ILM MIND MAP 2

Modul I Love Maths

TOPIC :1 NUMBER SYSTEM
SUBTOPIC: COMPLEX NUMBER

Complex number must be written in the form of z  a  bi
Conjugate of complex number z*  a  bi
Modulus of complex number: z  a2  b2
Arguement of complex number:     

       tan1 b 
   tan1 b a z  1 2i
z  1 2i a
   tan1 b 
      a z  1 2i

   tan1 b 
a



z  1 2i IMPORTANT!

Polar form : z  rcos   i sin . Make sure  in
r  z   argz,
radian.

Equality of complex number

z1  z2
a  bi  x  yi

STEEPQ1:UmAaLkIeTsYureOzF1 CanOdMz2PinLEthXe fNorUmMoBf EaR bi .
STEP 1: make sure z1SaTnEdPz22 :inCtohme pfoarrem:of a  bi .  z1  z2

STEP 2: Compare :
real part of the right hand side= real part of the left hand side
iimmaaggiinnaarryy ppaarrtt ooff tthhee rriigghhtt hhaanndd ssiiddee== iimmaaggiinnaarryy ppaarrtt ooff tthhee lleefftt hhaanndd ssiiddee

By: Madam Nuraini Abdullah 1

Kolej Matrikulasi Pahang

Modul I Love Maths

TOPIC :2 EQUATIONS AND INEQUALITIES

SUBTOPIC: LOGARITHM

PROBLEMS: HOW TO SOLVE LOGARITHM EQUATION?

METHOD:

LOG

SAME BASE DIFFERENT BASE

MAKE IT AS SINGLE LOG CHANGE BASE

CHANGE INTO INDEX FORM/ SOLVE THE EQUATION
COMPARE

2log x  log 2  log 3x  4 log6 x3  2 logx 6  1

STEP 1: COLLECT LOG AT ONE SIDE STEP 1: CHANGE BASE DIFFERENT BASE

2log x  log 2  log 3x  4 log6 x3  2 log6 6  1 USE RULE
log6 x
loga bm  m loga b
STEP 2: MAKE IT AS SINGLE LOG STEP 2 : SOLVE THE EQUATION
Use Law Of
Logarithm: log a  log b  log ab 3 log 6 x  2 x 1
log6
a
log a  log b  log b Let u  log6 x

log x2  log 23x  4 3u  2  1
u
STEP 3 : CHANGE INTO INDEX FORM/
COMPARE 3u2  2  u

x2  23x  4 3u2  u  2  0

STEP 4 : SOLVE THE EQUATION u 13u  2  0
x2  6x 8  0
TIPS:Don’t forget u  1, u   2
x  4x  2  0 to check the final 3

x  4, x  2 answer!! log6 x  1, log6 x   2
3

2

x  6, x  63

By: Madam Nuraini Abdullah 2

Kolej Matrikulasi Pahang

Modul I Love Maths

TOPIC :2 EQUATIONS AND INEQUALITIES
SUBTOPIC: INDICES
PROBLEMS: HOW TO SOLVE INDICES EQUATION?

METHOD:

SOLVING INDICES

2 TERMS TAKING LOG TO SOLVE
MORE THAN 3 BOTH SIDES
COUNT TERMS
TERMS COMPARE INDEX
EXAMPLE MAKE IT AS AN
2x5  4
EQUATION
STEP 1: Count Term
12 9x 10  3x  25  0

STEP 2: Change to similar base STEP 1 : Count Term
2x5  22 1 23

STEP 3: Compare index STEP 2: Change to similar base

x5  2 32x 10  3x  25  0

STEP 4: Solve the equation STEP 3: Make it as an equation

x7 let u  3x
u2 10u  25  0

STEP 4: Solve the equation

u 5u 5  0

u5

3x  5

x log 3  log 5

x  1.47

By: Madam Nuraini Abdullah 3

Kolej Matrikulasi Pahang

Modul I Love Maths

TOPIC :2 EQUATIONS,INEQUALITIES AND ABSOLUTE VALUES
SUBTOPIC: SOLVING INEQUALITIES

INEQUALITIES

LINEAR QUADRATIC RATIO

5x  6 x2  5x  6 2x 3 1 Do not
x2  5x  6  0 x5 cross
6 S1:SIMPLFY  multiply
5  x 3x  2  0
x  S1:SIMPLFY S2:FACTOR 2x 3
S2:FACTOR x  3, x  2 S3:CN x5
 6  S3:CN  1  0
5 
, 2x  3x  5

x5  0

S4:GRAPH

x  8  0
x  5

23 Critical Number
(, 2] [3, )
Choose positive x  8, x  5
sign! Shade
right and left of S4:S-LINE
the graph.
x8
x5 , 5 5,8 8,  

