2007 Catholic Trial HSC Examinations Mathematics Solutions
Page 1
Catholic Schools Trial Examinations
2007 Mathematics Solutions
1a.
25 559 000 = 2.5559 × 107
2a.(iii)
= 2.6 × 107 (2 dec pl)
1b.
5 +2+
1
5 +2
0+5
∴
x −0
5 −2
×
5 −2
=
5 +2+
5 −2
5−4
=
5 +2+
5 -2
1d.
5x + sin x + c
1e.
x–2 ≤ 3
x ≤ 5
=-
3
x = -3
2a.(iv)
∴ D(-3, 0)
Grad AD =
(3a − 1)
(a − 2)
3(3a − 1) − 4(a − 2)
=
4
3
12
9a − 3 − 4a + 8
=
12
5a + 5
=
12
3
5
15 = -5x
=2 5
1c.
5
grad CD = -
y–4=
4−0
4
2
=
=
3+3
6
3
2
(x – 3)
3
3y – 12 = 2x + 6
∴ 2x – 3y + 18 = 0
2a.(v) C(0, -5) and AD: 2x – 3y + 6 = 0
per dist =
| 2(0) − 3(−5) + 6 |
22 + (−3)2
or –(x – 2) ≤ 3
or
x – 2 ≥ -3
=
21
13
x ≥ -1
∴ -1 ≤ x ≤ 5
Area = base × height
2a.(vi)
3 – 2:
= 42 units
o
2b.(ii) ∠ LMK = 42 (base ∠ s of isos. ∆ KML)
o
4y = 20
y=5
∠ LKM = 96 ( ∠ sum of ∆ )
2b.(iii) ∠ MKJ = 42o (alt ∠ s, JK||ML)
o
Subs in 1: 3x + 5 = 6
3x = 6 - 5
3x = 1
1
3
1
∴ x=
and y = 5
3
∠ JKM = 42 (base ∠ s of isos. ∆ JKM)
o
∠ MJN = 84 (ext ∠ of ∆ result)
dy
3a.(i)
= 4(ex + 3)3.ex
dx
x=
2a.(i)
BC =
36 + 16
=
52
= 2 13 units
2a.(ii)
4 +1
3−6
5
5
=
=−3
3
m =
= 4ex(ex + 3)3
f ’(x) =
3a.(ii)
(6 − 0)2 + (−1 + 5)2
=
13
2
6x – 2y = -8 …………..….. 2
2 × 1: 6x + 2y = 12 ………………. 3
21
= 2 13 ×
3x + y = 6 ……………..…. 1
1f.
units
=
3b.(i)
x2
∫ x3 − 5
dx =
=
tan x.2 x − x2. sec2 x
tan2 x
2 x tan x − x 2 sec2 x
tan2 x
1
3
3x 2
∫ x3 − 5
dx
1
ln(x3 – 5) + c
3
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2007 Catholic Trial HSC Examinations Mathematics Solutions
2
3b.(ii)
∫
e− x
dx = − e

