stpm trial 2009 matht2 q&a (pahang)
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8/14/2019 STPM Trial 2009 MathT2 Q&A (Pahang)
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CONFIDENTIAL*
PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN
PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN
PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN
PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN
PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN
PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN
PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN
PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN
PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN
PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN
PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN
JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
PEPERIKSAAN PERCUBAAN
SIJIL TINGGI PERSEKOLAHAN MALAYSIA
NEGERI PAHANG DARUL MAKMUR 2009
Instructions to candidates:
Answerall questions. Answers may be written in either English or Malay.
All necessary working should be shown clearly.
Non-exact numerical answers may be given correct to three significant figures, or one
decimal place in the case of angles in degrees, unless a different level of accuracy is
specified in the question.
Mathematical tables, a list of mathematical formulae and graph paper are provided.
This question paper consists of 6 printed pages.
954/2 STPM 2009
Three hours
MATHEMATICS T
PAPER 2
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CONFIDENTIAL* 2
Mathematical Formulae for Paper 2 Mathematics T :
Logarithms :
a
xx
b
b
alog
loglog =
Series :
)1(2
1
1
+==
nnrn
r
)12)(1(6
1
1
2 ++==
nnnrn
r
22
1
3)1(
4
1+=
=
nnrn
r
Integration :
=
dxdx
du
vuvdxdx
dv
u
cxfdxxf
xf+= )(ln)(
)('
ca
x
adx
xa+
=
+
1
22tan
11
c
a
xdx
xa
+
=
1
22
sin1
Series:
Nnwhere ++
++
+
+=+ ,
21)( 221 nrrnnnnn bba
r
nba
nba
naba
Coordinate Geometry :
The coordinates of the point which divides the line joining (x1 ,y1) and (x2 ,y2) in
the ratiom :n is
++
++
nm
myny
nm
mxnx 2121 ,
The distance from ),( 11 yx to 0=++ cbyax is
22
11
ba
cbyax
+
++
Maclaurin expansions
1,!
)1()1(!2
)1(1)1(2
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CONFIDENTIAL* 3
Mathematical Formulae for Paper 2 Mathematics T :
Numerical Methods :
Newton-Raphson iteration for 0)( =xf :
)('
)(1
n
n
nnxf
xfxx =+
Trapezium rule :
+++++ b
ann yyyyyhdxxf ])(2[
2
1)(
1210
n
abhrhafyr
=+= andwhere )(
Correlation and regression :
Pearson correlation coefficient:
( )( )( ) ( )
=22
yyxx
yyxxr
ii
ii
Regression line ofy onx :
y =a +bx
where( )( )( ) = 2i ii xx yyxxb
xbya =
Trigonometry
BAAABA sincoscossin)sin( = BABABA sinsincoscos)cos( =
BA
BABA
tantan1
tantan)tan(
=
AAAAA 2222 sin211cos2sincos2cos === AAA 3sin4sin33sin = AAA cos3cos43cos 3 =
+=2
BAcos
2
BAsin2BsinAsin
+= 2BA
sin2
BAcos2BsinAsin
+=2
BAcos
2
BAcos2BcosAcos
+2
BAsin
2
BAsin2BcosAcos
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CONFIDENTIAL* 4
1. A passenger in an aeroplane flying at a certain height , H , sees two towns , Pekan
town and Kuantan town, directly to the left of the aeroplane. The angles of depression
to the Pekan towns and Kuantan town , are and respectively where < . If thetwo towns are d meter apart, show ( )eccossinsindH . [3]2.
The points A , B and C lie on the circumference of a circle as shown in the diagram
above with ABC= 80 and ACD=50. The tangent to the circle at point A meets thechord CD produced at point T.
(a) Show that AC = AT. [4]
(b) Show that the length of the chord CD is equal to the radius of the circle. [4]
3. The forces ( )Nj3i5F1 + , ( )Nj6i4F2 and ( )Nj7i2F3 + act at a point.(a) Calculate the magnitude of the resultant force. [2]
(b) Using the scalar product, calculate the angle between the resultant force and
force, ( )Nj3i5F4 + . [4]4. (a) A certain substance evaporates at a rate which is proportional to the amount of
substance left. Given that the initial amount of the substance is A and the amount which
has evaporated at time t is x , write a differential equation to show the rate of
evaporation. [1]
(b) Solve the differential equation and sketch the graph ofx against t. [6]
(c) Given that it took 2ln seconds for half the amount to be evaporated, find how
long it takes for4
3of the initial amount to be evaporated. [4]
5. The following table show the mean and standard deviation of the marks of the male
and female students who sat for a semester test.
