1
MATHEMATICS
SPM Section A Paper 2
( Selected Topics )
Disediakan Oleh:
EN. LEE MEOW CHOON ( SMK Bukit Gambir, Ledang )
Topic
1 Set …………………………………….. 2 - 6
2 Lines and planes in 3-dimensions …………………………………. 7 - 10
3 Mathematical Reasoning - Induction 11
4 Mathematical Reasoning - Argument …………………………………. 12 - 14
5 Volume of solids 15 - 18
6 Area and perimeter of sector …………………………………. 19 - 21
7 Matrices 22 - 25
Q P
R
1 2 5 4
3 6
7
K 8 cm L
H J
E F
G D
6 cm
6 = 2 (1 )2 + 4 12 = 2 (2 )2 + 4 22 = 2 (3 )2 + 4 36 = 2 (4 )2 + 4
4
10 7
R
T
60
Q
P
O 14
14
14
7
21
−−−
=
6
3624
153606
1yx
2
1. [ Set ] 1.1 Basic set operation
P = { 1 2 3 4 } Q = { 3 4 5 } addition
P ∪ Q = { 1 2 3 4 5 } P ∪ Q = {1234} ∪ {345} = {12345}
P = { 1 2 3 4 } Q = { 3 4 5 } only the
common ones P ∩ Q = { 3 4 }
P ∩ Q = {1234} ∩ {345} = {34}
P = { 1 2 3 4 } Q = { 3 4 5 }
ξ = P ∪ Q = { 1 2 3 4 5 } P = { 1 2 3 4 }
P ′ = { 5 } P ′ = { 1234}′ = {5}
1.2 It is given that the universal set [ SPM2004/P1/Q32 ] ξ = { x : 1 < x < 12 , x is an integer} Set P = { 2 , 3 ,7 ,9 } Set Q = { x : x is a prime number } and Set R = { x : x is a multiple of 4 }. The elements of the set ( P ∪ R ) ′ ∩ Q are
A 5, 11 B 1,5, 11 C 2,3 11 D 2,3,9 ξ = { 1 2 3 4 5 6 7 8 9 10 11 12 } P = { 2 3 7 9 }
Q = { 2 3 5 7 11 } R = { 4 8 12 }
( P ∪ R ) ′ ∩ Q ={ ( 2379) ∪ (48 12)}′ ∩ ( 2357 11) = { 234789 12 }′ ∩ ( 2357 11)
= {1 56 10 11} ∩ ( 2357 11) = { 5 11}
1.3 It is given that the universal set. ξ = { x : 19 < x < 31 , x is an integer } and [ SPM2006/P1/Q29 ] Set R = { x : x is a number such that the sum of its two digits is an even number } Find set R ′ .
A {20,22,24,26,28} B { 21,23,25,27,29 } C {19,21,23,25,27,29 } D {21,23,25,27,29,30 }
ξ = { 19 ,20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31 } R = { 19, 20, 22, 24, 26, 28, 31 } R ′ = { 21, 23, 25, 27, 29, 30 }
+
x
Not P, outside P
3
1.4. Strategy in solving 1.4.1(a) Shade on the Venn diagram provided the region for (i) P ∪ Q ∩ R
Steps 1.4(a)(i) 1.4(a)(ii) 1.4(a)(iii)
1.Assign each region with a number
`
2.A > N
P ∪ Q ∩ R ={12}∪{234}∩{45} ={ 1234 } ∩{45} = { 4 }
P ∪ Q ∩ R ={1245} ∪ {2356} ∩ {4567} = { 123456 } ∩ {4567} = {456}
P ∪ Q ∩ R = {12} ∪ {234} ∩ {4} = { 1234 } ∩ {4} = {4}
3. Shade the regions
`
1.4(b) Shade on the Venn diagram provided the region for P ∪ Q ∩ R′
Steps 1.4(b)(i) 1.4(b)(ii) 1.4(b)(iii) A> N P ∪ Q ∩ R′
={12} ∪ {234} ∩ {45} ′ = { 1234 } ∩ {123} = { 123 }
P ∪ Q ∩ R′ ={1245} ∪ {2356} ∩ {4567} ′ = { 123456 } ∩ {123} = { 123}
P ∪ Q ∩ R′ = {12} ∪ {234 }∩ {4} ′ = { 1234 } ∩ {123 } = { 123 }
3. Shade the regions
Q P
R
P
Q
R R
P
Q R
1 2 1 2
5 4
1
2 3 4 5
3 6
7 4 3
P
Q
R R
P
Q R
1 2
1
2 3 4 5
Q P
R
1 2 5 4
3 6
7 4 3
P
R
P
Q
R R
1 2 3 4 5
Q
1 2 5 4
3 6
7
P
Q R
1
2
4 3
4
1.4(c) Shade the region defined by P ∪ ( Q′ ∩ R)
Steps 1.4(c)(i) 1.4(c)(ii) 1.4(c)(iii) 1.Assign each regions with a number