+ -+

Its important to identify the -5 8
types of Inequalities, either
LINEAR / QUADRATIC /RATIO NOTE!!  (5, 8] Choose negative
-5 is a denominator. sign!
to avoid mistakes. Cannot be included ,need
to use open bracket.

By: Madam Nuraini Abdullah 4

Kolej Matrikulasi Pahang

Modul I Love Maths

INEQUALITIES

TOPIC : 2 EQUATIONS,INEQUALITIES AND ABSOLUTE VALUES
SUBTOPIC: SOLVING ABSOLUTE VALUES INEQUALITIES

ABSOLUTE VALUES INEQUALITIES

LINEAR QUADRATIC

5x  6 x2  7x  6  6 S1:DEFINE
S2: SIMPLIFY
x  6 A x   6 x2  7x  6  6 x2  7x  6  6 S3: FACTOR
5 N 5 x2  7x 12  0 S4: CRITICAL NUMBER
x2  7x  0 O S5: GRAPH
D x(x  7)  0 R x  3x  4  0
S6:ANSWERS
x  0, x  7 x  3, x  4 S7:NOM LINE

 6 6
5 5

 6 6 -7 0 -4 -3
5 5
, C1 :

(, 7] [0, ) O -C5 2 : [4,83]
R

-7 -4 -3 0 S8:FINAL
ANSWER
(, 7]4, 3 [0, )

By: Madam Nuraini Abdullah 5

Kolej Matrikulasi Pahang

Modul I Love Maths
RATIO

1 5
x 1

1 5 A x 1  5 S1:DEFINE
x 1 1 S2: SIMPLIFY

x 1 1  5 0 N x 1 1  5  0 S3: FACTOR
 D 
S4: CRITICAL
1  5 x  1  0 1 5 x 1  0 NUMBER

x 1 x 1

1 5x  5  0 1 5x  5  0
x 1 x 1

Divide /Multiply 5x  4  0 5x  6  0
with Negative 1 if x 1 x 1
x is negative.
5x  4  0
**CHANGE THE x 1
SIGN!!
x   4 , x  1 x   6 , x  1
5 5 S5: S-LINE

5x  4 , 1  1,  4   4 ,   5x  4  ,  6  6 , 1  1,  
x 1  5  5  x 1 5  5
+
+ -+ +-

1  4  6 1
5 5
S7:NOM LINE
4 AND S6:ANSWERS
5 C1 & C2
(, 1)   , )  (,  6    1, )
5 

 6 1  4 S8:FINAL
5 5 ANSWER

(,  6    4 , ) 6
5  5

By: Madam Nuraini Abdullah

Kolej Matrikulasi Pahang

Modul I Lo

SUMMARY

CHAPTER 1

TYPES BASIC Solve..
OF 1.INDICES
apply 1.INDICES
NUMBE Ex: Simplify 2n-6(4n-3) *count.. …
2.LOG 2terms
COMPLEX 3terms
NUMBERS Ex:
-in form of…….. a)9x+2-3x=8
Ex: b)2x-4=3x+2
c)ex=5
a) =
2.LOG
MODULUS
** BASE
z  a2 b2
apply Same Diff
ARGUMENT 
1)………….. 1)…

2)………….. 2)

3)……………. 3)…

Ex :2 ln x =ln (6-x)+

3logx3 +log3 3 x

3.SURD 3.SURD Ex:
*use expansion a)√ +
POLAR FORM 5 (a+b)2= b)√ +
= (a-b)2=
z cos  i sin  c) 2x
√2 − 7
By: Madam Nuraini Abdullah
Kolej Matrikulasi Pahang

