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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 2008
Instructions to candidates:
Answer all questions.
Answers may be written in either English or Bahasa Malaysia.
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.
950/1, 954/1 STPM 2008
Three hours
MATHEMATICS S
PAPER 1
MATHEMATICS T
PAPER 1
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
CONFIDENTIAL*
2
Mathematical Formulae for Paper 1 Mathematics T / Mathematics S :
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
duvuvdx
dx
dvu
cxfdxxf
xf+=∫ )(ln
)(
)('
ca
x
adx
xa+
=
+
−
∫1
22tan
11
ca
xdx
xa+
=
−
−
∫1
22sin
1
Series:
N n where ∈++
++
+
+=+
−−− ,21
)( 221 nrrnnnnnbba
r
nba
nba
naba LL
1,!
)1()1(
!2
)1(1)1( 2
<++−−
++−
++=+ xxr
rnnnx
nnnxx
rn whereL
LL
Coordinate Geometry :
The coordinates of the point which divides the line joining (x1 , y1) and (x2 , y2) in the
ratio m : n is
+
+
+
+
nm
myny
nm
mxnx 2121 ,
The distance from ),( 11 yx to 0=++ cbyax is
22
11
ba
cbyax
+
++
Numerical Methods :
Newton-Raphson iteration for 0)( =xf :
)('
)(1
n
n
nnxf
xfxx −=
+
Trapezium rule :
∫ +++++≈−
b
ann yyyyyhdxxf ])(2[
2
1)( 1210 L
n
abhrhafyr
−=+= and where )(
Trigonometry :
BAAABA sincoscossin)sin( ±=±
BABABA sinsincoscos)cos( m=±
BA
BABA
tantan1
tantan)tan(
m
±=±
AAAAA2222 sin211cos2sincos2cos −=−=−=
AAA3sin4sin33sin −=
AAA cos3cos43cos 3−=
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
CONFIDENTIAL*
3
1. Show that 07422=−−− xyy is a parabola.
Sketch the parabola and label its vertex and focus. [4 marks]
2. Given the complex numbers iz 321 += and iz 1252 +−= , verify that 1z is one
of the square roots of 2z .
Hence, simplify )*)(( 21 zz in terms of 1z . [5 marks]
3. Given the matrix
−=
16
2
k
kM , find the set of values of k for which the
inverse of M exists. [5 marks]
4. Given xy sin= , show that 014 4
2
23
=++ ydx
ydy . [5 marks]
5. Determine the solution set of the inequality 25
<+ x
x. [4 marks]
Hence, deduce the set of values of x that satisfy the inequality
25
<+ x
x. [2 marks]
6. The parametric equations of a curve are
2
,ln teytx == ( t > 0 )
(a) Express dx
dy in terms of t . [4 marks]
(b) Find the equation of the tangent to the curve at the point t = 1. [4 marks]
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
CONFIDENTIAL*
4
7. Expand y+1
1 as a series in ascending powers of y up to the term in 3y .
[2 marks]
Hence, by taking 2xxy += in your series, show that for small x ,
32
2 16
7
8
1
2
11
1
1xxx
xx+−−≈
++ [4 marks]
By taking 5
1=x , find the value of 31 correct to four decimal places.
[4 marks]
8. The points A(a , 0), B(-a , 0) and C(h , k) are the vertices of triangle ABC.
(a) State the perpendicular bisector of AB. [1 marks]
(b) Find the equation of the line which divides BC equally and also
perpendicular to BC. [4 marks]
Hence, (i) show that the centre of the circle that passes through the points A,
B and C is
−+
k
akh
2,0
222
, [2 marks]
(ii) determine the general equation of circle that passes through the
points (–2 , 0) , (2 , 0) and (1 , 1) [3 marks]
9. The polynomial qxpxxxf +++= 7)( 23 , where p, q are constants has a factor
of x – 1 . Given f '(1) = 20, find the values of p and q . [4 marks]
If RxQx
xf+=
+)(
1
)( ,
(a) state the value of R, [1 marks]
(b) show that Q(x) is positive for all real values of x. [5 marks]
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
CONFIDENTIAL*
5
10.(a) Express )21)(34(
17
xx
x
+−
+in partial fractions. [4 marks]
(b) The graph of )21)(34(
17
xx
xy
+−
+= is as shown below.
