m-MT KEDAH 2010
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KEDAH STPM TRIAL EXAMINATION 2010 MATHEMATICS T PAPER 1 CHU/SMKK
1. Find the solution set of the inequality 121 xx [4 marks]
2. Given point A(2,k) lies on the curve 032 33 xyyx , find the value of k.
Find also the gradient and equation of the normal to the curve at point A. [6 marks]
3. Express )2)(1(
43
rrr
r
rU in partial fractions. [3 marks]
Hence or otherwise, find
(a)
n
rr
U
1
[2 marks]
(b)
1rr
U [2 marks]
4. Three points have coordinates A ( 2 , 9 ) , B ( 4 , 3 ) and C ( 2 , −5 ). The line through
C with gradient 2
1 meets the straight line AB produced at D.
Find (a) the coordinates of D [3 marks] (b) the equation of the line through D perpendicular to the line 5y – 4x = 17 [3 marks] 5. Given that 3 + 2i, 5 – i and 4 – 6i are the first three terms of a geometric progression.
Find (a) the common ratio, [2 marks] (b) the fifth term, [3 marks] (c) the sum of the first 6 terms of this geometric progression. [2 marks]
6. Evaluate 4
0
2 2cos
dxxx . Give your answer in terms of . [7 marks]
7. The parametric equations of a curve are x = 4t , y = t
4 , where the parameter t takes
all non-zero values. The points A and B on the curve have parameters t1 and t2 respectively,
(a) Write down the coordinates of the midpoint of the chord AB in terms of
t1 and t2. [1 mark]
(b) Given that the gradient of AB is 2, show that t1t2 = 2
1 [3 marks]
(c) Find the coordinates of the points on the curve at which the gradient of the normal is 2
1.
[4 marks] 8. Functions f and g are defined by
f : 10 ,,222 xRxxxx and
g : 21 ,,1
2
xRx
x
xx respectively.
(a) Determine the range and inverse function of f. [4 marks]
(b) Given function h = gf, determine the range of h. [3 marks] (c) State with reason whether h has an inverse function. [1 mark]
9. (a) The matrices A and B are given by
421
134
432
A .
615
14415
9410
B .
State with reason whether matrix A is singular.
Find the matrix AB, and hence, deduce 1A . [5 marks]
(b) Using the result in (a), solve the system of linear equations
.
.334
,4132
,2 42
yzx
zyx
xzy
[5marks]
10. A curve is defined parametrically by x = 2t – 1, y = t3 and P is the point on the curve
when t = 2.
(a) Obtain an expression for dx
dy in terms of t and calculate the gradient
of the curve at P. [3 marks]
(b) Find 2
2
dx
yd in terms of t. [3 marks]
(c) Determine a Cartesian equation of the curve, expressing your answer in the form y = f(x). [3 marks] (d) Find the x and y intercepts. [1 mark]
11. Show that
(a) 2
0,1sin
dxxx . [2 marks]
(b) 2
0
2
4
1sin
xdx . [Hint: Use identity cos 2A = 1 – 2 sin2
A] [3 marks]
Find the area of the region bounded by the x-axis, the curve y = x – sin x and the line
x = .2
1 [3 marks]
Hence, show that the volume of the solid generated when the region bounded by the
x-axis, the curve y = x – sin x and the line x = 2
1 is rotated through 360
o about the
x-axis is . 48624
1 324 unit [4 marks]
12. Given p(x) = 1 2346 xbxaxx , where a and b are real constants. If (2x – 1)
is a factor of p(x) and (x – 1) is a factor of p‘(x),
(a) Find the values of a and b, factorise p(x) completely, and hence solve the equation p(x) = 0. [8 marks]
(b) Given that
)(132)13()( xqxxxp , find )(xq . Sketch the graph of )(xq and
determine the range of )(xq when 5,0x . [7 marks]
KEDAH STPM TRIAL EXAMINATION 2010 MATHEMATICS T PAPER 2 CHU/SMKK
1. By writing sin as cos ( 12
), solve the equation cos 3 = sin , for 0 < < . Give your answers
as multiples of . [4marks] 2. Three points P, Q, and R have position vectors 6i + 7j, 4j and 4i + 3j respectively. A is the point lying on
the line QR between Q and R such that 2
3
AR
QA , and B is the point on QR produced such that
2
3
BR
QB . Find the position vectors of A and B and evaluate the scalar product PA
PB
. Hence,
deduce the relationship between PA
and PB
. [7 marks]
3. P, Q, and R are three points on the horizontal ground with Q lies to the north of P and the bearing of R
from P is 060o. The angle of elevation of the point T at the top of a vertical tower located at Q from P
and R are each. Point S lies on PR such that PS : SR = 1 : 2. If the angle of elevation of T from S is ,
show that tan : tan = 3 : 7 [7marks]
4. Find the particular solution of the differential equation
ey dydx
1 = ln x ,
given that y = ln 2 when x = 1. [8 marks] 5. Two intersecting circles AOBC and APBQ are shown in the diagram below.
