2014 st josephs geelong spec maths

Specialist Maths
• Problems that achieved <60% for Year 2010-2013

The MEDIAN score is C+. 49% of students get B
or greater
21% of students get better than 33/40. Losing
any marks will have a big impact on your Study
Score. This was about the same as previous
years.
23/40 is the MEDIAN grade.

MEDIAN is C+: 43/80 B: 51/80
A: 65/80 A+:
70/80

Exam 1 Assessors Report
Hand writing and general setting out of work a
concern.
Use logical connectors between lines.

Short Answer 2012
• CALCULATOR FREE

6 + x
x2 + 4
= 6
x2 + 4
+ x
x2 + 4
6
x2 + 4
+ x
ò x2 + 4 ×dx
= 3×Tan-1 x
2
æè ç
öø ÷
+ 1
2
×ln x2 + 4 +c
Most students who recognised to split the fraction into two parts performed
quite well.

Not very well done. Average 0.9/3. Many student omitted cos(x)=0
as a possible solution. Many students only gave solutions for 0≤x≤2π.

Real coefficients so look for conjugate pair
Third solution must have real root.
Many students attempted long division and were unable to find the correct
solutions.

Many students incorrectly put the solution z=3 at z=2 confusing the
question with a root of a single complex number and not an expression.

Fr
N
mg
T
T sinq = mg
T = 100g

Sign errors were common with students omitting du/dx=-1.

Quite well done once students recognized the correct expression
for a.

Conjugate is a reflection in the Re axis
Multiply by i is a rotation of pi/4 anti-clockwise, in this case 3 times
This is the same result as C.

Perpendicular bisector of origin and z=-2i
Line z=-i. This is B
Note: Im(z) must be R, C is incorrectly represented.
Use calculator program or other to show y(0.2)=1.1995
In this region cos(x) is concave down OR gradient is decreasing
Answer will be an over estimate: B

Correct option is E
The options B,C, D, E cannot be
correct.
MUST be A.

dx
dt
= x Þ dt
= 1
x
dx
t = ln x +c
when t = 3, x = -1
c = 3
t = ln x + 3
t - 3 = ln x Þx = et-3

Area = p
8
2 ( )
2 - r1
r2
Area = p
8
22 -12 ( ) = 3p
8
units2
z = cis p
12
æè ç
öø ÷
Re(z) = 0, when Arg (z) = p
2
or -p
2
Þzn " n = 6 or - 6
Generalsoln n = 6 +12k ,k ÎJ
z = ±i

0 = -1
100
145t -t2 ( )+ 25
t = 20 sec
Question 3

19
ò pt
17 tan-1
0
6
æè ç
öø ÷
×dt
120 ´ 25
-1
100
145t -t2 ( )+ 25 æè ç
öø ÷
×dt
20
ò
0
Need to have completed an earlier
section to get first part. Most errors
were because of incorrect terminals
3637m use CAS
Don’t round each answer
this may lead to errors, add
separately and then round.

400 km
When directly above P, i component is
zero, which makes sin=0. When sin=0,
cos=1. 6800-6400=400 km

a(t) = r&&(t) = -1.32p 2 ×6800sin(p (1.3t - 0.1))i -1.32p 2 ×6800cos(p (1.3t - 0.1))j
a(t )^v(t)as
v(t)´a(t) = 0
where -Acosq sinq +Acosq sinq = 0
speed =1.3´p ´ 6800
= 27,772 m /s
x = 6800sin(p (1.3t - 0.1))
y= 6800cos(p (1.3t - 0.1))- 6400
circlemoved¬6400
Þx2 + (y+ 6400)2 = 68002

Use Pythagora’s Thm and a substitution of the time function:
x2 +y2 = 10002
(6800sinq )2 + (6800cosq - 6400)2 = 10002
where q =p (1.3t - 0.1)
-p £q £ 2p
givesq = -0.139, 0.139, 6.144
t = 0.04, 0.11, 1.58,...
t=1.58 is outside the domain, only two answers.

