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BENGKEL
KECEMERLANGAN AKADEMIK
MATEMATIK (1449) SPM 2016
SMK MUKAH
20 SEPTEMBER 2016
0900 – 1030
Disediakan oleh: Panitia Matematik
1 Mathematics SPM
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9 1
0
1
1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
1
9
2
0
2
1
2
2
2
3
2
4
2
5
2
6
Then,
KNOWLEDGE = 11 + 14 + 15 + 23 + 12 + 5 + 4 + 7 + 5 = 96%
HARDWORK = 8 + 1 + 18 + 4 + 23 + 15 + 18 + 11 = 98%
Both are important, but fall just short of 100%
But,
ATTITUDE = 1 + 20 + 20 + 9 + 20 + 21 + 4 + 5 = 100%
2 Mathematics SPM
ANALISIS JAWAPAN SPM
Tahun A B C D
2005 8 11 11 10
2006 10 10 10 10
2007 6 15 12 11
2008 10 11 9 10
2009 9 10 11 10
2010 10 9 10 11
ANALISIS JAWAPAN SPM
Tahun A B C D
2011 11 10 10 9
2012 11 8 10 11
2013 9 10 11 10
2014 10 10 11 9
2015 9 9 11 11
2016 8 – 11
Mathematics SPM5
Question 1
KETAKSAMAAN LINEAR
[ 3 marks]
Mathematics SPM6
Symbol Definition Line
> Lebih besar daripada Garis putus-putus
------------------
< Kurang daripada
 Lebih besar dan sama
dengan
Garis solid
 Kurang daripada dan
sama dengan
Mathematics SPM7
Example 5 : (SPM Nov 2005)
On the graph in the answer space, shade the
region which satisfies the three inequalities
y  2x + 10 , x < 5 and y  10 .
Answer :
x < 5
√ K1
√ K2
Mathematics SPM8
Example 6 :
State the three inequalities that satisfied the shaded
region in the graph below.
Answer :
(i) 3y  x + 12 ; (ii) y  2x + 4 ; (iii) x < 2
√ N1 √ N1 √ N1
Mathematics SPM9
LATIHAN: (SPM Jun 2016)
Pada graf tersebut, lorek rantau yang memuaskan
ketiga-tiga ketaksamaan
y  x + 8 , y  x and y  6 . (3 markah)
Answer :
Mathematics SPM10
JAWAPAN: (SPM Nov 2016)
Mathematics SPM11
Question 2
Persamaan Linear
[ 4 marks]
Example 7 : (SPM Nov 2015)
Calculate the value of x and of y that satisfy the following
simultaneous linear equations:
82
445


yx
yx
12 Mathematics SPM
 
6
366
441040
44285
)1(int)3(
)3(28:)2(
)2(82
)1(445







y
y
yy
yy
osubstitute
yxfrom
yx
yx



Method 1 : Substitution method
√ K1
√ N1
 
4
628
:)3(int6



x
oysubstitute
√ N1√ K1
13 Mathematics SPM
Method 2 : Elimination method
4
123:)1()3(
)1(445
)3(1642
2)2(:)2(
)2(82
)1(445







x
x
yx
yx
from
yx
yx




√ K1
√ N1
6
122
824
)2(int4




y
y
y
oxsubstitute
√ K1 √ N1
14 Mathematics SPM
     
     
     
6,4,
6
4
36
24
6
1
8541
8442
6
1
8
4
51
42
1425
1
8
4
21
45








































































yxso
y
x
y
x
Method 3 : Kaedah Matriks
√ K2
√ N1 √ N1
15 Mathematics SPM
Check answer : Calculator Scientific
16 Mathematics SPM
MODE  5: EQN  1 :
anX+bnY=cn
Example 8 :
Diberi jumlah 50 tiket telah dijual dengan harga RM 2080
dalam satu konsert. Jika harga satu tiket ialah RM35
ataupun RM50, cari bilangan tiket bagi harga RM35 dan
bilangan tiket bagi harga RM50 yang telah dijual.
Penyelesaian :
Katakan x = bilangan tiket bagi harga RM35
y = bilangan tiket bagi harga RM50
20805035
50


