stpmmathst kedah&scheme 2010

8/8/2019 STPMmathsT KEDAH&Scheme 2010

1/10

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 gradient2

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 iand 4 6iare 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 ttakes

all non-zero values. The points A and B on the curve have parameters t1and t2 respectively,

(a) Write down the coordinates of the midpoint of the chord AB in terms oft1 and t2. [1 mark]

(b) Given that the gradient of AB is 2, show that t1t2=21 [3 marks]

(c) Find the coordinates of the points on the curve at which the gradient of the normal is2

1.

[4 marks]

8. Functions fand gare 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 hhas an inverse function. [1 mark]


2/10

9. (a) The matrices A and Bare given by

421

134

432

A .

615

14415

9410

B .

State with reason whether matrix A is singular.

Find the matrix AB, and hence, deduce1A . [5 marks]

(b) Using the result in (a), solve the system of linear equations.

.334

,4132

,242

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 tand calculate the gradient

of the curve at P. [3 marks]

(b) Find2

2

dx

ydin terms of t. [3 marks]

(c) Determine a Cartesian equation of the curve, expressing your answer in the formy = f(x). [3 marks]

(d) Find the xand yintercepts. [1 mark]

11. Show that

(a) 20 ,1sin

dxxx . [2 marks]

(b) 202

4

1sin

xdx . [Hint: Use identity cos 2A = 1 2 sin 2 A] [3 marks]

Find the area of the region bounded by the x-axis, the curve y = x sin xand 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 xand the line x= 2

1is rotated through 360

oabout the

x-axis is .48624

1 324unit [4 marks]

12. Given p(x) = 1234

6 xbxaxx , where aand bare 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 aand b, factorise p(x) completely, and hence solve theequation 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]


3/10


4/10

(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 is35

. [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 >21 and X >

43 respectively, find P(B) and P(B|A).

[ 7 marks]


5/10

MARKING SCHEME FOR MATHEMATICS T PAPER 1

1 ( x + 1 )2 ( 2x + 1 )2 M1

3x2

+ 2x 0 M1x ( 3x + 2 ) 0 M1

x = ( ,3

2 ] [ 0, ) A1

2. 03233 xyyx

8 + 2k3+ 6k = 0

k3

+ 3k + 4 = 0 M1

(k + 1)(k2

k + 4) = 0k = 1 A1

03363 22 dx

dyxy

dx

dyyx

2

2

2yx

yx

dx

dy

M1

x = 2, y = 1;4

3

22

14

dx

dyA1

Equation of normal : y + 1 = )2(3

4

x . M1

y =3

11

3

4x A1

3. Let21)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

rA1

(a)

n

rrU

1

(31

21

12 )

+(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

nnn

rrU

=2

1

1

2

2

5

nn

=)2)(1(2

)95(

nn

nnA1

(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 B1D = (6,3) A1

(b) y + 3 = 4

5( x 6 ) B1M1

5x + 4y 18 = 0

5. (a) i

i

i

i

i

i

r 23

23

23

5

23

5

M1= 1i A1

(b) 4

T (4 6i) x (1i) = 2 10i B1

5T = ( 2 10i) x (1i) M1

= 12 8i A1

(c)6

T = (12 8i)(1i) = 20 + 4i

6S = (3 + 2i) + (5i) + (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

ttM1

2)( 1221

21

tttt

ttM1


6/10

21

21

tt

t1t2 =2

1A1

(c)dt

dx= 4 an

dt

dy= 4t

-2Both 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) = af(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

xxM1

= 1 +32

1

2 xx

x = 0, h(x) = 1 +3

4

3

1 both

x = 1, h(x) = 1 + =2

3B1

range of h= },2

3

3

4:{ Ryyy A1

(c) hhas an inverse function, his 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