x = 1
2
(u +v)t
10 = 1
2
(0 + 6)t
t = 10
3
sec
3.3 sec scores zero. Do not
give approximate answers
unless specified.
u = 10m /s x =ut + 1
2
at2
v= -6 = 10t - 4.9t2
a = -9.8m /s2 t = 2.5 sec
x = -6m (asit falls)
t =

u = ? s=ut+ 1
2
at2
v= -6 =u × 22
3
- 4.9 22
3
æè ç
öø ÷
2
a = -9.8ms-2 u = 35.1s
x = -6 m
t = 10 3+ 4 = 22
3sec

dv
dt
= - 196 -v2
10
dt
dv
= -10
196 -v2
Þt = -10 ×Sin-1 v
14
æè ç
öø ÷
+c
whent = 0, v= 7
0 = -10 ×Sin-1 1
2
æè ç
öø ÷
+c
c = 5p
3
5p
30
- t
10
= Sin-1 v
14
æè ç
öø ÷
v= 14sin p
6
- t
10
æè ç
öø ÷
or 14 cos p
3
+ t
10
æè ç
öø ÷
v = 0, 0 = 14sin p
6
- t
10
æè ç
öø ÷
t = 5p
3
sec

dx
dt
= 14 ×Sin p
6
- t
10
æè ç
öø ÷
x = 140 ×Cos p
6
- t
10
æè ç
öø ÷
+c
whent = 0, x = 0
Þc = -70 3
x = 140 ×Cos p
6
- t
10
æè ç
öø ÷
- 70 3
whent = 5p
3
5p
3
ò
x = 18.8m OR x = 14 ×Sin p
6
- t
10
æè ç
öø ÷
×dt = 18.8m
0

short answer 2011
• Calculator free

y = kx ×e2x dy
dx
= k(2x +1)×e2x d 2y
dx2 = 4k(x +1)×e2x
Þ4k(x +1)- 2k(2x +1)+ 5kx = 15x + 6
Þ4kx + 4k - 4kx - 2k + 5kx = 15x + 6
k = 3
Many students did not use the product rule to find dy/dx.

z =
2cis -p
3
æè ç
öø ÷
2cis 3p
4
æè ç
öø ÷
Arg (z) = -p
3
- 3p
4
= -13p
12
= 11p
12

åF = 0
considerhorizontalonly
T1
sin30 = T2
sin60
T1
= T2
3
T2
=
T1
3
ÞT2 ≤ 98
∴ T1 ≤ 3 × 98
ΣF = 0 vertical only
T1 cos 30 +T2 cos60 ≥ mg
98 3 ×
3
2
+ 98 ×
1
2
≥ mg
m ≤ 196
g
= 20 kg

y = cosec2 p x
6
æè ç
öø ÷
4
3
= cosec2 p x
6
æè ç
öø ÷
± 3
2
= sin p x
öø ÷ p x
6
æè ç
6
= p
3
, 2p
3
, 4p
6
, 5p
6
,...
x = 2,4,8,10
ptsÞ 2, 4
3
æè ç
öø ÷
, 4, 4
3
æè ç
öø ÷
, 8, 4
3
æè ç
öø ÷
, 10, 4
3
æè ç
öø ÷
Many students miss the last two points that come from the
negative sine angle.

z = x -iy
z = x +iy
1
z
= 1
x +iy
´ x -iy
x -iy
1
z
= x -iy
x2 +y2
Same angle but closer, as it will be
divided by magnitude of z. C

z3 = 27cis -p
2
æè ç
öø ÷
z = 3cis -p
6
æè ç
öø ÷
other solns± 2p
3
P4
, P8
, P12

A Perpendicular vectors
B Equal vectors
C Perpendicular diagonals
D Diagonals same length
E Vectors equal and opposite
direction

- dx
dy
=
2(y −1)
(x −1)
− dy
dx
=
(x −1)
2(y −1)
dy
dx
+
(x −1)
2(y −1)
= 0
Quite a difficult question. Only 25% correct.

When x=0, gradient is constant and positive. Omit E
When y=0, gradient is constant and negative. Omit C
When y=x, m is negative. Omit D
When y=-x m is positive. Omit B
Leaves A

t = 0, xo = 5
xn+1 = xn +h× f'(tn )
Þ x1
= xo +h× f' t0
( )
x1
= 5 - 0.5 ×1.0 = 4.5
t = 0.5, xo = 4.5
x2
= 4.5 - 0.5 × 10
9.5
= 3.97
Could also use CAS program to find the answer.
Be careful with the negative gradient.

