Download 1 (a) HCF of 360 and 84 = 4 × 3 = 12 (b) k = 3 × 5

1
(a) HCF of 360 and 84 = 4  3 = 12
(b) k = 3  52 = 75
2
(a) 2.8466
2.84646

2.84615
 Answer : 2.8 4 6
2.846846

 
2.84 6
2.8 4 6

2
9
3
11
13
(b) simple interest = 12500 × 12 × 0.002
= $ 18.75
(a) the greatest possible value of p  q = (–4)  (–1) = 4
(b) the smallest possible value of
𝑝 +𝑞
𝑞
𝑝
9
= the smallest possible value of ( 𝑞 + 1) = −1 + 1 = –8
(c) the smallest possible value of ( p 2  q 2 ) = 0 2 + (–1)2 = 1
4
(a) The cost of the handbag in $ = 23 000  64.1 = $ 358.81  $ 359
(b) The difference in the price between the two countries in $
= 414.40 – 358.81 = $ 55.59  $ 56
5
6
7
8
(a)
(i) 0.002177359
(ii) 0.002
(b)
– 17134913.82  – 17130000
= – 1.713 × 107
(a)
1 : 250 000 = 1 cm : 250 000 cm
= 1 cm : 2.5 km
= (1  28) cm : (2.5  28) km
= 28 cm : 70 km
Actual Distance = 70 km
(b)
(1cm)2 : (2.5 km)2 = 1 cm2 : 6.25 km2
37.5
37.5
= (1  6.25) cm2 : (6.25  6.25) km2
= 6 cm2 : 37.5 km2
 map area = 37.5 km2
(a)
When x = 0, y = –18,
(b)
y=
(a)
ac + 2bc – 3 ad – 6 bd = c (a + 2b) – 3d (a + 2b)
= (c – 3d) (a + 2b)
(b)
2x2 – 32 = 2(x2 – 16) = 2(x – 4)(x + 4)
3𝑥
2
– 18,
 P is (0, –18)
 gradient = 3/2
[Turn over
9
Depth of liquid
(d cm)
d2
d1
d1
0
2
4
6
8
Time (minutes)
10
12
10
(a)
Bearing of B from C = 86 + 50 = 136°
11
(b)
(a)
Area = 2 × 5 × 4 × 𝑠𝑖𝑛50° = 7.6604 ≈ 7.66 𝑚2
Difference = 24o C – (–12o C) = 36o C
(b)
(i)
1
(8 – 5)
the temperature at 0800 = – 12o + 36𝑜 × (13.5 − 5)
= 0.705 o C
 0.7 o C
(ii) the time when the temperature is 0o C
12 𝑜
= 0500 + (13.5 – 5) × 36 𝑜
= 0500 +
17
6
170
= 0500 + 60
= 0500 + 2 hr 50 min
= 0750
12 (a) & (b)
E
F
S
𝐹 ∪𝑆
𝐹 ∪ 𝑆 ′ 𝑜𝑟 𝐹 ′ ∩ 𝑆′
13
y  2 5 y  2x
1

(a) sub x = − 8 into
3
3x  1
(c)(i)
(ii)
y2

3
5 y  2( 
1
)
8
1
) 1
8
11
11
3
 y   15 y 
8
4
4
Page | 18
3( 
AHS Prelim 2010 Sec 4 Mathematics Paper 1
131
y2
8
16
y
131
(b)
y  2 5 y  2x

3
3x  1
( y  2)(3x  1)  3(5 y  2 x)
y(3x  1)  6 x  2  15 y  6 x
y(3x  1)  15 y  2
y(3x  16)  2
2
y
or
3x  16
14 (a) & (b)
2
16  3x
Q
C
A
B
(c)
(d)
15 (a)
The point of intersection is the centre of the given circle.
