seminar math tamb

SEMINAR KECEMERLANGAN SPM

2014

Matematik Tambahan

Mohd Ridwan b. Anang @ Talib

SMK Ulu Sapi, Telupid.

Penyelesaian Masalah

FahamiMasalah

Rancang Strategi

Semak Jawapan

Pelaksanaan Strategi

•Pilih strategi yang sesuai•Pilih formula yang betul

•Topik/Subtopik ? •Maklumat yang diberi•Apa yang perlu dicari

•Buat pengiraan•Melakar graf

•Membina jadual

•Adakah jawapan boleh diterima akal?

• Adakah jalan/strategi lain yang boleh diguna?

FORMAT KERTAS 1

Ujian Subjektif: Jumlah soalan: Jumlah Markah : Masa : Tahap Kesukaran : Peralatan Tamb. :

Soalan pendek 25 soalan 80 2 jam R (15) , S(7-8), T(2-3) Kalkulator saintifik, Set

geometri, Jadual taburan kebarangkalian.

FORMAT KERTAS 2

Soalan Subjektif Bil. Soalan : A (6), B (4/5), C (2/4) Jumlah Markah : 100 Masa : 2 jam 30 minit Tahap Kesukaran : R (6) , S(4-5), T(4-5) Peralatan Tambahan : Kalkulator saintifik, Set geometri, Jadual taburan kebarangkalian

Kunci Mencapai Kecemerlangan

Baca soalan dengan teliti Ikut arahan Mula dengan soalan yang mudah/pilihan Tunjukkan semua jalan kerja dengan jelas Pilih formula yang betul+(Gunakannya

dengan betul !!!) Berikan jawapan dalam bentuk termudah Jawapan akhir betul kepada 4 angka bererti. (atau ikut arahan dalam soalan)

3.142

Kunci Mencapai Kecemerlangan

Tulis simbol matematik dengan betul Semak Jawapan! Peruntukkan masa yang sesuai untuk setiap soalan

Kertas 1 : 3 - 7 minit untuk setiap soalanKertas 2 : Bhg. A : 8 - 10 minit untuk setiap soalan Bhg. B : 15 minit untuk setiap soalan Bhg. C : 15 minit untuk setiap soalan

Kesilapan Biasa Calon…

cxxdxx 4346.3 2

4. sin x = 300 ,

1500

2. y = 3x2 + 4x

y = 6x + 4

1. The Quadratic equation 3x2 - 4x

+ 5= 0

dxdy

x = 300 , 1500

31

PQ

AB5. 1

3

AB

PQ

Kesilapan Biasa Calon …• f ' (x) diinterpretasi sebagai f – 1(x) atau

sebaliknya

• x2 = 4 x = 2

• x2 = 4 x = ± 2

Kesilapan Biasa Calon…

PA : PB = 2 : 3

maka 2PA = 3 PB

Sebenarnya, …

PA : PB = 2 : 3

3 PA = 2 PB

32

PBPA

Kesilapan calon……

32 PA2 = 22 PB2

9 PA2 = 4 PB2

2222 )()(2)()(3

2222 23

Kesilapan biasa calon…loga x + loga y = 0, loga xy = 0 maka xy = 0

Sepatutnya… xy = a0 = 1

Kesilapan biasa…

loga (x – 3) = loga x – loga 3 2x x 2y = 1 x + y = 1

2x x 2y = 20

2x + y = 20

x + y = 0

Fungsi

F4

f : x x - 3 , g : x 3x , cari gf(1).1. Diberi

f(x) = x – 3, g(x) = 3xgf (1) = g [ f(1) ] = g [-2] = -6

Jika f -1(x) = y

Maka x = f (y)

x = 3 – 2y

Cara 1

2

31 xxf

23 xy

T4 BAB 1

Fungsi Songsang

2. Diberi f (x) = 3 – 2x, find f -1.

Cara 2

Jika x = 3 – 2f -1(x)

Maka 2f -1(x) = 3 – x

231 xxf

F4

T4 BAB 1

3. Diberi f:x 2 – x dan gf:x 2x – 2. Cari fg.

Ingat : Cari g dahulu !

