ST. STEPHEN’S GIRLS’ COLLEGE
Mid-Year Examination 2017 – 18
Form 5
MWC, WMC, YRK, SCHL
136 students
Mathematics
Paper I
Time allowed: 1 hour 30 minutes
Question/Answer Paper
SOLUTION
Please read the following instructions very carefully.
Class
1. Write your class, class number, name and division (if
applicable) in the spaces provided on this cover.
Class No.
Name
2.
This paper carries 70 marks. Attempt ALL questions.
Write your answers in the spaces provided in this
Question/Answer Paper.
Division
3. Graph paper and supplementary answer sheets will be
supplied on request. Write your class, class number
and name on each sheet, and fasten them with string
INSIDE this paper.
4. Unless otherwise specified, all working must be
clearly shown.
5. Unless otherwise specified, numerical answers should
either be exact or correct to 3 significant figures.
6. The diagrams in this paper are not necessarily drawn
to scale.
Page 1
Marker’s Use Only
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F.5
1.
Mathematics Paper I (Mid-Year 2017-18)
Solve 4 x  9(2 x )  22  0 .
(4 marks)
4 x  9(2 x )  22  0
(2 2 ) x  9(2 x )  22  0
(2 x ) 2  9(2 x )  22  0
By substituting 2 x  u into the above equation, we have
u 2  9u  22  0
(u  2)(u  11)  0
u  2 or u  11
Since 2  u , we have
2 x  2 or 2 x  11 (rejected)
x
2 x  21
x 1
∴
2.
2
 x  7  1.
x7
Solve
2
x7
(4 marks)
 x 7 1
2  ( x  7)  x  7
( 2  x  7) 2  ( x  7)
( x  5) 2  x  7
x 2  10 x  25  x  7
x 2  9 x  18  0
( x  6)( x  3)  0
x  6 or
∴
x  3
By checking, the solution of the equation is x = –6.
Page 2 of 11
F.5
3.


Mathematics Paper I (Mid-Year 2017-18)
Solve the following equations.
(a) 2 3x 1  54
(b) log10(3x  25)  2
4
3
(c) 48 ( x  1)  3 = 0 (x > 0)
(6 marks)
(a) 3x + 1 = 27
x+1=3
 x=2
(b) 3x + 25 = 100
3x = 75
x = 25
4
(c) 48 ( x  1) 3  3 = 0
4
( x  1) 3 = 16
x–1=8
x=9
2
4.

1 
1


Solve log 3  x    4 log 3  x    3  0 .
3 
3



(4 marks)
2

1 
1
1



By substituting log 3  x    u into log 3  x    4 log 3  x    3  0 ,
3 
3
3




we have
u 2  4u  3  0
(u  3)(u  1)  0
u  3 or u  1
1

Since log 3  x    u , we have
3

1
1


log 3  x    3
or log 3  x    1
3
3


1
1
x   3 3
or
x   3 1
3
3
1 1
1 1
x 
or
x 
3 27
3 3
10
2
x
or
x
27
3
Page 3 of 11
F.5
5.
Mathematics Paper I (Mid-Year 2017-18)
Let f(x) = 4x3 – 5x2 – 18x + c, where c is a constant. When f(x) is divided by x + 2, the
remainder is –25.
(a) Is x – 3 a factor of f(x) ? Explain your answer.
(3 marks)
(b) Someone claims that all the roots of the equation f(x) = 0 are rational numbers.
Do you agree ? Explain your answer.
(2 marks)
(a)
f (2)  25
4(2) 3  5(2) 2  18(2)  c  25
c  9
f (3)
 4(3) 3  5(3) 2  18(3)  9
0
Thus, x–3 is a factor of f(x).
(b)
f ( x)  0
4 x 3  5 x 2  18 x  9  0
( x  3)(4 x 2  7 x  3)  0
( x  3)( x  1)(4 x  3)  0
3
x  3 , x  1 or x 
4
3
Note that  1 , 3 and
are rational numbers.
4
Thus, the claim is agreed.
Page 4 of 11
F.5
6.
Mathematics Paper I (Mid-Year 2017-18)
The force between two particles at a distance r units in a solid is F units. It is given that F
is the sum of two parts, one part varies inversely as r2 and the other part varies inversely as
r4. When r = 1, F = – 5 and when r = 2, F = – 2.
(a) Find the value of F when r = 3.
(b) Find the value of r when F = 0.
(6 marks)
(a)
h
k
 4
2
r
r
 5  h  k