ACB = 90o
DBE = 36o/2 = 18o
DFE = 36o
FEC = DBE + DFE (ext  of BEF)
= 18o + 36o = 54o
A
(b)
H
F
D
36
C
G
E
B
(c)
Construct lines FG and GH
2 more isosceles triangles can be drawn
HFG = (18 + 54)o = 72o
FHG = 72o
HGC = 18o + 72o = 90o
No further triangles can be drawn.
[Turn over
y = (x + 3)(x – 4)
y = – (x + 3)(x – 4)
16 (a)
and
(There are other possible answers)
y
y = – (x + 3)(x – 4)
12.25
12
–3
–
12
– 12.25
(b)
17 (a)
4
O
x
y = (x + 3)(x – 4)
x = 0.5
 CBA = 180o – CDA (s in opp segment)
= 180o – 79o
= 101o
(b)
 FCB = 180o – BAC – CBA (s sum of ABC)
= (180 – 21 – 79)o
= 80o
(c)
CFE = FEB (alt. s, CF // EB)
= ½  FCB ( at centre = 2at circumference)
= 40o
𝐴𝑋 = 𝐴𝐵 + 𝐵𝑋
1
6
2
=
+ 3
=
0
4
18 (a) (i)
𝐴𝑋
(ii)
=
42 + 42 =
4
4
32 ≈ 5.66 𝑢𝑛𝑖𝑡𝑠
𝐵𝐸 = 𝐵𝐴 + 𝐴𝐸
3 4
−2
1
=
+ 4
=
−4
4
−1
(iii) 𝐸𝐷 = 𝐸𝐴 + 𝐴𝐷
3 4
6
= −4
+
0
4
3
=
−3
1
=3
−1
Since 𝐵𝐸 𝑎𝑛𝑑 𝐸𝐷 have common vector
collinear.
Page | 20
AHS Prelim 2010 Sec 4 Mathematics Paper 1
1
from same point E, therefore B, E and D are
−1
(b)
𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐸
𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐷
=
=
 𝐴𝑟𝑒𝑎
19 (a)(i)
1
𝐴𝐵×𝐵𝐸𝑠𝑖𝑛 ∠𝐴𝐵𝐷
2
1
𝐴𝐵×𝐵𝐷𝑠𝑖𝑛
∠𝐴𝐵𝐷
2
1
𝐵𝐷
𝐵𝐸
1
4
𝐵𝐷
=
𝐵𝐷
𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐸
𝑜𝑓 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚 𝐴𝐵𝐶𝐷
=4
=
𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐸
2×𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐷
=
1
8
Median =13.5 min
(ii)
Interquartile range = 16 – 12 = 4 min
(iii) No of competitors = 18
4
(iv) Percentage = 32 × 100 = 12.5%
(b)
The lower quartile or median or upper quartile for the 2008
competition is higher than
the 2009 competition. This shows that the quality of the competitors in the 2009 batch
seems to be better than the 2008 batch.
The inter-quartile range for the 2008 competition is wider than the 2009 competition.
This shows that the 2009 competitors are relatively more competitive.
20 (a)
(b)
(c)
(d)
p5 = 52 + 62 + 302 = 25 + 36 + 900 = 961
k = (s + 1)2
r = 5256
w (w + 1) = 5256 = 72 × 73
 w = 72
pn = n2 + (n + 1)2 + [n(n +1)]2
= n2 + n2 + 2n + 1 + n2(n + 1)2
= 2n2 + 2n + 1 + n2(n2 + 2n + 1)
= 2n2 + 2n + 1 + n4 + 2n3 + n2
= n4 + 2n3 + 3n2 + 2n + 1.