f(x) =2 - x , gf(x) = 2x-2Jika f(x) = u

Maka u = 2 – x atau x = 2 - u g(u) = 2(2-u) – 2

= 2-2u g(x) = 2-2x

fg(x) = f(2-2x) = 2 - (2-2x)

= 2x

F4

Persamaan KuadratikBina persamaan kuadratik yang mempunyai punca-punca – 3 dan ½ .

x = – 3 , x = ½ (x+3) (2x – 1) = 0 2x2 + 5x – 3 = 0

F4

Persamaan Kuadratik

02

acx

abx

x2 – ( HTP) x + (HDP) = 0

ax2 + bx + c = 0

ac

HTP = HTP = ab

F4

Persamaan Kuadratik ax2 + bx + c = 0 mempunyai

1. Dua punca berbeza jika

2. Dua punca yang sama jika

3. Tiada punca jika

b2 -

4ac

b2 -

4ac b2

- 4ac

> 0

< 0

= 0

Persamaan Kuadratik: Jenis-jenis Punca

F4

Fungsi Kuadratik : Ketaksamaan KuadratikCari julat nilai x jika x(x – 4) ≤ 12

x (x – 4) ≤ 12 x2 – 4x – 12 ≤ 0 (x + 2)(x – 6) ≤ 0

– 2 ≤ x ≤ 6

6 x -2

F4

Selesaikan:

x2 > 4

Back to BASIC

x> ±2???x2 – 4 > 0

R.H.S must be O !

(x + 2)(x – 2) > 0

x < -2 or x > 2

– 2 2

F4

1. Selesaikan persamaan serentak dan berikan jawapan anda kepada 3 tempat perpuluhan.

x + y =1

x2 + 3y2 = 8a

acbb2

42

Persamaan Serentak

F4

32(x – 1) . 3 (– 3x) = 1 2x – 2 – 3x = 1

– x = 3 x = – 3

Selesaikan .. 1

271.9 1

xx

Betul ke ???

5. INDEKS & LOGF4

32(x – 1) . 3 (– 3x) = 1 32x – 2 +(– 3x) = 30

– x – 2 = 0 x = – 2

Selesaikan 1

271.9 1

xx

F4 5. INDEKS & LOG

atau… 9x-1 = 27x

32(x – 1) = 3 3x 32x – 2 = 33x

2x – 2 = 3x x = – 2

Selesaikan 1

271.9 1

xx

F4 5. INDEKS & LOG

Selesaikan

2x + 3 = 2x+2

2x + 3 = 2x . 22

x = 0

2x + 3 = 4 (2x )

Dalam bentuku + 3 = 4u

3 = 3(2x )

1 = (2x )

F4 5. INDEKS & LOG

Selesaikan persamaan , beri jawapan anda betul kepada 2 tempat perpuluhan. [ 4 markah]

F4

9 (3x) = 32 + (3x)

8 (3x) = 32 3x = 4 x = 1.26

lg 4lg 3

x

23 32 3x x 5. INDEKS & LOG

Selesaikan:22x . 5x = 0.05

4x . 5x = 201

20x = 201

x = – 1

ambm = (ab)m

F4 5. INDEKS & LOG

pp 43

3

log1)

log3log(2

Diberi bahawa log3 p = m dan log4 p = n. Cari logp 36 dalam sebutan m dan n.

= 2log p 3 + logp 4

logp 36 = logp 9 + logp 4

nm12

Menukar asas logaritma

logaa =1

F4 5. INDEKS & LOG

Note to candidates:

Solutions to this question by scale drawing will not be accepted.

Coordinate Geometry

Note to candidates:

A diagram is usually given (starting from SPM 2004). You SHOULD make full use of the given diagram while answering the question.

Coordinate Geometry

Note to candidates:

Sketch a simple diagram to help you using the required formula correctly.

