                                                                                       only 2
           2 3 2 15
      x                            y  20.736                                                       inflexions
             5   5
           2 3      2 15
      x                  y  20.736
             5        5
      Check for changes of concavity
      From above, no change at (0, 0) but
                            2 15                 
      there is a change at  
                                      ,  20.736 
                                                  
                                5                
                     2 15                 
      Inflexions at  
                              ,  20.736 
                        5                 


                                                     y                                         2       1 for shape
                                               10                                                      of curve.

                                                         Max (0, 0)                                    1 for
                                                                                                       points
         -3           -2             -1                             1       2             3    x
              6                                                                     6                  shown
                                                                                                       correctly.
                                              -10

                                    -2 15                 2 15       
                                       5 – 20.7             5 – 20.7 
                                                                     
                                                                     
                                                                     
                                              -20




                                              -30

                      Minimum                                               Minimum
                      ( -2 , -32 )                                          ( 2 , -32 )
Question 6        Trial HSC Examination- Mathematics                               2008
Par Solution                                                                         Mark   Comment
t                                                                                    s
(b)                                                y                                 2      1 mark for
                f''(x) is positive where f'(x) is increasing                                x
                                                                                            coordinate
                                                                                            s and 1
                                        (-1, 4)                                             mark for
                                                                                            nature of
                                                                                            both.


                                                                                    x
               -4                                                  2



        f''(x) is zero where f'(x) is stationary


                                          f''(x) is negative where f'(x) is decreasing



    Stationary points on y occur where f '( x)  0 i.e. at
    x  4
    here f ''( x) is positive  min turning point at x  4
     and x  2
    here f ''( x) is negative  max turning point at x  2
Question 7    Trial HSC Examination- Mathematics   2008
Part Solution                                       Marks Comment
(a)    x 2  8 y                                   1     1 for answer
i)
       x 2  4  2  y
       a2
                             y




                                      x




       Focus is (0, -2)
ii)    Directrix is y  2                           1     1 for answer
iii)   x  8 y
        2                                           1     1 for sub

       x 2   8   64
                    2


       8 y  8  8   64
         8, 8  lies on the parabola.
iv)    Chord through (0, -2) and (-8, -8)           2     1 for correct
            8  2 6 3                                   gradient
       m                 
            8  0 8 4
       y  y1  m  x  x1                               1 for
                                                          equation
                        3
       y   2          x  0
                        4
           3
       y     x2
           4
       3x  4 y  8  0
v)     x 2  8 y                                   2     1 for
                                                          derivative
              x2
       y
              8                                           1 for
              x                                           equation
       y  
              4
                       8
       At A y           2
                        4
       y   8   2  x   8 
       y  8  2 x  16
       y  2x  8
(b)   Substitute y  x 2  3x into y  5 x  x 2                         2   1 for x
i)                                                                           substitution
      5 x  x 2  x 2  3x
      2x2  8x  0
      2x  x  4  0                                                        1 for points
      x  0, y  0
       x  4, y  4
      Intersect at (0, 0) and (4, 4).
ii)                  y                                                   3




                                                             (4,4)




                                                                     x




                                                                             1 for correct
                                       
                   4                        4
      Area            5 x  x 2 dx            x 2  3xdx                   integral stated
                 0                      0