a =v× dv
dx
if v= 2 1- x2
dv
= -2x
dx
1- x2
= 2 1- x2
a =v× dv
dx
1
´ -2x
1- x2
= -4x

let z = x +iy
z = x -iy
Þx +iy= a(x -iy)
lety= x
Þx +ix = a(x -ix)
LHS = RHS iff a = i

Area =2(1/4 circle - triangle)
A = 2 1
×1×1 æè ç
4
p (1)2 - 1
2
öø ÷
= p
2
-1

v = òa×dt
v(t) = -9.8tˆk+c
whent = 0,v= 35î + 5ˆj+ 24.5ˆk = c
v(t) = 35î + 5ˆj+ 24.5 - 9.8t ( )ˆk
r(t) = òv×dt
when t = 0,r(t) = 0Þc = 0
r(t) = 35t î + 5t ˆj+ 24.5t - 4.9t2 ( )ˆk
Not very well done for a easy question. Many answer lacking care and details. Marks
were lost for omitting the tilda’s on the vectors.

when dist ˆ k = 0
Þ24.5 = 4.9t
t = 5.0sec
r(5) = 175î + 25ˆj+ 0ˆk
asholeis200î
position from hole = 25î + 25ˆj
25î + 25ˆj = 35m

åF = ma
T - 3g sinq = 3a
2gsin(2q ) -T = 2a

2gsin(2q )-T = 2a
T - 3g sinq = 3a
ÞT = 2g sin(2q )- 2a = 3a+ 3g sinq
ÞT = 2g sin(2q )- 3g sinq = 5a
but sin(2x) = 2sin(x)cos(x)
4g sin ( q )cos( q )- 3g sinq = 5a
g sin q a =
( )
5
(4 cos(q )- 3)
Þa = 0
4 cos(θ)− 3 = 0
θ = cos−1 3
4
⎛⎝ ⎜
⎞⎠ ⎟
= 41.40

F = ma å5a = 2g sin2q - 3g sinq -mN
5a = 2g sin60 - 3g sin30 -m 3g cosq
5a = 2g 3
2
- 3g 1
2
-m 3g 3
2
10a = 2g 3 - 3g -m 3g 3
a = 0.2ms-2

Conc. = mass
volume
= x
10 +10t volume starts at 10 and
increases by 10 every min
dx
dt
= Input -Output
dx
dt
= 20e-0.2t -10 x
10 +10t
dx
dt
= 20e-0.2t - x
1+t
Þdx
dt
+ x
1+t
= 20e-0.2t
must show dx/dt=Input-Output

dx
dt
=
20(t2 + 7t + 31)e-0.2t
(t +1)2 - 600
(t +1)2
Use CAS don’t put x as variable
instead of t

dx
dt
=
20(t2 + 7t + 31)e-0.2t
(t +1)2 - 600
(t +1)2
& x(t) = 600
t +1
-
100e-0.2t (t + 6)
t +1
if dx
+ x
t +1
dt
= 20e-0.2t
Þ
20(t2 + 7t + 31)e-0.2t
(t +1)2 - 600
(t +1)2 + 1
t +1
æè ç
öø ÷
600
t +1
-
100e-0.2t (t + 6)
t +1
æ
è ç
ö
ø ÷
Þ
(20t2 +140t + 620)e-0.2t
(t +1)2 + -
e-0.2t (100t + 600)
(t +1)2
æ
è ç
ö
ø ÷
Þ
(20t2 + 40t + 20)e-0.2t
(t +1)2 =
20(t2 + 2t +1)e-0.2t
(t +1)2
= 20e-0.2t
Initial cond.
x(0) = 600 - 600 = 0
thisneedstobeshown.

Must have turning point. I would
have end point, it needed to be
shown in the correct square. Marks
were lost if the graph did not show
correct concavity.

(t +1)2 − 100e−0.2t (t + 6)
(t +1)2
0
10∫
= 600
⋅dt = 51.6g

AB = -a +b
= -î - 2ˆj- 2ˆk
( ) + -î + 3ˆj+ 4ˆk
( )
= -2î +ˆj+ 2ˆk
( )
a ×b = a b cosq
-2 + 2 + 4 = 3× 3cosq
Þcosq = 4
9
Must show step 2 to get mark.

Area = 1
2
a×b ×sinq
if cosq = 4
9
sinq = 65
9
Area = 1
2
3× 3× 65
9
= 65
2
units2
Many students were unable to start this problem. Either
ignoring the instruction hence, or unable to find sin(x)
given cos(x)=4/9.