ove Maths

Y

CHAPTER 2 BOTH MODULUS

|6 − 2 | > |5 |
**squaring both
sides

INEQUALITY

B LINEAR A LINEAR
A 7-6x < 8x B |7 − 2 | > 5
S S 7-2x>5x OR ………….
I X >…. O
C L QUADRATIC
QUADRATIC U | − 2| ≤ + 5
*coef x2 must be + x-2≤x2+5 AND…….
T X2-x+7≥ 0 ……………
*change sign > /<
E Graph 1 Graph 2
7-6x2 < 3x-4
G 6x2-3x…….0 RATIO
4
ferent Graph 1 2 − 3 ≥ 4 − 3
…………… ………………OR…………..
)………….. RATIO
…………… *coef x2 must be + S-line 1 S-line 2
+ln 3 *change sign > /<
*use table@no line +/- Combine: Shaded7
x =10/3 *if quadratic cannt be
factorized need to use CTS
+ 3 = 7 − 4 & the value is always +
+ 1 − √ − 3=2
−3 − − 1
x1 x1  x 2 ( + 2)( − 3) ≥ 0

S-line 1

Modul I Love Maths

TOPIC :5 FUNCTION AND GRAPHS QUADRATIC CUBIC
SUBTOPIC: SKETCHING GRAPH
BASIC GRAPH

LINEAR

y  mx  c y  x2 y  x3
ABSOLUTE VALUE SURD RECIPROCAL

y x y x y  1
EXPONENTIAL LOG x

y  ex y  ln x

By: Madam Nuraini Abdullah 8

Kolej Matrikulasi Pahang

Modul I Love Maths

There are 6 graph
movement

MOVEMENT QUADRATIC SURD
f (x)
y  x2 y x

Original 0 Domain: 0,  0
function Range: 0, 
f (x)  a Domain: , 
Move: Range:0,  y  x 5
+ upward
- downward y  x2  4

4 -5

Domain: ,  Domain: 0, 
Range: 4,  Range: 5,

f (x  a) y   x  52 y x2

Move: Domain: 5 Domain: -2
- right Range: Range:
+left -1
y   x  32 1 y  x 1 4
f (x  a)  b
Move: -3 4
- right
+left Domain: Domain: 1
Move: Range: Range:
+ upward
- downward y  x2 y x
 f (x)

Reflection on Domain: Domain:
x axis Range: Range:

f (x) y   x2 y  x

Reflection on Domain: Domain:
y axis Range Range:

By: Madam Nuraini Abdullah 9

Kolej Matrikulasi Pahang

Modul I Love Maths

MOVEMENT MODULUS RATIO
f (x)
y x y  1
x

Original Domain: Domain:
function Range:

Range:

f (x)  a y  2x  4 y  1  5
x
Move:
+ upward Domain:
- downward Range:

Domain:
Range:

f (x  a) y  x3 y  4

Move: x 3
- right
+left Domain:
Range:
Domain:
Range:

f (x  a)  b y x2 4 y  x 1 4  2
Move: 
- right Domain:
+left Range: Domain:
Move:
+ upward
- downward

Range:

 f (x) y x y   1  1
x 2

Reflection on Domain:
x axis Range:

Domain:
Range:

f (x) y  x y  1

2 x

Reflection on Domain: Domain:
y axis Range Range:

By: Madam Nuraini Abdullah 10

Kolej Matrikulasi Pahang

Modul I Love Maths

MOVEMENT EXPONENTIAL LAWN
f (x)
y  ex y  ln x

Original Domain: Domain:
function Range: Range:
y  ln x  5
f (x)  a y  ex  4

Move: Domain: Domain:
+ upward Range: Range:
- downward y  ln(x  2)
f (x  a) y  ex2

Move: Domain: Domain:
- right Range: Range:
+left y  ln(x 1)  2
y  ex3 1
f (x  a)  b Domain:
Move: Domain: Range:
- right Range: y   ln x
+left
Move: y  ex Domain:
+ upward Range:
- downward y  ln(x)
 f (x)

Reflection on Domain:
x axis Range:

f (x) y  ex

Reflection on Domain: Domain:
y axis Range Range:

By: Madam Nuraini Abdullah 11

Kolej Matrikulasi Pahang

Modul I Love Maths
SKETCHING GRAPH OF EXPONENTIAL

y  ex
STEP 1: Quadrant
LOOK AT
The sign of x=
The sign of y=
Choose the quadrant to get the starting point!