8
6
4
2
-2
-5 5 10
Show that the area of the region bounded by the curve )21)(34(
17
xx
xy
+−
+= ,
the line x = 3
1− and x =
2
1 is ( )3ln92ln19
6
1+ . [7 marks]
11.(a) The matrix A is defined by
−
−
−
=
1104
180
120
A .
Find (i) the matrix 2A , [1 mark]
(ii) the matrix B if IAAB 672+−= where I is the 3×3 identity
matrix. [2 marks]
Show that OIAB =− 24 where O is the 3×3 zero matrix.
Hence, deduce the inverse of matrix A. [4 marks]
(b) The points P(0 , 1), Q(0 , 4) and R(2 , 5) lie on the curve given by the equation
02222=+−−+ cbyaxyx .
Write a system of equations in the form of matrices that involve matrix A to
represent the above information.
Hence, find the values of a, b and c. [5 marks]
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
CONFIDENTIAL*
6
12. The function f is defined by
≥−
<−
−
=
1,22
1,1
1
)(
xx
xx
x
xf .
(a) Find )(1
limxf
x−
→ and )(
1
limxf
x+
→ . Hence, determine whether f is
continuous at x = 1. [4 marks]
(b) Sketch the graph of the function f . [3 marks]
Based on your graph,
(i) state the range of function f , [1 marks]
(ii) explain why the inverse of function f is not a function, [2 marks]
(iii) suggest the largest possible domain for function f so that
the inverse of f is a function. [1 marks]
(iv) sketch the graph of )(
1
xfy = . [3 marks]
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
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 2008
Instructions to candidates:
Answer all 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 7 printed pages.
954/2 STPM 2008
Three hours
MATHEMATICS T
PAPER 2
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
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
duvuvdx
dx
dvu
cxfdxxf
xf+=∫ )(ln
)(
)('
ca
x
adx
xa+
=
+
−
∫1
22tan
11
ca
xdx
xa+
=
−
−
∫1
22sin
1
Series:
N n where ∈++
++
+
+=+
−−− ,21
)( 221 nrrnnnnnbba
r
nba
nba
naba LL
Coordinate Geometry :
The coordinates of the point which divides the line joining (x1 , y1) and (x2 , y2) in the
ratio m : 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
<++−−
++−
++=+ xxr
rnnnx
nnnxx
rn whereLL
L
...!
...!2
12
+++++=r
xxxe
rx
( )( )
11...,1
...32
1ln
132
≤<−+−
+−+−=+
+
xr
xxxxx
rr
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
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 L
n
abhrhafyr
−=+= and where )(
Correlation and regression :
Pearson correlation coefficient:
( )( )
( ) ( )∑∑
∑−−
−−=
22
yyxx
yyxxr
ii
ii
Regression line of y on x :
y = a + b x
where b = ( )( )
( )∑∑
−
−−
2
xx
yyxx
i
ii
xbya −=
Trigonometry
sin ( A ± B ) = sin A cos B ± cos A sin B
cos ( A ± B ) = cos A cos Bm sin A sin B
tan ( A ± B ) = BA
BA
tantan1
tantan
m
±
cos 2A = cos 2 A − sin 2 B = 2 cos 2 A − 1 = 1 − 2 sin 2 A
sin 3A = 3 sin A − 4 sin 3 A
cos 3A = 4 cos 3 A − 3 cos A
sin A + sin B = 2 sin
+
2
BAcos
−
2
BA
sin A − sin B = 2 cos
+
2
BAsin
−
2
BA
cos A + cos B = 2 cos
+
2
BAcos
−
2
BA
cos A − cos B = − 2 sin
+
2
BAsin
−
2
BA
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
CONFIDENTIAL*
4
1. By using the factor formulae, find all values of x, where °≤≤° 3600 x , which
satisfy the equation cos x − 2 cos2 x + cos3 x = 0. [4 marks]
2. (a) Express 3 cos x + sin x in the form of r cos (x − θ) where r > 0 and
0 <θ < 2
π, find r and θ. [3 marks]
(b) If y =1xsinxcos3 ++++++++
ππππ where
6
π < x <
2
π, show that
3
π< y<
2
π.
