2 UNIT MATHEMATICS
2016 HSC EXAMINATION
SECTION I
1 sin θ > 0 and cos θ < 0.
∴ θ is in the 2nd quadrant.
1
B
2
C
3 Equation is:
!
y = 3 " (x " 2) 2
!
2
y " 3 = "(x " 2)
∴ Parabola is concave down with vertex at (2,3).
or
Equation is:
!
y = 3 " (x " 2) 2
= 3 " (x 2 " 4 x + 4)
= 3 " x2 + 4x " 4
2
= "x + 4 x "1
And:
" = b 2 # 4ac
= (4) 2 # 4(#1)(#1)
!
= 16 # 4
= 12
∴ Parabola is concave down with intercept at y = –1 and real roots.
3
B
4 Odd functions have rotational symmetry about the origin.
!
4
A
5 y = ln (cos x)
dy
"sin x
=
dx
cos x
= – tan x
5
B
6 f(x) = tan 3x
! Period = "
3
6
A
7
A
2 P(win prize) =
=
!
6
30
1
5
7 We have:
Length = r"
!
7 = 5"
" = 75
and so:
Area = 12 r 2"
= 12 # 5 2 # 75
!
= 17 12 square units
!
1
8 When cos 2x ≥ 0, we have
cos2x = 1
2x = 0°, 360°, 720°
x = 0°,180°, 360°
When cos 2x < 0, we have
"(cos2x) = 1
cos2x = "1
!
2x = 180°, 540°
x = 90°, 270°
∴ x = 0°, 90°, 180°, 270° or 360°
∴ There are 5 solutions.
or
!
8
D
9
C
10
D
∴ There are 5 solutions.
9
#
!
2
"3
x +1 dx = 12 bh + 12 bh
= 12 × 2 × 2 +
= 6 12
1
2
×3×3
!
!
10 4 + log2x = 4 × 1 + log2x
! = 4 × log!2 + log x
2
2
! = log 24 + log x
2
= log2(24 × x)
= log216x
2
2
or
Let N = 4 + log2x
So:
N " 4 = log 2 x
2 N "4 = x
2N ÷ 24 = x
N
2 ÷16 = x
2 N = 16x
log 2 2 N = log 2 16x
N log 2 2 = log 2 16x
N = log 2 16x
∴ 4 + log2x = log216x
!
SECTION
II
QUESTION 11
(a) (x – 3)2 + (y + 2)2 = 4
This represents a circle with centre (3,–2) and radius 2 units.
x +2
3x " 4
(3x " 4)(1) " (x + 2)(3)
dy
=
dx
(3x " 4) 2
(3x " 4) " 3(x + 2)
=
!
(3x " 4) 2
3x " 4 " 3x " 6
!
!
=
(3x " 4) 2
10
!
= "
(3x " 4) 2
(b) y =
!
!
3
(c)
!
!
!
(d)
!
"
1
0
(2x +1) 3 dx =
[
1
8
(2x +1) 4
1
]
0
= [ 18 (2(1) + 1)4] – [ 18 (2(0) + 1)4]
= [ 18 (3)4] – [ 18 (1)4]
= 10 18 – 18
!
= 10
!
!
! points !
(e) Solving for
of intersection:
! "!4 x
# y = "5
$
2
% y = 3 " 2x " x
!
!
!
!
x "2 ≤ 3
When x – 2 ≥ 0 (or x ≥ 2):
x "2 # 3
x #5
∴ 2 ≤ x ≤ 5.
When x – 2 < 0 (or x < 2):
"(x " 2) # 3
"x + 2 # 3
"x # 1
x $ "1
∴ –1 ≤ x < 2
∴ Solution is –1 ≤ x ≤ 5.
"5 " 4 x = 3 " 2x " x 2
x 2 " 2x " 8 = 0
(x " 4)(x + 2) = 0
x = 4 or –2
When x = 4,
y = "5 " 4(4)
= "21
When x = –2,
y = "5 " 4("2)
=3
∴ Points of intersection are (–2,3) and (4,–21).
(f) y = tan x
! dy
= sec2x
dx
When x = "8 ,
dy
= sec 2 "8
dx
1
=
2
!
cos "8
1
=
2
(cos "8 )
1
=
(cos22 12 °) 2
1
=
2
(0.9238795325)
= 1.171572875
∴ The gradient of the tangent at x =
"
8
is approximately 1.17.
4
!
!
" x%
(g) sin $ ' = 12
#2&
x
= 30° or 150°
2
x = 60° or 300°
!
