Answer Technique MM 2

7/29/2019 Answer Technique MM 2

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ANSWERING TECHNIQUES:SPM MATHEMATICS


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Paper 2

Section A


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Simultaneous Linear equation (4 m)

Simultaneous linear equations with two unknowns can be solved

by (a) substitution or (b) elimination. Example: (SPM07-P2) Calculate the values of p and q that

satisfy the simultaneous :

g + 2h = 1

4g 3h = 18

g + 2h = 1

4g 3h = 18

: g = 1 2h

into: 4(1 2h) 3h = 18 4 8h 3h = 18

11h = 22

h = 2

When h = 2, from:

g = 1 2(2)

g = 1 4

g = 3

Hence, h = 2

and g = 3

1

2

1


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Simultaneous Linear equation

Simultaneous linear equations with two unknowns can be solved

by (a) elimination or (b) substitution. Example: (SPM04-P2) Calculate the values of p and q that

satisfy the simultaneous :

p2q =13

3p + 4q = 2

p2q =13

3p + 4q = 2

2: p4q = 26

+: 4p = 24

p = 6

When p = 6, from:

(6)2q = 13

2q = 313

2q = - 10q = - 5

Hence, p = 6

and q = - 5

1

2

1


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Solis geometry (4 marks) Include solid geometry of cuboid, prism, cylinder, pyramid, cone and

sphere.

Example : (SPM04-P2) The diagram shows a solid formed by joininga cone and a cylinder. The diameter of the cylinder and the base of thecone is 7 cm. The volume of the solid is 231 cm3. Using = 22/7,calculate the height , in cm of the cone.

4 cm

Let the height of the cone be tcm.

Radius of cylinder = radius of cone= 7/2 cm (r)

Volume of cylinder = j2t

= 154 cm3

Hence volume of cone = 231154 = 77 cm3

= 77

t=

t= 6 cm

42

7

7

222

t

2

2

7

7

22

3

1

2

7

2

22

7377

tcm

7/2 cm

Rujuk rumus yang

diberi dalam kertas

soalan.

1

2

1


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Perimeters & Areas of circles (6 m) Usually involve the calculation of both the arc and area of

part of a circle.

Example : (SPM04-P2) In the diagram,PQ andRSare the

arc of two circles with centre O. RQ = ST= 7 cm andPO =14 cm.

Using = 22/7 , calculate

(a) area, in cm2, of the shaded region,

(b) perimeter, in cm, of the whole diagram. 60 O

T

R S

P

Q


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Perimeters & Areas of circles

60O

T

R S

P

Q

(a) Area of shaded region= Area sectorORSArea ofDOQT

=

= 346 98

= 248 cm2

2217

22

4

1 1414

2

1

(b) Perimeter of the whole diagram

= OP+ arcPQ + QR + arcRS+ SO

= 14 + + 7 + + 21

= 346 98

= 248 cm2

217

2224

1 147

222360

60

Formula

given in exam

paper.

Formula given

.

2

1

2

1


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Mathematical Reasoning (5 marks)

(a) State whether the following compound statement is true or false

76and12553

Ans: False 1


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Mathematical Reasoning(b) Write down two implications based on the following compound

statement.

4.ifonlyandif643 xx

Ans: Implication I : If x3 = -64, then x = -4

Implication II : If x = -4, then x3 = -64

(c) It is given that the interior angle of a regular polygon of n sides

is

Make one conclusion by deduction on the size of the size of theinterior angle of a regular hexagon.

180

21

n

Ans:

120

1806

21hexagonregularaofangleinteriorThe

2

2


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The Straight Line ( 5 or 6 marks)Diagram shows a trapezium PQRS drawn on a Cartesian plane. SR

is parallel to PQ.

Find

(a) The equation of the

straight line SR.