[3 marks]
3. B
In the figure above, OAB is an equilateral triangle with OA = 1 unit. M is the mid-
point of AB and P divides the line segment OA in the ratio 2:1 . Q is a point on OB
such that PQ intersects OM at G and PG: GQ = 4:3 . Given that →
OA = a and →
OB = b,
(a) Find →
OM in terms of a and b. [1 mark]
(b) Given that OQ : QB = k : ( l − k ) .
(i) Find →
OG in terms of k, a and b.
(ii) Use ABOG o to find k and show that →
PQ = a3
2b
2
1−−−− . [5 marks]
(iii) Using results from (ii) above, find →
PQ . [4 marks]
4. Two forces of magnitude P Newton each act on a particle. The resultant force
on the particle is 6 Newton. When the magnitude of one of the forces acting on the
particle is increased by 4 times, the angle between the two forces remains unchanged
but the resultant force is increased to 18 Newton. Calculate the value of P and the
angle between the forces. [7 marks] Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
CONFIDENTIAL*
5
5 C
An octagon ABCDEFGH is inscribed in a circle, centre O, as shown in the figure
above. K and J are the feet of the perpendiculars from O to the sides AB and DE
respectively. Chords AD and BE intersect at I.
Given that AB = BC = GH = HA = 3 units, and CD = DE = EF = FG = 2 units.
(a) Show that E, O, and A are collinear. [2 marks]
(b) Show that OK = 21
EB and OJ = 21
AD. [4 marks]
(c) Prove that ∆DCB is congruent to ∆DIB. [4 marks]
6. A cylindrical container has a height of 200 cm. The container was initially
full of a chemical but there is a leak from a hole in the base. When the leak is noticed,
the container is half-full and the level of the chemical is dropping at a rate of 1 cm per
minute. It is required to find for how many minutes the container has been leaking.
To model the situation it is assumed that, when the depth of the chemical remaining is
x cm, the rate at which the level is dropping is proportional to x .
Show that differential equation is . x 0.1 dt
dx−−−−==== Solve the differential equation,
and hence show that the container has been leaking for about 80 minutes (rounded up
to one significant figure). [13 marks]
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
CONFIDENTIAL*
6
7. The average weekly incomes in RM, of households in 11 states of Malaysia are
given below.
255.80 252.00 270.60 398.40
362.30 297.20 266.80 261.70
247.50 259.10 220.60
(a) Find the median and upper and lower quartiles. [3 marks]
(b) Draw a box and whiskers plot to represent the data. [2 marks]
(c) Describe the skewness of this distribution and identify a possible outlier.
[4 marks]
8. The table below shows the age distribution of accidents victims under the age of 16
in a country during one calendar year.
Age ( x , years ) Frequency
5 ≤ x < 8 99
8 ≤ x < 11 143
11 ≤ x < 13 130
13 ≤ x < 14 78
14 ≤ x < 15 84
15 ≤ x < 16 91
(a) Draw a histogram to represent the data. [2 marks]
(b) Calculate estimates of the mean and standard deviation of this distribution.
[5 marks]
(c) Calculate estimates of the median and mode. [4 marks]
(d) State which measure of location, mean or median would be chosen to represent the
data. Give a reason for your choice. [2 marks]
9. A Personal Identification Number (PIN) consists of 4 digits in order, each of which
is one of the digits 0 , 1 , 2 , 3 , …, 9. Irene has difficulty remembering her PIN. She
tries to remember her PIN and writes down what she thinks it is. The probability that
the first digit is correct is 0.8 and the probability that the second digit is correct is
0.86. The probability that the first two digits are both correct is 0.72. Find
(a) the probability that the first digit is correct and the second digit is incorrect.
[2 marks ]
(b) the probability that the second digit is incorrect given that the first digit is
incorrect. [2 marks ]
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
CONFIDENTIAL*
7
10. A bag consists four green apples and three red apples. A boy takes out apples one
by one randomly , without replacement until he gets a red apple. When he gets a red
apple, he stops. Let X represents the number of apples he takes out.
(a) Tabulate the probability distribution of X. [3 marks]
(b) Calculate E(X) and Var(X) . [3 marks]
11. The lifespan T ( in working hours ) of a certain type of drill used in oil exploration
drilling machinery is a continuous random variable with probability density function
defined by :
f ( t ) = >
−
otherwise
te t
0
0 µλ
where λ and µ are constants.
Drilling is planned to take place continuously for one six-hour shift each day.