** P ∪ ( Q′ ∩ R ) = { 45 } ∪ ({15} ∩ {12}) = {45 } ∪ ( 1 ) = { 145 }
P ∪ ( Q′ ∩ R ) = {1267} ∪ ({ 123} ∩ {2347}) = {1267} ∪ ( 23 ) = {12367 }
P ∪ ( Q′ ∩ R ) ={12} ∪ ( (1) ∩ (4) ) = {12} ∪ { } = {12}
3. Shade the regions
3. Shade the regions
P
R
P
R
P
Q
R R
P
Q R
Q
Q P
Q
R R
P
Q R
P
R
P
Q
R R
P
Q R
Q 5
3 2 1
4 1 2
3 4 5 6
7 1
2 3
4
P
R
Q P
Q
R R
P
Q R
5 4 3 2 1
1 2
3 4
5 6 7
1
2 3
4
5
1.4(d) Shade the region defined by P ∪ ( Q′ ∩ R) ′
Steps 1.4(d)(i) 1.4(d)(ii) 1.4(d)(iii) A > N P ∪ ( Q′ ∩ R) ′
P ∪ ( Q′ ∩ R) ′ P ∪ ( Q′ ∩ R) ′
1&3. Assign and Shade the regions
A > N P ∪ ( Q′ ∩ R ) ′ ={45} ∪ ( {15}∩ {12})′ = {45} ∪ ( {1})′ = {45} ∪ {2345} = { 2345 }
P ∪ ( Q′ ∩ R ) ′ ={1267}∪ ( {123} ∩{2347)}′ = {1267}∪ { 23 }′ = {1267}∪ { 14567 } = { 124567}
P ∪ ( Q′ ∩ R ) ′ = {12} ∪ ({1} ∩ {4})′ = {12} ∪ ({})′ = {12} ∪ { 1234} = {1234}
1&3. Assign and Shade the regions
P
Q
R R
Q P
R
P
Q R
P
Q
R R
Q P
R
P
Q R
7 4 5 1 6
2
5 4
2 1
3 3
1
2 3
4
6
1.5. The Venn diagrams below shows the sets of P, Q and R. Given that ξ = P ∪ Q ∪ R. In each diagram below, shade the region of
(a) Q ∪ R
(b) P ′ ∩ R (c) P ∩ ( Q ∪ R) ′
(a) Q ∪ R = { 34} ∪ {123} = {1234}
(b) P ′ ∩ R = { 1} ∩ {123}
= {1}
(c) P ∩ ( Q ∪ R) ′ = { 2345}∩({34}∪ {123})’ = { 2345}∩({1234})’ = { 2345}∩({5}) = {5}
1.6. The diagrams below show the set P, Q and R. Given that the universal set ξ = P ∪ Q ∪ R. Shade the region of (a) P ∩ R
(b) P ′ ∩ R ′ (c) P ∩ R′ ∩ Q
(a) P ∩ R
(b) P ′ ∩ R ′ (c) P ∩ R′ ∩ Q
P
Q R
P
Q
R P
Q R
P Q
R P Q
R
P Q
R
P
Q R
P
Q R
P
Q R
1 2
3 4
5
1 2
3 4
5
1 2
3 4
5
P Q
R P Q
R
P Q
R
7
1.7. The diagrams below show the Venn diagrams with the universal set ξ = P ∪ Q ∪ R . Shade the region of
(a) P′ ∪ Q
(b) P ∩ Q ∩ R ′
1.8. For each of the following Venn diagrams, the universal set ξ = A ∪ B ∪ C . Shade the region of
(a) A ∩ B ∪ C
(b) B′ ∩ C (c) ( A ∪ C ) ′ ∩ B
1.9.[SPM2004] 1.10. [SPM 2006]
(a) Shade 'BA∩
(a) Shade QP∩'
(b) Shade 'CBA ∩∪
(b) Shade RQP ∩∪ )'(
R P Q R P Q
A B
C
A B A
B
C C
A B C
P
Q R
A B
C
P
Q R
8
1.11 1.12 (a) Shade UTS ′∩∪
(a) Shade V ′ ∩ W
(b) Shade UTS ′∩∪
(b) Shade (V ∩ W ) ′
S
T U
V
W
ξ
S T
U V
W
ξ
9
1.[SPM2004] 2. [SPM 2006]
(a) Shade 'BA∩
A = { 1 , 2 } B’= { 1 , 4 }
'BA∩ = { 1 } A ∩ B ′ = ( 12 ) ∩ ( 14) = {1}
(a)
Shade QP∩'
P’ = { 1, 3, 4 } Q = { 1 , 2 , 3 }
QP∩' = { 1 , 3 } P ′ ∩ Q = {134} ∩ {123} = {13}
(b) Shade 'CBA ∩∪
A = { 1 , 2 , 5, 6 } B = { 4, 5, 6 , 7 }
A∪ B = { 1, 2, 4, 5, 6, 7 } C′ ={ 1, 6, 7 }
'CBA ∩∪ ={ 1, 6, 7 }
A ∪ B ∩ C ′ = {1256} ∪ {4567} ∩ {167 } = { 124567 } ∩ {167 }
= { 167 }
(b) Shade RQP ∩∪ )'(
P = { 2 } Q′= { 4 }
P ∪ Q′ = { 2 4 } R = { 3, 4 }
RQP ∩∪ )'( ={ 4 }
P ∪ Q ′ ∩ R = { 2 } ∪ {4} ∩ { 34} = { 24 } ∩ { 34}
= { 4 }
A B C
A B
C
P
Q R
P
Q R
1
A B C
1 2 3 4
A B C
1 2 3 4 P
Q R 1 2 3 4
P
Q R 1 2 3 4
Step 1
Step 2
Step 3
A B
C
1 P
Q R 1 2 3 4
2 3
4 5 6 7
P
Q R 1 2 3 4
Step 1
Step 2
Step 3
A B 1
2 3
4 5 6 7
10
