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ANSWERING TECHNIQUES: SPM MATHEMATICS. Paper 2. Section A. 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 :

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Answering techniques spm mathematics
ANSWERING TECHNIQUES:SPM MATHEMATICS


Paper 2

Paper 2

Section A


Simultaneous linear equation 4 m
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

1

1

  • Hence, h = 2

    and g = 3

2


Simultaneous linear equation
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 :

    ½p – 2q =13

    3p + 4q = 2

  • When p = 6, from :

    ½ (6) – 2q = 13

    2q = 3 – 13

     2q = - 10

    q = - 5

  • ½p – 2q =13 

    3p + 4q = 2 

  •   2: p – 4q = 26 

  •  + : 4p = 24

  • p = 6

1

1

  • Hence, p = 6

    and q = - 5

2


Solis geometry 4 marks

t cm

7/2 cm

4 cm

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 joining a cone and a cylinder. The diameter of the cylinder and the base of the cone is 7 cm. The volume of the solid is 231 cm3. Using  = 22/7, calculate the height , in cm of the cone.

  • Let the height of the cone be t cm.

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

  • Volume of cylinder = pj2t

  • = 154 cm3

  • Hence volume of cone = 231 – 154 = 77 cm3

  • = 77

  • t =

  • t = 6 cm

1

Rujuk rumus yang diberi dalam kertas soalan.

2

1


Perimeters areas of circles 6 m

S

R

Q

T

60

O

P

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 and RS are the arc of two circles with centre O. RQ = ST = 7 cm and PO = 14 cm.

    Using  = 22/7 , calculate

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

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


Perimeters areas of circles

S

R

Q

T

60

O

P

Perimeters & Areas of circles

Formula given in exam paper.

  • (a) Area of shaded region

    = Area sector ORS –Area of DOQT

  • =

  • = 346½ – 98

  • = 248½ cm2

2

1

  • (b) Perimeter of the whole diagram

  • = OP + arc PQ + QR + arc RS + SO

  • = 14 + + 7 + + 21

  • = 346½ – 98

  • = 248½ cm2

2

Formula given .

1


Mathematical reasoning 5 marks
Mathematical Reasoning (5 marks)

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

Ans: False

1


Mathematical reasoning
Mathematical Reasoning

(b) Write down two implications based on the following compound statement.

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

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

2

(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 the interior angle of a regular hexagon.

Ans:

2


The straight line 5 or 6 marks
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.

Ans:

1

1

1

1


The straight line
The Straight Line

Diagram 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


Graphs of functions 6 marks
Graphs of Functions (6 marks)

Diagram shows the speed-time graph for the movement of a particle for a period of t seconds.


Graphs of functions
Graphs of Functions

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

Ans: 20 m s-1

1

(b) Calculate the rate of change of speed, in m s-1, of the particle in the first 4 seconds.

1

Ans:

1

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

Calculate the value of t.

Ans:

2

1


Probability 5 or 6 marks
Probability (5 or 6 marks)

Diagram shows three numbered cards in box P and two cards labelled with letters in box Q.

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


Probability 5 or 6 marks1
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,

1

1

1

(b) A card with a number which is multiple of 3 or the card labeled R is picked.

1

1


Lines and planes in 3 dimensions 3m

E

M

F

H

G

D

A

C

8 cm

B

M

15 cm

4 cm

θ

A

E

Lines and planes in 3-Dimensions(3m)

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:

1

(b) Calculate the angle between the line AM and the plane ADEF.

Ans:

1

1


Matrices
Matrices

  • This topic is questioned both in Paper 1 & Paper 2

  • Paper 1: Usually on addition, subtraction and multiplication of matrices.

  • Paper 2: Usually on Inverse Matrix and the use of inverse matrix to solve simultaneous equations.


Matrices objective question
Matrices (objective question)

  • Example 1: (SPM03-P1)

5(-2) + 14

3(-2) + 44


Matrices 6 or 7 marks
Matrices (6 or 7 marks)

  • Example 2: (SPM04-P2)

  • (a) Inverse Matrix for

    is

Inverse matrix formula is given in the exam paper.

1

Hence, m = ½ , p = 4.

