SEKOLAH BERASRAMA PENUHBAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH / KLUSTERKEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN PERCUBAANSIJIL PELAJARAN MALAYSIA 2008
Kertas soalan ini mengandungi 15 halaman bercetak
3472/2 2008 Hak Cipta SBP [Lihat sebelah SULIT
For examiner’s use only
Question Total MarksMarks
Obtained1 32 33 34 35 36 37 38 39 310 411 412 413 414 315 216 317 418 319 320 321 322 323 324 425 3
TOTAL 80
MATEMATIK TAMBAHAN
Kertas 1 Dua jam
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1 This question paper consists of 25 questions. 2. Answer all questions. 3. Give only one answer for each question. 4. Write your answers 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-part
of a question are shown in brackets. 9. A list of formulae is provided on pages 2 to 3. 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
the examination .
Name : ………………..……………
Form : ………………………..……
3472/1Matematik TambahanKertas 1Ogos 20082 Jam
http://tutormansor.wordpress.com/
SULIT 3472/1
The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.
ALGEBRA
1 2 4
2
b b acx
a
2 am an = a m + n
3 am an = a m - n
4 (am) n = a nm
5 loga mn = log am + loga n
6 loga n
m = log am - loga n
7 log a mn = n log a m
8 logab = a
b
c
c
log
log
9 Tn = a + (n-1)d
10 Sn = ])1(2[2
dnan −+
11 Tn = ar n-1
12 Sn = r
ra
r
ra nn
−−=
−−
1
)1(
1
)1( , (r ≠ 1)
13 r
aS
−=∞ 1
, r <1
CALCULUS
1 y = uv , dx
duv
dx
dvu
dx
dy +=
2 v
uy = ,
2vdx
dvu
dx
duv
dy
dx−
= ,
3 dx
du
du
dy
dx
dy ×=
4 Area under a curve
= ∫b
a
y dx or
= ∫b
a
x dy
5 Volume generated
= ∫b
a
y 2π dx or
= ∫b
a
x 2π dy
3472/1 2008 Hak Cipta SBP [ Lihat sebelahSULIT
5 A point dividing a segment of a line
( x,y) = ,21
++
nm
mxnx
++
nm
myny 21
6 Area of triangle =
)()(2
1312312133221 1
yxyxyxyxyxyx ++−++
1 Distance = 221
221 )()( yyxx −+−
2 Midpoint
(x , y) = +
221 xx
, +
221 yy
3 22 yxr +=
4 2 2ˆ
xi yjr
x y
+=+
GEOMETRY
http://tutormansor.wordpress.com/
SULIT 3472/1
STATISTIC
3472/1 2008 Hak Cipta SBP [ Lihat sebelahSULIT
3
1 Arc length, s = rθ
2 Area of sector , L = 21
2r θ
3 sin 2A + cos 2A = 1
4 sec2A = 1 + tan2A
5 cosec2 A = 1 + cot2 A
6 sin 2A = 2 sinA cosA
7 cos 2A = cos2A – sin2 A = 2 cos2A - 1 = 1 - 2 sin2A
8 tan 2A = A
A2tan1
tan2
−
TRIGONOMETRY
9 sin (A ± B) = sinA cosB ± cosA sinB
10 cos (A ± B) = cosA cosB sinA sinB
11 tan (A ± B) = BA
BA
tantan1
tantan
±
12 C
c
B
b
A
a
sinsinsin==
13 a2 = b2 + c2 - 2bc cosA
14 Area of triangle = Cabsin2
1
1 x = N
x∑
2 x = ∑∑
f
fx
3 σ = N
xx∑ − 2)( = 2_2
xN
x−∑
4 σ = ∑
∑ −f
xxf 2)( =
22
xf
fx−
∑∑
5 m = Cf
FNL
m
−+ 2
1
6 1
0
100Q
IQ
= ×
7 1
11
w
IwI
∑∑=
8 )!(
!
rn
nPr
n
−=
9 !)!(
!
rrn
nCr
n
−=
10 P(A ∪ B) = P(A)+P(B)- P(A ∩ B)
11 P (X = r) = rnrr
n qpC − , p + q = 1
12 Mean µ = np
13 npq=σ
14 z = σ
µ−x
http://tutormansor.wordpress.com/
Blank Page
3472/1 2008 Hak Cipta SBP [Lihat sebelah
SULIT
4
http://tutormansor.wordpress.com/
SULIT 3472/1
Answer all questions.