−x
0
π
π
dy π
( ) = -sin( )ecos 2
dx 2
2
2

0
= -1.e0 = -1
-2
= -(e
= -(
0
–e )
1
∴ grad of normal = 1
- 1)
e2
=1-
Now, y(
2
π
) = ecos 2
= e0 = 1
e2
π
∴ y – 1 = 1(x -
6
∴ y–1 =x -
= 19
p=
π
1
p2 = 82 + (3 3 )2 – 2(8)(3 3 ).cos
3c.
Page 2
19
4d.
3d.
tan θ =
∴θ =
and θ lies in 2nd quad.
4a.
a = 15
4
5
a
1−r
15
=
= 24
1−r
π
2
π
2
3
5π
11π
,
6
6
α+β=
∴ 4αβ
2
−9
= -3
3
= 4.
4b.(i) In ∆ s ABC and DCE
∠ ACB = ∠ DEC (alt ∠ s equal, AC||ED)
∠ ABC = ∠ DCE (alt ∠ s equal, AC||ED)
∴ ∆ ABC ||| ∆ DCE (2 ∠ s equal)
∴
x
8
=
12 + x
22
22x = 8(12 + x)
22x = 96 + 8x
14x = 96
96
6
=6
14
7
6
6
∴ BC = 12 + 6
= 18 cm
7
7
dy
4c.
= -sin x . ecos x
dx
x=
1
.-3
3
2
5a.(i) f(x) = x – x – 8x – 3
f ‘(x) = 3x2 – 2x – 8 = 0
(3x + 4) (x - 2) = 0
−4
,2
3
−4
−4 3
−4 2
−4
Now f(
) =(
) -(
) - 8(
)
3
3
3
3
14
=3
27
x=
and f(2) = (2)3 - (2)2 - 8(2)
Let CE = x
4b.(ii)
1
3
= -4
3
9
3
=
24
8
αβ =
+ 4 α2β = 4αβ (α + β )
15 = 24 – 24r
r=
, 1)
)
15 = 24(1 – r)
24r = 9
2
For 3x2 + 9x + 1 = 0
4e.
S∞ =
24r = 24 – 15
π
3 tanθ + 1 = 0
−1
By pythag, BC = 4
∴ cos θ = -
∴ (
= -7
−4
14
stat pts at (
,3
) and (2, -7)
3
27
Now, f ‘’(x) = 6x – 2
∴ f ‘’(
−4
)<0
3
and f ‘’(2) > 0
∴ max (
−4
14
,3
)
3
27
∴ min (2, -7)
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2007 Catholic Trial HSC Examinations Mathematics Solutions
Page 3
1
4
5a.(ii)
∑
6a.(i)
w2 = 0 + 1 +
2 +
3 + 2
w =0
=3+
2 +
3
6b.(i) 1 + 1 + 3 + 9 + …
n=5
5
∴ 1 – 2(5) + 2 = 23
n = 6 ∴ 1 – 2(6) + 26 = 53
n = 7 ∴ 1 – 2(7) + 27 = 115
∴ 115 is a term
6b.(ii) 1 + 1 + 3 + 9 + 23 + 53 + 115 + …
separate into arithmetic : 1 - 2n
5a(iii) f ‘(x) = (3x + 4) (x - 2) < 0
−4
<x< 2
3
∴
P(select B ) =
5b.(i)
: -1, -3, -5, …
a = -1, d = -2, n = 25
and geometric: 2
n
: 2, 4, 8, …
∴ S25 =
1
2
1
4
3
×
×
2
11
10
3
=
55
= 67 108 237
6b.(iii)
67 108 237 ÷ 2 500 000
= 26.8 drops/cm2
= 1 – P(no green from B)
1
7
6
×
×
2
11
10
6c.(i) y: 2.773; 5.727; 9.888; 15.346; 22.181
6c.(ii)
89
110
0.5
[2.773 + 22.181 +
3
4[5.727 + 15.346] + 2[9.888]]
5c.(i) Subs (2, 16) in y = 20 – 2x
16 = 20 – 2(2) yes
16 = 2(2)3 yes
∴ (2, 16) is point of intersection.
2
∫
0
∫
θ = 1.8 radians
(20 − 2 x ) dx
2
reflex ∠ = 2π - 1.8 radians)
1
6d.(ii) A =
2
2
 x4 
10
=   + 20 x − x 2 

2
 2 
0
=
22.5 = 12.5θ
(note: diagram is not accurate :
10
2 x3 dx +
= 21.504
6d.(i) l = rθ
Subs (2, 16) in y = 2x3
5c.(ii) A =
250m2 = 250 × 10 000
= 2 500 000cm2
5b.(ii) P(at least 1 green from B)
=
25
1(225 − 1)
[2(-1) + 24 × (-2)] +
2
2 −1
= -625 + 67 108 862
∴ P(B then 2 greens) =
=1-
a = 1, r = 2, n = 25
θ (r12 – r22)
=
24
- 0 + 20(10) – (10)2 - 20(2) + (2)2
2
1
1.8 (92 – 3.52)
2
= 61 875cm2
= 8 + 200 – 100 – 40 + 4
= 72
2
∴ 72 units
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2007 Catholic Trial HSC Examinations Mathematics Solutions
Page 4
P = 106 × ekt
7a.(i)
dP
= k(106 × ekt)
dt
A3(1.0075)33 = 178967.7174
= kP
A3 = 178967.7174 ÷ (1.0075)33
6
6
10k
1.7 × 10 = 10 × e
7a.(ii)
= 139858.1858
1.7 = e10k
10k
ln(e
But A3 = P(1.005)3 - 12 060.10
) = ln1.7
∴ P = (139858.1858 + 12 060.10) ÷ (1.005)3
10k = ln1.7
k =
= $149 662.11
ln 1.7
10
8a.
= 0.053 [3 dec pl]
7a.(iii)
1 [use calc to verify]
8b.(i)
x = 42 – 4loge (4 – 1) + 5
= 16 – 4loge3 + 5
∴ t = 27
6
= 21 – 4loge3
27k
∴ P = 10 × e
= 4 180 000 [3 sig figs]
7a.(iv)
1(1.007533 − 1
]=0
1.0075 − 1
= A3(1.0075)33 – 4800[
8b.(ii)
8.5 × 106 = 106 × ekt
v = 2t -
4
=0
t −1
2t(t – 1) – 4 = 0
8.5 = ekt
2t2 – 2t – 4 = 0
ln(ekt)
= ln8.5
t2 – t – 2 = 0
kt
= ln8.5
(t – 2) (t + 1) = 0
ln 8.5
t =
k
t = 2, as t ≥ 0
as v = 2t – 4(t – 1)-1
8b.(iii)
= 40.3
a = 2 + 4(t – 1)-2
∴ during 2020
=2+
7b.(i) A1 = P(1.005) – 4000
7b.(ii) A1 = P(1.005) – 4000
8b.(iv)
Subs t = 4 and t = 2 into x:
= [P(1.005) – 4000] × 1.005 – 4000
x(4) = 42 – 4loge (4 – 1) + 5
2
= P(1.005) – 4000[1 + 1.005]
= 21 – 4loge3
A3 = A2 × 1.005 – 4000
x(2) = 22 – 4loge (2 – 1) + 5
2
= P(1.005) – 4000[1 + 1.005] × 1.005
– 4000
= P(1.005)3 – 4000[1 + 1.005 + 1.0052]
=9
∴ distance = 21 – 4loge3 – 9
= 12 – 4 loge3
1
1
8c.(i) As y =
,1–x=
1− x
y
1
x=1y
7b.(iii) A4 = A3 × 1.075 – 4800
∴ A5 = A4 × 1.075 – 4800
= [A3 × 1.075 – 4800] × 1.0075 – 4800
2
= A3(1.0075) – 4800[1 + 1.0075]
∴ A36 = A3(1.0075)
=0
Now when t > 1, a > 0
A2 = A1 × 1.005 – 4000
33
4
(t − 1)2
2
∴ Vy =
{a = 1, r = 1.0075, n = 33}