3. 4. (a) Shade UTS ′∩∪ S = { 2 , 3 , 4, 5 } T = { 1 , 2 5 6 } S ∪ T = { 1 , 2 3, 4 , 5 , 6 } U ′ = { 1 , 2 , 3 } UTS ′∩∪ ={ 1 , 2 , 3 } S ∪ T ∩ U ′ = {2345} ∪ {1256} ∩ { 123 } = { 123456 } ∩ {123} = { 123 }
(a) Shade V ′ ∩ W
V ‘ = { 1 4 } W = { 3 , 4 }
V ′ ∩ W = { 4 } V ′ ∩ W = { 14 } ∩ { 34} = { 4 }
S
T U
V
W
ξ
S
T U
1 2
3 4
5 6
7
V
W 2
1
3 4
S
T U
1 2
3 4
5 6
7
V
W 2
1
3 4
Step 1
Step 2
Step 3
11
(b) Shade ( UTS ′∩∪ S = { 3 , 4, 5 } T = { 1 , 2 , 3 } S ∪ T = { 1 , 2 3, 4 , 5 } U ′ = { 1 , 5 } UTS ′∩∪ ={ 1 5 } S ∪ T ∩ U ′ = {345} ∪ {123 } ∩ {15}
= { 1234 5} ∩ {15 } = {15 }
(a) Shade (V ∩ W ) ′
V = { 2 , 3 } W = { 3 , 4 }
V ∩ W = { 3 } (V ∩ W ) ′ = { 1 , 2 4 } ( V ∩ W ) ′ = ( {23} ∩ {34} ) ′ = { 3 } ′ = { 124}
S T
U V
W
ξ
V
W 2
1
3 4
S T
U
1 2 3 4 5
6
S T
U
1 2 3 4 5
6
V
W 2
1
3 4
Step 1
Step 2
Step 3
ξ
12
2. Angle between Line and Plane 3-D 2.1 (a) Line AR with plane ABCD
Step 1 Write line AR and plane ABCD AR
A BCD 2 Write the common point in the middle
A
3 Write the other remaining point/s in the front
R A
4 Choose point on the plane that is nearest to the remaining point. (Shade the plane, use arrow to denote the relative position)
R A C
2.1 (b) Line BS with plane CDSR
Step 1 Write the line BS and plane CDSR
B S CD S R
2 Write the common one in the middle S
3 Write the other remaining point/s in the front
B S
4 Choose point on the plane that is nearest to the remaining point. (Shade the plane, use arrow to denote the relative position)
B S C
B A
C D
P Q
R S
C D
P Q
R S
B A
13
2.2 Angle between plane and plane 2.2(a) Plane ARS with plane ABCD
Step 1 Write plane ARS with plane ABCD ARS
A BCD 2 Write the common one in the middle
A
3 Write the other remaining point/s at the front
R/S A
4 Choose a point that is nearer to the middle point
R/S A
5 Choose point on the plane that is nearest to the remaining point. (Shade the plane, use arrow to denote the relative position)
R/S A D
2.2 (b) Plane BRS with plane CDSR
Step 1 Write plane BRS with plane CDSR B R S
CD R S
2 Write the common one in the middle R/S
3 Write the other remaining point/s at the front
B R/S
4 Choose a point that is nearer to the middle point
B R/S C
5 Choose point on the plane that is nearest to the remaining point. (Shade the plane, use arrow to denote the relative position)
B R/S C
B A
C D
P Q
R S
B A
C D
P Q
R S
14
2.3. Determine the length of the sides or the angles. (a) (b) (c) (d) (e)
(f) (g) (h) (i) (j) (k) (l)
(m)
3
4 10 8 5
13
8=O 13=H
7=O
do
4=A
12=H
5=A
eo
fo
13=O
10=A
15=O 17=H
7=A
12=H
12=H
6=O
h o i o
j o
g o
14=H 5=A k o
14=A
9=O
l o
c=12 b=6
a=5
m o
10=O 13=A
SOH CAH TOA
a2+b2= c2
15
(a) (b) (c) (d) (e)
(f) (g) (h) (i) (j) (k) (l)
(m)
3
4 10 8 5
13
8=O 13=H
7=O
do
4=A
12=H
5=A
eo
fo
13=O
10=A
15=O 17=H
7=A
12=H
12=H
6=O
h o i o
j o
g o
14=H 5=A k o
14=A
9=O
l o
c=12 b=6
a=5
m o
10=O 13=A
SOH CAH TOA
OH SOH Sin d = O/H = 8/13 d = 37.98
AH CAH Cos e = A/H = 5/12 e = 65.38
