1. PROGRAM DIDIK CEMERLANG AKADEMIK
SPM
MODEL SPM QUESTIONS ( PAPER 1 )
ORGANISED BY:
JABATAN PELAJARAN NEGERI PULAU PINANG
ADDITIONAL
MATHEMATICS
MODULE 19
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2. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 2
3472/1 NO. KAD PENGENALAN
Additional
Mathematics
Paper 1 ANGKA GILIRAN
2 Hours
JABATAN PELAJARAN NEGERI PULAU PINANG
ADDITIONAL MATHEMATICS
Paper 1
Two Hours
JANGAN BUKA KERTAS SOALAN INI
SEHINGGA DIBERITAHU
1. Tuliskan angka giliran dan nombor kad
pengenalan anda pada ruang yang
disediakan.
2. Calon dikehendaki membaca arahan
di halaman 2.
Kod Pemeriksa
Questions Marks Actual Marks
1 2
2 3
3 3
4 3
5 3
6 3
7 3
8 3
9 3
10 2
11 4
12 4
13 4
14 2
15 2
16 3
17 4
18 4
19 4
20 4
21 3
22 4
23 3
24 3
25 4
Total
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3. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 3
INFORMATION FOR CANDIDATES
1. This question paper consists of 25 questions.
2. Answer all questions.
3. Give only one answer for each question.
4. Write your answer clearly in the spaces provided in the question paper.
5. Show your working . It may help you to get marks .
6. If you wish to change your answer , cross out the work that you have done. Then write down the
new answer.
7. The diagrams in the questions provided are not drawn to scale unless stated.
8. The marks allocated for each question and sub-figure mathematical tables is provided .
9. You may use a non-programmable scientific calculator .
10.A booklet of four-figure mathematical tables is provided.
11.You may use a non –programmable scientific calculator.
12.This question paper must be handed in at the end of examination.
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4. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 4
Answer all questions
1 . Diagram 1, the function h maps x to y and the function k maps y to z.
DIAGRAM 1
Determine
(a) hk –1
(– 4),
(b) kh –1
(– 2). [2 marks]
Answer : (a) ___________
(b) ___________
2. The function p–1
is defined as p–1
(x) =
x
x
4
3
, x ≠ 4.
Find
(a) p(x),
(b) p(5). [3 marks]
Answer : (a) ___________
(b) ___________
x y z
4
3
2
h
k
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5. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 5
3. The following information refers to the functions h and g.
Find f (x). [3 marks]
Answer : ___________
4. The straight line y = t intercept with the curve y = 3 + 5x – 4x2
at the two points A and B .
Find the range of values of t . [3 marks]
Jawapan : ___________
5. Solve the quadratic equation
5
2
3
1
2 2
x
x
Give your answer correct to three decimal
places . [3 marks]
Answer : x = ___________
g (x) = 4 – 3 x
fg (x) = 2 x + 5
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6. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 6
6. Diagram 2 shows the graph of a quadratic functions g(x) = 4 – 2(x + h)2
where h is constant.
DIAGRAM 2
The curve y = g(x) has maximum point (3, k) , where k is a constant .State
(a) the value of h,
(b) the value of k,.
(c) the equation of the axis of symmetry .
[3 marks]
Answer : (a) h = ___________
(b) k = ___________
(c) ___________
7. Solve the equation .
)
4
)(
8
(
1
64 1
3
x
x
x
[3 marks]
Answer : x = ___________
y = g(x)
●
(3,k)
x
y
O
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7. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 7
8. Solve the equation log2 (x – 1) + 2 = 2 log2 x [3 marks]
Answer : x = ___________
9. Given that log2 x = p and log2 y = r, express
2
3
2
32
log
y
x
in term p and r.
[3 marks]
Answer : ___________
10. The sum of the first n term of an arithmetric progression is given by Sn= 5n – 3. Find the
fifth term . [2 marks]
Answer : ___________
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8. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 8
11. The first three terms of an arithmetic progression are –
2
1
, 1, –2 , …
Find
(a) the common ratio ,
(b) the sum of the first 10 terms after the 5th term .
[ 4 marks]
Answer : (a) ___________
(b) ___________
12. The sum of the first n terms of an arithmetric progression 47, 44, 41, … is –1325. Find
(a) the common difference of the progression ,
(b) the value of n. [4 marks]
Answer : (a) ______________
(b) n = ___________
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9. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 9
13. Diagram 3 shows a straight line graph of log10 y againts log10 x.
The variables x and y are related by equation y = p x q
, where p and q are contants .
DIAGRAM 3
(a) Calculate the value of p and q,
(b) Find the value of y if x = 10. [4 marks]
Answer : (a) p = ___________
q = ___________
(b) y =___________
14. The following information refers to the equations of two straight lines , AB and CD , which
parallel to each other.
Express p in terms q. [2 marks]
Answer : p =___________
AB : 2y = p x + q
CD : 3y = (q + 1) x + 2
Where p and q are constants
log10 y
log10 x
2
4 O
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10. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 10
15. Given that B point is (–8, 5) and O is origin.
(a) Express
OB in terms of
~
i and
~
j .