[Turn over
AHS 2010 Prel EM P2 Answers
1ai)
1b)
2ai)
2bi)
3aii)
3biii)
4ci)
5ii)
6i)
75 km/h
1aii)
1c)
2aii)
2bii)
3bi)
4a)
4cii)
5iii)
6ii)
1 h 45 min
1aiii) 71 km/h
7bi))
8a)
9bi)
–3 ½
4/(5 – y)
8
26
2aiii) 60
9
2biii) 2
0≤𝑥≤6
15.0 cm
20%
3bii) $ 560
40%
7 hrs 55 min
4bii) 1 hr 15 min
2, 3, 5, 7
5i)
3.38 cm
½<x9
34.1o
11.6 cm
6vi) Shop B
6 2
35.80 35.80
6iv)
96.30 96.00
4 3
35.20 34.70
5 1
25.30 25.50
The elements in R represent the total cost of buying the CDs and paper at shop A or shop
B for Andrew, Betty and Charles respectively.
The elements in T represent the total spent at Shop A or Shop B by Andrew, Betty and
Charles.
1/5
7bii) 5/13
7c)
119/1235
135o
8b)
112.5o
8d)
10.6 cm2
9 cm
9biii) 12 10 cm
10i) 60o
10ii)
10v)
11d)
176 m
64.5o
16.7
6iii)
6v)
5
10iii) 19.8o
11a) –3.25
11e) –0.6 , 4
Anglican High School Preliminary Examinations 2010
11
10iv) 144.5o
11c) 3.25
Mathematics Paper 2
AHS 2010 Prel EM P2 Solution
1
(a)
A train travelled a distance of 375 km in 5 hours from Station A to Station B. After
taking a break of 45 minutes, it then travelled a further distance of 157.5 km at an
average speed of 90 km/h.
Calculate:
i)
its average speed for the first part of the journey,
[1]
ii) the time taken, in hours and minutes, for the second part of the journey, [1]
iii) its average speed for the whole journey.
[3]
(b) Solve the equation
1
32−𝑥 =
3
[2]
35𝑥 +1
(c) Express the following as a single fraction in its simplest form.
5+𝑦
25 − 𝑦
3
−
2
[3]
𝑦 −5
Answer:
1a i) average speed for the first part of the journey
ii)
time taken for the second part of the journey
iii) average speed for the whole journey
=
= 375  5
= 75 km/h
= 157.5  90
= 1¾ h
= 1 h 45 min
375  157.5
3
3
5  1
4
4
= 532.5  7.5
= 71 km/h
Q1b)
32−𝑥 =
3
2−𝑥
=
1
3
3
5𝑥
1
3−( 3 + 3)
2–x=−
2𝑥
35𝑥 +1
5𝑥
=−
x=−
3
7
−
1
3
3
7
2
Anglican High School Preliminary Examinations 2010
12
Mathematics Paper 2
5+𝑦
Q1c)
25 − 𝑦
2 −
3
𝑦 −5
=
5+𝑦
5−𝑦 (5+𝑦)
=
=
=
=
2
(a)
+
3
5−𝑦
5 + 𝑦 + 3(5+𝑦)
5−𝑦 (5+𝑦)
20 + 4𝑦
5−𝑦 (5+𝑦)
4(5 + 𝑦)
5−𝑦 (5+𝑦)
4
5−𝑦
A soccer club has 150 members. X is the set of strikers. Y is the set of defenders.
The letters a, b and c in the Venn diagram represent the number of members in
each of the subsets of X and Y. The letter d represents the numbers of members
who are neither strikers nor defenders. Given that n(X) = 90 and n(Y) = 68, find
(i)
the value of b if d = 0,
[1]
(ii)
the value of d if b = c,
[1]
(iii) the largest possible number of members who are neither strikers nor
defenders.
[1]
2a i)

When d = 0, 90+ 68 – b = 150
b=8
X
a
ii) When b = c, b = c = (68÷ 2) = 34
Therefore, a – b + b + c + d = 150
90 + 34 + d =150
d = 26
Y
b
c
d
iii) Largest possible value of members who are neither strikers nor
defenders is when Y becomes a proper subset of X.
Therefore d + a = 150
Largest value of d = 150 – 90 = 60
Anglican High School Preliminary Examinations 2010
13
Mathematics Paper 2
2
(b)
The table below shows the number of pens own by each student in a class.