Coordinate Geometry

6. Coordinate Geometry6.2.2 Division of a Line Segment

Q divides the line segment PR in the ratio PQ : QR = m : n

nmP(x1, y1) R(x2, y2)Q(x, y)

●

n

m

R(x2, y2)

P(x1, y1)

Q(x, y)

nmmyny

nmmxnx 2121 ,Q(x, y) =

6. Coordinate Geometry (Ratio Theorem)

The point P divides the line segment joining the point M(3,7) and N(6,2) in the ratio 2 : 1. Find the coordinates of point P.

nmmyny

nmmxnx 2121 ,P(x, y) =

●1

2

N(6, 2)

M(3, 7)

P(x, y)

12)2(2)7(1,

12)6(2)3(1

311,

315

115,3

=

=

P(x, y) =

6. Coordinate Geometry

m1.m2 = –1 P

Q

R

S

Perpendicular lines :

6. Coordinate Geometry(SPM 2006, P1, Q12)

Diagram 5 shows the straight line AB which is perpendicular to the straight

line CB at the point B.

The equation of CB is y = 2x – 1 .Find the coordinates of B. [3 marks]

mCB = 2

mAB = – ½

Equation of AB is y = – ½ x + 4

At B, 2x – 1 = – ½ x + 4

x = 2, y = 3

So, B is the point (2, 3).

x

y

O

A(0, 4)

C

Diagram 5B

●

●

●y = 2x – 1

6. Coordinate Geometry

Given points P(8,0) and Q(0,-6). Find the equation of the perpendicular bisector of PQ.

34

43

)4(34)3( xy

mPQ=

mAB=

Midpoint of PQ =

(4, -3)

The equation :

4x + 3y -7 = 0

K1

K1

N137

34

xyor

P

Q

x

y

O

TASK : To find the equation of the locus of the moving point P such that its distances from the points A and B are in the ratio m : n

(Note : Sketch a diagram to help you using the distance formula correctly)

6 Coordinate Geometry

6. Coordinate GeometryFind the equation of the locus of the moving point P such that its distances from the points A(-2,3) and B(4, 8) are in the ratio 1 : 2.(Note : Sketch a diagram to help you using the distance formula correctly)

A(-2,3), B(4,8) and m : n = 1 : 2 Let P = (x, y)

●

2

1

B(4, 8)

A(-2, 3)

P(x, y)2 2

2 2 2 2

12

2

4

4 ( 2) ( 3) ( 4) ( 8)

PAPBPA PB

PA PB

x y x y

3x2 + 3y2 + 24x – 8y – 28 = 0

6. Coordinate Geometry

Find the equation of the locus of the moving point P such that its distance from the point A(-2,3) is always 5 units. (≈ SPM 2005)

●

5

A(-2, 3)

P(x, y)

● A(-2,3) Let P = (x, y)

is the equation of locus of P.2 2 4 6 12 0x y x y

2 2 2( 2) ( 3) 5x y

6. Coordinate GeometryFind the equation of the locus of point P which moves such that it is always equidistant from points A(-2, 3) and B(4, 9).

A(-2, 3)●

B(4, 9)●

Locus of P

● P(x, y)

Constraint / Condition :

PA = PB

PA2 = PB2

(x+2)2 + (y – 3)2 = (x – 4)2 + (y – 9)2

x + y – 7 = 0 is the equation of

locus of P.

Note : This locus is actually the perpendicular bisector of AB

Solutions to this question by scale drawing will not be accepted.(SPM 2006, P2, Q9) Diagram 3 shows the triangle AOB where O is the origin. Point C lies on the straight line AB.

(a) Calculate the area, in units2, of triangle AOB. [2 marks](b) Given that AC : CB = 3 : 2, find the coordinates of C. [2 marks](c) A point P moves such that its distance from point A is always twice its distance from point B.

(i) Find the equation of locus of P, (ii) Hence, determine whether or not this locus intercepts the y-axis.

