maths t paper 2 jit sin 2011
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−1−
KOD KERTAS : 954
08.09.2011
SMJK JIT SIN
BUKIT MERTAJAM
PEPERIKSAAN PERCUBAAN
2011
UPPER SIX
MATHEMATICS T
PAPER 2
THREE HOURS
Arahan kepada calon
JANGAN BUKA KERTAS SOALAN INI SEHINGGA ANDA DIBENARKAN
BERBUAT DEMIKIAN
Instruction to candidates.
Answer all the questions.
The marks for questions or parts of questions are given in brackets [ ].
All necessary working must be shown clearly.
For questions which do not require answers in any specified form, the answer
may contain , e, surds, or fractions.
Kertas soalan ini terdiri daripada 5 halaman bercetak
*Kertas soalan ini SULIT sehingga peperiksaan kertas ini tamat
−2−
Answer all questions.
1. Find the values of x, where 0 x , which satisfy the equation
02
x3sin2xsin
2
xsin2 . [4 marks]
2. The diagram below shows a prism with uniform cross section. It has a rectangle
base PQRS of length 4 cm and width 3 cm. Another identical rectangle PQTU is
inclined at 30o to the base PQRS.
Find
(a) the inclination of the line PT to the horizontal plane, [3 marks]
(b) the angle between planes PTS and PQRS. [2 marks]
3. The position vectors of the points P, Q, R and S, relative to an origin, are 2pi + 5j,
−4i − (q −1)j, (x + 3)i + j and (x + 5)i + 7j respectively.
(a) Determine the values of p and q for which PQRS is a parallelogram. [3 marks]
(b) Determine the value(s) of x for which PQRS is a rhombus. [3 marks]
4. The position vectors of three points P, Q and R, relative to an origin O, are b3a2 ,
ba4 , and b3a5 respectively. L, R and S are the midpoints of PQ, QN and PN
respectively. Find, in terms of a and b , the position vectors of L, N and S. The point M
lies on PR such that PRMR . Express OM in terms of , a and b . The point K
lies on QS such that QSKS . Express OK in terms of , a and b . Hence, find the
position vector of T in terms of a and b if T is the intersecting point between PR and
QS. [10 marks]
P
T
R
Q
S
3
4
U
3 30
o
−3−
5. Find the general solution of the differential equation
,2y3ydx
dyxcos 22 where 0 < y < 1.
Hence, express y explicitly in terms of x for which y = 0.5 when x = 0. [6 marks]
6. The diagram below shows the circumscribed circle of the quadrilateral ABGE. The
tangent to the circle at B meets the line AE extended to C. AFGD and BFE are straight
lines.
(a) Show that CEGD is a cyclic quadrilateral. [4 marks]
AFGD is the angle bisector of the angle BAE. Show that
(b) the triangles BDG and BFG are congruent, [3 marks]
(c) the triangles ADC and AEG are similar, [3 marks]
(d) ADAGACAE . [2 marks]
7. Class U6P2 of SMJK JIT SIN consists of 16 boys and 10 girls. In the beginning of
the year, four officers are to be chosen at random as “Class Monitor”, “Assistant Class
Monitor”, “Treasurer” and “Cleaniness Officer”. Find, giving your answers correct to
three significant figures, the probability that
(a) all the four officers are boys, [2 marks]
(b) two officers are girls and two are boys, [2 marks]
(c) the “Treasurer” and “Cleaniness Officer” are girls given that the two monitors
are of opposite sex. [3 marks]
A B
C
D
E
F
G
−4−
8. The rate of change of water temperature is described by the differential equation
)(kdt
do
where is the water temperature at time t measured in hours, o is the surrounding
temperature, and k is a positive constant.
A boiling water at C100o is left to cool down in a room temperature of o30 C. The
water takes 1 hour to decrease to the temperature of o70 C. Show that k = 4
7ln .
[5 marks]
When the water reaches o60 C, the water is placed in a freezer at − o4 C to be frozen to
ice. Find the time required, from the moment the water is put in the freezer until it
becomes ice at o0 C. [5 marks]
9. A nutrition drink „Milo‟ is packed in packets of two sizes. The mass in each small
packet may be regarded as an normal variable with mean 1005 g and standard deviation
8 g while the mass in each large packet is another independent normal variable with
mean 2008 g and standard deviation 12 g.