1. A function f maps the elements from set { 3,4,9,16,25}P m
to set Q = { 0, 3, 8, 15, 24 } as shown below in ordered pairs :
{ ( 3m , 0), (4, 3), (9, 8), (16, 15), (25, 24) }
(a) Write down the function notation for f.
(b) State the value of m.
(c) State the codomain.
[ 3 marks ]
Answer : ( a )……………………..
( b )...…………………...
( c )..................................
2. Given that function : 3, 0f x ax a and 2 : 16f x x b , find the value of a and of
b.
[ 3 marks ]
3472/1 2008 Hak Cipta SBP [ Lihat sebelahSULIT
5 3
2
3
1
For examiner’s use only
http://tutormansor.wordpress.com/
Answer : a =…………………….
b =..............................
3. Given that function : 3 , 0k
f x xx
, and 1 4
: ,f x x hx h
, find the value of h
and of k.
[ 3 markah ]
Answer : h =.........…………………
k =.....................................
4. Given m and -3 are the roots of a quadratic equation 22 4 1x x k , find the value of m and of k.
[ 3 marks ]
3472/1 2008 Hak Cipta SBP [Lihat sebelah
SULIT
6
For examiner’s use only
3
4
3
3
http://tutormansor.wordpress.com/
SULIT 3472/1
Answer : m =.........………
k =.....................5. Given that quadratic equation x(3x – p) = 2x – 3 has no roots, find the range of values
of p. [ 3 marks ]
Answer : .................................
___________________________________________________________________________
6. Diagram 1 shows the graph of a curve qpxxf ++= 2)(2)( , where p and q are constants.
Diagram 1
Given the straight line 3−=x is the axis of symmetry of the curve and parallel to they-axis. Find
(a) the value of p,
(b) the value of q,
(c) the turning point of the curve )(xf .
[3 marks]
Answer : (a) .……........................
3472/1 2008 Hak Cipta SBP [ Lihat sebelahSULIT
7
3
5
3
6
For examiner’s use only
x
qpxxf ++= 2)(2)(
• 8−O3−=x
f(x)
http://tutormansor.wordpress.com/
(b) ……........................
(c)..................................
7. Find the range of the values of x for which )52(2)12(3 +≤− xxx .[3 marks]
Answer : ..................................
8. Solve the equation 23
2
18 −
+ =x
x .
[3 marks]
Answer : ...................................
9. Solve the equation xx 2log1)25(log 55 =−− .[3 marks]
3472/1 2008 Hak Cipta SBP [Lihat sebelah
SULIT
8
3
7
3
8
For examiner’s use only
http://tutormansor.wordpress.com/
SULIT 3472/1
Answer : ....................................
10. Given that 2log2log 927 =− qp , express p in terms of q.
[4 marks]
Answer : ....……………...………..
11. An arithmetic progression has 9 terms. The sum of the first four terms is 24 and the
sum of all the odd number terms is 55. Find
a) the first term and the common difference,
b) the seventh term.
[4 marks]
3472/1 2008 Hak Cipta SBP [ Lihat sebelahSULIT
9
3
8
3
9
4
10
For examiner’s use only
4
11
http://tutormansor.wordpress.com/
Answer: ( a )…...…………..….......
( b).....................................
12. For a geometric progression, the sum of the first two terms is 30 and the third
term exceeds the first term by 15. Find the common ratio and the first term
of the geometric progression.
[4 marks]
Answer: …...….………..….......___________________________________________________________________________
13. Diagram 2 shows the graph of xy against 2x .
The variables x and y are related by the equation x
kxy +=2 , where k is a
constant. Find the value of h and of k.[4 marks]
3472/1 2008 Hak Cipta SBP [Lihat sebelah
SULIT
10
4
12
Forexaminer’s use
only
(0 , h)
2x
(7 , 2)
O
xy
Diagram 2
http://tutormansor.wordpress.com/
SULIT 3472/1
Answer : h = …………………….
k = ….……………….....
