# teknik menjawab matematik spm

35
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 T 60 Q P O 14 14 14 7 21 = 6 36 2 4 15 3 60 6 1 y x

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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}

`

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 }

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}

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]

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 (V ∩ W ) ′

S

T U

V

W

ξ

S T

U V

W

ξ

9

1.[SPM2004] 2. [SPM 2006]

A = { 1 , 2 } B’= { 1 , 4 }

'BA∩ = { 1 } A ∩ B ′ = ( 12 ) ∩ ( 14) = {1}

(a)

P’ = { 1, 3, 4 } Q = { 1 , 2 , 3 }

QP∩' = { 1 , 3 } P ′ ∩ Q = {134} ∩ {123} = {13}

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 }

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

18

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

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

=

6

366

1yx

=

169

121

yx

=

6

366

1yx

−−

=

169

121

yx

−−

=

6

36606

1yx

−−

=

169

5121

yx

−−

=

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

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/

4.5 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