Student Number of students Mean Standard deviation
Male 80 52. 9 5. 3
Female 100 61. 4 4. 1
Calculate the mean and standard deviation of the marks of all the students.[6]
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CONFIDENTIAL* 5
6. Two bowlers , Adam Lambert and Kris Allen , take turns to throw, with bowler
Adam Lambert taking the first throw. The probabilities of bowlers Adam Lambert and
Kris Allen scoring a strike are3
2and
5
4respectively. Find the probabilities of bowler
Adam Lambert scoring a strike first. [3]
7. There are one or two flowers on the faces of 50 cents stamps. 90% of all these 50 cents
stamps have two flowers while the rest of the stamps have single flower.
From the stamps which have single flower, 95% of these stamps have a flower at the
centre of the stamps while the rest have a flower on the left side of the stamps.
(a) By using a suitable approximation, determine the probability that between 5 and 15
stamps inclusive have one flower , out of a random sample of 100 pieces of 50 cents
stamps. [5]
(b) By using a suitable approximation, determine the probability that less than 3 stamps
have only one flower on the left side of the stamp out of a random sample of 100 pieces
of 50 cents stamps. [4]
8. X is a continuous random variable with a probability density function,fdefined as
( )
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CONFIDENTIAL* 6
10. (a) Prove that 2
tan2sin6sin4sin2
2sin6sin4sin2 =+
.
Hence, find the value of 15tan2 , leaving your answer in surd form. [5](b) Solve the equation
2cos
2sin6sin4sin2
2sin6sin4sin2 =+
for 3600 . [5]
11. (a) At noon, a man A is travelling at a speed of 20 km/h in the direction of N 25 W.Another man , B is 2 km to the north of A and travelling at the speed of 15 km/h in the
direction of N 60 W. Find the magnitude and the direction of the velocity of A relativeto B. Hence, find the shortest distance between man A and man B . [9]
(b) Find the course man A must travel in order to intercept B if man A maintains hisspeed but changes its course. [4]
12. (a) A machine is used to fill up bottles with a mean volume of 550 ml. Suppose that
the volume of water delivered by the machine follows a normal distribution with mean , ml and standard deviation 4 ml. Find the range of values of mean , , if it is requiredthat not more than 1% of the bottles contain less than 550 ml. [4]
(b) The mass of a box of chocolate cereal is distributed normally with mean 100 g and
standard deviation 2 g.
(i) Calculate the probability that three boxes of chocolate cereal chosen at
random, each has a mass less than 98 g. [2]
(ii) Calculate the probability that three boxes of chocolate cereal chosen at
random, have a total mass exceeding 305 g. [2]
(iii) Calculate the probability that out of three boxes of chocolate cereal chosen at
random, exactly two have a mass greater than 98 g while the mass of the remaining box
is greater than 105 g. [3]
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Marking Scheme
PEPERIKSAAN PERCUBAAN STPM NEGERI PAHANG 2009
Mathematics T Paper 2 ( 954 / 2 )
Question Scheme Marks
1
ABK ,AK
Hsin = AK = sin
H M1
AKP , Sine Rule :
( )sin dsinAK , ( )
sin
d
sin
sin
H
( )eccossinsindH M1
A1
Question Scheme Marks
2. (a) ADC = 180 80 ( Opposite angles of a cyclic= 100 quadrilateral are supplementary)TDA = 180 100 ( Angles on a straight line )= 80TAD = ACD = 50 ( Angles in the alternate segment)
ATC , DTA or CTA = 180 ( 80 + 50)= 50
{ The sum of angles in a triangle is 180 )
Since CTA = ACT = 50, ATC is an isoscelestriangle because the base angles are the same.AC = AT
M1
M1
M1
A1
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2. (b) Join A and C to the centre ,
then OA=OC=radius
AOC = 2 ABC= 160
( Angle at the center = 2 angle at the circumference.)-------------------------------------------------------------------
OCA = ( )1601802
1= 10
OAC is an isosceles triangle ( OA = OC = radius )-------------------------------------------------------------------
ODC, OCD =OCA + ACD= 50+10= 60 OCD =ODC = 60OCD is an isosceles triangle ( OA = OD = radius )
COD= 180 ( 60 + 60 ) = 60Since OCD =ODC = COD = 60 , thereforeOCD is an equilateral triangle andCD = OC = OD = radius of circle.