Given that O is the centre of the circle APBQ and ACO = 30.
(a) Find APC. [4 marks]
(b) Show that APC is an equilateral triangle. [3 marks] (c) Prove that BQ is parallel to PA. [3 marks]
6. Two boats P and Q are able to give light signal to one another. The light signal can be seen at a
maximum distance of 10 km. Initially, boat P is at a distance of 2.5 km due east of boat Q. At 1200 noon,
the two boats start to move with boat P moves towards east with a speed of 5 km h-1
and boat Q moves
in the direction north-east with a speed of 10 km h1.
(a) Find the magnitude and direction of the velocity of boat Q relative to boat P. [4 marks] (b) Calculate the shortest distance between the two boats and the time, to the nearest
minutes, when the two boats are at the shortest distance. [5 marks]
Q
C
P
O
B
A
(c) Find the time after which the two boats are unable to see the light signal from one another.
[5 marks]
7. Two bags each contains 8 discs which are indistinguishable apart from their colour. The first bag
contains 3 red and 5 black discs and the second, 6 red and 2 black discs. A disc is chosen at random
from the first bag and placed in second. Then, after thoroughly mixing, a disc is taken from the second
bag and placed in the first. Find the probability that the first bag still contains exactly 3 red discs.
[4 marks]
8. A continuous random variable X is distributed normally with mean and variance 2. Find the
value of if the probability that X lies within the range of 9.8 from the mean is [4 marks]
9. The mean and variance of the four numbers 2, 3, 6, 9 are 5 and 7.5 respectively. Two numbers m and n
are added to this set of four numbers, such that the mean is increased by 1 and the variance is
increased by 2.5 . Find m and n. [ 7 marks]
10. The binomial variable X represents the number of eggs laid each year by a certain species of
birds where E(X) = 4 and Var(X) = 43
. Find P(X = 6).
Hence, find the probability that four or more eggs hatched in a year given that the probability that any
egg hatched is 35
. [9 marks]
11. The following data shows the number of books borrowed from a school library for the past 26 days.
61 72 83 57 78 80 67 20 85 70 54 62 76 60 48 75
52 62 72 52 46 83 54 74 82 69
(a) Display the above data in an ordered stemplot. [2 marks]
(b) Find the median and interquartile range. [4 marks]
(c) Draw a boxplot to represent the above data. [3 marks]
(d) State the type of distribution of the above data. Justify your answer. [2 marks]
12. The continuous random variable X has probability density function given by
otherwise0,
1,x1for),xk(1f(x)
2
where k is a constant.
(a) Find the value of k. [ 3 marks]
(b) Sketch the graph of f(x) and hence state the value of E(X). [ 2 marks]
(c) Determine Var (X). [ 3 marks]
If A and B are the events represented by X > 2
1 and X >
4
3 respectively, find P(B) and P(B|A).