[6 marks]

x

y

O

A(-3, 4)

Diagram 3C

●

●

●

B(6, -2)

(SPM 2006, P2, Q9) : ANSWERS

9(a)

= 9

0 6 3 01 1 0 24 0 0 6 00 2 4 02 2

x

y

O

A(-3, 4)

Diagram 3C

●

●

●

B(6, -2)

3

2

9(b) 2( 3) 3(6) 2(4) 3( 2),3 2 3 2

12 2,5 5

K1

N1

nmmyny

nmmxnx 2121 ,

Use formula correctly

N1K1

Use formulaTo find area

(SPM 2006, P2, Q9) : ANSWERS

√

AP = 2PB

AP2 = 4 PB2

(x+3)2 + (y – 4 )2 = 4 [(x – 6)2 + (y + 2)2

x2 + y2 – 18x + 8y + 45 = 0

N1

K1 Use distance formula

K1Use AP = 2PB

x

y

O

A(-3, 4)

C

●

●

●

B(6, -2)

2

1

P(x, y)●

AP = 2 2[ ( 3)] ( 4)x y

9(c) (i)

(SPM 2006, P2, Q9) : ANSWERS

9(c) (ii) x = 0, y2 + 8y + 45 = 0

b2 – 4ac = 82 – 4(1)(45) < 0

So, the locus does not intercept the y-axis.

Use b2 – 4ac = 0or AOM

K1

K1 Subst. x = 0 into his locus

N1√ (his locus & b2 – 4ac)

Given that A(-1,-2) and B(2,1) are fixed points . Point P moves such that the ratio of AP to PB is 1 : 2. Find the equation of locus for P.

2 AP = PB

x2 + y2 + 4x + 6y + 5 = 0

K1

J14[ (x+1)2 + (y+2)2 ] = (x -2 )2 + (y -1)2

6. Coordinate Geometry : the equation of locus

N13x2 + 3y2 + 12x + 18y + 15 = 0

2222 )1()2()2()1(2 yxyx

F4

Marks f6-10 12

11-15 2016-20 2721-25 1626-30 1331-35 1036-40 2Total 100

From a given set of data,(e.g. The frequency distribution of marks of a group of students)

Students should be able to find ….

• the mean, mode & median• Q1, Q3 and IQR• the variance & S.Deviations

• Construct a CFT and draw an ogive• Use the ogive to solve related problems

Statistics

F4

To estimate median from Histogram

F5

10

20

30

40

50

60

70

80

0.5 20.5 40.5 60.5 80.5 100.5Modal age = 33.5

Age

Number of people

33.5

Graph For Question 6(b)

F4 CHAPTER 8

8. CIRCULAR MEASURE

‘Radian’ ‘Degrees’

S = rθ (θ must be in RADIANS) A = ½ r2 θ

Always refer to diagram when answering this question.

θ

F4

8. CIRCULAR MEASUREDiagram shows a sector of a circle OABC with centre O and radius 4 cm. Given that AOC = 0.8 radians, find the area of the shaded region.

C

A

B O0.8c

Area of sector OABC

= ½ x 42 x 0.8= 6.4 cm 2

= ½ x 42 x sin 0.8= 5.7388 cm2

Area of triangle OAC

Area of shaded region

= 6.4 – 5.7388 = 0.6612 cm2

K1

N1

K1

K1

In radians !!!!

DIFFERENTIATION :

2

( (3) (3 1)(4)(4 5

5)

4 )dy xxdx x

2)54(11

x

dxdy

5413

xxyGiven that , find

d udx v

F4

9 Differentiation : The second derivative

Given that f(x) = x3 + x2 – 4x +

5 , find the value of f ” (1)

f’ (x) = 3x2 + 2x – 4 f” (x) = 6x + 2

f” (1) = 8

F4

9 Differentiation : The second derivative

Given that , find the value of g ” (1) .

g’ (x) = 10x (x2 + 1)4

F4

52( ) 1g x x

g’’ (x) = 40x (x2 + 1) 3 . 2x

Ya ke ??

g’ (x) = 10x (x2 + 1)4

F4- 9

Given that , find the value of g ” (-1) .

52( ) 1g x x

d uvdx

g’’ (x) = 10x . 4(x2 + 1) 3.2x +(x2+1)4. 10g’’ (-1) = 10(-1) . 4[(-1)2 + 1] 3 +[(-1)2+1)4. 10

= 800 Mid-year, Paper 2

Given that y = 2x3 – x2 + 4, find the value of at the point (2, 16). Hence, find the small increment in x which causes y to increase from 16 to 16.05.

dxdy

Differentiation : Small increments

K1

K1

N1

= 6x2 – 2x

= 20 , x = 2

F4

Progressions : A.P & G.P

A.P. : a, a+d, a+2d, a+3d , …….. Most important is “d”

F5

G.P. : a, ar, ar2, ar3, ……..