yx
yx
17 Mathematics SPM
     
22,28,
22
28
330
420
15
1
2080
50
135
150
135501
1
2080
50
5035
11






















































yxmaka
y
x
y
x
Method 3 : Kaedah Matriks
√ K2
√ N1 √ N1
18 Mathematics SPM
Mathematics SPM19
Question 4
Lines and Planes in 3-Dimensions
[ 3 marks]
WON Technique
Mathematics SPM20
Example 12 :
Given that V and W are midpoints of UT and PS.
Name the angle between line VQ and the base PQRS.
WON Technique
Mathematics SPM21
Step 1 : Arrange the line and the plane in two rows.
Then draw 3 boxes.
VQ
PQRSW
Step 2 : Find out the same alphabet. Slash the same
alphabet.
VQ
PQRSW
WON Technique
Mathematics SPM22
Step 3 : Write V in the first box.
VQ
PQRSW
Step 4 : Look at V in the diagram. Choose the slash
alphabet
nearest to V. Write in the centre box.
VQ
PQRSW
V
V Q
WON Technique
Mathematics SPM23
Step 5 : Look at V in the diagram.
Choose the non-slash alphabet nearest to V.
Write in the last box.
VQ
PQRSW
 VQW
V Q W
Mathematics SPM24
LATIHAN : (SPM Jun 2016)
Diagram 4 shows a right pyramid with height 7cm.
M is the midpoint of AB. Given AB = 4cm and BC
= 6cm.
(a) On the diagram, draw the orthogonal projection
of the line ME and the base ABCD.
(b) Calculate the angle between the line ME and
the
plane ABCD. (3
Mathematics SPM25
JAWAPAN : (SPM Jun 2016)
Mathematics SPM26
Question 7
Gradient and Area under a Graph
[ 5 ~ 6 marks]
(I) Distance-time Graph
Mathematics SPM27
(a) Find distance or period of time, when object
stationary
(b) Gradient = Speed = Rate of change of distance
(c) Average speed =
(II) Speed-time Graph
Mathematics SPM28
(a) Find speed or period of time, when object at ‘uniform speed’
(b) Gradient = Acceleration = Rate of change of speed
(c) Average speed =
* (d) Total distance travelled = Area under the graph
Mathematics SPM29
LATIHAN : (SPM Jun 2016)
Diagram 9 shows a speed time-graph for the movement of
two particles, A and B, for a period of 30 seconds. The graph
PR represents the movement of particle A and the graph
PQR represents the movement of particle B. Both particles
start at the same point and move along the same route.
Diagram 9
Mathematics SPM30
Example 15 : (SPM Jun 2016)
(a) State the uniform speed, in ms-1, of particle B.
(b) Calculate the rate of change of speed, in ms-2, of particle
B.
12 ms-
1
2
2
1
24
12
306
012







ms
Particle A
Particle B
(6,
12)
(30,
0)
√ N1
√ K1
√ N1
Mathematics SPM31
Example 15 : (SPM Jun 2016)
(c) Find the difference between the distance, in m, travelled
by
particle A and particle B for the period of 30 seconds.
Distance travelled by particle A = Area under the graph
PR
=
12
30
m180
1230
2
1


√ K1
Lakarkan bentuk geometri
Mathematics SPM32
Example 15 : (SPM Jun 2016)
(c) Find the difference between the distance, in m, travelled
by
particle A and particle B for the period of 30 seconds.
Distance travelled by particle B = Area under the graph
PQR
=
12
30
m216
12)306(
2
1