Number of pens
Number of students
0
1
1
3
2
x
3
7
4
5
(i) If the mode of the distribution is 3, find the range of values of x.
(ii) If the median is 3, what is the largest value of x.
(iii) If the mean is 2.9, find the value of x.
2bi)
ii)
5
2
[1]
[1]
[2]
0≤𝑥≤6
1+3+𝑥+7+5+2+1
2
19 + 𝑥
𝑥
≤ 7+5+2
≤ 28
≤9
Therefore, largest value of x is 9.
iii)
0+3+2𝑥+21+20+10
= 2.9
1+3+𝑥+7+5+2
54+2𝑥
18+𝑥
= 2.9
54 + 2𝑥 = 52.2 + 2.9𝑥
0.9𝑥 = 1.8
𝑥=2
Anglican High School Preliminary Examinations 2010
14
Mathematics Paper 2
3
(a)
B
The diagram shows the arcs AB and EF
of a circle, centre O, with radius r cm.
E
BCX, OP and EDY are straight lines that
perpendicular to the line segment AXPYF.
Given that AX =XY =YF = 5cm,
CDYX is a rectangle with CX = 4 cm.
i)
Show that r = 8.5 cm
[1]
ii)
Calculate the arc AB.
[4]
O
C
A
X
P
D
Y
Solution
(a)
i)
r = 𝐴𝑃2 + 𝑂𝑃2
= (5 + 2.5)2 + 42
= 8.5 cm
ii)
(shown)
COA =  OAP
4
= tan – 1
7.5
= 28.07o
 BOC = cos –1
2.5
8.5
= 72.89o
AOB = AOC +  COB
= 28.07o + 72.89o
= 100.96o
Arc AB =
100.96𝑜
360 𝑜
 2  (8.5)
= 14.97
 15.0 cm
Anglican High School Preliminary Examinations 2010
15
Mathematics Paper 2
F
3
(b)
The cost price of a particular model of camera is $300. Shop A advertises its sale
as “usual price $450 now $360”. Shop B sells the same model of camera at $420
after 25% discount.
Calculate
3b)
i)
i)
the discount given by shop A as a percentage,
[2]
ii)
the original marked price of the camera at shop B,
[1]
iii)
the profit gained by shop B as a percentage.
[2]
Discount =
450−360
450
 100%
= 20 %
ii)
Original marked price =
420
75%
= $ 560
iii)
Percentage of profit =
420−300
300
 100%
= 40%
4
(a)
It takes 3 hours and 10 min for 5 workers to paint a hall. Given that all the workers
work at the same rate, how long will 2 workers take to paint the same hall?
[Leave your answer in hours and minutes]
4(a)
=
No. of workers
5
2
: No. of hours
1
:
3
=
5
=
2
:
=
2
:
=
2
:
5
Anglican High School Preliminary Examinations 2010
[2]
6
19
:

6
95
5
2
12
475
60
7 hours 55 min
16
Mathematics Paper 2
3(b)
Mr Hai takes x minutes to carve a swan out of a block ice and Mr Li takes 25 more
minutes to carve an identical sculpture. If both of them work together, it only takes
30 minutes to complete carving an identical sculpture.
i)
Form an equation involving x, and show that it simplifies to
x2 – 35x – 750 = 0.
ii)
[2]
Determine the time taken by Mr Li to carve the sculpture on his own.
[3]
1
4(b)
𝑥
+
1
=
𝑥 + 25
2𝑥+25
𝑥(𝑥 + 25)
=
1
30
1
30
60 x + 750 = x2 + 25x
x2 – 35x – 750 = 0
(shown)
(x + 15)( x – 50) = 0
x = – 15 (rej)
or
x = 50
Time taken by Mr Li = 50 + 25
= 75 min
= 1 hr 15 min
4
(c)
4 (c) i)
i)
Solve – 4 < 2x – 5  13
[2]
ii)
Write down all the prime numbers which satisfies this condition.
