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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 :
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
(a) Area of shaded region
= 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
AD
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
AD
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
AD
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
AD
8.3253
12
3
1
1
THE ENDGOD BLESS
&Enjoy teaching