!
= "3 or 53"
!
QUESTION 12
(a) (i)
Equation of BC:
y " y1 y 2 " y1
=
x " x1 x 2 " x1
y " 4 1" 4
=
x "2 6 "2
y " 4 "3
=
x "2 4
4(y " 4) = "3(x " 2)
4 y "16 = "3x + 6
3x + 4 y " 22 = 0
(ii) AD =
!
!
!
!
(iii)
!
!
(b) (i)
=
ah + bk + c
a2 + b2
3.1+ 4.0 " 22
32 + 4 2
3 + 0 " 22
=
25
"19
=
5
19
=
units
5
BC2 = (x1 – x2)2 + (y1 – y2)2
= (2 – 6)2 + (4 – 1)2
= (–4)2 + (3)2
= 25
BC = 5 units
∴ Area of ∆ABC = 12 bh
= 12 " 5 " 195
= 9 12 square units
!
∠BAO = x
∠BOA = x!
! – 2x
∠ABO = 180
∠CBO = 180 – (180 – 2x)
= 180 – 180 + 2x
= 2x
(given)
(equal angles opp equal sides)
(angle sum of ∆)
(supplementary angles)
5
(ii) ∠BCO = 2x
∠BOC = 180 – 2x – 2x
= 180 – 4x
Now:
"AOB + "BOC + "COD = 180
x + (180 # 4 x) + 87 = 180
267 # 3x = 180
#3x = #87
x = 29
(equal angles opp equal sides)
(angle sum of ∆)
(supplementary angles)
(c) Let α be the angle in the triangle opposite side 8 cm.
20 2 +15 2 # 8 2
! cos " =
2 $ 20 $ 15
561
=
600
" = 20.77185505
% 21°
Since the tile is square, we have:
θ = 90 – 21
= 69°
!
(d) (i)
(ii)
y = xe3x
dy
= x × 3e3x + e3x × 1
dx
= e3x(3x + 1)
"
2
0
e 3x (3 + 9x) dx =
"
2
0
3e 3x (3x +1) dx
2
!
3x
= 3 " 0 e (3x +1) dx
!
!
!
!
[
3x
= 3 xe
]
2
0
3(2)
= 3[(2)e ] – 3[(0)e3(0)]
= 3[2e6] – 3[0]
= 6e6
QUESTION 13
(a) (i)
y = 4x3 – x4
dy
= 12x2 – 4x3
dx
d2y
2
2 = 24x – 12x
dx
!
For turning points,
dy
= 0.
dx
12x 2 " 4 x 3 = 0
4 x 2 (3 " x) = 0
!
So:
!
!
or:
2!
4x = 0
x =0
3" x =0
x=3
6
!
d2y
When x = 0, y = 0 and
=0
dx 2
∴ Horizontal point of inflection at (0,0).
d2y
When x = 3, y = 27 and
= –36 < 0.
dx 2
!
∴ Maximum turning point at (3,27).
d2y
(ii) For points of inflection,
= 0.
dx 2
!2
24 x "12x = 0
12x(2 " x) = 0
So:
12x = 0 !
x =0
! or:
2"x =0
x =2
! When x =, y = 0.
d2y
When x = –0.1,
2 = –2.52 < 0
dx
!
d2y
When x = 0.1,
= 2.28 > 0
dx 2
∴ Horizontal
point of inflection at (0,0).
!
When x = 2, y = 16.
d2y
When!x = 1.9,
= 2.28 > 0
dx 2
d2y
When x = 2.1,
= –2.52 < 0
dx 2
∴ Point of inflection at (2,16).
!
!
7
(b) (i)
Equation of parabola is:
x 2 " 4 x = 12y + 8
x 2 " 4 x + 4 = 12y + 8 + 4
(x " 2) 2 = 12y +12
2
(x " 2) = 12(y +1)
2
(x " 2) = 4.3.(y +1)
∴ Vertex is (2,–1) and focal length is 3 units
(ii) Focus is (2,2)
!
M(t) = Ae–kt
When t = 0, M = 10.
10 = Ae "k(0)
10 = Ae 0
A = 10
∴ M(t) = 10e–kt
(ii) When t = 163, M = 5.
5 = 10e "k(163)
!
e "163k = 0.5
"163k = ln0.5
ln0.5
k=
"163
= 0.00425243669
= 0.0043
∴ M(t) = 10e–0.0043t
(c) (i)
(d) Shaded area = Area under curve – Area of triangle
!