OA TOA Tan f = O/A = 7/4 f= 60.26
OA TOA Tan g = O/A = 13/10 g = 52.43
OH SOH Sin h = O/H = 15/17 h = 61.93
AH CAH Cos i = A/H =7/12 i = 54.31
OH SOH Sin j = O/H = 6/12 j = 30.00
AH CAH Cos k = A/H =5/14 k = 69.08
OA TOA tan l = O/A = 9/14 l = 32.74
OA TOA Tan m = O/A = 10/13 m = 37.57
16
2.4 Diagram shows a pyramid VJKLM The base JKLM is a horizontal rectangle. Q is the midpoint of JM. The apex V is 8 cm vertically above the point Q. Calculate the angle between the line KV and the base JKLM. [4 marks] SPM 2004 ( KV )( JKLM Q) KV J K LMQ V K Q
Q
J
V
K
L M
10 cm
12 cm
Diagram 2
K Q
V
5 5
13
8=O T0A tan VKQ= O/A Tan VKQ = 8/13 =31.61 = 31º 36º 27.01 = 31º 36º
13=A
13
8
17
2.5 Diagram 1 shows a prism with horizontal square base HJKL . Trapezium EFLK is the uniform cross-
section of the prism. The rectangular surface DEKJ is vertical while the rectangular surface GFLH is inclined.
Calculate the angle between the plane DLH and the base HJKL. [4 marks] SPM 2003 DLH K LH J D L/H J TOA tan DHJ = O/A = 6/8 DHJ = 36.87 = 36º 52º 11.63 = 36º 52º
K 8 cm
L
H J
E F
G D
6 cm
Diagram 1
K 8 cm
L
H J
E F
G D
6 cm
Diagram 1
8
6
8= A
6=O
D
H J
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3. [ Mathematical Induction ] 3.1. Make a general conclusion by induction for the following sequences. (a) 5 = ( 0 ) + 5 6 = ( 1 ) + 5 7 = ( 2 ) + 5 8 = ( 3 ) + 5 . . . . . . . . . . . . . . ( ) + 5 ( n ) + 5 n = 0,1,2 . . .
(b) 3 = 2 ( 2 ) – 1 5 = 2 ( 3 ) – 1 7 = 2 ( 4 ) – 1 9 = 2 ( 5 ) – 1 . . . . . . . . . . . . . . 2 ( ) – 1 2 ( n ) – 1 n = 2,3,4 . . .
(c) 3 = ( 0 ) 2 + 3 4 = ( 1 ) 2 + 3 . 7 = ( 2 ) 2 + 3 12 = ( 3 ) 2 + 3 . . . . . . . . . . . . . . ( ) 2 + 3 ( n2 ) + 3 n = 0,1,2 . . .
Write the unchanged. Replace the sequence with n write first 3 n follows by
(d) –3 = 4(0) – 3 1 = 4(1) – 3 5 = 4(2) – 3 9 = 4(3) – 3 . . . . . . . . . . . . . . 4n – 3 n=0,1,2,……
(e) 6 = 2 (1 )2 + 4 12 = 2 (2 )2 + 4 22 = 2 (3 )2 + 4 36 = 2 (4 )2 + 4 . . . . . . . . . . . . . . 2(n)2+ 4 n= 1,2,3,……..
(f) 12 = 2 (3 ) 2 – 6 26 = 2 (4 ) 2 – 6 44 = 2 (5 ) 2 – 6 66 = 2 (6 ) 2 – 6 . . . . . . . . . . . . . . 2x2 – 6 x= 3,4,5,……
(g) – 1 = 7 – (2)3
– 20 = 7 – (3)3
– 57 = 7 – (4)3 – 127 = 7 – (5)3 . . . . . . . . . . . . . . 7 – n3 n = 2,3,4,…..
(h) 35 = 4 ( 3 ) 2 – 1 63 = 4 (4 ) 2 – 1 99 = 4 (5 ) 2 – 1 143 = 4 (6 ) 2 – 1 . . . . . . . . . . . . . . 4 (n )2 – 1 n=3,4,5,……
(i) 1 = 2 1 – 1 3 = 2 2 – 1 7 = 2 3 – 1 15 = 2 4 – 1 . . . . . . . . . . . . . . 2 n – 1 n=1,2,3…
(j) 10 = 3 x 2 0 + 7 13 = 3 x 2 1 + 7 19 = 3 x 2 2 + 7 31 = 3 x 2 3 + 7 . . . . . . . . . . . . . . 3 (2)n + 7 n=0,1,2,……
(k) 19 = 42 + 3 67 = 43 + 3 259 = 44 + 3 1027 = 45 + 3 . . . . . . . . . . . . . . 4 n + 3 n=2,3,4,…
(l) 4 = 3 x 1 + 1 = 3 (1)2 + 1 13 = 3 x 4 + 1 = 3 (2)2 + 1 28 = 3 x 9 + 1 = 3 (3)2 + 1 49 = 3 x 16 + 1 = 3 (4)2 + 1 . . . . . . . . . . . . . . 3 x n2 + 1 n = 1,2,3, ….