2

1

8

4

113

15

PQm

1321

42

19

82

19SR,linestraighttheofEquation

xy

xy

xy

Ans: 1

1

1

1


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The Straight LineDiagram shows a trapezium PQRS drawn on a Cartesian plane. SR

is parallel to PQ.

Find

(b) The y-intercept of

the straight line SR

Ans: The y-intercept

of SR is 13.1


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Graphs of Functions (6 marks)Diagram shows the speed-time graph for the movement of a particle

for a period of t seconds.


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22

4

1220

seconds4firstin theparticletheofspeedofchangeofrateThe

ms

Graphs of Functions(a) State the uniform speed, in m s-1, of the particle.

Ans: 20 m s-1

(b) Calculate the rate of change of speed, in m s-1, of the particle

in the first 4 seconds.

Ans:

(c) The total distance travelled in t seconds is 184 metres.

Calculate the value of t.

Ans:

seconds10

20020

184802064

184420420122

1

t

t

t

t

1

1

1

2

1


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Probability (5 or 6 marks)Diagram shows three numbered cards in box P and two cards

labelled with letters in box Q.

2 3 6 Y R

P Q

A card is picked at random from box P and then a card is picked atrandom from box Q.


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Probability (5 or 6 marks)By listing the sample of all the possible outcomes of the event,

find the probability that

(a) A card with even number and the card labeled Y are picked,

3

1

2

1

3

2

card)P(Ynumber)P(Evencard)YandnumberP(Even

(b) A card with a number which is multiple of 3 or the card

labeled Ris picked.

6

5

3

1

2

1

3

2

card)R3ofP(multiplecard)P(R3)ofP(multiplecard)Ror3ofP(multiple

1

1

1

1

1


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Lines and planes in 3-Dimensions(3m)

M

D

H

G

C

B

F

8 cm

E

A

Diagram shows a cuboid. M is

the midpoint of the side EH and

AM = 15 cm.

(a) Name the angle between the

line AM and the plane ADEF.

Ans: EAM(b) Calculate the angle between

the line AM and the plane

ADEF.

A

M

E

15 cm

4 cm

Ans:

'2815

'2815

15

4sin

EAM

1

1

1


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Matrices

This topic is questioned both in Paper 1 &

Paper 2

Paper 1: Usually on addition, subtractionand multiplication of matrices.

Paper 2: Usually on Inverse Matrix and the

use of inverse matrix to solve simultaneousequations.


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Matrices (objective question)

Example1: (SPM03-P1)

4

2

43

15

43

15

4

2

5(-2) + 14

3(-2) + 44

166

410

10

6


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Matrices (6 or 7 marks)

Example2: (SPM04-P2)

(a) Inverse Matrix for

is

65

43

35

6 p

m

Inverse matrix

formula is given

in the exam

paper.

1

65

43

35

)4(6

5)4()6(3

1

35

46

2

1

Hence, m = ,p = 4. 2

1


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Matrices Example2: (SPM04-P2) (contd)

(b) Using the matrix method , find the value of x and y that

satisfy the following matrix equation:

3x4y = 1

5x6y = 2

Change the simultaneous equation into matrix equation:

Solve the matrix equation:

2

1

65

43

y

x

2

1

65

43

65

43

65

4311

y

x

2

1

35

46

2

1

y

x

23)1()5(

24)1()6(

2

1

y

x

11

14

2

1

y

x

2

15

7

y

x

Maka,x = 7,y = 5

1

1

2

1


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Paper 2

Section B


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Graphs of functions(12 marks) This question usually begins with the calculation of two to

three values of the function.( Allocated 2-3 marks)

Example: (SPM04-P2)

y = 2x24x3

Using calculator, find the values ofkand m: Whenx = - 2,y = k.

hence, k = 2(-2)24(-2)3

= 13

Whenx = 3,y = m.hence, m = 2(3)24(3)2

= 3

Usage of calculator:

Press 2 ( - 2 ) x2 - 4

( - 2 ) - 2 = .Answer 13 shown on

screen.