14. In Diagram 3 below, the equation of straight line LM is −6
x1
4=y
. The points
L and M lie on the x-axis and y-axis respectively.
Find the equation of a straight line which is perpendicular to LM and passes
through the point M.
[3 marks]
Answer : .…………………
15. In Diagram 4, OA a
and OB b
. On the same square grid , draw the line that
represents the vector 3OC a b
.
[ 2 marks ]
3472/1 2008 Hak Cipta SBP [ Lihat sebelahSULIT
11
4
13
3
14
For examiner’s use only
x
y
0
M
L
Diagram 3
O
A
B
http://tutormansor.wordpress.com/
16. Given (2, 5), (3,4)A B and ( , )C p q . Find the values of p and q such that
~ ~
2 9 5AB BC i j
.
[3 marks]
Answer: ……..…….…………... ___________________________________________________________________________
17. Solve the equation 3sin2cos4 xx for 3600 x .
[4 marks]
Answer: …...…………..….......18. Diagram 5 shows a sector BOC of a circle with centre O.
3472/1 2008 Hak Cipta SBP [Lihat sebelah
SULIT
12
2
15
4
16
4
17
For examiner’s use only
3
16
For examiner’s use only
B
Diagram 4
http://tutormansor.wordpress.com/
SULIT 3472/1
Given the perimeter of the sector BOC is 24.5 cm, calculate
(a) the radius of the sector
(b) the area of the sector.[3 marks]
Answer: a)...………………………
b) ……………………….___________________________________________________________________________
19. Given 3)35(
2)(
xxf
, find the value of (2)f .
[3 marks]
Answer:………………………
3472/1 2008 Hak Cipta SBP [ Lihat sebelahSULIT
13
3
19
3
18
For examiner’s use only
O
C
0.45 rad
Diagram 5
http://tutormansor.wordpress.com/
20. Find the equation of normal to the curve xxy 53 at the point when 1−=x .
[ 3 marks ]
Answer: …...…………..….......___________________________________________________________________________
21. Given that ∫ −=2
5
6)( dxxf , calculate dxx
xf∫
−
5
2 4)( .
[3 marks]
Answer: ……………………..
22. It is given that the sum of 5 numbers is α . The mean of the sum of the square is 200
and the variance is 3 2β . Express α in terms of .β
[3 marks]
3472/1 2008 Hak Cipta SBP [Lihat sebelah
SULIT
14
3
20
3
21
3
22
http://tutormansor.wordpress.com/
SULIT 3472/1
Answer: ……………………..
23. A pack of present containing three pens are chosen from 4 blue pens and 5 black pens. Find the probability of getting at least one black pen in the pack of present.
[3 marks]
Answer: …...…………..….......
___________________________________________________________________________
24.
Diagram 6
Diagram 6 shows six cards of different letters.
(a) Find the number of possible arrangements beginning with P.
(b) Find the number of these arrangements in which the vowels are separated.
[ 4 marks ]
Answer: ( a )……………………………
( b )……………………………
3472/1 2008 Hak Cipta SBP [ Lihat sebelahSULIT
15
For examiner’s use only
4
24
3
23
P O L I T E
http://tutormansor.wordpress.com/
25. X is a random variable which is normally distributed with mean µ and standard
deviation of 6. Find
(a) the value of µ, if the z-value is 1.5 when x = 42,
(b) P( X < 47 ).
[3 marks ]
Answer: ( a)…...…………..….......
( b )...................................
