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
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