6
The difference = 216 – 180 = 36 m
√ K1
√ N1
Lakarkan bentuk geometri
Mathematics SPM33
Question 8
Solid Geometry ( Volume )
[ 4 marks]
Mathematics SPM34
Example 16 : (SPM Nov 2007)
Diagram 6 shows a solid, formed by joining a cylinder to a
right prism. Trapezium AFGB is the uniform cross-section of
the prism.
AB = BC = 9 cm. The height of the cylinder is 6 cm and its
diameter is 7 cm.
Diagram 6
Calculate the volume, in cm3, of the solid.
[Use ]7
22

9 cm
9 cm
6 cm
7 cm
Mathematics SPM35
Example 16 : (SPM Nov 2007)
(i) Volume of
cylinder
231
6
2
7
7
22
2
2








 hr
(ii) Volume of prism
 
756
98129
2
1








 Ah
(iii) Volume of the solid = 231 +
756
= 987 cm3
√ K1 √ K1
√ N1
√ K1
Mathematics SPM36
LATIHAN : (SPM Jun 2016)
A container contains 1386 cm3 of water. Then of the water
is poured into the right prism container as shown in Diagram
3. Trapezium ABCD is the uniform cross section of the prism.
Find the depth, in cm, of the water. (3 marks)
[Use ]
7
22

4
3
Mathematics SPM37
JAWAPAN : (SPM Jun 2016)
Mathematics SPM38
Question 10
Matrices
[ 6 marks]
Mathematics SPM39
Inverse matrix
Let , then
.







dc
ba
A
No Inverse matrix
If a matrix has no inverse, then ad – bc = 0
.
ad – bc =
0










ac
bd
bcad
A
11
Mathematics SPM40
LATIHAN: (SPM Jun 2016)
(a) It is given that is the inverse matrix of .
Find the value of p and of q.
;






 q
p
5
3






35
47
p =  4 q = 7
√ N1 √ N1
Mathematics SPM41
Example 18 : (SPM Jun 2016)
(b) Write the following simultaneous linear equations as a
matrix
form :
7x + 4y = 5
5x + 3y = 3
Hence, using matrix method, calculate the value of x and
of y.
Answer :


































































4
3
3755
3453
3
5
75
43
5437
1
3
5
35
47
y
x
y
x
y
x
y
x
 x = 3 , y =  4
√ K1
√ K1
√ N1√ N1
Mathematics SPM42
Question 11
Probability II
[ 5 ~ 6 marks]
Mathematics SPM43
LATIHAN: (SPM Jun 2016)
Diagram 11 shows two boxes, P and Q. Box P contains five
tokens labelled with letter and box Q contains three tokens
labelled with number. Two tokens are picked at random. The
first token is picked from box P and the second token is
picked from box Q.
Diagram 11
Mathematics SPM44
JAWAPAN: (SPM Jun 2016)
(a) List all the elements in the sample space.
6 8 9
K (K , 6) (K , 8) (K , 9)
E (E , 6) (E , 8) (E , 9)
L (L , 6) (L , 8) (L , 9)
A (A , 6) (A , 8) (A , 9)
H (H , 6) (H , 8) (H , 9)
n(S) = 15 √ P2
Mathematics SPM45
JAWAPAN: (SPM Jun 2016)
(b) By listing down all the possible outcomes of the event, find
the
probability that
(i) a token labelled with letter A and a token labelled with
an
even number are picked,
X = { (A, 6), (A, 8) }
P(X) =
15
2
√ K1
√ N1
6 8 9
K (K , 6) (K , 8) (K , 9)
E (E , 6) (E , 8) (E , 9)
L (L , 6) (L , 8) (L , 9)
A (A , 6) (A , 8) (A , 9)
H (H , 6) (H , 8) (H , 9)
√ √
Mathematics SPM46
JAWAPAN: (SPM Jun 2016)
(b) By listing down all the possible outcomes of the event, find
the
probability that
(ii) a token labelled with a vowel or a token labelled with
number 9 is picked.6 8 9
K (K , 6) (K , 8) (K , 9)
E (E , 6) (E , 8) (E , 9)
L (L , 6) (L , 8) (L , 9)
A (A , 6) (A , 8) (A , 9)
H (H , 6) (H , 8) (H , 9)
X = { (K, 9), (E, 6), (E, 8), (E, 9), (L, 9), (A, 6), (A, 8), (A,
9), (H, 9)}
P(X) =
15
9
√ K1
√ N1
√
√
√
√
√
√
√
√
√
Kem akademik  sept 16
Kem akademik  sept 16
Kem akademik  sept 16