(a) If the mean life of the drill is 20 hours, find the values of the constant λ and µ.
[5 marks]
(b) If a new drill is used for a shift, what is the probability that it will fail during the
shift ? [2 marks]
12. (a), A machine packs sugar into bags marked 2 kg. The weight produced by the
machine is normally distributed. The standard deviation of the measures produced by
the machine is 0.03 kg . At least 95% of the bags must be over 2 kg in weight , find
the limit of the value of µ, mean of the measures produced by the machine.
[3 marks]
(b) When a telephone call is made in the country “ABC” , the probability of getting
the intended number is 0.95.
(i) Ten independent calls are made. Find the probability of getting eight or
more of the intended numbers . [2 marks]
(ii) Three hundred independent calls are made. Find the probability of failing
to get the intended number on at least ten but not more than twenty of the calls.
[3 marks]
(iii) Four hundred independent calls are made. For each call the probability of
getting “number unobtainable” is 0.004. Find the probability of getting “number
unobtainable” less than three times. [3 marks]
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
Marking Scheme
PEPERIKSAAN PERCUBAAN STPM NEGERI PAHANG 2008
Mathematics T Paper 1 (954/1)/ Mathematics S Paper 1 (950/1)
1.
)24( )1(
84 )1(
74 2
0742
2
2
2
2
+=−
+=−
+=−
=−−−
xy
xy
xyy
xyy
This is a standard equation of parabola, therefore the curve is a parabola. A1
x
y
-3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
[4]
2. (2 + 3i)2 = -5 + 12i M1
It is verified that 2 + 3i is one of the square roots of -5 + 12i. A1
(2 + 3i)* = 2 – 3i B1
(2 – 3i) ( 2 + 3i)2 = 13 1z M1A1 [5]
3. M =
−16
2
k
k
Determinant of M is M = k( k – 1) – 12 M1
= k2 – k – 12 A1
The inverse for M exists if M ≠ 0,
k2 – k – 12 ≠ 0 M1
(k – 4) (k + 3) ≠ 0
k ≠ -3 , k ≠ 4 A1
∴The set of values of m for which the inverse of M exists is
{ 4,3,: ≠−≠∈ kkRkk } A1 [5]
(y-1)2 = 4(x + 2)
M1
A1
D1 : correct
curve
x
y
(-2,1) (-1,1)
D1 : both vertex
and focus
curve
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
4. y2 = sin x
xdx
dyy cos2 = M1
xdx
dy
dx
ydy sin22
2
2
2
−=
+ M1A1
xy
x
dx
ydy sin
2
cos22
2
2
2
−=
+
xy
x
dx
ydy sin
4
sin122
2
2
2
2
−=
−+
2
2
4
2
2
2
12 y
y
y
dx
ydy −=
−+
44
2
23 214 yy
dx
ydy −=−+ M1
014 4
2
23
=++ ydx
ydy A1 [5]
5. 025
<−+ x
x
05
210<
+
−−
x
xx M1
05
10<
+
−−
x
x A1
∴{x : x > -5 , x < -10} M1A1
10,5 −<−> xx M1
Hence, {x : 0≥x } A1 [6]
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6. tdt
dx 1= and
2
2 tte
dt
dy= M1A1
2
2
221
2 tt
et
t
te
dx