# "x &
)1
= + 2 cos % ( dx – 12 bh
*0
$4'
1
)4 2
# "x &,
= +
sin % (. – 12 × 1 × 1
"
$ 4 '- 0
*
!
!
)4 2
&
# " &, # 4 2
= +
sin % ( . – %
sin (0) ( – 12
$ 4 '- $ "
* "
'
!
$4 2
'
!
1 ' $4 2
= &
#
# 0 ) – 12
)– &
2( % "
% "
!(
4
!
!
=
– 0 – 12
"
8
"
!
=
#
!
2" 2!"
8 "#
=!
square units
!
2#
!
!
8
QUESTION 14
(a) The increase in area can be approximated by looking at the difference in the heights.
Using Simpson's rule:
x
y
W
P
–2
0
1
0.0
–1
0.40
4
1.6
0
0.5
2
1.0
1
0.40
4
1.6
2
0
1
0.0
4.2
1
Area ≈ 3 × 1 × 4.2
≈ 1.4 square metres
(b) (i)
!
Let An be the number of insects remaining after the nth day.
Each day the number of insects is reduced by 35%. This means that 65% of the starting
number of insects will remain.
After 1 day, number of insects remaining before new insects are produced
= 0.65 × 100000
After 1 day, number of insects remaining after new insects are produced
A1 = 0.65 × 100000 + 5000
After 2 days, number of insects remaining before new insects are produced
= 0.65 × [0.65 × 100000 + 5000]
After 2 days, number of insects remaining after new insects are produced
A2 = 0.65 × [0.65 × 100000 + 5000] + 5000
= 0.65 × 0.65 × 100000 + 0.65 × 5000 + 5000
= 100000(0.65)2 + 5000(0.65) + 5000
(ii) After 3 days, number of insects remaining before new insects are produced
= 0.65 × [100000(0.65)2 + 5000(0.65) + 5000]
= 100000(0.65)3 + 5000(0.65)2 + 5000(0.65)
After 3 days, number of insects remaining after new insects are produced
A3 = 100000(0.65)3 + 5000(0.65)2 + 5000(0.65) + 5000
∴ After n days, number of insects remaining after new insects are produced
An = 100000(0.65)n + 5000(0.65)n–1 + ... + 5000(0.65)2 + 5000(0.65) + 5000
2
n "1
= 100000(0.65)n + 5000 [1+
0.65
+ 0.65
+ ...+
0.65
!#
##
#
#"#
##
##
$]
GP: a =1, r =0.65, n terms
# a(1 " r n ) &
= 100000(0.65) + 5000 %
(
$ 1" r '
#1(1 " 0.65 n ) &
!n
= 100000(0.65) + 5000 %
(
$ 1 " 0.65 '
#1 " 0.65 n &
!
= 100000(0.65)n + 5000 %
(
$ 0.35 '
(iii) After 14 days, (when
! n = 14),
#1 " 0.6514 &
14
A14 = 100000(0.65) + 5000 %
(
$ 0.35 '
!
n
= 240.3183829 + 14251.38309
= 14491.70147
∴ After 14 days there are approximately 14 500 insects remaining.
!
9
(c) (i)
!
!
Let y be the length of the enclosure.
Since the area is 720 m2,
A = lb
720 = xy
720
y=
x
Now, the total length of fencing is:
l=x+x+x+x+x+y
720
= 5x +
!
x
(ii) l = 5x + 720x–1
dl
= 5 – 720x–2
dx
! d 2l
= 1440x–3
dx 2
dl
d 2l
For minimum length,
= 0 and
> 0.
dx
dx 2
5 " 720x "2 = 0
720x "2 = 5
"2
1
!x = 144 !
x 2 = 144
x = 12
When x = 12,
720
l = 5(12) +
12
!
= 60 + 60
= 120
d 2l
"3
= 1440(12)
dx 2
= 56
!
>0
∴ The minimum length of fencing required is 120 metres.
(d) Factorising:
! x5 – 1 = x5 – 15
= (x – 1)(x4.10 + x3.11 + x2.12 + x1.13 + x0.14)
= (x – 1)(x4 + x3 + x2 + x1 + x0)
= (x – 1)(x4 + x3 + x2 + x1 + 1)
Therefore:
5
4
3
2
lim x #1 lim (x #1)(x + x + x + x +1)
x "1 x #1 = x "1
x #1
lim
4
3
= x "1 (x + x + x 2 + x +1)
= 14 +13 +12 +1+1
=5
(e) log 2 + log 4 + log 8 + ... + log 512 = log 2 + log 22 + log 23 + ... + log 29
= log 2 + 2 log 2 + 3 log 2 + ... + 9 log 2
!