[ 7 marks]
MARKING SCHEME FOR MATHEMATICS T PAPER 1
1 ( x + 1 )2 ( 2x + 1 )
2 M1
3x2 + 2x 0 M1
x ( 3x + 2 ) 0 M1
x = (− ,3
2 ] [ 0, ) A1
2. 032 33 xyyx
8 + 2k3+ 6k = 0
k3 + 3k + 4 = 0 M1
(k + 1)(k2
− k + 4) = 0
k = −1 A1
03363 22 dx
dyxy
dx
dyyx
2
2
2yx
yx
dx
dy
M1
x = 2, y = − 1; 4
3
22
14
dx
dy A1
Equation of normal : y + 1 = )2(3
4x …. M1
y = 3
11
3
4x A1
3. Let 21)2)(1(
43
r
C
r
B
r
A
rrr
r
)1()2()2)(1(43 rcrrBrrrAr B1
1,1,2 CBA B1
2
1
1
12
)2)(1(
43
rrrrrr
r A1
(a)
n
rrU
1
(3
1
2
1
1
2 )
+(4
1
3
1
2
2 )
+(5
1
4
1
3
2 ) M1
+(nnn
1
1
1
2
2
)
+(1
1
1
1
2
nnn)
+(2
1
1
1
2
nnn)
2
1
1
2
2
112
1
nn
n
rrU
= 2
1
1
2
2
5
nn
= )2)(1(2
)95(
nn
nn A1
(b) 4622
925n
nlim
1
nn
n
rrU
=
2
462
95
lim
nn
n
n
M1
= 2
5 A1
4. (a) y = 2
1x – 6 B1
y = − 3x+15 B1 D = (6,−3) A1
(b) y + 3 = −4
5( x – 6 ) B1M1
5x + 4y − 18 = 0
5. (a) i
i
i
i
i
ir
23
23
23
5
23
5
M1
= 1 – i A1
(b) 4T (4 – 6i) x (1 – i) = − 2 – 10i B1
5T = (− 2 – 10i) x (1 – i) M1
= − 12 – 8i A1
(c) 6T = (−12 – 8i)(1 – i) = − 20 + 4i
6S = (3 + 2i) + (5 – i) + (4 – 6i) + (−2 –10i) +
(−12 – 8i) + (−20 + 4i) M1 = − 22 – 19i A1
6. 4
0
2 2cos
dxxx =
4
0
2sin4
0
2sin2
2
1
dxxxxx M1M1
= 4
04
2sin4
0
2cos2
14
0
2sin2
2
1
xxxxx M1M1
=
0
42sin
4
10
42cos
42
10
42sin
2
42
1
A1A1
= 4
1
32
2
A1
7. (a)
2
44
,2
44 2121 tttt
)
11(2),(2
2121
tttt B1
(b) 244
44
12
12
tt
tt M1
2)( 1221
21
tttt
tt M1
21
21
tt
t1t2 = 2
1 A1
(c) dt
dx = 4 an
dt
dy= −4t
-2 Both correct B1
dx
dy =
2
1
t
2
1
t = − 2 M1
t = 2
1 A1
Points are (2 2 , 4 2 ) and (−2 2 , −4 2 ) A1
8 (a) Range of f = {y : y }21 , yR B1
Let f1
(x) = a f(a) = x
a2
− 2a + 2 = x
a2
− 2a + (2 – x) = 0 M1
)1(2
)2(422 2 xa
11 xa M1
21 , ,11:1 xRxxxf A1
(b) h(x) = g( )222 xx
= 122
2222
2
xx
xx=
32
422
2
xx
xx M1
= 1 + 32
12 xx
x = 0, h(x) = 1 + 3
4
3
1 both
x = 1, h(x) = 1 + ½ = 2
3 B1
range of h = },2
3
3
4:{ Ryyy A1
(c) h has an inverse function, h is a one-to-one function B1
9. (a) )38(4)116(3)212(2 A
05 M1
A is not singular because 05 A . A1
615
14415
9410
421
134
432
AB
I5
500
050
005
M1A1
5
6
5
11
5
14
5
43
5
9
5
42
615
14415
9410
5
11A
A1
(b)
2
3
1
421
134
432
z
y
x
B1
2
3
1
615
14415
9410
5
1
z
y
x
M1
2
3
1
615
14415
9410
5
1
20
55
40
5
1 A1
4
11
8
A1
x=8, y=−11, z=4 A1
10. (a) dt
dx = 2 and
dt
dy= 3t
2 M1
2
3 2t
dx
dy M1
t = 2 , dx
dy = 6 A1
(b) dx
dtx
t
dx
yd
2
62
2
M1
= 2
1
2
6x
t M1
= 2
3t A1
(c) 2
1
xt M1
y =
3
2
1
x M1
= 318
1x A1