To calculate the next

value, change 2 to 3.

2


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Graphs of functions

To draw graph

(i) Must use graph paper.

(ii) Must follow scale givenin the question.

(iii) Scale need to be

uniform.(iv) Graph needs to be

smooth with regularshape.

Example: (SPM04-P2)

y = 2x24x3


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Graphs of functions Example: (SPM04-P2)

Draw y = 2x24x3

To solve equation

2x2

+x23 = 0,2x2 +x + 4x4x3 -20 = 0

2x24x3 = - 5x + 20

y = - 5x + 20

Hence, draw straight line

y = - 5x + 20

From graph find values of x

4

1

1

2


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Plans & Elevations(12 marks) NOT ALLOW to sketch.

Labelling not important.

The plans & elevations can be drawn from any angle.(except when it becomes a reflection)

Points to avoid:

Inaccurate drawing e.g. of the length or angle.

Solid line is drawn as dashed line and vice versa.

The line is too long.

Failure to draw plan/elevation according to given scale.

Double lines.

Failure to draw projection lines parallel to guiding lineand to show hidden edges.


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Plans & Elevations (3/4/5 marks)

X

H

G

D E

LK

M

J

F

N

4 cm

6 cm

3 cm


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Statistics (12 marks) Use the correct method to draw ogive, histogramand

frequency polygon.

Follow the scale given in the question.

Scale needs to be uniform.

Mark the points accurately. The ogive graph has to be a smooth curve.

Example(SPM03-P2) The data given below shows theamount of money in RM, donated by 40 families for a

welfare fund of their children school.


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Statistics 40 24 17 30 22 26 35 1923 28 33 33 39 34 39 2827 35 45 21 38 22 27 35

30 34 31 37 40 32 14 28

20 32 29 26 32 22 38 44

Upper

boundaryAmount

(RM)

Frequency Cumulative

Frequency

11 - 15

16 - 20

21 - 25

26 - 30

31 - 35

36 - 40

41 - 45

1

3

6

10

11

7

2

0

1

4

10

20

31

38

40

10.5

15.5

20.5

25.5

30.5

35.5

40.5

45.5

To draw an ogive,

Show the Upper

boundary column,

An extra row to indicate

the beginning point.

3


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Statistics Ogif bagi wang yang didermakan

0

5

10

15

20

25

30

35

40

45

0 10 20 30 40 50

Wang (RM)

Kekerap

anLonggokanThe ogive drawn is

a smooth curve.

Q3 4

d) To use value from graph to solve question given (2m)


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Combined Transformation (SPM03-P2)

(a) RReflection in the liney = 3,

Ttranslasion

Image ofHunder(i) RT

(ii) TR

10

x-6 -4 -2 2 4 6 8

2

4

6

8

y

O

P

N M

L

G

H

JK

CB

AD

E F

4

2

2

2


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Combined Transformation(12 marks) (SPM03-P2)

(b) V mapsABCD toABEF

V is a reflection in the lineAB.

W mapsABEFto GHJK.

W is a reflection

in the linex = 6.

10

x-6 -4 -2 2 4 6 8

2

4

6

8

y

O

P

N M

L

G

H

JK

CB

AD

E F

2

2


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(b) (ii) To find a transformation that is equivalent to twosuccessive transformations WV.

Rotation of 90 anti clockwise about point (6, 5).

10

x-6 -4 -2 2 4 6 8

2

4

6

8

y

O

G

H

JK

CB

AD

3


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(c) Enlargement which mapsABCD toLMNP.

Enlargement centered at point (6, 2) with a scale factor of3.

AreaLMNP

= 325.8 unit2

Hence,

AreaABCD

= 36.2 unit210

x-6 -4 -2 2 4 6 8

2

4

6

8

y

O

P

N M

L

CB

AD

8.3253

1

2

3

1

1


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THE ENDGOD BLESS

&Enjoy teaching
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Answer Technique MM 2

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