TEKNIK MENJAWAB MATEMATIK SPM 2009

MOHD NAZAN BIN KAMARUL ZAMAN SMK. KOTA KLIAS, BEAUFORT

1 Solve the quadratic equation 2

6

33 2

x

xx

01253

012233

12233

)6(233

26

33

2

2

2

2

2

xx

xxx

xxx

xxx

x

xx

1 mark 33

4

0)()(

01253 2

xx

xx

3x = - 4

x – 3 = 0

x - 3

3x + 4 = 0

3x + 41 mark

1 mark1 mark

2 Calculate the value of d and of e that satisfy the following simultaneous linear equations: 8d − 9e = 5 2d − 3e = −1

(i)(ii)

(ii) x 4 2d(4) – 3e(4) = -1(4)

8d – 12e = - 4 (iii)

8d = 32

0 – 3e = -9

(iii) – (i)

33

9

e

e

Substitute e = 3 to (i)

8d – 9(3) = 5

8d – 27 = 5

8d = 5 + 27

8d – 9e = 5 (i)

48

32

d

d

1 mark

1 mark

1 mark

1 mark

3 Diagram 1 shows a right prism. The base JKLM is a horizontal rectangle. The right angled triangle UJK is the uniform cross section of the prism.

J K

LM

U

T

9cm

12cm

5cm

DIAGRAM 1Identify and calculate the angle between the plane UJL and the plane UJMT .

J

LM

U

T

J

LM

U

T

L

J

LM

12cm

5cm

MJL tan MJL =

12

5

atau setara 22.61o atau 22o 37’

(1m)

(1m)

(1m)

MJL @ LJM

4 In Diagram 2, O is the origin, point P lies on the y-axis. Straight line PR is parallel to the x-axis and straight line PQ is parallel to straight line SR. The equation of straight line PQ is 2y = x + 12.

0 x

y

P

Q

R

S (6 , −1)●

DIAGRAM 2

(a) State the equation of the straight line PR . (b) Find the equation of the straight line SR and hence, state its x-intercept

a) P = y-intercept for PQ and PR

find the y-intercept

2y = x + 12 , y-intercept x = 0

2y = 0 + 12

y = 6 (1m)

b) SR is parallel to straight line PQ

mSR = mPQ

2y = x + 12

2

1

62

12

12

m

cmxy

xy

xy

4

31

31

)6(2

11

c

c

c

c

cmxy

2

1

2

1

S (6, -1) x = 6 , y = -1

y = mx + c

y = x - 4

x-intercept, y = 0

0 = x - 4

x = 8 1(m)

1(m)

1(m)

2

1

6

)1(

x

y

1(m)

1(m)

p2 – 3p + 2 = 0

5 (a) Determine whether the following sentence is a statement or non-statement.

(b) Write down two implications based on the following

sentence:

= 7 if and only if

(c) Make a general conclusion by induction for a list of numbers 8, 23, 44, 71, … Which follows the following pattern:

8 = 3(2)2 – 4 23 = 3(3)2 – 4 44 = 3(4)2 – 4 71 = 3(5)2 – 4

x 49x

QP

R

QP

R

6.The Venn diagram in the answer space shows sets P, Q and R such that the universal set ξ = P Q R

On the diagram in the answer space, shadea)the set Q’.b)the set P (Q R)

Answer:

i viviiiii

Q = ii, iii, iv

Q’ = i, v

P = i, ii

Q R = ii, iii, iv , v

P ∩ (Q R) = ii

i iviiiii v

7. The inverse matrix of

a) Find the value of m and n

b) Write the following simultaneous linear equations as a matrix equation:

2x + 3y = -4 - 4x – 3y = 2

hence, using the matrix method, calculate the value of x and y

2

331is

34

32

nm

24

33

6

1

2

331

24

33

126

1

24

33

)4(3)3(2

11 a)

nm

ac

bd

bcad

m = 6 and n = 4

b) Write the following simultaneous linear equations as a matrix equation:hence, using the matrix method, calculate the value of x and y

2x + 3y = -4 - 4x – 3y = 2

21

2

1

2

4

24

33

6

1

2

4

34

32

1

yandx

y

x

y

x

CAB

CBA

y

x

8. Diagram shows the speed- time graph of a particle for a period of 26s

0 10 14 18

u

Time (s)

Speed (m s-1)

26

35

50

a) Stat the duration of time, in s, for which the particle moves with uniform speed.

b) Calculate the rate of change of speed, in m s‾² , in the first 10 secondsc) Calculate the value of u, if the total distance travelled for the last 12

seconds is 340 m.