dy== M1A1
When t = 1,
x = 0 , y = e M1
edx
dy2=∴ A1
Equation of tangent, y - e = 2e (x – 0) M1
y = 2ex + e A1 [8]
7. 2
1
)1(1
1 −+=
+y
y
= L+
−
−
−
+
−
−
+−32
!3
2
5
2
3
2
1
!2
2
3
2
1
2
11 yyy M1
= L+−+−32
16
5
8
3
2
11 yyy A1 [2]
When y = x + x2,
...)(16
5)(
8
3)(
2
11
1
1 32222
2++−+++−=
++xxxxxx
xx M1A1
= ......)(16
5)2(
8
3
2
1
2
11 34322
++−+++−− xxxxxx M1
32
16
7
8
1
2
11 xxx +−−≈ A1 [4]
Let x = 5
1,
31
5
25
31
1
25
1
5
11
1==
++
B1
32
5
1
16
7
5
1
8
1
5
1
2
11
31
5
+
−
−≈∴ M1A1
5648.531 = ( 4 d.p. ) A1 [4]
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8. (a) x = 0 B1 [1]
(b) Midpoint of BC =
−
2,
2
kah B1
mBC = ah
k
+
Equation of line
−−
+−=−
22
ahx
k
ahky B1M1
k
akhx
k
ahy
2
222−+
+
+−= A1 [4]
(i) x = 0 -----------------------------(1)
k
akhx
k
ahy
2
222−+
+
+−= ----------(2)
Centre of circle is
−+
k
akh
2,0
222
M1A1 [2]
(ii) radius = 5
centre = (0, -1) B1
equation of circle
222 )5()1()0( =++− yx M1
x2 + y
2 + 2y – 4 = 0 A1 [3]
9. f(x) = x3 + px
2 + 7x + q
f(1) = 1 + p + 7 + q = 0
p + q = -8 M1
f '(x) = 3x2 + 2px + 7
f '(1) = 3 + 2p + 7 = 20 M1
p = 5 A1
q = -13 A1 [4]
f(x) = x3
+ 5x2
+ 7x -13
R = f(-1) = -16 B1
Q(x) = x2 + 6x + 13 B1
= (x + 3)2 + 4 M1A1
Since (x + 3)2 > 0, xxQ ∀>∴ 0)( M1A1 [6]
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10. Let x
B
x
A
xx
x
2134)21)(34(
17
++
−≡
+−
+
17 + x ≡ A(1 + 2x) + B(4 – 3x) M1
Substituting x = 2
1− :
=
2
11
2
33B M1
B = 3
Substituting x = 3
4 :
=
3
11
3
55A
A = 5 A1(both A&B)
Thus xxxx
x
21
3
34
5
)21)(34(
17
++
−=
+−
+ A1(stating)
Area = ∫ − +−
+2
1
3
1)21)(34(
17dx
xx
x B1
= dxxx∫ −
++
−
2
1
3
1
21
3
34
5 M1
=2
1
3
1
)21ln(2
3)34ln(
3
5
−
++−− xx M1A1
=
+−−+
−
3
1ln
2
35ln
3
52ln
2
3
2
5ln
3
5 M1A1
=
−+
−
3
1ln
2
32ln
2
3
2
5ln5ln
3
5
= 3ln2
32ln
2
32ln
3
5++
= 3ln2
32ln
6
19+
= )3ln92ln19(6
1+ A1 [11]
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11(a) (i)
−−
−−
−−
=
−
−
−
−
−
−
13784
7544
164
1104
180
120
1104
180
120
B1
(ii) B = A2 – 7A + 6I
=
+
−
−
−
−
−−
−−
−−
100
010
001
6
1104
180
120
7
13784
7544
164
M1
=
−
−
−
0832
044
682
A1
(iii) AB – 24I =
−
100
010
001
24
2400
0240
0024
= O M1A1
AB = 24I
A-1
AB = 24 A-1
I
24
1B = A
-1 M1
−
−
−
=−
03
1
3
4
06
1
6
14
1
3
1
12
1
1A A1 [7]
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(b) 1- 2b + c = 0
6 – 8b + c = 0
-4a – 10b + c = 0 B1B1 (all correct)
B1B0 (any two
correct)
2b – c = 1 B0B0 ( others )
8b – c = 16
4a + 10b – c = 29
=
−
−
−
29
16
1
1104
180
120
c
b
a
M1
−
−
−
=
29
16
1
0832
044
682
24
1
c
b
a
M1
=
4
2
25
A = 2, b = 2
5, c = 4 A1 [5]
12. (a)1
1lim)(lim
11 −
−=
−−→→ x
xxf
xx
= )1(
1lim
1 −−
−−
→ x