(a) Find the probability that the mass of one randomly chosen large packet is less than
2000 g. [2 marks]
(b) Find the probability that the mass of one randomly chosen large packet exceeds the
total mass of two randomly chosen small packets. [3 marks]
(c) Find also the probability that the mass of one randomly chosen large packet is less
than twice the mass of one randomly chosen small packets. [3 marks]
(d) Three large packets and six small packets are chosen at random.
Find the probability that the average mass of all the nine packets chosen lies between
1330 g and 1345 g. [4 marks]
10. Rectangular wooden doors produced by a manufacturer have flaws which occur at
random at the rate of 0.15 per square metre. Find the probability that a wooden door
with dimension 2 m by 1 m will contain exactly one flaw. [3 marks]
For 50 such doors produced, find, using a suitable approximation, the probability that at
least 40 of them will contain no flaw. [4 marks]
−5−
11. The continuous random variable X has probability density function
f(x) =
,2
1x,
x
m
,2
1x0,x21
,0x,0
3
where m is a constant.
(a) Determine the value of m. [3 marks]
(b) Find the cumulative distribution function of X. [3 marks]
(c) If 2P(X > n) = 3P(X n), find the exact value of n. [3 marks]
12. The lifespans of 100 new tyres manufactured by a tyre company are shown in the
following table.
Lifespans,
x (km) x 69500 x 74500 x 79500 x 84500 x 89500 x 94500
Cumulative
frequency 0 8 30 67 88 100
Prepare a frequency table for the above grouped data. [2 marks]
(a) Draw a histogram of the grouped data. Comment on the shape of the frequency
distribution. [3 marks]
(b) Calculate estimates of the mean, median and mode of the lifespans of the new tyres.
Use your calculations to justify your statement about the shape of the frequency
distribution. [7 marks]
~~ The End ~~
−6−
No Working/Answer Partial
marks
Total
marks
1 0xsin)
2
xcosxsin2(2 [use formula sum to product]
0)12
xcos4(xsin
sin x = 0, 4
1
2
xcos
0, 2.64 (or 0.839 rad),
M1
M1
A1
A1
4
2
(a) TW = 1.5 cm, TP = 5 cm
sin TPW = 5
5.1 or 0.3
TPW = 17.46o or 17
o27‟ (or 17.5
o)
B1 (both)
M1
A1
5
(b) tan TXW = 4
5.1 or 0.375
TXW = 20.56o or 20
o33‟ (or 20.6
o)
M1
A1
3 (a) PQ = (−4 − 2p)i + (− q − 4)j, SR = − 2i − 6j
(−4 − 2p)i + (− q − 4)j = − 2i − 6j
−4 − 2p = −2, −q − 4 = −6
p = −1, q = 2
B1 (both)
M1
A1 (both)
6
(b) PR = (x + 5)i − 4j, QS = (x + 9)i + 8j
0QSPR
(x + 5)(x + 9) − 32 = 0
x = −1, −13
M1
M1
A1
P
T
R
Q
S
3
4
U
3 30
o
W
X
30o
−7−
4 baOL
2
ONOQOR
b5a14ON
ba8OS
OR)1(OPOM
b)63(a)35(OM
OS)1(OQOK
ba)128(OK
b)63(a)35( ba)128(
= 3
1 or =
3
1
ba4OT
B1
B1
B1
M1
A1
M1
A1
M1
A1
A1
10
5
xcos
dx
2y3y
dy22
xtandyy2
1
y1
1
cxtan)y2ln()y1ln(
c0tan5.02
5.01ln
y = 3e
3e2xtan
xtan
or equivalent
B1
M1A1
M1
M1
A1
6
6
(a) Let DBG = ,
12
A B
C
D E
F
G
Q N
R 2
1
P
L
1 − T
1
1
1
S
1
1 1 −
−8−
Statement Reason
ABD = 90o,
BAG = DBG =
BDC is a tangent to the
circle at B
Alternate segment theorem