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Kem akademik sept 16

  • 1. BENGKEL KECEMERLANGAN AKADEMIK MATEMATIK (1449) SPM 2016 SMK MUKAH 20 SEPTEMBER 2016 0900 – 1030 Disediakan oleh: Panitia Matematik 1 Mathematics SPM
  • 2. A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0 2 1 2 2 2 3 2 4 2 5 2 6 Then, KNOWLEDGE = 11 + 14 + 15 + 23 + 12 + 5 + 4 + 7 + 5 = 96% HARDWORK = 8 + 1 + 18 + 4 + 23 + 15 + 18 + 11 = 98% Both are important, but fall just short of 100% But, ATTITUDE = 1 + 20 + 20 + 9 + 20 + 21 + 4 + 5 = 100% 2 Mathematics SPM
  • 3. ANALISIS JAWAPAN SPM Tahun A B C D 2005 8 11 11 10 2006 10 10 10 10 2007 6 15 12 11 2008 10 11 9 10 2009 9 10 11 10 2010 10 9 10 11
  • 4. ANALISIS JAWAPAN SPM Tahun A B C D 2011 11 10 10 9 2012 11 8 10 11 2013 9 10 11 10 2014 10 10 11 9 2015 9 9 11 11 2016 8 – 11
  • 6. Mathematics SPM6 Symbol Definition Line > Lebih besar daripada Garis putus-putus ------------------ < Kurang daripada  Lebih besar dan sama dengan Garis solid  Kurang daripada dan sama dengan
  • 7. Mathematics SPM7 Example 5 : (SPM Nov 2005) On the graph in the answer space, shade the region which satisfies the three inequalities y  2x + 10 , x < 5 and y  10 . Answer : x < 5 √ K1 √ K2
  • 8. Mathematics SPM8 Example 6 : State the three inequalities that satisfied the shaded region in the graph below. Answer : (i) 3y  x + 12 ; (ii) y  2x + 4 ; (iii) x < 2 √ N1 √ N1 √ N1
  • 9. Mathematics SPM9 LATIHAN: (SPM Jun 2016) Pada graf tersebut, lorek rantau yang memuaskan ketiga-tiga ketaksamaan y  x + 8 , y  x and y  6 . (3 markah) Answer :
  • 12. Example 7 : (SPM Nov 2015) Calculate the value of x and of y that satisfy the following simultaneous linear equations: 82 445   yx yx 12 Mathematics SPM
  • 14. Method 2 : Elimination method 4 123:)1()3( )1(445 )3(1642 2)2(:)2( )2(82 )1(445        x x yx yx from yx yx     √ K1 √ N1 6 122 824 )2(int4     y y y oxsubstitute √ K1 √ N1 14 Mathematics SPM
  • 15.                   6,4, 6 4 36 24 6 1 8541 8442 6 1 8 4 51 42 1425 1 8 4 21 45                                                                         yxso y x y x Method 3 : Kaedah Matriks √ K2 √ N1 √ N1 15 Mathematics SPM
  • 16. Check answer : Calculator Scientific 16 Mathematics SPM MODE  5: EQN  1 : anX+bnY=cn
  • 17. Example 8 : Diberi jumlah 50 tiket telah dijual dengan harga RM 2080 dalam satu konsert. Jika harga satu tiket ialah RM35 ataupun RM50, cari bilangan tiket bagi harga RM35 dan bilangan tiket bagi harga RM50 yang telah dijual. Penyelesaian : Katakan x = bilangan tiket bagi harga RM35 y = bilangan tiket bagi harga RM50 20805035 50   yx yx 17 Mathematics SPM
  • 18.       22,28, 22 28 330 420 15 1 2080 50 135 150 135501 1 2080 50 5035 11                                                       yxmaka y x y x Method 3 : Kaedah Matriks √ K2 √ N1 √ N1 18 Mathematics SPM
  • 19. Mathematics SPM19 Question 4 Lines and Planes in 3-Dimensions [ 3 marks]