x
x
= -1 M1
)22(lim)(lim11
−=++
→→
xxfxx
= 0 A1
Since )(lim)(lim11
xfxfxx
+−→→
≠ , )(lim1
xfx→
does not exist. M1
So f is not continuous at x = 1. A1
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(b)
(i) ),0[}1{ ∞∪−∈y B1
(i) From the graph , f is many to one, so it is not one to one, M1
therefore the inverse is not a function. A1
(iii) }1:{ ≥xx A1
(iv)
x
y
1
– 1
D1 : line y = 2x – 2
D1 : line y = –1
D1 : correct and
x
y
x = 1
D1 : correct
D1 : correct
Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
1
Marking Scheme
PEPERIKSAAN PERCUBAAN STPM NEGERI PAHANG 2008
Mathematics T Paper 2 (954/2)
1. °°°°≤≤≤≤≤≤≤≤====++++ 360x0,0x3cosx2cos2-xcos
0=x2cos2-xcosx2cos2 factor formula [M1]
0=)2-xcos2(x2cos [A1]
2
2=xcosor,0=x2cos [A1]
oooooo315,45=xor,630,450,270,90=x2
oooo
315,225,135,45=x.ie [A1 all correct]
2(a) )-xcos(xsinxcos3 θθθθr ≡≡≡≡++++
xsinsinrxcoscosr θθθθθθθθ ++++≡≡≡≡
3=θcosr,1=θsinr (can be implied ) [B1 both]
3
1=θtan
2221+)3(=r [M1 either one]
6
π=θ 2=r [A1 both]
(b) 1+)
6
π-xcos(2
π=
1+xsin+xcos3
π=y [M1]
36-x0
2x
6
ππππππππππππππππ<<<<<<<<⇒⇒⇒⇒<<<<<<<<Q
3cos)
6-xcos(0cos
ππππππππ>>>>>>>> [M1]
2
1)
6-xcos(1 >>>>>>>>
ππππ 1)
6-xcos(22 >>>>>>>>
ππππ 21)6
-xcos(23 >>>>++++>>>>ππππ
2
1
1)6
-xcos(2
1
3
1<<<<
++++
<<<<ππππ
21)
6-xcos(2
3
ππππ
ππππ
ππππππππ<<<<
++++
<<<< [A1]
3(a) 2
b+a=OM [B1]
kb=OQanda
3
2=OP)i((b)
4+3
)kb(4+)a3
2(3
=OG [M1]
7
kb4+a2=OG∴ [A1]
0=a)-(b•7
4kb+2a
0=AB.•OGsocollinear,areMandGO,Since(ii)
[M1]
0=)1(2-)2
1)(k4-2(+)1(k4
2
1=60cos)1)(1(=b•a
0=a2-b•a)k4-2(+bk4
22
22
o
2
1=k [A1]
b2
1=OQ
a3
2b
2
1PQ −−−−==== [A1]
2
1=60cos)1)(1(=b•a)c( o (relax on dot) [B1]
)a3
2b
2
1()a
3
2b
2
1(PQ
2
−−−−
−−−−==== o
[B1]
22
22
2
)1(9
4)
2
1(
3
2)1(
4
1
a9
4ba
3
2 b
4
1
)a3
2b
2
1()a
3
2b
2
1(PQ
++++−−−−====
++++••••−−−−====
−−−−
−−−−==== o
[M1]
36
13=
6
13=PQ [A1]
3
300
B
4300
a
b
2 1
k
1-k
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2
4. Correct arrows & labels (angles &magnitude) in parallelogram or triangle of forces
[correct form/shape & ≥2 labels in either diagram D1] [all correct D1]
)180cos()P(P2PP6 222 αααα−−−−°°°°−−−−++++==== ⇒⇒⇒⇒ ααααcosP2P236 22++++==== (eqn1) substitute into either eqn [M1]
)180cos()P4(P2P4P18 222 αααα−−−−°°°°−−−−++++==== ⇒⇒⇒⇒ ααααcosP8P17324 22++++==== (eqn2) both eqn correct [A1]
Eqn2 −−−− 4 ×××× eqn 1 , 2P9180 ==== , N472.420P ======== , solving eqn [M1] , ans for P [A1]
From eqn 1 , ααααcos)20(2)20(236 ++++====
°°°°°°°°°°°°====
−−−−====
−−−− 7.95@73.95@'449540
4cos 1αααα answer for αααα [A1]
5(a)Arcs ED, DC, CB and BA make up half of a whole circle, therefore
ECA is a semicircle. [B1]