(m) 6 = 3 x 2 = 3 ( 3 – 1 ) 12 = 4 x 3 = 4 ( 4 – 1 ) 20 = 5 x 4 = 5 ( 5 – 1 ) 30 = 6 x 5 = 6 ( 6 – 1 ) . . . . . . . . . . . . . . n x (n -1) n = 3,4,5,…… 6 = 3 x 2 = ( 2 + 1 ) (2) 12 = 4 x 3 = ( 3 + 1 ) (3) 20 = 5 x 4 = ( 4 + 1 ) (4) 30 = 6 x 5 = ( 5 + 1 ) (5) . . . . . . . . . . . . . . n = (n+1)xn n = 2,3,4,......
19
4. Mathematical Reasoning 4.1 Anatomy of a statement.
Front Centre Back Example All Some
- - All sets have subsets. Some triangles are isosceles.
********* if and only if +++++++++ If ** ** then ++ ++ If ++ ++ then ** **
True(1)/ False (0) And (x)
True(1)/ False (0) 2+3 = 5 and 23 = 6 (1) x 0 = 0
True(1)/ False (0) Or (+)
True(1)/ False (0) – 4 > –5 or (2+5)2 > (25)2 ( 1) + 0 = 1
If **** ( antecedent)
Then +++++ (consequence)
Converse If +++++ then ****
4.2. Argument 4.2.1 Premise 1 : All even numbers are divisible by 2 Premise 2 : 12 is an even number Conclusion : 12 is divisible by 2. 4.2.2 Premise 1 : If the mean of m numbers is n then the sum is mn. Premise 2 : The mean of 6 numbers is 7 Conclusion : The sum is 42 . 4.2.3 Premise 1 : All parallelograms have two pairs of parallel lines. Premise 2 : ABCD is a parallelograms Conclusion : ABCD have two pairs two pairs of parallel lines. 4.2.4 Premise 1 : All even number is divisible by 2. Premise 2 : 223 is not divisible by 2 Conclusion : 223 is not an even number . 4.2.5 Premise 1 : If the perimeter of a square is 40 cm then the area of the square is 100 cm2 Premise 2 : the area of a square is not 100 cm2 Conclusion : The perimeter of the square is not 40 cm. 4.2.6 Premise 1 : If xo is within 90o and 270 o , then cos x o < 0 Premise 2 : cos x > 0 Conclusion : xo is not within 90o and 270 o 4.2.7 Premise 1 : If P ( x, y ) lies on x-axis ; then y = 0 Premise 2 : y ≠ 0 Conclusion : P( x, y ) do not lie on x-axis
20
4.3.1 Premise 1 : If m is a factor of 12 then m is a factor of 36. Premise 2 : 4 is a factor of 12. Conclusion : 4 is a factor of 36 4.3.2 Premise 1 : If perimeter of a square is 12 cm, then its area is 16 cm2. Premises 2 : Perimeter of square PQRS is 12 cm Conclusion : Area of the square PQRS is 16 cm2. 4.3.3 Premise 1 : If √ x = y , then y 2 = x . Premise 2 : √ 16 = 4 Conclusion : 4 2 = 16 4.3.4 Premise 1 : If the product of the gradients of two lines is – 1, then they are perpendicular to each other. Premise 2 : Line PQ and RS are not perpendicular to each other. Conclusion : The product of gradient line PQ and RS is not – 1 . 4.3.5 Premise 1 : If p > q , then p + 4 > q + 4 . Premises 2 : P +4 < q +4 Conclusion : p < q . 4.3.6 Premise 1 : If m is a negative number then 2 m < 0. Premises 2 : 2m > 0; 2m is not < 0 Conclusion : m is not a negative number. 4.3.7 Premise 1 : If the radius of a circle is 5 cm then the area is 25 π cm2 Premises 2 : The area of the circle R is not 25 π cm2. Conclusion : The radius of circle R is not 5 cm. 4.3.8 Premise 1 : If the sum of 4 number is 100 then the average is 25 Premise 2 : The average of 4 numbers is not 25. Conclusion : The sum of the 4 numbers is not 100.