END OF QUESTION PAPER
3472/1 2008 Hak Cipta SBP [Lihat sebelah
SULIT
16
3
25
For examiner’s use only
http://tutormansor.wordpress.com/
SULIT 3472/1
3472/1 2008 Hak Cipta SBP [ Lihat sebelahSULIT
17
http://tutormansor.wordpress.com/
SULIT3472/1Additional MathematicsPaper 1Ogos 2008
SEKOLAH BERASRAMA PENUHBAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH / KLUSTERKEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN PERCUBAAN SPM TINGKATAN 5
2008
ADDITIONAL MATHEMATICS
Paper 1
MARKING SCHEME
This marking scheme consists of 7 printed pages
http://tutormansor.wordpress.com/
PAPER 1 MARKING SCHEME 3472/1
Number Solution and marking schemeSub
MarksFull
Marks1 (a) 1)( −= xxf
(b) 4=m
(c) {0,3,8,15,24}
1
1
1
3
2
4−=a and 9−=b
4−=a or 9−=b
162 =a or - ba −=− 33
bxxfa −=− 163)]([
3
B2
B1
3
34−=k and 3−=h
4−=k or 3−=h
3)(1
−−=−
x
kxf
3
B2
B1
3
4 5=m and 31=k
−−=−+
2
4)3(m or
2
1)3(
+−=− km
0142 2 =+−− kxx
3
B2
B1
3
548 <<− p
0)4)(8( <−+ pp
0)3)(3(4)2( 2 <−−− p
3
B2
B1
3
6
(a) 3=p(b) 26−=q(c) )26,3( −−
1
1
13
2
http://tutormansor.wordpress.com/
Number Solution and marking schemeSub
MarksFull
Marks7
26
5 ≤≤− x
0)2)(56( ≤−+ xx
01076 2 ≤−− xx
3
B2
B1
3
81−=x
22
1)3(3 +−=+ xx
2
1)3(3
2+x or 22 +−x
3
B2
B1
3
9
12
5
512 =x
xx
25
25 =− or equivalent
3
B2
B1
3
103729qp =
1236
2
3 3loglog =q
p or
126
2
3=q
p
3log12log6log2 333 =− qp or
3log22
log2
3
log3
33 =−qp
27log
log
3
3 p or
9log
log2
3
3 q or 2
3 3log
4
B3
B2
B1
4
3
http://tutormansor.wordpress.com/
Number Solution and marking schemeSub
MarksFull
Marks 11 (a) 3=a and 2=d
2464 =+ da or 55205 =+ da
(b) 15
)2(637 +=T
2
B1
2
B1
4
12 12=a
2
3=r
30
15
)1(
)1( 2
=+−ra
ra
30)1( =+ ra or 15)1( 2 =−ra
3021 =+TT or 1513 =−TT
4
B3
B2
B1
4
13 3−=k
2
3−=h
7
2
2
1 h−=
22
2 kxxy +=
4
B3
B2
B1
4
14 y =
2
3− x – 4 or 832 −−= xy
)0(2
3))4(( −−=−− xy
Gradient LM = 3
2 or )0,6(L and )4,0( −M
3
B2
B1
3
4
http://tutormansor.wordpress.com/
Number Solution and marking schemeSub
MarksFull
Marks 15
2 2
16 11,1 =−= qp (both)
−=
−−
14
8
82
62
q
p
−
=
−
−
−
−
5
9
4
32
5
2
4
3
q
p or equivalent
3
B2
B1
3
17 °°= 96.298,04.241x or '0'0 57298,3241 and °= 90x
°= 90x or 1sin =x or 8
7sin −=x
0)7sin8)(1(sin =+− xx or
07sinsin8 2 =−− xx
03sin)sin21(4 2 =++− xx
4
B3
B2
B1
4
18 (a) r = 10 cm
2r + 0.45r = 24.5
(b) Area = 22.5 cm2
2
B1
13
1918
4'
))2(35(
18)2(
−=f
)3()35(6)( 4' −−−= −xxf
3
B2
B1
3
5
http://tutormansor.wordpress.com/
Number Solution and marking schemeSub
MarksFull
Marks
20 92 −−= xy
gradient of normal = 2
1−
53 2 +−= xdx
dy
3
B2
B1
3
21
8
33
6 - 5
2
2
8
x
dxx
dxxf∫ ∫−5
2
5
2 4)(
3
B2
B1
3
22α = 5000 - 75 2β
3 2β = 200 -
2
5
α
5
α=x or 22 )(200 x−=σ
3
B2
B1
3
23
21
20
1 -
××
7
2
8
3
9
4
1 - P (all blue pens) or any combination
3
B2
B1
3
6
http://tutormansor.wordpress.com/
Number Solution and marking schemeSub
MarksFull
Marks24
(a) 120
5P5
(b) 144
3P3 or 4P3
2
B1
2
B1
4
25 (a) 51
5.142 =−
σµ
(b) 0.3284
2
B1
1
3
7
http://tutormansor.wordpress.com/
Top Related