Most important is “ r ” !!

Progressions : G.P - Recurring Decimals

SPM 2004, P1, Q12Express the recurring decimal 0.969696 … as a fraction in the simplest form.

9699

x = 0. 96 96 96 … (1) 100x = 96. 96 96 ….. (2)(2) – (1) 99x = 96 x = = 32

33

F5

Usual Answer : S10 – S5 = ……. ???

Correct Answer : S10 – S4

ProgressionsGiven that Sn = 5n – n2 , find the sum from the 5th to the 10th terms of the progression.

Back to basic… …

Ans :-54

F5

Y

X

1. Table for data X and Y

2. Correct axes and scale used

3. Plot all points correctly4. Line of best fit5. Use of Y-intercept to determine value of constant6. Use of gradient to determine another constant

1

1-2

1

1

2-4

Linear LawF5

Y

X

Bear in mind that …......

1. Scale must be uniform 2. Scale of both axes may defer : FOLLOW given instructions !

3. Horizontal axis should start from 0 !4. Plot ……… against ……….

Linear Law

Vertical Axis Horizontal Axis

F5

0 2 4 6 8 10 12 x

0.5

1.0

1.5

2.5

2.5

3.0

3.5

4.5Y

x

x

x

x

x

x

Linear lawF5

Read this value !!!!!

dxx 4)13(.1

dxx 4)32(.2

dx

x 4)13(1.3

dxx 4)13(

2.4

=

=

=

=

cx

15)32( 5

cx

15

)13( 5

cx

3)13(9

1

cx

3)13(9

2

INTEGRATION F5

INTEGRATION SPM 2003, P2, Q3(a) 3 marks

Given that = 2x + 2 and y = 6 when x = – 1, find y in terms of x.

dydx

Answer: = 2x + 2

y =

= x2 + 2x + c

x = -1, y = 6: 6 = 1 + 2 + c c = 3

Hence y = x2 + 2x + 3

dydx

(2 2)x dx

F5

INTEGRATION SPM 2004, K2, S3(a) 3 marks

The gradient function of a curve which passes through

A(1, -12) is 3x2 – 6 . Find the equation of the curve.

Answer: = 3x2 – 6

y = = x3 – 6x + cx = 1, y = – 12 : – 12 = 1 – 6 + c c = – 7

Hence y = x3 – 6 x – 7

dydx

2(3 6)x dx Gradient Function

F5

22 34

A

BGiven that OA = 2i + j and OB = 6i + 4j, find the unit vector in the direction of AB

AB = OB - OA = ( 6i + 4j ) – ( 2i + j )

= 4i + 3j

l AB l =

= 5

Unit vector in the direction of AB = )34(51 ji

Vectors : Unit Vectors

K1 N1

K1

F5

Parallel vectors

Given that a and b are parallel vectors, with a = (m-4)i +2 j and b= -2i + mj. Find the the value of m.

a = k b

(m-4) i + 2 j = k (-2i + mj)

m- 4 = -2k

mk = 2

1

2

a = b

m = 2

K1

N1

K1

F5

Prove that tan2 x – sin2 x = tan2 x sin2 x

xkosx

2

2sin sin 2x

xkosxxkosx

2

222 sinsin

xx 22 sintan

xkosxkosx

2

22 )1(sin

tan2 x – sin2 x = K1

N1

K1

5 TRIGONOMETRIC FUNCTIONSF5

Solve the equation 2 cos 2x + 3 sin x - 2 = 0

sin x ( -4 sin x + 3 ) = 0

sin x = 0 ,

2( 1 - 2sin2 x) + 3 sin x - 2 = 0

-4 sin2 x + 3 sin x = 0

sin x =43

x = 00, 1800, 3600 x = 48.590, 131.410

K1

N1

K1

N1

5 TRIGONOMETRIC FUNCTIONSF5

5 TRIGONOMETRIC FUNCTIONS (Graphs)

(Usually Paper 2, Question 4 or 5) - WAJIB !

