# WESTERN REGION by nw17D34

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```									WESTERN REGION

2008
TRIAL HSC
EXAMINATION

Mathematics

SOLUTIONS
Question 1    Trial HSC Examination- Mathematics                2008
Part Solution                                                    Marks Comment
(a)    1      
 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
 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
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

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

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
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
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
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
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
PQ 2  478202
PQ  692 km (nearest km)
(d)   First find QPR                                      2     2 for
ii)   sin QPR sin110                                           complete
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
ii)   ABC  180  156                   (cointerior  on || lines AE and BC) 1     1 for
 24

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

x                 when t = 0
4 0  1
8                                                                           1 for
working
(b)               4t 2  t  8                                                      1 for
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
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
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
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

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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