2


Matrices1
Matrices

  • Example 2: (SPM04-P2) (cont’d)

  • (b) Using the matrix method , find the value of x and y that satisfy the following matrix equation:

    3x – 4y =  1

    5x – 6y = 2

  • Change the simultaneous equation into matrix equation:

  • Solve the matrix equation:

1

1

1

Maka, x = 7, y = 5½

2


Paper 21

Paper 2

Section B


Graphs of functions 12 marks
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 = 2x2 – 4x – 3

  • Using calculator, find the values of k and m:

  • When x = - 2, y = k.

    hence, k = 2(-2)2 – 4(-2) – 3

    = 13

  • When x = 3, y = m.

    hence, m = 2(3)2 – 4(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


Graphs of functions1

Graphs of functions

  • To draw graph

    (i) Must use graph paper.

    (ii) Must follow scale given in the question.

    (iii) Scale need to be uniform.

    (iv) Graph needs to be smooth with regular shape.

  • Example: (SPM04-P2)

  • y = 2x2 – 4x – 3


Graphs of functions2

Graphs of functions

  • Example: (SPM04-P2)

  • Draw y = 2x2 – 4x – 3

  • To solve equation

    2x2 + x – 23 = 0,

    2x2 + x + 4x – 4x – 3 -20 = 0

    2x2 – 4x – 3 = - 5x + 20

    y = - 5x + 20

  • Hence, draw straight line

    y = - 5x + 20

    From graph find values of x

4

1

1

2


Plans elevations 12 marks
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 line and to show hidden edges.


Plans elevations 3 4 5 marks

N

H

M

J

3 cm

L

K

F

G

X

6 cm

E

D

4 cm

Plans & Elevations (3/4/5 marks)


Statistics 12 marks
Statistics (12 marks)

  • Use the correct method to draw ogive, histogram and 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 the amount of money in RM, donated by 40 families for a welfare fund of their children school.


Statistics

Cumulative

Frequency

Amount

(RM)

Frequency

11 - 15

1

1

15.5

4

3

16 - 20

20.5

6

10

21 - 25

25.5

10

26 - 30

20

30.5

31 - 35

11

31

35.5

36 - 40

7

38

40.5

41 - 45

2

40

45.5

40 24 17 30 22 26 35 19

23 28 33 33 39 34 39 28

27 35 45 21 38 22 27 35

30 34 31 37 40 32 14 28

20 32 29 26 32 22 38 44

Statistics

Upper

boundary

10.5

0

  • To draw an ogive,

  • Show the Upper

  • boundary column,

  • An extra row to indicate

  • the beginning point.

3


Statistics1
Statistics

The ogive drawn is

a smooth curve.

Q3

4

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


Combined transformation

y

L

P

8

6

G

D

A

4

H

2

C

M

N

B

F

K

E

J

x

O

-4

-6

-2

2

6

8

4

10

Combined Transformation

  • (SPM03-P2)

  • (a) R – Reflection in the line y = 3,

    T – translasion

  • Image of H under

    (i) RT

    (ii) TR

2

2


Combined transformation 12 marks

y

L

P

8

6

G

D

A

4

H

2

C

M

N

B

F

K

E

J

x

O

-4

-6

-2

2

6

8

4

10

Combined Transformation (12 marks)

  • (SPM03-P2)

  • (b) V maps ABCD to ABEF

  • V is a reflection in the line AB.

  • W maps ABEF

    to GHJK.

  • W is a reflection

    in the line x = 6.

2

2


Combined transformation1

y

8

6

G

D

A

4

H

2

C

B

K

J

x

O

-4

-6

-2

2

6

8

4

10

Combined Transformation

  • (SPM03-P2)

  • (b) (ii) To find a transformation that is equivalent to two successive transformations WV.

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

3


Combined transformation2

y

L

P

8

6

D

A

4

2

C

M

N

B

x

O

-4

-6

-2

2

6

8

4

10

Combined Transformation

  • (SPM03-P2)

  • (c) Enlargement which maps ABCD to LMNP.

  • Enlargement centered at point (6, 2) with a scale factor of 3.

  • Area LMNP

    = 325.8 unit2

  • Hence,

    Area ABCD

    = 36.2 unit2

3

1

1


THE END

GOD BLESS

&

Enjoy teaching


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