[1]
– 4 < 2x – 5  13
1 < 2x  18
½ < x 9
ii) Prime numbers are 2, 3, 5 & 7
Anglican High School Preliminary Examinations 2010
17
Mathematics Paper 2
5
The diagram shows a solid in which ABCD, DCFE and ABFE are rectangles. G is the
foot of perpendicular from D to AE. Given that AB = 5 cm, AD = 8 cm, DE = 5.5 cm
and angle DAE is 25°.
A
G
E
o
25
5 cm
5.5 cm
8 cm
D
B
F
Find,
(i)
5
C
the length of DG,
[1]
(ii) angle DCG,
[2]
(iii) the length of AE,
[4]
(i)
𝐷𝐺
8
= sin 25o
𝐷𝐺 = 8 × 𝑠𝑖𝑛25°
= 3.3809
≈ 𝟑. 𝟑𝟖 𝒄𝒎
ii) In ΔDCG, tan DCG =
𝐷𝐺
𝐷𝐶
=
3.3809
5
∠𝐷𝐶𝐺 = 34.065𝑜
≈ 𝟑𝟒. 𝟏°
Anglican High School Preliminary Examinations 2010
18
Mathematics Paper 2
iii) AG = 82 − 3.38092
= 7.250 cm
GE = 5.52 − 3.38092
= 4.338 cm
AE = AG + GE
= 7.250 + 4.338
= 11.588
 11.6
6
cm
Two shops A and B sell the same brand of compact discs (CDs) and printing paper. Each
box of CDs is sold at $3.70 and $3.80 at Shop A and Shop B respectively. Each ream of
paper is sold at $6.80 and $6.50 at Shop A and Shop B respectively. The prices of the
items are represented by the matrix P,
𝑷=
$3.70
$6.80
$3.80
$6.50
Andrew plans to buy 6 boxes of CDs and 2 reams of paper. Betty plans to buy 4 boxes of
CDs and 3 reams of paper. Charles plans to buy 5 boxes of CDs and 1 ream of paper.
(i)
Represent the quantities of CDs and papers to be bought by the 3 persons as
a matrix Q.
[1]
(ii)
Evaluate the product, R = QP
[2]
(iii)
State what the elements of R represent.
[1]
(iv)
Evaluate T = (1
[1]
(v)
State what the elements of T represent.
(vi)
If Andrew, Betty and Charles combine their purchase, which shop will give them a
better deal.
[
1
1) R.
Anglican High School Preliminary Examinations 2010
19
[1]
Mathematics Paper 2
6i) Q =
6
4
5
2
3
1
6 2
3.70 3.80
ii) 𝑹 = 4 3
6.80 6.50
5 1
6 × 3.7 + 2 × 6.8 6 × 3.8 + 2 × 6.5
= 4 × 3.7 + 3 × 6.8 4 × 3.8 + 3 × 6.5
5 × 3.7 + 1 × 6.8 5 × 3.8 + 1 × 6.5
35.80 35.80
= 35.20 34.70
25.30 25.50
iii) The elements in R represent the total cost of buying the CDs and paper at shop A
or shop B for Andrew, Betty and Charles respectively.
iv) 𝑻 = 1
1 1
35.80 35.80
35.20 34.70 = 96.30 96.00
25.3
25.5
v) The elements in T represent the total spent at Shop A or Shop B by Andrew, Betty
and Charles.
vi) Shop B
7
A bowl of sweets contains 12 chocolate nuggets, 13 mints and 15 toffees. Mr Chan takes
two sweets at random and eats them.
(a)
Draw a tree diagram to represent all the possible outcome.
(b)
Calculate the probability, as a fraction in its simplest form, that
(c)
[3]
i) a chocolate nugget and a mint was taken.
[1]
ii) none of the two sweets chosen were toffees.
[1]
A third sweet was randomly taken from the bowl. Calculate the probability that the
first two sweets drawn were of the same type and the third was a chocolate nugget.