collinearareAand,O,Ehence,180EOAo
====∠∠∠∠ [B1]
(b) EA is a diameter, therefore o90EBA ====∠∠∠∠ [B1]
Given OK is perpendicular to AB & K midpoint of AB
OK is parallel to EB [B1]
EB2
1=OK ( midpoint theorem ) [B1]
Similarly, AD2
1=OJ [B1]
(c) DB is common to both triangles. [B1]
DC=DE & CB=AB the corresponding arcs subtend equal angles at the
circumference [B1]
Hence αααα====∠∠∠∠====∠∠∠∠====∠∠∠∠ IBDEBDCBD & θθθθ====∠∠∠∠====∠∠∠∠====∠∠∠∠ IDBADBCDB [B1]
DIBDCB ∆∆∆∆∆∆∆∆ ≡≡≡≡ (by using ASA) [B1]
xk dt
dx.6 −−−−==== Rate of change [B1]
x =100 and 1dt
dx−−−−==== ⇒ k = 0.1.Solving for k [B1] [M1] [A1]
∴ x1.0dt
dxisDE −−−−==== For correct DE [A1]
dxx 2
1
∫∫∫∫−−−−
= ∫∫∫∫−−−− dt 1.0 ⇒⇒⇒⇒ 2
1
x2 = ct 1.0 ++++−−−−
Separation of variables and their integration [B1] [ M1]
for correct RHS ∫∫∫∫ ⇒⇒⇒⇒ 2
1
x2 [A1]
for correct LHS ( kt ±±±± ) [A1]
arbitrary constant C included
or equivalent statement of both pairs of definite integral
limits [B1]
0t,200x ======== ⇒⇒⇒⇒ 2002c ==== for evaluating C [M1]
2002t1.01002,100x ++++−−−−======== for evaluating t [M1]
)figure tsignificanone to ( min 80 about
for leaking been has container the Hence.utesmin84.82=t
For correct value 82.84@82.8 mins seen before rounding [A1]
2
2 2
2
2
3
3
3
C 2 3
3
α
αθ
θ
P N
P N
P N
P N
P N
18 N
4P N
6 N
180-α α
α
180-α
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3
7. (a) median = RM261.70 Q 1 = RM252.00 Q 3 = RM297.20 B1 B1 B1
7. (b) Box = D1 Whiskers = D1 * Must be drawn on graph paper.
Q 1 Q 2 Q 3 outlier
••••
Income (RM)
220 260 300 340 380 420
7.(c) Q 2 −−−− Q1 = 9.70 and Q 3 −−−− Q 2 = 35.50
Q 3 −−−− Q 2 > Q 2 −−−− Q1 , therefore the distribution is positively skewed. M1 A1
Upper boundary = Q 3 + 1.5 ( Q 3 −−−− Q1 ) = RM 365.00 , Q 3 −−−− Q1 = 45.20
Lower boundary = Q1 −−−− 1.5 ( Q 3 −−−− Q1 ) = RM 184.20 M1 ( find boundaries )
∴∴∴∴ The outlier is RM 398.40 A1
8. (a)
Age ( x , years ) Mid-point Frequency Class width Frequency density
5 ≤≤≤≤ x <<<< 8 6.5 99 3 33
8 ≤≤≤≤ x <<<< 11 9.5 143 3 47.4
11 ≤≤≤≤ x <<<< 13 * 12 130 2 65
13 ≤≤≤≤ x <<<< 14 13.5 78 1 78
14 ≤≤≤≤ x <<<< 15 14.5 84 1 84
15 ≤≤≤≤ x <<<< 16 15.5 91 1 91
Suitable Scale = D1
Correct Rectangles = D1
8. (b) mean = 625
5.7243
M1
mean = 11. 5896
= 11.59 @ 11.6 A1
variance = 625
75.89547 −−−−
2
625
5.7243
B1 for 89547.75
standard deviation =
2
625
5.7243
625
75.89547
− M1
= 2.992 or 2.99 A1
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4
8.(c)
median = 11 +
( )
143
22426252
1×
−
M1
= 11.986
= 11.99 or 12.0 A1
mode = 8 + 31344
44×
+ M1
mode = 10.3158
= 10.32 or 10.3 A1
