Add Maths Paper 1

PROGRAM DIDIK CEMERLANG AKADEMIK
SPM
MODEL SPM QUESTIONS ( PAPER 1 )
ORGANISED BY:
JABATAN PELAJARAN NEGERI PULAU PINANG
ADDITIONAL
MATHEMATICS
MODULE 19
http://mathsmozac.blogspot.com
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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

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.

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

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

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

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 : ___________

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 = ___________

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

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 = ___________

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

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 : ___________

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 = ___________

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

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

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

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

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

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

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

=
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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