[2]
Anglican High School Preliminary Examinations 2010
20
Mathematics Paper 2
7a)
11
C
39
13
C
12
M
39
T
15
39
40
C
12
39
12
M
13
M
39
15
40
T
39
15
12
40
C
39
13
T
M
39
14
T
39
12
bi) P(that to get chocolate and mint) =
40
13
×
bii) P(none of the two sweets are toffees) =
+
39
12
40
×
13
40
24
39
×
+
12
39
13
40
×
=
24
39
1
5
=
5
13
(c) P(two sweets of the same type and the third a chocolate nugget)
12
11
=
×
=
119
1235
40
39
×
10
38
+
13
40
×
12
39
×
12
38
+
15
40
Anglican High School Preliminary Examinations 2010
×
14
39
×
21
12
38
Mathematics Paper 2
8.
The diagram shows two regular octagons meet at points A and B.
P
A
Q
C
B
(a)
Find the size of each interior angle of the octagons.
[1]
(b)
Find  PAQ.
[2]
(c)
Show that
(d)
8
𝐴𝐵
𝐴𝐶
= 1.85, correct to 3 significant figures.
[3]
Hence or otherwise, find the area of the smaller octagon if the area of the bigger
octagon is 36.2 cm2.
(a)
Interior angle = 180o – 360o/8 = 135o
(b)
 CAB = (180 – 135)o/2 = 22.5o
[3]
 PAQ = 360o – 135o – (135o – 22.5o)
= 112.5o
Anglican High School Preliminary Examinations 2010
22
Mathematics Paper 2
𝐴𝐵
(c)
sin 135 𝑜
𝐴𝐵
𝐴𝐶
=
=
𝐴𝐶
sin 22.5𝑜
sin 135 𝑜
sin 22.5𝑜
= 1.8477
 1.85
(d)
The two octagons are similar objects,
area of the larger octagon
area of the smaller octagon
=(
𝐴𝐵 2
)
𝐴𝐶
 area of the smaller octagon =
= 1.852
36.2
1.85 2
= 10.577
 10.6 cm
y
9.
The diagram shows a circle, centre O and radius 3 cm.
An inscribed triangle ABC is
A
drawn with A and B are on
C
the y-axis.
AC is extended to meet the
D
E
x
O
x-axis at D and OD meets
BC at E.
B
(a)
Prove that triangles BOE and DOA are similar.
(b)
If OE = 1 cm,
(i)
find OD,
(ii)
show that AC =
[3]
[2]
3 10
5
[3]
cm,
(iii) hence find the exact length of DC.
Anglican High School Preliminary Examinations 2010
23
[2]
Mathematics Paper 2
Solution
9. (a) BOE = DOA = 90o (axes are perpendicular)
ACB = 90o ( rt  in a semicircle)
ABC = 180o – 90o – BAC ( sum of )
= 90o – BAC
ADO = 180o – 90o – BAC
= 90o – BAC
 ABC = ADO
By AAA property, triangles BOE and DOA are similar.
9
(b) (i)
Since triangles BOE and DOA are similar,
𝑂𝐷
𝑂𝐴
=
𝑂𝐵
𝑂𝐸
𝑂𝐷
3
=
3
1
 OD = 9 cm
(ii)
ABC is similar to EBO,
𝐴𝐶
𝐸𝑂
𝐴𝐶
1
=
𝐴𝐵
=
6
 AC =
=
𝐸𝐵
1+9
6 10
10
3 10
5
cm
DC = AD – AC
= 9 + 81 –
= 3 10 –
=
12 10
5
3 10
5
3 10
5
cm
Anglican High School Preliminary Examinations 2010
24
Mathematics Paper 2
10
T
N
O
N
15O
P

W
40O
R
200 m
O
280
150 m
O, P , W and R are points on a horizontal plane. A vertical memorial tower, OT, is due
north of P. The angle of elevation of a man at P to the top of the tower is 15°. He walks a
distance of 200 metres to point W. Given that the bearing of W from P is 040° and the
bearing of O from W is 280°, calculate
i)
∠PWO,
[2]
ii)
OP,
[2]
iii)
the height of the tower OT,
[2]
iv) the angle of depression of W from T.