y
1
2
– 4800[1 + 1.0075 + …
+ 1.007532] = 0
2

π ∫ 1 − 1  dy
=π
2
1
∫ 1 − y + y2
dy
1
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2007 Catholic Trial HSC Examinations Mathematics Solutions
2
Vy = π
8c.(ii)
∫
2
1
+
dy
y y2
1−
1
=
4 a2 + 4 − 2a + (8 + 3π )
hours
24
1
2
=π
Page 5
2
∫1 − y + y
−2
dy
−
1
4. (a2 + 4) 2 .2a − 2
dt
2
=
=0
da
24
9b.(ii)
1
1
4a(a + 4) 2 − 2
2

1
= π y − 2 ln x − 
y 1

1
= π[2 – 2ln2 - (1 – 0 - 1)]
2
24
−
g(-x) =
=
1
a2 + 4
= 0.357 [3 dec pl]
g(x) =
=0
4a(a2 + 4) 2 = 2
2a
= π[1.5 – 2ln2]
9a.(i)
−
2
2
=1
a2 + 4
2a =
x2 − 1
4a2 = a2 + 4
2
3a2 = 4
(− x )2 − 1
2
as a > 0,
x2 − 1
= g(x) ∴ even function
9a.(ii) All real except x = ± 1
9b.(i) amplitude = 3
9b.(ii)
a =
2
3
dt
Now check neighbourhood of
:
da
2
dt
(
- ∈) < 0
da
3
2
dt
(
) =0
da
3
2
dt
(
+ ∈) > 0
da
3
∴ minimum when a =
10a.(i)
2
3
dV
= -bt
dt
V=-
bt 2
+c
2
t = 0, V = 250 000,
9b.(iii) From the graph, 2 solutions
9c.(i) By Pythag, SX2 = a2 + 42
SX =
a2 + 4
XY = 4 – a
FY =
1
. 2π(2)
4
=π
Using Time =
D
S
∴ 250 000 = 0 + c
∴ c = 250 000
bt 2
∴ V= + 250 000
2
1
= 250 000 - bt2
2
1
10a.(ii) 85 000 = 250 000 - (0.431)t2
2
1
2
(0.431)t = 165 000
2
t2 = 765 661.2529
T =
a2 + 4
4−a
π
+
+
6
12
8
=
4 a2 + 4 + 2(4 − a) + 3π
24
t = 875.02
dV
Now,
= -bt
dt
= -(0.431) 875.02 = -377.1 L/min
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2007 Catholic Trial HSC Examinations Mathematics Solutions
Page 6
10b.(i) Using Pythag: BD2 = 152 + 152
BD = 15 2
∴ BD is 15 2 cm
10b.(ii) Now, ∠ BDC =
π
4
(diagonals of square bisect ∠ )
π
BE
Now in ∆ DBE, sin(θ +
)=
4
∴ BE = 15 2 sin(θ +
10b.(iii) in ∆ AGD,
15 2
π
4
)
AG
= cosθ
15
∴ AG = 15cosθ
AF
and in ∆ AFB,
= sinθ
15
∴ AF = 15sinθ
as FG = AG + AF
∴ FG = 15(sinθ + cosθ )
10b.(iii) As FG = BE,
π
15(sinθ + cosθ ) = 15 2 sin(θ +
∴ sinθ + cosθ = 2 sin(θ +
π
4
4
)
)
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