           8x  2 x
           4
                        2
                             dx
           0
                             4                                               1 for
         2 2 x3 
       4 x                                                                integration
                3 0
              128 
        64      0
               3 
        64       1                                                           1 for sub and
            21 u 2
         3       3                                                           correct
                                                                             answer
Question 8             Trial HSC Examination- Mathematics   2008
Part Solution                                                Marks Comment
(a)      P  Ae kt                                           1     1 for answer
i)
       dP
           Ae kt .k
       dt
           kAe kt
             kP
ii)    t = 1 was 147 200                                     2
              P  Aekt
       147200  Aek                  (i)
       t = 2 was 154 800
       154800  Ae 2 k                (ii)
                        2k
       154800 Ae
                                      (ii)  (i)
       147200 Ae k
       1.0516  e k                                                1 for k
       k  ln 1.0516 
       k  0.05
       147200  Ae0.05(1)
           147200                                                  1 for A
       A  0.05  139973
             e
iii)   When t  4                                            1     1 for answer
       P  Aekt
       P  139973e0.05(4)
          171197
iv)           P  Ae kt                                      1     1 for answer
         200000  139973e0.05t
         200000
                  e0.05t
         139973
           1.429  e0.05t
       ln 1.429   ln  e0.05t 
            0.05t  ln 1.429 
                      ln 1.429 
                  t
                         0.05
                    7.1
       t = 7 is start of 2012
       Population will reach 200 000 in 2012
Question 8                Trial HSC Examination- Mathematics                 2008
Part Solution                                                                 Marks Comment
(b)                                     A                                     2     1 each for 2
i)                                                                                  pairs of
                                                                                    equal angles
                                                                                    and reasons
                      Q                                    P




       C                                                                 B

      In APQ and ABC
      A is common
      AQP=ACB                  (Corresp  on || lines)
      APQ= ABC                 (Corresp  on || lines)
      APQ ||| ABC (Corresponding angles equal)
ii)   AP 1                                                                    2     2 for any
                ( P is midpoint of AB)                                             reasonable
      AB 2
      AP AQ                                                                         explanation
                (sides of similar triangle in same ratio)                          using ratio of
      AB AC                                                                         correspond
      AQ 1                                                                          sides
               (from above)
      AC 2                                                                           1 for partial
       Q is midpoint of AC.                                                        explanation

      
          3
(c)                                                                           2     1 for sub in
              g  x  dx 
                          1
                          6
                            12  4 8  2  0   4  3  5                     formula
       1
                                                                                    correctly
                          61
                        
                           6                                                        1 evaluate
                             1                                                      correctly
                         10
                             6
(d)                                                                           1     1 for answer
      
        5

               n 2  1   22  1   32  1   42  1   52  1
       n2

                       3  8  15  24
                       50
Question 9             Trial HSC Examination- Mathematics                      2008
Part Solution                                                                   Marks Comment
(a)    y  2 x2  2                                                             3     1 for
                                                                                      formula used
             x dy
               6
       V         2
                                                                                      correctly
               0

              y2
             2 dy
               6
                                                                                      1 for
                                                                                    integration
               0
                         6
              y2 
             y
             4   0                                                                  1 for
                                                                                      evaluation
              36           
             6    0  
              4            
          15 u 3
(b)    P WWW   0.8  0.8  0.6  0.384                                       1     1 answer
i)
ii)    P  LLL   0.2  0.2  0.4  0.016                                      1     1 answer
iii)   P  at least 1 win   1  P  LLL                                      1     1 answer
                              1  0.016
                              0.984
(c)    A6  20000  6M                                                          1     1 answer
i)
ii)    A7   20000  6M 1.01  M                                              1     1 answer
       A8   20000  6M 1.01  M  1.01  M
                                   
            20000  6M 1.01  1.01M  M
                              2


            20000  6M 1.012  M 1  1.01
iii)   A9   20000  6M 1.013  M 1  1.01  1.012                          2     1 for
                                                                                      developing
       An   20000  6M 1.01n 6  M 1  1.01  1.012  .......1.01n 7           series further
       A36   20000  6M 1.0130  M 1  1.01  1.012  .......1.0129 
                                                                                      1 for result
                                                                                      for A36
Question 9         Trial HSC Examination- Mathematics                    2008
Part Solution                                                             Marks Comment
iv)   Since repaid after 36 months A36  0                                2
       20000  6M 1.0130  M 1  1.01  1.012  .......1.0129   0
      M 1  1.01  1.012  .......1.0129    20000  6M 1.0130
      Need to evaluate 1  1.01  1.012  .......1.0129
      Geometric series with a  1, r  1.01, n  30
                        a  r n  1
                  S30 =
                           r 1
                        11.0130  1
                      =
                            1.01  1
                      =34.785
            34.785M   20000  6M 1.0130
           34.785M
                      20000  6 M
            1.0130
           34.785M
      6M             20000
            1.0130
           34.785 
      M 6          20000
           1.0130 
             31.8M  20000
                       20000
                  M
                        31.8
                      $629 (nearest dollar)
Question 10   Trial HSC Examination- Mathematics   2008
Part Solution                                       Marks Comment
a) i)   V  2  3 cos t  sin t                     1     1 for answer
        dV
             3 sin t  cos t
         dt
ii)     When t  0                                  1     1 for answer
        dV                                                or for correct
             3 sin 0  cos 0                             value of
         dt
                                                          derivative.
         1
         the tank is emptying at this time.