L1 L2

a) 14 – 10 = 4 s

b) The rate of change of speed = gradient

12

12

xx

yy

2

3@

10

15

100

3550

c) Distance = area under a graph

340 = L1 + L2 uLuL 82

124)35(

2

11

45

82

14)35(

2

1340

u

uu

9. On the graph in the answer space, shade the region which satisfies the three inequalities 0and3,3

2

1 yyxxy

y

x0

32

1 xy

3 yx

y = 0 equal to x- axis

10. Diagram 9 shows two boxes , P and Q . Box P contains four cards labeled with letters and box Q contains three cards labeled with numbers.

TSEB 764

Two cards are picked at random, a card from box P and another card from box Q .

a)List the sample space and the outcomes of the events .

b) Hence , find the probability that(i) a card labeled with letter E and a card labelled with an even number are picked

(ii) a card lebelled with letter E or a card labelled with an even number arepicked

a) {(B, 4), (B, 6), (B, 7), (E, 4), (E, 6), (E, 7), (S, 4), (S, 6), (S, 7), (T, 4), (T, 6), (T, 7)} Notes : 1. Accept 8 correct listings for 1 mark

b) i) {(E, 4), (E, 6)}

ii) {(E, 4), (E, 6), (E, 7), (B, 4), (B, 6), (S, 4), (S, 6), (T, 4), (T, 6)}

6

1@

12

2

4

3@

12

9

1(m)

1(m)

2(m)

1(m)

1(m)

9 cm

V

16 cm

Diagram 3

22 7

10 cm

6. Diagram 3 shows a solid formed by joining a cone and cylinder. The height from vertex V to the base of cylinder is 16 cm anddiameter of cylinder is 10cm.By using = , calculate the volume of solid.

9 cm

V

16 cm

10 cm

Calculate the volume of solid.Answer:

V. of cone = j2t13

= (5)2 (16 – 9) 13

= 550 3

Volume of cylinder = j2t

= (5)2 (9)

= 4950 7

Volume of solid = 550 3

= 18700 21

+ 4950 7

= 890 cm31021

K1

K1

K1

N1

PX O

Q

R

Y

S4 cm

3 cm

Diagram 4

7. Diagram 4 shows a quadrant PQR and sector of a circle OYS, both with centre O.

PXO and OYR are straight lines and YOS = 600. OP = 7cm Use = , Calculate

a) The area of the shaded region. b) The perimeter of the whole diagram.

22 7

7 cm

600

2

PX O

Q

R

Y

S4 cm

3 cmRajah 4

7 cm

600

Calculate the area of shaded region.Answer:

Area of sectorArea of circle

= Angle at centre 3600

A. sector OPQ = r2 Angle at centre 3600

= (7)2 900

3600

A. of sector OYS = r2 Angle at centre 3600

= (4)2 600

3600

= 77 2

= 176 21

A. triangle OXY = 3 4 12

= 6

Area of shaded region

= + – 6 77 2

176 21

= 40.88 cm2

K1

K1

N1

PX O

Q

R

Y

S4 cm

3 cmRajah 4

7 cm

600

Perimeter seluruh rajahJawapan

panjang lengkok lilitan bulatan

= sudut pusat 3600

lengkok PQR = 2j sudut pusat 3600

= 2(7) 900

3600

lengkok YS = 2j sudut pusat 3600

= 2(4) 600

3600

= 11

= 8821

perimeter seluruh rajah = 11 + 3 + + 4 + 7 8821

K1

K1

N1 = 29.19 cm