= log 2 × (1 + 2 + 3 + ... + 9)
= 45 log 2
∴ a = 45 and b = 2
10
QUESTION 15
(a) For C1:
Equation of curve is:
x2 + y2 = 4
y2 = 4 " x2
#
=π#
Volume = π
!
!
!
0
"2
0
"2
y 2 dx
4 " x 2 dx
[
3
= π 4 x " 13 x
]
0
"2
= π[4(0) – 13 (0)3] – π[4(–2) – 13 (–2)3]
= π[0 – 0] – π[–8 + 83 ]
= 163" cubic units
!
For C2:
!
!
Equation of curve is:
!
x2 y2
!
+
=1
9
4
2
y
x2
=1"
4
9
2
= 1 " 19 x
y 2 = 4(1 " 19 x 2 )
= 4 " 49 x 2
"
=π#
Volume = π
!
!
[
3
0
3
0
y 2 dx
4 " 49 x 2 dx
3
= π 4 x " 274 x
]
3
0
= π[4(3) – 274 (3)3] – π[4(0) –
!
= π[12 – 4] – π[0 – 0]
= 8π cubic units
!
∴ Total volume = 163" + 8π
!
= 403" cubic!units
4
27
(0)3]
!
!
11
(b) (i)
If the game ends before the fourth roll, then the game ends on either the 1st, 2nd or 3rd
roll. An 8 must be rolled on the 1st, 2nd or 3rd roll.
P(8 on 1st roll) = 18
P(8 on 2nd roll) = 78 " 18
P(8 on 3rd roll) = 78 " 78 " 18
∴ P(game ends before fourth roll) = 18 + 78 " 18 + 78 " 78 " 18
!
2
= 18 + 78 " 18 + ( 78 ) " 18
!
49
= 18 + 647 + 512
!
169
! ! = 512 !
(ii) We need the probability of the
game
ending before the nth roll to be greater than 43 .
!
!
!
1
7
1
7 2
1
7 n #2
1
3
+
"
+
"
+
...+
"
>
(
)
(
)
8
8
8
8
8
! 8 ! !8 4
1
7
7 2 !
7 n #2
> 43
8 1+ 8 + ( 8 ) + ...+ ( 8 )
!
2
n #2
1+ 78 + ( 78 ) + ...+ ( 78 ) > 6
[
]
2
n #2
+ ( 7 ) + ...+ ( 78 ) > 5
!#8##"##
#
$
7
8
GP:a = 78 , r = 78 , n #2 terms
a(1 # r n )
>5
1# r
7
7 n #2
)
8 (1 # ( 8 )
>5
7
1# 8
7
7 n #2
)
8 (1 # ( 8 )
>5
1
8
(1 # ( 78 ) n #2 ) > 58
1 # ( 78 ) n #2 > 57
#( 78 ) n #2 > # 27
( 78 ) n #2 < 27
log( 78 ) n #2 < log 27
(n # 2)log 78 < log 27
log 27
n #2 >
log 78
n # 2 > 9.381786139
n > 11.381786139
n = 12,13,14,...
∴ The probability that the game ends before the 12th roll will be greater than
7
8
!
!
12
3
4
.
(c)
(i)
ABCD is a square
CD || BA
∠CFB = ∠ABF
∠FCB = ∠BAT
∴ ∆FCB is similar to ∆BAT
(given)
(opp sides of square)
(alt angles in || lines)
(angles in square)
(two pairs of equal corr angles)
(ii)
Let α = ∠STA
∠TAS = 180 – α – 90 (angle sum of ∆)
= 90 – α
∠DAB = 90°
∠BAE = 180 – 90 – (90 – α)
=α
∴ ∠STA = ∠EAB
∠TSA = ∠AEB
∴ ∆TSA is similar to ∆AEB
(iii) From (i):
FC CB
=
BA AT
x
1
=
1 AT
1
AT =
x
!
13
(angle in square)
(supp angles)
(as shown)
(given)
(two pairs of equal corr angles)
From (ii):
TS
=
AE
h
=
y
!
AT
BA
AT
1
h
AT =
y
Therefore:
h 1
=
y x
y
h=
x
QUESTION 16
!
