−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