M1
M1
M1
A1
Question Scheme Marks
3. (a) Resultant force, RF = ( )j3i5 + + ( )j6i4 + ( )j7i2 + = ( )j4i7 +
Magnitude of the resultant force,
RF =22 47 + = 65 = 8.062 N
B1
B1
3. (b)
4R
4R
FF
FF
cos
= ,( ) ( )
j3i5j4i7
j3i5j4i7
cos ++
= ( ) ( )2222
3547
3457cos +
=
3465
47cos =
'131
M1
M1
M1
A1
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Question Scheme Marks
4. (a) Amount which has evaporated at time t = x
Amount which has not evaporated at time t = A x( )xA
dt
dx , ( )xAkdt
dx B14. (b) Interpret initial condition t=0 ,x = 0
( ) = t0x0 dtkdxxA 1 separate variables M1( )] [ ] t0x0 tkxAln = correct integration
Substitute upper limits & lower limits
M1
M1
tkA
xAln
tkeA
xA =
M1
( )tke1Ax A1Graph
D1
4. (c) ( )tke1Ax ,Substitute , 2lnt = ,2
Ax = , M1
k = 2 A1( )t2e1Ax , Substitute A4
3x = M1
2lnt = A1
Question Scheme Marks
5 Mean =FM
FM
nn
xx
++
, Mean =( ) ( )
10080
4.611009.5280
+
Mean =180
61404232 +, Mean =
180
10372= 57.622
M1
A1
Male : ( ) [ ]222M nx + ( ) [ ]222
M 9.523.580x + = 226 120Female : ( ) [ ]222F 4.611.4100x + = 378 677
B1
B1
Standard deviation =
22
F
2
M
180
10372
180
xx
+
=
2
180
10372
180
604797
= 6.2978
M1
A1
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Question Scheme Marks
6
P(A) + P(A K A ) + P(A K A K A ) + =
3
2+
3
2
5
1
3
1+
3
2
5
1
3
1
5
1
3
1+
=
15
11
3
2
Infinite Geometric Series
=7
5
M1
M1
A1
Question Scheme Marks
7. (a) X represents the number of stamps with one flower
X B ( 100 , 0.1 )Normal Approximation : mean=10 , variance = 9
( )15X5P =
3
105.15Z
3
105.4P
= ( )833.1Z833.1P = ( )0334.021 = 0. 9332
B1
B1
M1 Standardize
M1 Continuity
correction
A1
7. (b) Y represents the number of stamps with only one flower
on the left side. 005.005.01.0p = Y B ( 100 , 0. 005 )Poisson Approximation : mean, = 0.5
)3X(P < = )0X(P = + )1X(P = + )2X(P = =
+
2
5.05.01e
25.0
= 1.625 5.0e = 0.9856
B1
B1
M1
A1
Question Scheme Marks
8. (a)
1dxe
4
12
xk
0
= , 1e2
1k
0
2
x =
Integration M1
k = 2 ln 3 A1
8. (b)
3ln2x0 < , dxe4
1)x(F 2
xx
0Integration
x
0
2
x
e2
1)x(F
= ,
1e
2
1)x(F 2
x
M1
The cumulative distribution function is
>
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Question Scheme Marks
10. (a)LHS,
2sin6sin4sin2
2sin6sin4sin2
+ ( )( )2sin6sin4sin2 2sin6sin4sin2 ++= ( )( )2cos4sin24sin2 2cos4sin24sin2 +=
( )( )2cos14sin2 2cos14sin2 += [ ][ ]1cos21 sin211 22
=
2
2
cos2
sin2= 2tan Q.E.D
Factor
formulae M1
Double
Angle M1
A1
Substitute 15 into
2sin6sin4sin2
2sin6sin4sin2tan2 +
=
= 30sin90sin60sin230sin90sin60sin2
15tan2
2
11
2
32
2
11
2
32
+
=
332
332
+=
( )( )332 332 += ( )( )332 332 347
Values of
sin andcosM1
A1
10. (b)
2cos2sin6sin4sin2
2sin6sin4sin2 =+
, 2costan2 = 1cos2
cos
sin 22
2
242 coscos2sin 242 coscos2cos1
4
1cos4 = , 8409.0cos
2.3278.212,2.147,8.32
M1
M1
cos
sintan =
M1 Basic
Identity
M1
A1
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Question Scheme Marks
11. (a)
2
BAV = ( )( ) 35cos152021520 22 BAV = ( )( ) 35cos152021520 22 BAV =11.5546 km/h
The magnitude of the velocity of
A relative to B is 11.6 km/h .
D1 arrows
D1
directions/angles
M1
A1
35sin5546.11
sin
15
5546.1135sin15
sin=
'848 , '82325'848 or 23.13The direction of the velocity of A relative to B
is N 23 8 E.
M1
A1
d = '823sin2 d = 0.7857 km
The shortest distancebetween A and B is 0.786 km
D1 diagram
M1
A1
11. (b)
120sin20
sin
15
20
120sin15sin
=
'3040 '3040120180
'3019 In order to intercept B , man A must travel in the
direction of N 40 30 W . or (bearing 319 30)
D1 arrows
D1
directions/angles
M1
A1
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Question Scheme Marks
12 (a) ( ) 01.0550XP < , 01.04
550ZP 12
300305ZP
= ( )443.1ZP > = 0.0745
M1 Standardize
A1
(iii) 3 ( )98XP > ( )98XP > ( )105XP > = 3 2
2
10098ZP
>
>
2
100105ZP
= 3 ( )]21ZP ( )5.2ZP > = 3 [ ]28413.0 0.00621= 0.0132
M1 Standardize
M1
A1
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