Upload
khai83
View
170
Download
13
Embed Size (px)
DESCRIPTION
Teknik menjawab Matematik SPM SMK. Kota Klias
Citation preview
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