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ANSWERING TECHNIQUES: SPM MATHEMATICS

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### Paper 2

### Paper 2

ANSWERING TECHNIQUES:SPM MATHEMATICS

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 :
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 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.
- 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

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 circlesFormula 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)

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

Ans: False

1

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)

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

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)

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

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)

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 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

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

- 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 (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

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

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 = .

Answer 13 shown on screen.

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.

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.

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

StatisticsUpper

boundary

10.5

0

- To draw an ogive,
- Show the Upper
- boundary column,
- An extra row to indicate
- the beginning point.

3

Statistics

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

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