h k

 2  4  16
h  9, k  4
F
when r  3,
F
(b)
9 4
77
 4  
2
81
3
3
when F  0,
9 4

0
r2 r4
2
r
3
7.
It is given that f(x) is the sum of two parts, one part varies directly as x2 and the other part
varies directly as x. Suppose that f(1) = –1 and f(2) = –8.
(a) Factorize f(x).
(b) Find the coordinates of the vertex of the graph y = f(x) + 3.
(6 marks)
(a)
f( x)  hx 2  kx
 1  h  k

 8  4 h  2 k
h  3, k  2
f( x)   x(3x  2) or x(2  3x)
(b)
y  3x 2  2 x  3
2
1 1

 3 x 2  x     3
3
9 9

2
1  10

 3 x   
3
3

 1 10 
vertex   ,

3 3 
Page 5 of 11
F.5
8.
Mathematics Paper I (Mid-Year 2017-18)
Geologists claim that the energy E released from an earthquake at Richter scale R is given
by
E  a  10bR ,
where a and b are constants.
(a) Express log E as a linear function of R.
(b) Figure 1 shows the graph of log E against R.
log E
Figure 1
Slope = 1.5
4.8
R
0
Using the information in Figure 1, find the value of b and estimate the value of a
correct to the nearest integer.
(c) It is known that the 2008 Sichuan earthquake was measured at Richter scale 8 while
the 2011 Japan earthquake was measured at Richter scale 9.2. Using (b), find the ratio
of the energy released from the Japan earthquake to the energy released from the
Sichuan earthquake in the form n : 1, where n is correct to the nearest integer.
(6 marks)
(a) log E = bR + log a
(b) b = slope of the graph = 1.5
log10 a = vertical intercept = 4.8

a = 104.8 = 63096 (nearest integer)
(c) From (b), E = 63096101.5R
63096  101.5(9.2)
The required ratio 
: 1  101.5(1.2) : 1  63 : 1 (nearest integer)
1.5(8)
63096  10
Page 6 of 11
F.5
9.
Mathematics Paper I (Mid-Year 2017-18)
5  cos x
.
6 sin x cos x
(a) Show that 6 cos 2 x  cos x  1  0 .
5  cos x
(b) Hence, solve tan x 
for 0  x  360 

6 sin x cos x
(Give your answers correct to 1 decimal place if necessary.)
It is given that tan x 
(2 marks)


marks)
(a)
5  cos x
6 sin x cos x
6 sin x cos x tan x  5  cos x
tan x 
 sin x 
6 sin x cos x
  5  cos x
 cos x 
6 sin 2 x  5  cos x
6(1  cos 2 x)  cos x  5  0
6 cos 2 x  cos x  1  0
(b)
5  cos x
6 sin x cos x
2
6 cos x  cos x  1  0
(from (a))
tan x 
(2 cos x  1)(3 cos x  1)  0
cos x  
1
1
or cos x 
2
3
1
When cos x   ,
2
x  180  60 or 180  60
i.e. x  120
or 240
1
When cos x  ,
3
x  70.5
or 360  70.5
i.e. x  70.5 (cor. to 1 d. p.) or 289.5 (cor. to 1 d. p.)
 x = 70.5 (cor. to 1 d.p.), 120 , 240 or 289.5 (cor. to 1 d.p.)
Page 7 of 11
F.5
Mathematics Paper I (Mid-Year 2017-18)
10. In ABC, AD is the angle bisector of BAC, where D is a point on BC. AB = 6 cm, AC = 5 cm.
It is given that the area of ABC is 12 cm2.
(a) Find BAC.
(3 marks)
(b) Find the length of AD if AD is shorter than 3 cm.
(2 marks)
(Give your answers correct to 3 significant figures.)
(a)
1
(5)(6) sin BAC  12
2
BAC  53.13010235° or 126.8698976°
=53.1° or 127°, cor. to 3 sig. fig.
(b) Let AD  x cm.
1
126.868976 1
126.868976
(5)( x) sin(
)  (6)( x) sin(
)  12
2
2
2
2
x = 2.44, cor. to 3 sig. fig.
 AD = 2.44 cm
Page 8 of 11
F.5
Mathematics Paper I (Mid-Year 2017-18)
11. At 1 p.m., a ship sails at a speed of 30 km/h in a direction of 045° from point A. At
3:30 p.m., the ship arrives at point B. The distance between lighthouse H and point A is
88 km and the bearing of point B from the lighthouse H is 300. It is given that A, B and H
are on the same horizontal plane as shown in Figure 2.
B
N
Figure 2
H
N
A
(a) Find the distance between point B and lighthouse H.
(3 marks)
(b) The ship keeps sailing in the same direction after leaving point B and reaches point C
where BH = HC.
(i) Find the compass bearing of point C from lighthouse H.
(ii) Ken claims that the ship can reach point C before 4:30 p.m. on the same day. Do
you agree? Explain your answers.
(4 marks)
(Give your answers correct to 3 significant figures if necessary.)
(a) AB = (2.5  30) km = 75 km
ABH = 45° + 60° = 105°
AH
AB