I would consider an easy question but only 50% correct

Many students got the answer as 1/6 instead of-1/6

Note there is a typo in first part but does not
change the answer

Need to identify the Difference of Perfect Squares.
1ò
area = - (x2 -1) x +1
-1
×dx
1ò
area = - (x -1)(x +1) x +1
-1
×dx
-1
ò ×dx
area = (x -1)(x +1)
3
2
1
letu = x +1and x -1 = u - 2
terms: x = -1Þu = 0
x = 1Þu = 2
0ò
area = (u - 2)(u)
3
2
2
×du
3
2 æ
5
2 - 2u
0ò
area = u
è ç
ö
ø ÷
2
×du
5
2 é
area = 2
7
7
2 - 4
u
5
u
ë ê
ù
û ú
0
2
area = (0) - 2
5
2 æè ç
7
(2)
7
2 - 4
5
(2)
öø ÷
é
ë ê
ù
û ú
area = 4
5
(2)
5
2 - 2
7
(2)
7
2 = 256 2
7
- 128 2
5
area = 32 2
35

Not A, C and D
B has the same co-efficient for x2 and y2. Asymptote y=x
Only E will have non-perpendicular asymptote.

Graph the function. There are 4
intersections. Only option is E

There is a complex coefficient so the conjugate root
theorem does not apply. Minimum would be 3.

Gradient is independent of t: NOT A
Gradient for positive V is negative: NOT B or D
As V increases Abs(V) increases NOT E
Must be C
V = e-t
dV
dt
= -e-t = -V
C
OR

F = ma = m
d 1
v2 æè ç
2
öø ÷
dx
Þm ´ 1
2
v2 = òF(x)×dx
Þv2 = 2
m
òF(x)×dx
v= 2
m
òF(x)×dx ÞD

m = 80
4 - 3
from previous
åF = 0
mg sinq +mN = 20g
mg sinq +mmg cosq = 20g
m = 20 -m sin30
m cos30
= 0.07735

åF = ma
ma = mg sinq -mN
ma = mg sinq -mmg cosq
a = g sinq -mg cosq
a = 9.8sin30 - (0.077)(9.8)cos30 = 4.244ms-2
v=u +at
v= 3´ 4.244 = 12.7ms-1

p +1
2
,0
⎛
⎝ ⎜
⎞
⎠ ⎟
- p +1
2
,0
⎛
⎝ ⎜
⎞
⎠ ⎟
Note : ArSin(1) = p
2

y = Arsin(2x2 -1)
Þx2 = 1
2
(siny+1)
b
ò
Area =p x2 ×dy
a
0ò
Area =p 1
2
(siny+1)×dy
-p
2

y = Sin-1 (2x2 -1)
let u = 2x2 -1Þdu
dx
= 4x
y= Sin-1 (u)Þdy
= 1
du
1-u2
dy
dx
= dy
du
du
dx
= 4x
1-u2
= 4x
1- (2x2 -1)
2 = 4x
4x2 1- x2 ( )
= 2x
1- x2 ( )
and a = 1

Use CAS show: z1 = -10 -10i
z1
= 10 2

öø ÷ z = 200
z3 = 200
1
2cis -3p
4
æè ç
öø ÷æ
è ç
1
2cis -3p
4
æè ç
ö
ø ÷
1
3
z = 200
1
6cis -p
æè ç öø ÷
4
also 200
1
6cis -p
4
+ 2p
3
æè ç
öø ÷
= 200
1
6cis 5p
12
æè ç
öø ÷
200
1
6cis -p
4
+ 4p
3
æè ç
öø ÷
= 200
1
6cis 13p
12
æè ç
öø ÷
= 200
1
6cis -11p
12
æè ç
öø ÷

u = 2 10cis(-a )
w = 10cis(p / 2 -a )
Þ u = 2 10cis a ( )
iw = 10cis(p -a ), rotatebyp / 2
u
iw
=
2 10cis a ( )
10cis(p -a )
= 2cis(2a -p )

z1 =
(u +w)u
iw
( ) = Arg (u +w) +Arg u
Arg z1
æè çöø ÷
iw
Þ -3p
4
= Arg (u +w)+ 2a -p
Arg (u +w) = p
4
- 2a

2014 st josephs geelong spec maths

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