yx

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

2M1

2

32t

dx

dy M1

t= 2 ,dx

dy= 6 A1

(b)dx

dtxt

dx

yd

2

6

2

2

M1

=2

1

2

6xt

M1

=2

3tA1

(c)2

1x

t M1

y =3

2

1

xM1

= 3

18

1

x A1

(d) y intercept =8

1

xintercept = 1 Both correct A1


7/10

2

5

6

5

20

y

x

11. (a) 2020

2

0coscossin

xdxxxxdxx

= 20sincos

xxx M1

= 10sin02

sin2

cos2

A1

(b) 2

0

220 2cos12

1

dxxxdxins M1

=2

0

2sin2

1

2

1

xx M1

=4

)0sin2

10(sin

2

1

22

1

A1

Area = 20 sin

dxxx M1

=2

0

2

cos

2

x

xM1

= )0cos0(2

cos8

2

22

18

unit

A1

Volume = 202

sin

dxxx B1

= 2022

sinsin2

dxxxxx

=

2

0

22

0

2

0

2sinsin2

xdxxdxxdxx M1

=

4)1(2

3

2

0

3

xM1

=4

224

24

48624

1 24 unit3

A1

12.a) 012

12

2

13

2

14

2

16

2

1

bap

M112 ba M1

01)1(22)1(3

3)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

2B1

Shape of graph D1

5

62

5

255q(5),5

x M1

107

5,0x ,

107,

5

6qR A1


8/10

P R

Q

T

60

MARKING SCHEME FOR MATHEMATICS T PAPER 2

1. cos 3 = cos (12)

cos 3 cos (12) = 0 B1

2sin

4

1 sin

4

12 =0 M1

sin

4

1 = 0 or sin

4

12 = 0

4

1 = or

4

12 = 0, M1

= 4

3or =

8

1, 8

5

=

8

1,

4

3,

8

5. A1

2. ARQA 32

OAOROQOA 32

OQOROA5

2

5

3 M1

= 345

3 + j4

5

2

= j5

17

5

12 A1

BRQB 32

OBOROQOB 32

OQOROB 23 M1

42343 j12 A1

j5

18

5

18 OPOAPA

j66 OPOBPB

66518

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

= QR2

+ SR2 2(QR)(SR) cos 60

o.

QS2 = ( htan

)2 + ( 2h3tan

)2

2(h

tan )(

2h

3tan ) cos 60

oM1

QS =7 h

3 tan A1

tan =h

QSM1

tan =3 tan

7

tan : tan = 3 : 7 A1

4. dxxdyey )ln1( M1

dx

xxxxxye

1ln + c M1A1

cxxey ln A1

2ln,1 yx , c = 2 M1A1

2ln xxey M1Particular 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


9/10

5

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 5M1

= 73o41

The direction of VQP is at N 16o19 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

= s2

= (2.071t 2.5)2

+ (7.071t)2

2sdsdt

= 4.142(2.071t 2.5) + 100t M1

For shortest distancedsdt

= 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 kmThe shortest distance s = 2.399 km or 2.40

km A1

(c) | PQ

| = 10 M1

102

= (2.071t 2.5)2

+ (7.071t)2

100 = 54.288t2 10.355t + 6.25

54.288t2 10.355t 93.75 = 0

t = 1.413

t = 1 hour 25 min A1

The time is 1325 A1

Method 2

VQP = VQVP

VQP2

= 102

+ 52 2(10)(5) cos 45

o. M1

VQP = 7.37 km h1. A1

sin 5

=sin 45

o

7.37

= 28o40, or 28.7

o

The direction of VQP is at N 16o20 E or N 16.4

oE A1

(b) The shortest distance between the two boats = d

sin 73o40 =

d2.5

M1

d = 2.399 km or 2.40 km A1

Time taken t =2.5 cos 73

o40'

7.37M1

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'

10M1

sin = 0.2399

= 13o53 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

1A1

9. New mean = 6

66

9632

nmM1

16 nm .(1) A1

New variance = 7.5+2.5=10

1066

813694 222

nm

M1

14622 nm (2) A1Solving (1) and (2):

VQ = 10 km

Vp = 5 km

4

4

VQP

o

2.5 km

d

PQ

73o40

10 km

T


10/10

14616 22 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

4M1(Both)

p =3

2and n = 6 A1

X B( 6,3

2)

P(X= 6) = (3

2)6

M1

=64729

A1

P( Y 4) =64729

[P ( Y = 4 ) + P ( Y = 5 ) + P ( Y = 6)] M1

=64729

[6C4(

35

)4

(25

)2

+6C5 (

35

)5(25

)1

+ (35

)6]

M1M1

=64729

[9723125

+291615625

+64

15625A1

= 0.0478 A1

11.

(a)

b)

Median = 68 B1

Q1 = 54, Q3 = 76

Interquartile range = 76 54 B1B1

= 22 B1

(c)

R1( shape + his Qs)

(d) Skewed negative. Reason : Q2Q1 > Q3Q2

12

(a) X is random: 111

1

2

dxxk

13

1

1

3

xxk M1

1

3

1)1(

3

11

3

k A1

8

3k A1

b)

R1

By symmetry : E(X)=0. B1

(c) Var(X) = 011

1

22

dxxkx M1

=

1

1

53

53

xx

k 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 5B1

Key : 5|7 means 57 B1

54 68 76

854620

outlier

x

218

3xy

f(x)

8

3

stpmmathst kedah&scheme 2010

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