21
4.4 Write two implications from each of the following compound statements.
Compound Statement Answer a) 10 a = 1 if and only if a = 0
Implication 1 : I f 10 a = 1 then a = 0 Implication 2 :If a = 0 then 10 a = 1
b) x3 = –64 if and only if x = –4 Implication 1 : I f x3 = –64 then x = –4 Implication 2 : I f x = –4 then x3 = –64
c) Abu will be punished if and only if he is late to school
Implication 1 : I f Abu will be punished then he is late to school Implication 2 : I f Abu is late to school then he will be punished
d) x + 3 = –7 if and only if x – 8 = –18
Implication 1 : I f x + 3 = –7 then x – 8 = –18 Implication 2 : I f x – 8 = –18then f x + 3 = –7
e) BA⊂ if and only if ABA =∩
Implication 1 : If BA⊂ then ABA =∩ Implication 2 : If ABA =∩ then BA⊂
f) y2 – 4y = –4 if and only if y = 2
Implication 1 : If y2 – 4y = –4 then y = 2 Implication 2 : If y = 2then y2 – 4y = –4
g) k is a perfect square if and only if k is an integer
Implication 1 : If k is a perfect square then k is an integer Implication 2 : If k is an integer then k is a perfect square
h) m is a negative number if and only if m3 is a negative number
Implication 1 : if m is a negative number then m3 is a negative number Implication 2 : If m3 is a negative number then m is a negative number
i) 10 –1 =z1
if and only if z =10 Implication 1 : If 10 –1 =z1
then z =10
Implication 2 : If z =10 then 10 –1 =z1
then z =10
j) 5=m if and only if 52 = m Implication 1 : If 5=m then 52 = m
Implication 2 : If 52 = m then 5=m then
22
5. Volume of solids. 5.1. The basic Main Sub/ Derivative ½ Self
combined With others
V = (8)(6)(15) = 720
V = 31 (8)(6)(15)
= 240
V = (722 ) (7)2(15)
= 2310
V= (31 )(
722 )(7)2(15)
= 770
Cross-sectional area x depth ( ½ )(4)(3) [6] = 36
Cross-sectional area x depth ( ½) ( 3+5)(4) [6] = 96
Cross-sectional area x depth ( ½ )( 3+5)(4)[6] = 96
(
34 ) (22/7)(21)3
=38808
( ) 1940421722
34
21 3 =
8 6
15
8 6
15
7
7
15 15
4 6
3 5
4 3
6
3
5
4 6
21 21
23
5.2.The marking Scheme 2
Diagram below shows a solid cylinder with diameter 8 cm and height 10 cm. A cone with radius 4 cm and height 7 cm is taken out of the solid.
Calculate the volume, in cm3, of the remaining solid. (Use π = 227
)
[4 marks]
A n s w e r
2 2
2 2
3
22 1 224 10 or 4 77 3 722 1 224 10 4 77 3 7
502.857 117.333385.53cm
× × × × ×
× × − × × ×
−
K1 K1 K1 N1
Case 1 Volume of the cylinder – Volume of the cone
K 1
5.3. Strategy in solving Step 1 : Label each length, assign unknown with x Step 2 : Draw separate diagram for different solid. Step 3 : Substitute into the formulae. Step 4 : Involving addition ( 2 different solids ) or subtraction (Original – take away) 5.4 Finding the volume of a combined solid. 5.4.1 [SPM 2003]
Vol. of pyramid = 31 (7)(10)(9) = 210
Vol. of semi cylinder = ½ (722 )(3.5)2 (10)=192.5
Total = 210+192.5 = 402.5
D
E
F
G
7 10
9
Volume of cylinder or cone
Big - small
4
10 7
10 7
9
10 5.3
27=
10
10
7
24
5.4.2 [SPM2004]
Vol. of cylinder = (722 )(3.5)2 (4) = 154
Vol. of Cone = (31 )(
722 )(3.5)2 x =
215.269 x
Total = 154 + 21
5.269 x = 231,
x = 6 5.4.3[SPM2005]
Vol. of cone = = (31 )(
722 )(9)2 (14) =1188
Vol. of cylinder = (722 )(3)2 (7) = 198
The cylinder is removed, find the volume. Vol remains = 1188–198 = 990
9
14
3
4
7
x
Total volume= 231 cm3 Find the height of the cone.
7
3
4
x
5.327=
7
9
3
14
5.327=
25
5.4.4 [SPM 2006] Total volume=584 cm3
Vol. of pyramid = 31 (6)(14)(8) = 224
Vol. of prism = ½ ( 10 + 14 )( x ) = 72 x Total Volume = 584 224 + 72x = 584; x = AF = 5 5.4.5
(a) Volume of the right pyramid (31 )(10)(10)(12)=400cm3]