iii)                              dV                3     2 if final
        Full (or empty) when         0                   value of t is
                                  dt
        dV                                                given.
            0
        dt
                                                          1 if equation
         3 sin t  cos t  0                              formed
          3 sin t  cos t                                 correctly but
                                                          not solved
        sin t     1
                                                         correctly
        cos t       3
                  1                                       1 if an
        tan t                                            incorrect
                   3
                                                          equation is
           7 13
         ,        ,     ...                              formed and
           6 6 6                                          is solved
        As tank is initially emptying,                    correctly.
        second value corresponds to when full.
                                    7
        Tank is first full when t 
                                     6                    1 for finding
        V  2  3 cos t  sin t                           volume from
                      7       7                         value of t
         2  3 cos       sin
                       6        6
        4
(b)   D is common                                          2   1 for
i)                                DA                    B       statement of
      C  B  90o
                                                                similarity
      E  A (coresponding angles on|| lines)
      ABD ||| ECD (equiangular)
       x  h 2t                                                 1 for finding
               t                                              required
         x      2
                                                                expression
      x  h  tx
      tx  x  h
      x(t  1)  h
             h
      x
           t 1
(b)         1                     1                         2   1 for using
      V    2t  .  h  x     2  .x
                    2                   2
ii)         3                     3                             correct
                                                                formulae
         1        2        h  1         2  h 
         2t  .  h            2 .        
         3               t 1  3            t 1 
         1        2  ht     1      2  h                     1 for
         2t  .            2 .        
         3           t 1  3           t 1                 algebraic
                                                                manipulation
              2  h  3
                         t  1
       1
         2 .                                               to achieve
       3         t 1                                         result
       4  h 
        .          t  1  t  t  1
                                  2

       3  t 1 
        4 h  2
              t  t  1
        3 
iii)   Sum of radii and height =12                                3   1 for
       2  h  2t  12                                                expressing V
       h  10  2t                                                    in terms of t
                                                                      or h
               4 h  2
         V             t  t  1
               3 
            4 
                  10  2t   t  t  1
                                                                      1 for
                                2

             3                                                      derivative
            4 
                10t  10t  10  2t  2t  2t 
                          2                3 2

            3                                                        1 for finding
            4  2
                   8t  8t  2t  10 
                                                                      values for
                                  3
                                                                      max case. If
            3 
                                                                      only found
             4  2
       V         8t  8t  2t  10 
                                     3
                                                                      the correct
              3                                                     value of t ,
       dV  4                                                       give the
                    16t  8  6t 
                                       2

        dt  3                                                       mark.
       dV
            0
        dt                                                            Accept any
       16t  8  6t 2   0                                          reasonable
                                                                      rounding of
            b  b 2  4ac                                            values which
       t
                 2a                                                   indicate
                                                                      correct
           16    16   4  6 8 
                        2

                                                                     working has
                     2  6                                          been done.
                                                                      i.e. don’t
         16  448
                                                                     deduct for
             12                                                      error in
        0.43 or 3.10 ( Ignore negative value)                       rounding.
       d 2V  4 
                 16  12t 
       dt 2  3 
       When t = 3.10
       d 2V  4                          4 
                 16  12  3.10          21.2 
              3                          3 
          2
       dt
        88.7
        d 2V
             0
         dt 2
        V is a maximum

       V 
            4 
            3 
                   
                 8  3.10   8  3.10   2  3.10   10
                            2                         3
                                                              
        218.2

								
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