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