  • 20. WON Technique Mathematics SPM20 Example 12 : Given that V and W are midpoints of UT and PS. Name the angle between line VQ and the base PQRS.
  • 21. WON Technique Mathematics SPM21 Step 1 : Arrange the line and the plane in two rows. Then draw 3 boxes. VQ PQRSW Step 2 : Find out the same alphabet. Slash the same alphabet. VQ PQRSW
  • 22. WON Technique Mathematics SPM22 Step 3 : Write V in the first box. VQ PQRSW Step 4 : Look at V in the diagram. Choose the slash alphabet nearest to V. Write in the centre box. VQ PQRSW V V Q
  • 23. WON Technique Mathematics SPM23 Step 5 : Look at V in the diagram. Choose the non-slash alphabet nearest to V. Write in the last box. VQ PQRSW  VQW V Q W
  • 24. Mathematics SPM24 LATIHAN : (SPM Jun 2016) Diagram 4 shows a right pyramid with height 7cm. M is the midpoint of AB. Given AB = 4cm and BC = 6cm. (a) On the diagram, draw the orthogonal projection of the line ME and the base ABCD. (b) Calculate the angle between the line ME and the plane ABCD. (3
  • 25. Mathematics SPM25 JAWAPAN : (SPM Jun 2016)
  • 26. Mathematics SPM26 Question 7 Gradient and Area under a Graph [ 5 ~ 6 marks]
  • 27. (I) Distance-time Graph Mathematics SPM27 (a) Find distance or period of time, when object stationary (b) Gradient = Speed = Rate of change of distance (c) Average speed =
  • 28. (II) Speed-time Graph Mathematics SPM28 (a) Find speed or period of time, when object at ‘uniform speed’ (b) Gradient = Acceleration = Rate of change of speed (c) Average speed = * (d) Total distance travelled = Area under the graph
  • 29. Mathematics SPM29 LATIHAN : (SPM Jun 2016) Diagram 9 shows a speed time-graph for the movement of two particles, A and B, for a period of 30 seconds. The graph PR represents the movement of particle A and the graph PQR represents the movement of particle B. Both particles start at the same point and move along the same route. Diagram 9
  • 30. Mathematics SPM30 Example 15 : (SPM Jun 2016) (a) State the uniform speed, in ms-1, of particle B. (b) Calculate the rate of change of speed, in ms-2, of particle B. 12 ms- 1 2 2 1 24 12 306 012        ms Particle A Particle B (6, 12) (30, 0) √ N1 √ K1 √ N1
  • 31. Mathematics SPM31 Example 15 : (SPM Jun 2016) (c) Find the difference between the distance, in m, travelled by particle A and particle B for the period of 30 seconds. Distance travelled by particle A = Area under the graph PR = 12 30 m180 1230 2 1   √ K1 Lakarkan bentuk geometri
  • 32. Mathematics SPM32 Example 15 : (SPM Jun 2016) (c) Find the difference between the distance, in m, travelled by particle A and particle B for the period of 30 seconds. Distance travelled by particle B = Area under the graph PQR = 12 30 m216 12)306( 2 1   6 The difference = 216 – 180 = 36 m √ K1 √ N1 Lakarkan bentuk geometri
  • 33. Mathematics SPM33 Question 8 Solid Geometry ( Volume ) [ 4 marks]