F5

1. Sketch given graph : (4 marks) (2003) y = 2 cos x , (2004) y = cos 2x for (2005) y = cos 2x , (2006) y = – 2 cos x ,

0 2x

0 00 180x

0 2x 32

0 2x

Find the number of four digit numbers exceeding 3000 which can be formed from the numbers 2, 3, 6, 8, 9 if each number is allowed to be used once only.

No. of ways = 4 . 4. 3. 2 = 96

3, 6, 8, 9

F5 PERMUTATIONS AND COMBINATIONS

Vowels : E, A, IConsonants : B, S, T, R Arrangements : C V C V C V C

No. of ways = 4! 3 ! = 144

Find the number of ways the word BESTARI can be arranged so that the vowels and consonants alternate with each other

[ 3 marks ]

F5

Two unbiased dice are tossed. Find the probability that the sum of the two numbers obtained is more than 4.Dice B, y

4

1

5

6

2

3

Dice A, x2 3 4 51 6

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

n(S) = 6 x 6 = 36

Constraint : x + y > 4

Draw the line x + y = 4

We need : x + y > 4

P( x + y > 4) = 1 – 366

= 65

F5

The Binomial Distributionrnr

rn qpCrXP )()()(

r = 0, 1, 2, 3, …..n

n = Total number of trials

q = probability of ‘failure’

p = Probability of ‘success’

r = No. of ‘successes’

p + q = 1

F5

PROBABILITY DISTRIBUTIONS

Mean = npVariance = npq

The NORMAL Distribution

F5

PROBABILITY DISTRIBUTIONS

Candidates must be able to … determine the Z-score

Z =

use the SNDT to find the values (probabilities)

x

z

f(z)

0 0.50

z

f(z)

0 1.5z

z

f(z)

0-1.5 1

= – 1

f(z)

0 1

–

T5

Index Numbers

• Index Number =

• Composite Index =

• Problems of index numbers involving two or more basic years.

1000

1 HHI

wwI

I_

F4

Solution of Triangles• The Sine Rule• The Cosine Rule• Area of Triangles• Problems in 3-Dimensions.• Ambiguity cases (More than

ONE answer)

Motion in a Straight Line

Initial displacement, velocity, acceleration... Particle returns to starting point O... Particle has maximum / minimum velocity.. Particle achieves maximum displacement... Particle returns to O / changes direction... Particle moves with constant velocity...

Motion in a Straight Line

Question involving motion of TWO particles. ... When both of them collide / meet ??? … how do we khow both particles are of the same

direction at time t ??? The distance travelled in the nth second. The range of time at which the particle returns …. The range of time when the particle moves with

negative displacement Speed which is increasing Negative velocity Deceleration / retardation

Linear ProgrammingTo answer this question, CANDIDATES must be able to ..... form inequalities from given mathematical information

draw the related straight lines using suitable scales on both axes

recognise and shade the region representing the inequalities

solve maximising or minimising problems from the objective function (minimum cost, maximum profit ....)

Linear Programming

y ≤ 2x12. The ratio of the quantity of Q (y) to the quantity of P (x) should not exceed 2 : 1

x ≥ y + 1011. x must exceed y by at least 10

y - 2x >1013. The number of units of model B (y) exceeds twice the number of units of model A (x) by 10 or more.

x + y > 4010. The sum of x and y must exceed 40x + y ≥ 509. The sum of x and y is not less than 50

3x - 2y ≥ 188. The minimum value of 3x – 2y is 18 x + 2y ≤ 607. The maximum value of x+ 2y is 60

y ≥ 356. The minimum value of y is 35 x ≤ 1005. The maximum value of x is 100

y ≥ 2x4. The value of y is at least twice the value of xx ≤ y3. x is not more than y

x ≤ 802. x is not more than 80 x ≥ 101. x is at least 10

KetaksamaanMaklumat

THANK Q & Selamat maju jaya !