4
(a) v = 2 –
t +1
(t +1)(0) " (4)(1)
a =–
(t +1) 2
4
! =
(t +1) 2
x = 2t – 4 ln (t + 1) + C
!
(i) When t = 0,
4
v =2"
!
0 +1
4
=2"
1
= "2
∴ The initial velocity of the particle is –2 m/s.
(ii) Particle is stationary when v = 0.
4
2"
=0
!
t +1
4
2=
t +1
2(t +1) = 4
2t + 2 = 4
2t = 2
t =1
When t = 1,
4
a=
(1+1) 2
!
4
= 2
2
=1
∴ When the particle is stationary the acceleration is 1 m/s2.
!
14
(iii) When t → ∞,
t+1→∞
4
→0
t +1
4
2–
→2
t +1
∴ As t increases indefinitely, v approaches 2 m/s.
!
!
(iv) Since the particle is stationary at t = 1, it changes direction at that time.
When t = 0,
x = 2(0) – 4 ln (0 + 1) + C
= 0 – 4 ln 1 + C
= 0 – 4(0) + C
=C
When t = 1,
x = 2(1) – 4 ln (1 + 1) + C
= 2 – 4 ln 2 + C
When t = 7,
x = 2(7) – 4 ln (7 + 1) + C
= 14 – 4 ln 8 + C
= 14 – 4 ln 23 + C
= 14 – 4 × 3 ln 2 + C
= 14 – 12 ln 2 + C
Distance travelled from t = 0 to t = 1
= (2 " 4 ln2 + C) " (C)
= 2 " 4 ln2 + C " C
= 2 " 4 ln2
= 4 ln 2 – 2
!
Distance travelled from t = 1 to t = 7
!
= (14 "12ln2 + C) " (2 " 4 ln2 + C)
!
= 14 "12ln2 + C " 2 + 4 ln2 " C
= 12 " 8ln2
= 12 – 8 ln 2
!
∴ Total distance travelled = 4 ln 2 – 2 + 12 – 8 ln 2
!
= (10 – 4 ln 2) metres
!
15
200
1+19e "0.5t
dy (1+19e "0.5t )(0) " (200)("9.5e "0.5t )
=
dt
(1+19e "0.5t ) 2
"0.5t
1900e
!
=
(1+19e "0.5t ) 2
(ii) When t = 0,
200
y=
"0.5(0)
1+19e
!
200
=
0
1+19e
200
=
1+19
= 10
As t → ∞,
–0.5t → –∞
e–0.5t → 0
!
1 + 19e–0.5t → 1
200
"0.5t → 200
1+19e
∴ The range of the function is 10 ≤ y < 200.
(iii) Rearranging the equation for y gives:
200
y=
!
1+19e "0.5t
"0.5t
y(1+19e ) = 200
y +19ye "0.5t = 200
19ye "0.5t = 200 " y
200 " y
e "0.5t =
19y
# 200 " y &
"0.5t = ln %
(
$ 19y '
# 200 " y &
t = "2 ln %
(
$ 19y '
# 200 " y & "1
= 2 ln %
(
$ 19y '
# 19y &
= 2 ln %
(
$ 200 " y '
(b) (i)
y=
!
16
And so:
$ (200 # y)(19) # (19y)(#1)
dt
19y '
=2 "&
÷
)
2
dy
200 # y (
(200 # y)
%
$19(200 # y) +19y 200 # y '
=2 "&
"
)
19y (
(200 # y) 2
%
$ 3800 #19y +19y
1 '
=2 "&
"
)
200 # y
19y (
%
$ 3800
1 '
=2 "&
"
)
% 200 # y 19y (
7600
=
19y(200 # y)
400
=
y(200 # y)
Therefore:
dy y(200 " y)
=
dt
400
!
y
=
(200 " y)
400
(iv) To find when the population is growing at its fastest rate, we look for the maximum
value of the rate.
Let R(y) be the rate of population growth.
!
y(200 " y)
R(y) =
400
200y " y 2
=
400
2
! R '(y) = (400)(200 " 2y) " 2(200y " y )(0)
400
400(200 " 2y)
!
=
160000
800(100 " y)
=
!
160000
100 " y
=
!
200
(200)("1) " (100 " y)(0)
!R"(y) =
200 2
"200
=
!
40000
1
=–
!
200
For a maximum rate we need R'(y) = 0 and R"(y) < 0 (which is true).
!100 " y
=0
200
!
100 " y = 0
y = 100
∴ When there is a population of 100 yabbies, the population is increasing at its fastest
rate.
!
17