34

# ANSWERING TECHNIQUES: SPM MATHEMATICS. Paper 2 Section A

Embed Size (px)

Citation preview

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 :

g + 2h = 14g 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 = 2and g = 3

1

2

1

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 =133p + 4q = 2

½p – 2q =13 3p + 4q = 2

2:p – 4q = 26 + : 4p = 24 p = 6

When p = 6, from : ½ (6) – 2q = 13 2q = 3 – 13 2q = - 10 q = - 5

Hence, p = 6and q = - 5

1

2

1

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.

4 cm

Let the height of the cone be t cm. Radius of cylinder = radius of cone= 7/2 cm (r) Volume of cylinder = j2t = 154 cm3

Hence volume of cone = 231 – 154 = 77 cm3

= 77 t = t = 6 cm

42

7

7

222

t2

2

7

7

22

3

1

2

7

2

22

7377

t cm

7/2 cm

Rujuk rumus yang diberi dalam kertas

soalan.

1

2

1

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. 60O

T

R S

P

Q

Perimeters & Areas of circles

60O

T

R S

P

Q

= Area sector ORS –Area of OQT = = 346½ – 98 = 248½ cm2

2217

22

4

1 1414

2

1

(b) Perimeter of the whole diagram = OP + arc PQ + QR + arc RS + SO = 14 + + 7 + + 21 = 346½ – 98 = 248½ cm2

217

222

4

1 14

7

222

360

60

Formula given in

exam paper.

Formula given .

2

1

2

1

Mathematical Reasoning (5 marks)(a) State whether the following compound statement is true or false

76 and 12553

Ans: False 1

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

statement.

4. ifonly and if 643 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 the

interior angle of a regular hexagon.

180

21

n

Ans:

120

1806

21hexagonregular a of angleinterior The

2

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.

2

18

4113

15

PQm

132

1

42

19

82

19 SR, linestraight theofEquation

xy

xy

xy

Ans: 1

1

1

1

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

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

for a period of t seconds.

22

4

1220seconds 4first in the particle theof speed of change of rate The

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

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 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,

3

12

1

3

2

card) P(Ynumber)P(Even card) Y andnumber P(Even

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

6

53

1

2

1

3

2

card) R3 of P(multiplecard) P(R3) of P(multiplecard) Ror 3 of P(multiple

1

1

1

1

1

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 cm4 cm

θ

Ans:

'2815

'281515

4sin

EAM

1

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 (objective question) Example 1: (SPM03-P1)

4

2

43

15

43

15

4

2

5(-2) + 14

3(-2) + 44

166

410

10

6

Matrices (6 or 7 marks)

Example 2: (SPM04-P2) (a) Inverse Matrix for

is

65

43

35

6 pm

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 = ½ , p = 4. 2

1

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 = 15x – 6y = 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 = 5½

1

1

2

1

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 = .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.

Plans & Elevations (3/4/5 marks)

X

H

G

D E

LK

MJ

F

N

4 cm

6 cm

3 cm

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 40 24 17 30 22 26 35 1923 28 33 33 39 34 39 2827 35 45 21 38 22 27 3530 34 31 37 40 32 14 2820 32 29 26 32 22 38 44

Upper boundary

Amount (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

Statistics Ogif bagi wang yang didermakan

0

5

10

15

20

25

30

35

40

45

0 10 20 30 40 50

Wang (RM)

Kek

erap

an L

ongg

okan

The ogive drawn is a smooth curve.

Q3 4

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

Combined Transformation (SPM03-P2) (a) R – Reflection in the line y = 3,

T – translasion

Image of H under

(i) RT

(ii) TR

10x

-6 -4 -2 2 4 6 8

2

4

6

8

y

O

P

N M

L

G

H

JK

C B

E F

4

2

2

2

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.

10x

-6 -4 -2 2 4 6 8

2

4

6

8

y

O

P

N M

L

G

H

JK

C B

E F

2

2

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).

10x

-6 -4 -2 2 4 6 8

2

4

6

8

y

O

G

H

JK

C B

3

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 10x

-6 -4 -2 2 4 6 8

2

4

6

8

y

O

P

N M

L

C B

8.3253

12

3

1

1

THE ENDGOD BLESS

&Enjoy teaching