BEG = BAG = angles subtended by same
segments are equal
ABG = 90o − ,
AGB = 90o
AEB = 90o
Sum of the interior angle of
a triangle is 180o
angles subtended by same
segments are equal
CEG = 90o − Complementary angle
CDG = 90o + Exterior angle of the
ABD
CDG + CEG
= (90o + ) + (90
o − )
= 180o
CEGD is a cyclic
quadrilateral
B1
B1
B1 (both)
B1
(b)
Statement Reason
DBG =
BGD = BGA = 90o AFGD is a straight line
EAG = BAG =
AFGD is the angle bisector
of the angle BAE
EBG = EAG =
FBG = EBG =
FBG = DBG =
angles subtended by same
segments are equal
BG is a common side of
the triangles BDG and
BFG
the triangles BDG and
BFG are congruent
SAS
B1
B1
B1
(c)
Statement Reason
ADC = AEG = 90o +
CAD = EAG = Common angle
ACD =AGE= 90o−2
Sum of the interior angle of
a triangle is 180o
B1
B1
−9−
(the third angle of a
triangle
ADC and AEG are
similar
All the corresponding
angles are equal
B1
(d)
Statement Reason
AC
AG
AD
AE
Since ADC and AEG
are similar
AEAC = AGAD
B1
B1
7 (a)
4
26
4
16
C
C OR
23
13
24
14
25
15
26
16
= 0.1217 (or 0.122)
M1
A1
6
(b) 4
26
2
10
2
16
C
CC OR
!2!2
!4
23
9
24
10
25
15
26
16
= 0.3612 (or 0.361)
M1
A1
(c) 23
8
24
9 OR
)225
10
26
16(
)223242526
891016(
= 0.1304 (or 0.130)
M1
A1
8
dtk
30
d
ckt30ln
70lnc
70lnk3070ln
k = 4
7ln
B1
M1
A1
M1
A1
11
)4)(4
7(ln
dt
d
dt
4
7ln
4
d
dt)4
7(ln4ln
64lnd
64lnt)4
7(ln4ln
t = 4.95 hrs or 4 hrs 57 min
B1
B1
M1
A1
M1
A1
−10−
9 (a) )
12
20082000Z(P)2000L(P
(normalize)
= 0.2525
M1
A1
12
(b) )0SSL(P)SSL(P 2121
)272
)2(0Z(P
= 0.4517
B1
M1
A1
(c) )0S2L(P)S2L(P
)20
)2(0Z(P
= 0.5398
B1
M1
A1
(d) 9
SSSSSSLLLT 654321321
)
27
272
3
40181345
T
27
272
3
40181330
(P
= R(−2.9406) − R(1.7854)
= 0.9613
B1
M1
M1
A1
10 X ~ Po(0.3)
P(X = 1) = )3.0(e 3.0
= 0.2222
B1
M1
A1
7
Y ~ B(50, 3.0e ) or Y ~ B(50, 0.7408)
np = 50(0.7408) = 37.04 > 5
nq = 50(0.2592) = 12.96 > 5
Using Normal approximation,
Y ~ N(37.04, 9.6008) approximate
P(Y 40) = P(Y > 39.5) continuity correction
= )6008.9
04.375.39Z(P
= P(Z > 0.7939)
= R(0.7939)
= 0.2136
B1
B1
M1
A1
11
(a) 1dxx
mdxx21
2
13
2
1
0
M1
9
−11−
12
xmxx
2
1
2
2
1
0
2
1x
1
2
m
4
1
2
1
2
12
m = 8
3
M1
A1
(b) For 0 t 2
1, F(t) =
t
0
t
0
dxx21dx)x(f
For t > 2
1, F(t) =
t
2
13
t
2
1
dxx8
3)
4
1
2
1(dx)x(f)
2
1(F
F(x) =
.2
1x,
x16
31
,2
1x0,xx
,0x,0
2
2
M1
M1
A1
(c) 2[1 − P(X n)] = 3P(X n)
P(X n) = 5
2 > F(
2
1) and F(
2
1) = P(X
2
1) =
4
1
2
1 =
4
1
P(X n) = F(n) = 1 − 2n16
3 =
5
2
since n > 2
1, n =
4
5
B1
M1
A1
12
Lifes
pans,
x
(×104
km)
6.95 < x
7.45
7.45 < x
7.95
7.95 < x
8.45
8.45 < x
8.95
8.95 < x
9.45
f 8 22 37 21 12
B1 (x)
B1 (f)
12
(a) Histogram : Frequency is proportional to the height on
graph paper
All correct
The distribution is symmetrical
D1
D1
B1
−12−
(b) 100
105.823x
4
mean = 82350 km
)5000(37
305079500M
median = 82202 km
Mode = 79500 + 50001615
15
mode = 81919 km
mean median mode
The distribution is symmetrical
(or almost symmetrical or slightly skewed or not skewed)
M1
A1
M1
A1
M1
A1
B1
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