[2]
From W, the man walks 150 metres to a point R on a bearing of 𝜃° on the horizontal
plane. At R, the angle of elevation of the man to the top of the tower is 10°.
Calculate
v)
angle OWR,
[3]
vi)
the value of 𝜃.
[1]
Solution
Anglican High School Preliminary Examinations 2010
25
Mathematics Paper 2
10 i)
ii)
∠𝑂𝑊𝑁 = 360 − 280
= 80°
∠𝑃𝑂𝑊 = 80° 𝑎𝑙𝑡 𝑎𝑛𝑔𝑙𝑒
∠𝑃𝑊𝑂 = 180 − 80 − 40 = 60°
𝑂𝑃
200
=
sin 60
𝑠𝑖𝑛 80
200
𝑂𝑃 = 𝑠𝑖𝑛60 ×
𝑠𝑖𝑛 80
= 175.87
≈ 𝟏𝟕𝟔𝒎
iii)
iv)
Height of Tower = 𝑂𝑃 𝑡𝑎𝑛15°
= 47.124
≈ 𝟒𝟕. 𝟏𝒎
𝑂𝑊
𝑠𝑖𝑛 40 𝑜
=
200
sin 80 𝑜
𝑂𝑊 = 200 × 𝑠𝑖𝑛40𝑜 ÷ sin 80𝑜
≈ 130.54
47.124
130.54
= 19.849° ≈ 19.8°
𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑑𝑒𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑜𝑓 𝑊 𝑓𝑟𝑜𝑚 𝑇 = 𝑡𝑎𝑛−1
v)
47.124
𝑂𝑅
= tan 10o
𝑂𝑅 = 47.124 ÷ 𝑡𝑎𝑛10°
 267.25
130.542 + 1502 − 267.252
cos ∠𝑂𝑊𝑅 =
2 × 130.54 × 150
∠𝑂𝑊𝑅 = 144.50 𝑜
≈ 𝟏𝟒𝟒. 𝟓°
vi) Value of  = 144.50o – 80o
= 64.50o
 64.5o
Anglican High School Preliminary Examinations 2010
26
Mathematics Paper 2
11. Answer the whole of this question on a sheet of graph paper.
The variables x and y are connected by the equation
𝑦 = 3 2𝑥 − 4.
Some corresponding values of x and y are given in the following table.
x
–2
–1
0
1
2
3
4
y
a
–2.5
–1
2
8
20
44
(a)
Find the value of a.
[1]
(b)
Using a scale of 2 cm to 1 unit, draw a horizontal x–axis for –2 ≤ x ≤ 4.
Using a scale of 2 cm to 5 units, draw a vertical y–axis for –5 ≤ y ≤ 45.
On your axes, plot the points given in the table and join them with a smooth curve.
[3]
(c)
Use your graph to find the solution of 2𝑥 = 9.
[2]
(d)
By drawing a tangent, find the gradient of the curve at the point (3, 20).
[2]
(e)
Use your graph to find the solutions of the equation 3 2𝑥 − 10𝑥 − 8 = 0.
[3]
Anglican High School Preliminary Examinations 2010
27
Mathematics Paper 2
11. (a)
𝑎 = 3 2−2 − 4 = –3.25
(b)
(c)
2𝑥 = 9
3(2𝑥 ) − 4 = 3 9 − 4
3(2𝑥 ) − 4 = 23
From the graph, when y=23, x= 3.25
(d)
(e)
gradient of the curve at the point (3, 20)  16.7
3 2𝑥 − 10𝑥 + 8 = 0
3 2𝑥 = 10𝑥 + 8
3 2𝑥 − 4 = 10𝑥 + 4
Insert 𝑦 = 10𝑥 + 4,
From the graph,
 x  – 0.6, 4
Anglican High School Preliminary Examinations 2010
28
Mathematics Paper 2