Teknik Menjawab Matematik Spm2009

TEKNIK MENJAWAB MATEMATIK SPM 2009

MOHD NAZAN BIN KAMARUL ZAMAN SMK. KOTA KLIAS, BEAUFORT

1

Solve the quadratic equation

3x 2 − 3x =2 x+6

3 x 2 − 5 x − 12 = 0 ( 3x + 4 ) ( x - 3 ) = 0

3 x − 3 x = 2( x + 6) 2

3x + 4 = 0

3 x − 3 x = 2 x + 12 2

x–3=0

3x = - 4

3 x − 3 x − 2 x −12 = 0 2

3 x 2 − 5 x − 12 = 0

3 x 2 −3 x =2 x +6

1 mark

4 x=− 3 1 mark

x =3 1 mark

1 mark

2

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

(ii) x 4

2d(4) – 3e(4) = -1(4) 8d – 12e = - 4 0 – 3e = -9

−9 e= −3 e=3

1 mark (iii) (i) 1 mark

8d – 9e = 5 (iii) – (i)

(i) (ii)

1 mark

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

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

5cm

M

M

L

12cm

L

J T

J

U

∠ MJL M

J

L

(1m)

5 12

tan ∠ MJL =

(1m)

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

R

P

0

●

S (6 , −1)

x

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

find the y-intercept 2y = x + 12 , y-intercept x = 0 2y = 0 + 12 y=6

1(m) (1m)

y = mx + c 1 − 1 = ( 6) + c 2 − 1= 3 + c c = − 1− 3 c=− 4

y − (−1) 1 = x −6 2

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

y = mx + c

x + 12 y= 2 1 y= x+6 2 y = mx + c m=

1 2

y=

1 x-4 2

1(m)

x-intercept, y = 0 0= 1(m)

1 x-4 2

x=8

1(m) 1(m)

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 ξ = P∪ Q∪ R On the diagram in the answer space, shade a)the set Q’. b)the set P (Q ∪R)

∩

Answer: R

P

i

Q

ii

iii

iv

R

P

v

Q

i

ii

iii

iv

∪

Q = ii, iii, iv

P = i, ii

Q’ = i, v

Q

R = ii, iii, iv , v

P ∩ (Q ∪ R) = ii

v

7. The inverse matrix of

3   2 1 − 3   is  m n  − 4 − 3

− 3  2

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

− 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 26s Speed (m s-1 ) 50 u 35

L1 0

10

14

L2 18

26

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‾² , 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 ≥ 0 2

y

1 y= − x + 3 2 0

x + y =3

y = 0 equal to x- axis

x

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

9 3 @ 12 4

1(m) 1(m)

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 π =22 , calculate the volume of solid. 7 V

16 cm 9 cm Diagram 3 10 cm

the volume of solid. V. of cone = π 1j2t 3 = π1 (5)2 (16 – 9) 3 550 = 3

V

Answer: Calculate

Volume of cylinder = π j2t = π (5)2 (9) 4950 = 7

Volume of solid

550 4950 = + 3 7 18700 = 21 10 = 890 cm3 21

K1

16 cm 9 cm 10 cm

K1

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 π = 7 , Calculate R a) The area of the shaded region. b) The perimeter of the whole diagram. Y Q 2

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 2 A. sector OPQ = × π r 3600 0 90 = × π (7)2 3600 K1 P 77 = 2 Angle at centre 2 A. of sector OYS = × π r 3600 0 60 = × π (4)2 3600 = 176 21 1 A. triangle OXY = × 3× 4 2 =6

R

Answer: Calculate

Y

Q

S

4 cm 600 Rajah 4 O

X 3 cm 7 cm Area of shaded region =

77 + 176 – 6 21 2

= 40.88 cm2 N1

K1

seluruh rajah panjang lengkok = sudut pusat 3600 lilitan bulatan sudut pusat lengkok PQR = × 2π j 0 360 0 90 = × 2π (7) K1 0 360 P = 11

R

Jawapan Perimeter

Y

Q

4 cm 600 X 3 cm 7 cm

sudut pusat × 2π j 0 360 0 60 = × 2π (4) 0 360 = 88 21 perimeter seluruh rajah = 11 + 3 + 88 + 4 + 7 21 = 29.19 cm N1 lengkok YS =

K1

Rajah 4 O

S