1 / 34

# ANSWERING TECHNIQUES: SPM MATHEMATICS - PowerPoint PPT Presentation

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 :

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Paper 2

Section A

• 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 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

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.

• 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

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.

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

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

Ans: False

1

(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

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

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

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

(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

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.

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

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

• 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.

• Example 1: (SPM03-P1)

5(-2) + 14

3(-2) + 44

• 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

• 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 2

Section B

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 = .

To calculate the next value, change – 2 to 3.

2

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 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)

• 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.

H

M

J

3 cm

L

K

F

G

X

6 cm

E

D

4 cm

Plans & Elevations (3/4/5 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.

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

The ogive drawn is

a smooth curve.

Q3

4

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

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

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

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

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

GOD BLESS

&

Enjoy teaching