  • 34. Mathematics SPM34 Example 16 : (SPM Nov 2007) Diagram 6 shows a solid, formed by joining a cylinder to a right prism. Trapezium AFGB is the uniform cross-section of the prism. AB = BC = 9 cm. The height of the cylinder is 6 cm and its diameter is 7 cm. Diagram 6 Calculate the volume, in cm3, of the solid. [Use ]7 22  9 cm 9 cm 6 cm 7 cm
  • 35. Mathematics SPM35 Example 16 : (SPM Nov 2007) (i) Volume of cylinder 231 6 2 7 7 22 2 2          hr (ii) Volume of prism   756 98129 2 1          Ah (iii) Volume of the solid = 231 + 756 = 987 cm3 √ K1 √ K1 √ N1 √ K1
  • 36. Mathematics SPM36 LATIHAN : (SPM Jun 2016) A container contains 1386 cm3 of water. Then of the water is poured into the right prism container as shown in Diagram 3. Trapezium ABCD is the uniform cross section of the prism. Find the depth, in cm, of the water. (3 marks) [Use ] 7 22  4 3
  • 37. Mathematics SPM37 JAWAPAN : (SPM Jun 2016)
  • 39. Mathematics SPM39 Inverse matrix Let , then .        dc ba A No Inverse matrix If a matrix has no inverse, then ad – bc = 0 . ad – bc = 0           ac bd bcad A 11
  • 40. Mathematics SPM40 LATIHAN: (SPM Jun 2016) (a) It is given that is the inverse matrix of . Find the value of p and of q. ;        q p 5 3       35 47 p =  4 q = 7 √ N1 √ N1
  • 41. Mathematics SPM41 Example 18 : (SPM Jun 2016) (b) Write the following simultaneous linear equations as a matrix form : 7x + 4y = 5 5x + 3y = 3 Hence, using matrix method, calculate the value of x and of y. Answer :                                                                   4 3 3755 3453 3 5 75 43 5437 1 3 5 35 47 y x y x y x y x  x = 3 , y =  4 √ K1 √ K1 √ N1√ N1
  • 43. Mathematics SPM43 LATIHAN: (SPM Jun 2016) Diagram 11 shows two boxes, P and Q. Box P contains five tokens labelled with letter and box Q contains three tokens labelled with number. Two tokens are picked at random. The first token is picked from box P and the second token is picked from box Q. Diagram 11
  • 44. Mathematics SPM44 JAWAPAN: (SPM Jun 2016) (a) List all the elements in the sample space. 6 8 9 K (K , 6) (K , 8) (K , 9) E (E , 6) (E , 8) (E , 9) L (L , 6) (L , 8) (L , 9) A (A , 6) (A , 8) (A , 9) H (H , 6) (H , 8) (H , 9) n(S) = 15 √ P2
  • 45. Mathematics SPM45 JAWAPAN: (SPM Jun 2016) (b) By listing down all the possible outcomes of the event, find the probability that (i) a token labelled with letter A and a token labelled with an even number are picked, X = { (A, 6), (A, 8) } P(X) = 15 2 √ K1 √ N1 6 8 9 K (K , 6) (K , 8) (K , 9) E (E , 6) (E , 8) (E , 9) L (L , 6) (L , 8) (L , 9) A (A , 6) (A , 8) (A , 9) H (H , 6) (H , 8) (H , 9) √ √
  • 46. Mathematics SPM46 JAWAPAN: (SPM Jun 2016) (b) By listing down all the possible outcomes of the event, find the probability that (ii) a token labelled with a vowel or a token labelled with number 9 is picked.6 8 9 K (K , 6) (K , 8) (K , 9) E (E , 6) (E , 8) (E , 9) L (L , 6) (L , 8) (L , 9) A (A , 6) (A , 8) (A , 9) H (H , 6) (H , 8) (H , 9) X = { (K, 9), (E, 6), (E, 8), (E, 9), (L, 9), (A, 6), (A, 8), (A, 9), (H, 9)} P(X) = 15 9 √ K1 √ N1 √ √ √ √ √ √ √ √ √