TEKNIK MENJAWAB MATEMATIK SPM2009

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Teknik menjawab Matematik SPM SMK. Kota Klias

Text of TEKNIK MENJAWAB MATEMATIK SPM2009

TEKNIK MENJAWAB MATEMATIK SPM 2009

MOHD NAZAN BIN KAMARUL ZAMAN SMK. KOTA KLIAS, BEAUFORT

1

Solve the quadratic equation

3 x 2 3 x =2 x +63 x 2 5 x 12 = 0 ( 3x + 4 ) ( x - 3 ) = 03x + 4 = 0 3x = - 4 x3=01 mark

3x 2 3x =2 x+6 3 x 3 x = 2( x + 6)2 2 2

3 x 3 x = 2 x + 12 3 x 3 x 2 x 12 = 0 3 x 2 5 x 12 = 01 mark

4 x= 31 mark

x =31 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) Substitute e = 3 to (i) 8d 9(3) = 5 8d 27 = 5 8d = 5 + 27 8d = 32 32 d= 8 d =4

(ii) x 4

2d(4) 3e(4) = -1(4) 8d 12e = - 4 8d 9e = 5

1 mark (iii) (i) 1 mark

(iii) (i)

0 3e = -9

9 e= 3 e=3

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.T

U

9cm

M

L

12cm J 5cm K

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

T U

M

5cm

L

M

L

12cm

J T U J

MJLM L

(1m)

tan MJL =

5 12

(1m)

J

atau setara 22.61o atau 22o 37 (1m)

L

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.y Q

P

R

0

S (6 , 1)

x

DIAGRAM 2

(a) (b)

State the equation of the straight line PR . 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 1(m) (1m)y (1) 1 = x 6 2

S (6, -1) x = 6 , y = -1y = mx + c 1 1 = ( 6) + c 2 1= 3 + c c = 1 3 c= 4

b) SR is parallel to straight line PQ mSR = mPQ 2y = x + 12

x + 12 y= 2 1 y= x+6 2 y = mx + c m= 1 21(m)

y = mx + c y=1 x-4 2

1(m)

x-intercept, y = 0 0=1 x-4 2

1(m) 1(m)

x=8

5

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

p2 3p + 2 = 0 (b) Write down two implications based on the following sentence:

x = 7 if and only if x = 49

(c) Make a general conclusion by induction for a list of numbers 8, 23, 44, 71, Which follows the following pattern: 8 23 44 71 = = = = 3(2)2 4 3(3)2 4 3(4)2 4 3(5)2 4

6.The Venn diagram in the answer space shows sets P, Q and R such that the universal set = P Q ROn the diagram in the answer space, shade a)the set Q. b)the set P (Q R)

Answer:P R Q P R Q

i

ii

iii

iv

v

i

ii

iii

iv

v

Q = ii, iii, iv Q = i, v

P = i, ii Q R = ii, iii, iv , v P (Q R) = ii

7. The inverse matrix of a) Find the value of m and n

3 2 1 3 is 4 3 m n

3 2

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

3 1 d b 1 = a) ad bc c a 2(3) 3(4) 4 1 3 3 = 2 6 + 12 4 1 3 m n m = 6 and n = 4 3 1 3 = 2 6 4

3 2

3 2

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 2 4 3 x 4 = 3 y 2 A B = C B = A 1 C x = y x = y 1 3 6 4 3 4 2 2

1 2 x =1 and y = 2

8. Diagram shows the speed- time graph of a particle for a period of 26sSpeed (m s-1 ) 50 u 35

L10 10 14 18

L226 Time (s)

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 , in the first 10 seconds c) Calculate the value of u, if the total distance travelled for the last 12 seconds is 340 m.

a) 14 10 = 4 s b) The rate of change of speed = gradient y y = 2 1 x 2 x1 50 35 = 0 10= 15 3 @ 10 2

c) Distance = area under a graph 340 = L1 + L2

1 1 340 = ( 35 + u ) 4 + 8 u 2 2 u = 45

1 L1 = ( 35 + u ) 4 2

L2 =

1 8 u 2

9. On the graph in the answer space, shade the region which satisfies the three 1 inequalities y x + 3 , x + y 3 and y 02y

1 y= x + 3 20 x

x + y =3

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.

B

E

S

T

4

6

7

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 are picked

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

2 1 @ 12 6ii) {(E, 4), (E, 6), (E, 7), (B, 4), (B, 6), (S, 4), (S, 6), (T, 4), (T, 6)}

1(m) 1(m)

9 3 @ 12 4

1(m) 1(m)

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 and diameter of cylinder is 10cm. By using =22 , calculate the volume of solid. 7 V

16 cm 9 cm Diagram 3 10 cm

the volume of solid. V. of cone = 12t j 3 = 1 (5)2 (16 9) 3 550 = 3

Answer: Calculate

V

K1

16 cm 9 cm 10 cm

Volume of cylinder = j2t = (5)2 (9) 4950 = 7 550 4950 = + 3 7 18700 = 21 10 = 890 cm3 21

K1

Volume of solid

K1

N1

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 22 Use = 7 , Calculate R a) The area of the shaded region. b) The perimeter of the whole diagram. Y Q2

4 cm 600 P X 3 cm 7 cm O

S Diagram 4

the area of shaded region. Area of sector = Angle at centre 3600 Area of circle Angle at centre A. sector OPQ = r2 3600 900 (7)2 = 3600 K1 P 77 = 2 Angle at centre A. of sector OYS = r2 3600 600 (4)2 = 3600 = 176 21 1 A. triangle OXY = 3 4 2 =6

Answer: Calculate

R Q Y S

4 cm 600 X 3 cm 7 cm Area of shaded region = 77 + 176 6 21 2N1

Rajah 4 O

K1

= 40.88 cm2

seluruh rajah panjang lengkok = sudut pusat 3600 lilitan bulatan sudut pusat lengkok PQR = 2 j 0 360 900 2 (7) = K1 0 360 P = 11 lengkok YS =

Jawapan Perimeter

R Q Y S

4 cm 600 X 3 cm 7 cm Rajah 4 O

sudut pusat 2 j 0 360 600 2 (4) = 3600 = 88 21 perimeter seluruh rajah = 11 + 3 + 88 + 4 + 7 21 = 29.19 cm N1

K1