(b) Total volume =775, find AM Find the height of AM = x
Vol. of prism = ( ½ ) (10+15)(10)(x) = 125x 400 + 125 x = 775; x = AM = 3.0 cm
D M N
P Q
V
10 cm
12 cm
10 cm
A B
C
15 cm
V
G
H E
F
B
C D
A 6
10
14
8
6 14
6 x
x
x
6
10
14
8
6 14
6 x
x
14 6
6
6
14
6 14
x
10
10
10
10
10
10
12 10
x
10 10
10
15
15 10
26
6. Perimeter and Area of a sector. 6.1. The basics
1. Perimeter of a circle = 2 π r
Area of a circle = π r 2
6.2.1 Angles 2.1 Equilateral triangle 6.2.2 Isosceles right triangle
6.2.3 Theorem Pythagoras. 6.2.4 Big - Small
7 )7(7222 ××=P
120o
40o
)7(7222
360120
1 ×××
=P
)7(7222
36040
2 ×××
=P
7
7
7 2)7(722
×=A
120o
40o
7
21 )7(
722
360120
××
=A
22 )7(
722
36040
××
=A
40o
7 5
9
120o 60o
30o
45o 135o
7
x
20)57()97( 22 =+++=x
12
5
120o
12
12 12
5
5122112
722
360120 2 ××−××=AreaShaded
27
6.3. Strategy in solving.
Step 1 Label each side/segment/angles Step 2 Draw separate diagram for each part of a question. Step 3 Each question must have angle, and length or radius.
The table below will guide you in solving.
θ r P/A P = Perimeter A =Area
45
15
( )157222
36045
×××
( )215722
36045
×××
6.4. Skill Practice 6.4.1 [SPM2003]
rs ×××=7222
360θ
(a) Perimeter = 14.67+11+14+7+7 =53.67
(b) Area of shaded region = 38.5+77=115.5
N M
P Q
T
O
60o 14 cm
7 cm 14 cm
45o
Smart 4 2
7 7
7 7
15
15
T 60o 7
7 7 7
θ
r
rs ×××=7222
360θ
90
7
77222
36090
×××=s
θ
r
rs ×××=7222
360θ
90 14 147222
36060
×××=s
14
N M
P Q
T
O
60o
60o
14 60o
90o
θ
r
2
722
360rA ××=
θ
60 14 214722
36060
××=A
θ
r
2
722
360rA ××=
θ
60 7 27722
36060
××=A
7 14
7 7
7
θ
r
2
722
360rA ××=
θ
90 7 27722
36090
××=A
28
6.4.2 [SPM2004] ½ (14)(14)
(a) Area of shaded region =346.5- 98 =248.5
(b) Perimeter = 33+21+14+14.67+7 =89.67
R
Q
P
T
S
O 60º
14 cm
7 cm 7 cm
14 14
14 14 21 21
90º 90º
θ
r
2
722
360rA ××=
θ
90 21 221722
36090
××=A
14 cm
7 cm
14 θ
r
rS ×××=7222
360θ
60 14 147222
36060
×××=S 90º
60º
θ
r
rS ×××=7222
360θ
90 21 217222
36090
×××=S
7
29
6.4.3 [SPM2005] RO = 2 OV, OV = 7cm and UOV∠ = 60°
(a) Perimeter = 3264721
388
322
=+++
(b) Area of shaded region = 1543
77777
616=+
−
S
T
R
W
O V
U
7 14
7 120o 60o
θ
r
2
722
360rA ××=
θ
120 14 214722
360120
××=A
θ
r
rS ×××=7222
360θ
120 14 147222
36060
×××=S
60o 7
7 7
180o 120o
14
14 7 θ
r
2
722
360rA ××=
θ
180 7 27722
360120
××=A
θ
r
2
722
360rA ××=
θ
60 7 27722
36060
××=A
30
6.4.4 [SPM2006] OMRN is a quadrant of a circle OM = MP = 7 cm and oPOQ 60=∠
(a) Perimeter = 7(4)+3.67+7.33+14.67=53.67
(b) Area of shaded region =12.83 +(102.67- 25.67)= 89.83
N
Q
O M P
60°
θ
r
rS ×××=7222
360θ
30 7 77222
36030
×××=S
7 7
7
7 30o
θ
r
rS ×××=7222
360θ
60 14 147222
36060
×××=S
7
60°
30° 7 7
7 60°
14
14
θ
r
2
722
360rA ××=
θ
30 7 27722
36030
××=A
60°
θ
r
2
722
360rA ××=
θ
60 7 27722
36060
××=A
θ
r
2
722
360rA ××=
θ
60 14 214722
36060
××=A
31
6.4.5 . [SPM2007]
OS = 21 cm and OP = 14 cm (a) Area of shaded area = 346.50 – 102.67= 243.83 (b) Total perimeter =21+33+7+29.33+14 = 104.33
R
S
T
60o
Q
P
O 14
14
14
7
21
21
21
60o
14
14
θ
r
2
722
360rA ××=
θ
90 21 221722
36090
××=A
θ
r
2
722
360rA ××=
θ
60 14 214722
36060
××=A
S
T
60o
Q
P
O 14
14
14
7
21
120o
θ
r
2
7222
360rS ×××=
θ
90 21 217222
36090
×××=s θ
r
2
7222
360rS ×××=
θ
120 14 147222
360120
×××=s
32
7. Matrices 7.1 Multiplication of matrices
Example 1 Example 2 Example 3 Example 4
( )
35
12 ( )
−5143
12
− 4
34213
− 53
124213