(d) y – intercept = 8
1
x – intercept = −1 Both correct A1
2
5
6
5
20
y
x
11. (a) 2
0
20
2
0coscossin
xdxxxxdxx
= 20sincos
xxx M1
= 10sin02
sin2
cos2
A1
(b) 2
0
22
02cos1
2
1
dxxxdxins M1
=2
0
2sin2
1
2
1
xx M1
= 4
)0sin2
10(sin
2
1
22
1
A1
Area = 2
0sin
dxxx M1
= 2
0
2
cos2
x
x M1
= )0cos0(2
cos8
2
22
18
unit
A1
Volume = 2
0
2sin
dxxx B1
= 2
0
22 sinsin2
dxxxxx
= 2
0
22
0
2
0
2 sinsin2
xdxxdxxdxx M1
=
4)1(2
3
2
0
3
x M1
= 4
224
24
48624
1 24 unit3 A1
12.a) 01 2
12
2
13
2
14
2
16
2
1
bap
M1 12 ba M1
01)1(22)1(33)1(24)1(' bap M1
2523 ba
13a , 7b A1A1
)13)(1)(1)(12()( xxxxxp A1
0)13)(1)(1)(12( xxxx
3
1 ,1 ,
2
1x A1
(b)
)2425(132)13()( xxxxxp M1
2425)( xxxq A1
5
62
5
25)(
xxq M1
Minimum point
5
6,
5
2 B1
Shape of graph D1
5
62
5
255q(5) ,5
x M1
107
5,0x ,
107,
5
6qR A1
P R
Q
T
S
60o
MARKING SCHEME FOR MATHEMATICS T PAPER 2
1. cos 3 = cos ( 12 )
cos 3 cos ( 12 ) = 0 B1
2sin
π
4
1θ sin
π
4
12θ =0 M1
sin
π
4
1θ = 0 or sin
π
4
12θ = 0
π
4
1θ = or
π
4
12θ = 0, M1
= π4
3 or = π
8
1, π8
5
= π8
1, π
4
3, π
8
5. A1
2. ARQA 32
OAOROQOA 32
OQOROA5
2
5
3 M1
= ji 345
3 + j4
5
2
= ji5
17
5
12 A1
BRQB 32
OBOROQOB 32
OQOROB 23 M1
jji 42343
ji12 A1
ji5
18
5
18 OPOAPA
ji 66 OPOBPB
jiji 665
18
5
18
PBPA
= 65
186x
5
18
M1
= 0. A1
PA is perpendicular to PB . B1
3.
Let h = TQ (height of vertical tower)
tan = h
PQ =
hQR
therefore PQ = QR = h
tan M1
( PQR is isosceles)
QPR = QRP = 60o.
( PQR is equilateral)
PR = h
tan M1(Implied)
and SR = 2h
3tan M1
QS2 = QR
2 + SR
2 2(QR)(SR) cos 60
o.
QS2 = (
h
tan )
2 + (
2h
3tan )
2
2(h
tan )(
2h
3tan ) cos 60
o M1
QS = 7 h
3 tan A1
tan = h
QS M1
tan = 3 tan
7
tan : tan = 3 : 7 A1
4. dxxdye y )ln1( M1
dx
xxxxxye
1ln + c M1A1
cxxey ln A1
2ln,1 yx , c = 2 M1A1
2ln xxey M1
Particular solution xxy ln2ln A1
5.(a)
ABO = ACO = 30 ( Angles subtended by the same arc) B1
BAO = ABO = 30(Base angles of isosceles ) B1
AOB =180 BAO ABO ( Sum of interior angles of )
= 120 M1
APC =
2
1 AOB(Angle at circumference is half angle at centre)
= 60 A1
5(b)
S
5
2
5 2 5
ACP = 180− AOB = 60 ( Opposite angles of cyclic
quadrilateral) B1
CAP=180 ACB APB ( Sum of interior angles of )
= 60
Since APC = ACP = CAP= 60 B1
APC is an equilateral triangle. B1
5(c)
QBC = CAP = 60 (Exterior angles of cyclic quadrilateral) B1
QBC = APC
B1
BQ is parallel to PA (Corresponding angles are equal) B1
6. (a) VP = 5i + 0j , VQ = 5 2 i + 5 2 j.
VQP = VQ VP.