sin ABH sin AHB
75 sin 105
sin AHB 
88
AHB  55.4096745°
BAH  180°  105° 55.4096745°
= 19.5903255°
BH
AH

sin BAH sin ABH
88 sin 19.5903255
km
BH 
sin 105
= 30.54658951 km
= 30.5 km, cor. to 3 sig. fig.
(b) (i)
(ii)
BH = HC
CBH = BCH = 75°
BHC = 180°  75°2 = 30°
 The compass bearing of point C from lighthouse H is N30°W.
AC
AH

sin AHC sin ACH
88 sin(55.4096745  30)
km
AC 
sin 75
= 90.81207832 km
Distance travelled by the ship from 1 p.m. to 4:30 p.m. = (3.5  30) km
= 105 km > AC
Yes, the claim is agreed.
Page 9 of 11
F.5
12.
Mathematics Paper I (Mid-Year 2017-18)
Figure 3.1 shows a rectangular A4 size paper card with AB = 21 cm and AD = 29.7 cm. F is a
point lying on AD such that CF  BD. BD intersects AC and CF at M and E respectively.
F
D
A
E
Figure 3.1
M
B
C
(a) Find CE, DE and EF.
(4 marks)
(b) The paper card described in (a) is folded along BD such that the plane ABD lies on the
horizontal ground as shown in Figure 3.2. It is given that AC = 23 cm.
C
A
Figure 3.2
D
B
(i) Find CDA.
(ii) Peter claims that the angle between the planes ABD and BCD is CDA, while
David claims that it should be CMA instead. Do you agree? Explain your
answers. If you do not agree, find the angle between the planes ABD and BCD,
correct to 3 significant figures.
(7 marks)
Page 10 of 11
F.5
Mathematics Paper I (Mid-Year 2017-18)
(a)
BC 29.7

DC
21
CDB  54.73698827°
tan CDB 
CE
DC
CE  21 sin 54.73698827° cm
= 17.14671978 cm
= 17.1 cm, cor. to 3 sig. fig.
sin CDB 
DE
DC
DE  21 cos 54.73698827° cm
= 12.12394328 cm
= 12.1 cm, cor. to 3 sig. fig.
cos CDB 
EF
DE
EF  12.12394328 tan (90°54.73698827°) cm
= 8.572485148 cm
= 8.57 cm, cor. to 3 sig. fig.
tan FDE 
(b) (i)
CD 2  DA2  CA 2
2CD DA
CDA  50.46153352°
= 50.5°, cor. to 3 sig. fig.
cos CDA 
(ii) Both claims are not correct, since CM (and MA) are not perpendicular to BD
and CD (and DA) are not perpendicular to BD.
The required angle is CEF.
DF  DE 2  EF 2
 14.84848485 cm
CF  CD 2  DF 2  2(CD)( DF ) cos CDA
 16.26262626 cm
CE 2  EF 2  CF 2
cos CEF 
2CE EF 
CEF  69.48532561°
= 69.5°, cor. to 3 sig. fig.
Thus, the required angle is 69.5°.
End of paper
Page 11 of 11