1.Fill in the grid. Row by column
5 3 2 1
3 –4 1 5 2 1
3 4 3 1 –2 4
2 1 1 5 3 1 –2 4
2.Multiplication of corresponding elements
2x5 + 1 x 3 2x3+1x1 2x(–4)+1x5 3x3+1x(4) (–2)x(3)+4x4
3x2+1x1 3x1+1x5 (–2)x2+4x1 (–2)x1+4x5
3. Simplify 10 + 3
6+1 –8+ 5
9 + 4 –6 + 16
6 + 1 3 + 5 –4 + 4 –2 + 20
( 13 ) ( 7 – 3 )
1013
18087
Question 1 Question 2 Question 3 Question 4
( )
43
12 ( )
−5143
12
− 4
34213
− 53
124213
1. Fill in the grid. Row by column
2. Multiplication of corresponding elements
3. Simplify
33
7.2 Simultaneous Equation with Matrix method
Example 1 Example 2
12532
=− yx ; 4x + 3y = 6
– 15y = 36
3312 =+ yx ;5x – 2y = 16
5x = 9
1. Donate away the denominator
2 x – 15 y = 36
6x + 1y = 9
2. Align the terms
2 x – 15 y = 36 4 x + 3 y = 6
6 x + 1y = 9 5 x – 2y = 16
3. Write in matrix equation
=
−6
3634152yx
yx
=
− 16
925
16yx
yx
=
−6
3634152
yx
=
− 16
925
16yx
4.Inverse
−=
6
3634152
Inverseyx
−
=
169
2516
Inverseyx
4.1 1/
=
6
361yx
4.1 1/ad
=
6
366
1yx
−
=
169
121
yx
4.2 1/ad–
−
=
6
366
1yx
−−
=
169
121
yx
4.3 1/(ad–cd)
−−
=
6
36606
1yx
−−
=
169
5121
yx
4.4 Change addresses
−−
=
6
362
3606
1yx
−−−
=
169
62
5121
yx
4.5 Change sign
−−−
=
6
3624
153606
1yx
−
−−−−
=
169
6512
5121
yx
5. Multiplication
−
=
6
3624
153661
yx
−
−−−
=
169
6512
171
yx
6.Simplify
+−+
=
12144
90108661
yx
+−−+−
−=
96451618
171
yx
−
=
132
198661
yx
−−
−=
5134
171
yx
−=
6613266
198
yx
−−−−
=
17571734
yx
−
=
2
3yx
−
=
3
2yx
7. Answer x = 3 , y = –2 x = 2 , y = –3
34
7.3 Practice on your own.
Question 1 Question 2
12532
=− yx ; 4x + 3y = 6 3312 =+ yx ; 5x – 2y = 16
1. Donate away the denominator
2. Align the terms
3. Write in matrix equation
4.Inverse
4.1 1/
4.1 1/ad
4.2 1/ad–
4.3 1/(ad–cd)
4.4 Change addresses
4.5 Change sign
5. Multiplication
6.
7. x = 3 , y = –2 x = 2 , y = –3
35
7.4 Solve the simultaneous equations with matrices methods.
7.4.1. M is a 2 x 2 matrix where M (4523
−−
) = (1001
) SPM2003
a) Find the matrix M b) Write the following simultaneous linear equation as a matrix equation 3 x – 2 y = 7 5 x – 4 y = 9 Hence , calculate the values of x and y using matrices. [6 marks]
7.4.2. a) The inverse matrix of
−−
6543
is m
−−
356 p
SPM2004
Find the value of m and of p. b) Using matrices, calculate the value of x and of y that satisfy the following , simultaneous linear equation: 3x – 4y = –1 5x – 6y = 2 [6 marks]
7.4.3. It is given that matrix P =
−3152
and matrix Q = k
− 213 h
SPM2005
such that PQ =
1001
.
a)Find the value of k and of h. b)Using matrices , find the value of x and of y that satisfy the following simultaneous linear equation: 2x – 5y = –17 x + 3y = 8 [7 marks]
7.4. 4. a)It is given that
n
21
21 is the inverse matrix of
−
−2143
. SPM2006
Find the value of n. b)Write the following simultaneous linear equations are matrix equation:
3u – 4v = –5 –u + 2v = 2 hence , using matrices , calculate the value of u and of v. [6 marks]
7.4.5. (a) The inverse matrix of
−
−4274
is
2272
r, find the value of r .
SPM2007 (b) Using matrices, find the value of x and y that satisfies the following
simultaneous equations. 4x – 7y = 15 – 2x + 4y = –8 [6 marks] Answer
7.4.1 (a)
−−
−=3524
21M (b) x = 5 , y = 4
7.4.2 (a) m = ½ p = 4 (b) x = 7 , y = 11/2 7.4.3 (a) k = 1/11 , h = 5 (b) x = – 1 , y = 3 7.4.4 (a) n = 3/2 (b) u = – 1 , v = ½ 7.4.5 (a) r = 1 (b) x = 2 , y = –1