=(5 2 5)i + 5 2 j. M1
| VQP |2 = (5 2 5)
2 + (5 2 )
2
|VQP | =7.368 km h1 or 7.37 km h1
. A1
tan = 5 2
5 2 5 M1
= 73o 41’
The direction of VQP is at N 16o 19’ E A1
(b) At time t, position vector of Q = OQ
OQ
= (0i + 0j) + (VQP)t
OQ
= (5 2 5)t i + (5 2 )t j. M1
Position vector of P = OP
OP
= 2.5i + 0j
PQ
= OQ
OP
PQ
= [(5 2 5)t 2.5] I + (5 2 )t j. M1
PQ
= (2.071t 2.5) I + 7.071t j .
| PQ
|2 = s
2 = (2.071t 2.5)
2 + (7.071t)
2
2s dsdt
= 4.142(2.071t 2.5) + 100t M1
For shortest distance dsdt
= 0
4.142(2.071t 2.5) + 100t = 0
t = 0.09536 hr
t = 5 min 43 sec A1
s2 = [2.071(0.095) 2.5}
2 + {7.071(0.095)}
2
s2 = 5.756 km
The shortest distance s = 2.399 km or 2.40
km A1
(c) | PQ
| = 10 M1
102 = (2.071t 2.5)
2 + (7.071t)
2 M1
100 = 54.288t2 10.355t + 6.25
54.288t2 10.355t 93.75 = 0 M1
t = 1.413
t = 1 hour 25 min A1
The time is 1325 A1
Method 2
VQP = VQ VP
VQP2 = 10
2 + 5
2 2(10)(5) cos 45
o. M1
VQP = 7.37 km h1. A1
sin
5 =
sin 45o
7.37
= 28o 40’, or 28.7
o
The direction of VQP is at N 16o 20’ E or N 16.4
o E A1
(b) The shortest distance between the two boats = d
sin 73o 40’ =
d2.5
M1
d = 2.399 km or 2.40 km A1
Time taken t = 2.5 cos 73
o40'
7.37 M1
t = 5 min 43 sec A1
The time when the two boats are at shortest
distance is 1206 A1
. (c) sin 2.5
= sin 73
o40'
10 M1
sin = 0.2399
= 13o 53’ A1
= 92o27’
QTsin 92
o27'
= 10
sin 73o40'
M1
QT = 10.41 km
t = 10.417.37
t = 1 hour 25min M1
The time is 1325 hr A1
7. Required Probability = P ( R1R2)+P(B1B2)
= 9
3x
8
5
9
7x
8
3 B1B1M1
= 2
1 A1
9. New mean = 6
66
9632
nm M1
16 nm ……….(1) A1
New variance = 7.5+2.5=10
1066
813694 222
nm
M1
14622 nm ……(2) A1
Solving (1) and (2):
VQ = 10 km
h1
Vp = 5 km
h1
4
5o
4
5o
VQP
o
2.5 km
d
P Q
73o40’
10 km
T
1461622 mm
0110322 2 mm
511 morm A1
5,11 nm or 11,5 nm
The two numbers are 5 and 11. A1
10.
X B( n, p)
np = 4 , np(1-p) =3
4 M1(Both)
p = 3
2 and n = 6 A1
X B( 6, 3
2)
P(X= 6) = (3
2)6 M1
= 64
729 A1
P( Y 4) = 64
729 [P ( Y = 4 ) + P ( Y = 5 ) + P ( Y = 6)] M1
= 64
729 [
6C4(
35 )
4 (
25 )
2 +
6C5 (
35 )
5(25 )
1 + (
35 )
6]
M1M1
= 64
729 [
9723125
+ 2916
15625 +
6415625
A1
= 0.0478 A1 11.
(a)
b)
Median = 68 B1
Q1 = 54, Q3 = 76
Interquartile range = 76 54 B1B1
= 22 B1
(c)
R1( shape + his Q’s)
R1( outlier)
R1 ( All correct)
(d) Skewed negative. Reason : Q2Q1 > Q3−Q2 B1B1
12
(a) X is random: 11
1
1
2
dxxk
13
1
1
3
xxk M1
13
1)1(
3
11
3
k A1
8
3k A1
b)
M1 h has an inverse function, h is a one-to-one function B1 A1
R1
By symmetry : E(X)=0. B1
(c) Var(X) = 01
1
1
22
dxxkx M1
=
1
1
53
53
xxk M1
= 5
2
15
16k A1
Stem Leaves
2 0
3
4 6 8
5 2 2 4 4 7
6 0 1 2 2 7 9
7 0 2 2 4 5 6 8
8 0 2 3 3 5
B1 Key : 5|7 means 57 B1
54 68 76
85 46 20
outlier
x
218
3xy
f(x)
8
3