(b) Find the unit vector in the direction of
OB . [2 marks]
Answer : (a) ___________
(b) ___________
16. Given that
OA = –2
~
i + 3
~
j ,
OB = 10
~
i + 6
~
j and R is a point on AB such that AR :
AB = 2 : 3. Find
(a)
AB ,
(b)
OR . [3 marks]
Answer : (a) ___________
(b) ___________
17. Solved the equation 3
cot
1 2
2
x
sek
x
for 0 ≤ x ≤ 360.
[4 marks]
Answer : x = ___________
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11. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 11
18. Diagram 4 shows the sector OPQ, centre O with a radius of 5 cm . Line PR is perpendicular to
the line OQ and QR = 1 cm.
[4 marks]
Find ( a ) POR in radians ,
( b ) the perimeter of the shaded region .
Answer : ( a ) ………………….
( b ) ………………….
19. Given that 2
)
2
5
(
4
)
( x
x
x
f
, find )
(
" x
f [4 marks]
Answer : ___________
RAJAH 4
Q
P O
R
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12. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 12
20. Given that x = t + 3t 2
and y = 2t – 1
(a) find
dy
dx
in term of t.
(b) If y decreases from 4.0 to 3.98, Find the corresponding small change in t .
[4 marks]
Answer : (a) ___________
(b) ___________
21. Given that 2
3
2
1
x
x
y
and )
(
4 x
f
dx
dy
with )
(x
f is a function in x .
Calculate the value of dx
x
f )
(
3
2
2
. [3 marks]
Answer : ___________
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13. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 13
22. A chess team consists that of 5 students . The team will be chosen from a group of 6 boys
and 4 girls . Calculate the number of teams that can be formed such that each team consists
of
(a) 4 boys,
(b) Not more than 2 girls.
[4 marks]
Answer : (a) ___________
(b) ___________
23. The mean of five numbers is m. The sum of the squares of the numbers is 720 and the
standard deviation is 9h2
. Express m in term of h.
[3 marks]
Answer: m = ___________
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14. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 14
24. Ali has 6 accessories of the item which consists of P, Q, R, S, T and U. He wants to arrange
4 of the items in a row . Find the probability that
(a) the arrangment is TUPS ,
(b) the arrangment does not incule P item
[3 marks]
Answer : (a) ___________
(b) ___________
25. X is a random variable of normal distribution with a mean of 10 and a variance 9 , find the
value of r such that P ( X < r ) = 0 . 9 7 5.
[4 marks]
Answer : ___________
END OF QUESTION PAPER
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15. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 15
Answers
1. ( a ) h k – 1
(– 4) = – 2
( b ) k h – 1
(– 2) = – 4
2 . ( a ) p – 1
( x ) =
x
x
4
3
p – 1
( y ) = y
y
4
3
x =
y
y
4
3
x ( 4 – y ) = 3 y
4 x – x y = 3 y
y ( x + 3 ) = 4 x
y =
3
4
x
x
p ( x ) =
3
4
x
x
, x ≠ –3,
( b ) 3
5
)
5
(
4
)
5
(
p
=
2
5
3 . g ( x ) = 4 – 3 x
y = 4 – 3 x
3 x = 4 – y
x =
3
4 y
g – 1
(x) =
3
4 x
f ( x ) = f g g – 1
( x ) = f g
3
4 x
= 2 5
3
4
x
=
3
2
23 x
4 . y = t , y = 3 + 5 x – 4 x 2
3 + 5 x – 4 x 2
= t
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16. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 16
4 x 2
– 5 x + t – 3 = 0
Intercept at two points , use b 2
– 4 a c > 0
( – 5 ) 2
– 4 ( 4 ) ( t – 3 ) > 0
1 6 t < 7 3
t <
16
73
5 . 5 ( 2 x – 1 ) = 3 ( x 2
– 2 )
3 x 2
– 10 x – 1 = 0
Use x =
a
ac
b
b
2
4
2
x =
)
3
(
2
)
1
)(
3
(
4
)
10
(
)
10
( 2
x =
6
112
10
x = 3.43, x = – 0. 0972
6 . g (x) = 4 – 2 ( x + h ) 2
x + h = 0 , x = – h ; y = 4
( – h , 4 ) = ( 3 , k )
(a) h = – 3
(b) k = 4
(c) symmetry Axis : x = 3