8. (d) Median because the distribution is negatively skewed. B1 B1
9. (a) A = the event that the first digit is correct P ( A ) = 0.8
B = the event that the second digit is correct P ( B ) = 0.86, P( A ∩∩∩∩ B ) =0.72
P( A ∩∩∩∩ B′′′′ ) = P ( A ) −−−− P( A ∩∩∩∩ B ) = 0.8 −−−− 0.72 = 0.08 M1 A1
9. (b) P ( B′′′′ A′′′′ ) = )'(
)''(
AP
BAP ∩=
)(1
)(1
AP
BAP
−
∪−=
[ ])(1
)()()(1
AP
BAPBPAP
−
∩−+−
= 8.01
]72.086.08.0[1
−
−+− =
2.0
06.0 = 0.3 M1 A1
10. (a) X = x 1 2 3 4 5 Total B3 if 5 values correct
P(X =
x) 7
3
7
2
35
6
35
3
35
1
1 B2 if 4 values correct
B1 if 3 values correct
P(X=1) =7
3 P(X=2) =
7
4××××
6
3 =
7
2 P(X=3) =
7
4××××
6
3 ××××
5
3 =
35
6
P(X=4) = 7
4××××
6
3 ××××
5
2××××
4
3 =
35
3 P(X=5) =
7
4××××
6
3 ××××
5
2××××
4
1 ×××× 1 =
35
1
10(b) E(X) = 1 ××××
7
3 + 2 ××××
7
2 + 3 ××××
35
6 + 4 ××××
35
3 + 5 ××××
35
1 = 2 B1
E(X 2 ) = 1 2××××
7
3 + 2 2
××××
7
2 + 3 2
××××
35
6 + 4 2
××××
35
3 + 5 2
××××
35
1 =
5
26
Var(X) = 5
26 −−−− ( 2 ) 2 =
5
6 their E(X 2 ) M1√ A1
11.(a)
dtet
∫∞
−
0
µλ = 1
∞
−
−
0
te
µ
µ
λ = 1 M1
0 + µ
λ = 1
λλλλ = µµµµ A1
Given E ( T ) = 20
dtet
∫∞
−
0
t µλ = 20 B1
dtet
∫∞
−
0
t λλ = 20 because µµµµ = λλλλ
[ ] ∞−−
0
tet
λ + dte
t
∫∞
−
0
λ
= 20 M1 “by part”
0 +
∞
−
−
0
1 te λ
λ = 20
0 +
−−
λ
10 = 20 ;
λ
1 = 20 ,
∴∴∴∴λλλλ = 0.05 and µµµµ = 0.05 A1 Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/
5
11.(b)
f ( t ) = >
−
otherwise
te t
0
0 05.0 05.0
P ( T < 6 ) = dtet
∫−
6
0
05.0 05.0
= [ ] 6
0
05.0 te
−− M1
= 1 −−−− 3.0 −e
= 0.2592 or 0.259 A1
12.(a) X ∼∼∼∼ N ( µµµµ , 0.03 2 ) and P ( X > 2 ) ≥0.95
−>
03.0
2 µZP ≥ 0.95 . M1 standardized
03.0
2 µ− ≤ −−−−1.645 B1 for ( −−−−1.645 )
µµµµ ≥ 2.049 kg or µµµµ ≥ 2.05 kg A1
12.(b) (i)
X represents the number of the intended calls obtained
X ∼∼∼∼ B ( 10 , 0.95 ) x = 0,1,2,…10
P ( X ≥≥≥≥ 8 ) = P ( X = 8 ) + P ( X = 9 ) + P ( X = 10 )
= 8
10C (0.95) 8 (0.05) 2 + 9
10C (0.95) 9 (0.05) 1 + 10
10C (0.95) 10 (0.05) 0 M1
= 0.9885 or 0.989 A1
(ii) Y represents the number of unsuccessful calls.
Y ∼∼∼∼ B ( 300 , 0.05 ) x = 0,1,2,…300
Normal Approximation : µµµµ = np = 15
σσσσ2 = n p q = 14.25 B1 ( both correct) may be implied
P ( 10 ≤≤≤≤ Y ≤≤≤≤ 20 ) =
−≤≤
−
25.14
155.20
25.14
155.9ZP M1√(continuity correction & standardise)
= P ( −−−−1.457 ≤≤≤≤ Z ≤≤≤≤ 1.457 )
= 1 −−−− 0.0725 −−−− 0.0725
= 0.855 A1
(iii) T represents the number of unobtainable calls .
T ∼∼∼∼ B ( 400 , 0.004 ) x = 0,1,2,…400
Poisson Approximation : λ = np = 1.6 B1 may be implied
P ( T <<<< 3 ) = P ( T = 0 ) + P ( T = 1 ) + P ( T = 2 )
= 6.1−e
++
!2
6.1
!1
6.1
!0
6.1 210
M1 √ their λ
= 0.783 A1
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