7 . 2 6 ( x+3 ) x ½
= 2 – 3x
. 2 – 2 ( x + 1 )
3 ( x + 3 ) = – 3 x + [ – 2 x – 2 ]
8 x = – 11
x =
8
11
8 . log 2 ( x – 1 ) + 2 = 2 log 2 x 2 = log 2 22
= log 2 4
log 2 4 ( x – 1 ) = log 2 x 2
4 x – 4 = x 2
x 2
– 4 x + 4 = 0
( x – 2 ) 2
= 0
x = 2
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17. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 17
9 . 2
2
3
2
2
2
3
2 log
log
32
log
32
log y
x
y
x
= y
x 2
2
5
2 log
2
log
3
2
log
= 5 + 3 p – 2 r
10 . S n = 5 n – 3
T n = S n – S n – 1
T 5 = S 5 – S 4
= [ 5 ( 5 ) – 3 ] – [ 5 ( 4 ) – 3 ]
= 22 – 17
= 5
11 . (a) r =
1
2
= – 2
(b) T 1 , T 2 , ……… T 5 , T 6 , ……….. T 15
the sum of = S 15 – S 5
S 15 =
1
)
2
(
1
)
2
(
2
1 15
= – 5461.5 ;
S 5 =
1
)
2
(
1
)
2
(
2
1 5
= – 5.5
S 15 – S 5 = – 5456
12 . 47, 44, 41, …………
( a ) d = 4 4 – 4 7 = – 3
( b ) S n = ]
)
1
(
2
[
2
d
n
a
n
)]
3
)(
1
(
)
47
(
2
[
2
n
n
= –1325
3 n 2
– 9 7 n – 2 6 5 0 = 0
( 3 n + 5 3 ) ( n – 5 0 ) = 0
n = 50
13 . y = p x q
( a ) log 10 y = log 10 p + q log 10 x
log 10 p = 2
p = 10 2
= 100
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18. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 18
q =
2
1
4
0
0
2
( b ) log 10 y = 2 + ½ log 10 x
x = 10 , log 10 y = 2 + ½ log 10 10
log 10 y = 2.5
y = 10 2.5
y = 316 . 2
Alternatif method
y = p x q
= 100 x ( 10 ) ½
= 316 . 2
14 . y =
2
2
q
x
p
y =
3
2
3
1
x
q
3
1
2
q
p
p = )
1
(
3
2
q
15 . ( a )
OB =
~
~
5
8 j
i
( b ) the unit vector in the direction of
OB =
89
5
8 j
i
16 . ( a )
AB =
AO +
OB
= –
3
12
6
10
3
2
( b )
OR =
1
2
)
(
1
)
(
2
OA
OB
OR =
3
2
1
6
10
2
1
2
1
OR =
5
6
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19. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 19
17.
x
2
cot
1
. + sek 2
x = 3
tan 2
x + sek 2
x = 3
tan 2
x + 1 + tan 2
x = 3
2 tan 2
x = 2
tan 2
x = 1
tan x = ± 1
x = 45 o
, 135 o
, 225 o
, 315 o
18 . ( a ) Cos POR =
5
4
= 0 . 8
POR = 0 . 6435 r
( b ) s PQ = j r
= 5 x 0.6435 = 3 . 2175
RQ = 1 cm , PR = 3 cm ( Pythagoras Teorem )
The perimeter of the shaded region = 3 + 1 + 3. 2175
= 7 . 2175
19 . f ( x ) = 2
)
2
5
(
4 x
x
u = 4x v = ( 5 – 2 x ) 2
4
dx
du
)
2
)(
2
5
)(
2
(
x
dx
dv
dx
du
v
dx
dv
u
dx
dy
)
4
(
)
2
5
(
)]
2
5
(
4
[
4
)
(
' 2
x
x
x
x
f
= 4 8 x2
– 160 x + 100
160
96
)
(
'
'
x
x
f
20 . (a) x = t + 3 t 2
y = 2 t – 1
t
dt
dx
6
1
2
dt
dy
)
6
1
(
2
1
2
1
)
6
1
(
t
t
dy
dt
dt
dx
dy
dx
(b) 4
98
.
3
y
= -0.02
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20. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 20
2
dt
dy
t
y
dt
dy
dt
dy
y
t
01
.
0
2
02
.
0
21 .
dx
x
f
dy )
(
4
dx
x
f
dy )
(
3
4
3
2
2
2
2
2
3
2
1
4
3
)
(
3
x
x
dx
x
f
=
12
4
1
12
4
1
4
3
=
2
1
22 . ( a ) 6
20
2
4
3
6
C
C
= 120
( b ) Total of player boys and girls = 6 + 4 = 10
The number of ways of selecting from 5 player = 252
5
10
C
The number of ways of selecting player consisting of 1 boy and 4 girls
= 6
1
6
4
4
1
6
C
C
The number of team is less than 2 boys = 2 5 2 – 6 = 2 4 6
23 . m
x ; 720
2
x ; 2
2
9h
2
2
2
x
n
x
9 h 2
= 2
5
720
m
m 2
= 144 – 9h2
m =
2
9
144 h
24 . ( a ) The probability of arrangement is TUPS
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21. Additional Mathematics ( Paper 1) SPM
Jabatan Pelajaran Pulau Pinang 21
=
360
1
3
1
4
1
5
1
6
1
( b ) The probability does not included the P accessory
=
4
6
4
5
C
C
=
3
1
25 . variance = 2
= 9
Standard deviation = = 3
mean = = 10
P ( X < r ) = 0 . 975
Z =
X
P
3
10
3
10 r
X
= 0 . 975
P
3
10
r
Z = 0 . 975
1 – P
3
10
r
Z = 0 . 975
P
3
10
r
Z = 1 – 0. 975
= 0 . 025
3
10
r
= 1 . 9 6
r = 1 5 . 8 8
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