STEP 2: Intercept
Find the y intercept When x=0, y=?
STEP 3: Asymptote
Find the asymptote?
STEP 4: Sketch
Don’t hit the asymptote

By: Madam Nuraini Abdullah 12

Kolej Matrikulasi Pahang

Modul I Love Maths

STEPS y  ex 1 y  log  x  3
LOOK AT
The sign of x= The sign of x= - The sign of x= +
The sign of y=
Choose the quadrant to get The sign of y= + The sign of y= +
the starting point!
Choose the quadrant to get the Choose the quadrant to get
starting point! the starting point!

 Starting point is at  Starting point is at

Quadrant 2 Quadrant 1

y y

x
x

STEP 2: Intercept When x=0, when x = 0, y = not exist
Find the intercept y=e0 +1=2 y when y = 0, x = 4
STEP 3: Asymptote
2 y
STEP 4: Sketch
Don’t hit the asymptote x x

y  ex 1 4
Horizontal Asymptote, y = 1
y  log  x  3
y
Vertical Asymptote,
x-3=0
x=3

y x=3

2
y=1
x

Sketch From left to right x
Don’t hit the asymptote 4
Sketch From top to bottom
y Don’t hit the asymptote
y

2 x
y=1 4
x

By: Madam Nuraini Abdullah 13

Kolej Matrikulasi Pahang

Modul I Lo

FIND INVERSE?? Show 1-1 f
f  f 1
Function METHOD 1   f 1(x)
f (x)  x  4 f (x1)  f (x2 )   f 1(
 f 1(
x1  4  x2  4
x1  x2 f  f 1(x
f  f 1(x)
METHOD 2 f 1(x) 
f 1(x) 
f(x) 4 HLT
f 1(x
4

f :x x  2,x  2 By using horizontal line test it
cut the graph at 1 point.Thus
f(x) is 1-1 function.

METHOD 1
f (x1)  f (x2 )

x1  2  x2  2

x1  2  x2  2
x1  x2

METHOD 2 f(x)

HLT

-2
By using

horizontal line test it cut the
graph at 1 point.Thus f(x) is 1-1
function.

By: Madam Nuraini Abdullah

Kolej Matrikulasi Pahang

ove Maths

1  x Graph f and f-1 Df, Rf Df-1, Rf-1

1(x)  x f(x) 4 Df  ,   Rf 1
  4  x 4 Rf  ,   Df 1
(x)  x  4
(x)  x  4 f-1(x)

x)  x f(x) Df  2 ,   Rf 1
)  x Rf  0 ,   Df 1
2x -2
 2  x2 -2 f-1(x)
x)  x2  2
x0

14

Function Show 1-1 Modul I Lo

g(x)  x2  2x, x  1 f

g(x) = ex + 3 METHOD 1
f (x1)  f (x2 )
ex1  3  ex2  3 f  f 1(x)
ex1  ex2 f  f 1(x)
ln ex1  ln ex2
x1  x2 e f 1 ( x)  3
e f 1 ( x)
METHOD 2
ln e f 1 ( x)
f(x) f 1(x)

HLT

By using horizontal line test it
cut the graph at 1 point.Thus
f(x) is 1-1 function.

By: Madam Nuraini Abdullah

Kolej Matrikulasi Pahang

ove Maths Sketch Graph f and f-1 Df, Rf Df-1, Rf-1

1  x

)x f(x) 4 Df  ,   Rf 1
  x 3
3 x Rf  3,   Df 1
)  x3 34
)  ln x  3 f-1(x)
)  ln x  3
x=3
Vertical asymptote

15

Function Show 1-1 Modul I Lo
f ( x)  ln(x  1)
f

f (x)  2x 1 , x  1
2

By: Madam Nuraini Abdullah

Kolej Matrikulasi Pahang

ove Maths Sketch Graph f and f-1 Df, Rf Df-1, Rf-1

1  x

16

Function Show 1-1 Modul I Lo

f

f (x)  x 1
x3

f(x) = 2+ e2x

f (x)  ln(x  3)

By: Madam Nuraini Abdullah

Kolej Matrikulasi Pahang

ove Maths Sketch Graph f and f-1 Df, Rf Df-1, Rf-1

1  x

17

Modul I Lo

BASIC GRAPH
FIND DOMAIN RANGE?

F

By: Madam Nuraini Abdullah

Kolej Matrikulasi Pahang

ove Maths

BASIC GRAPH LOG AND EXPONENTIAL
Find Domain Range?

FUNCTIONS FIND INVERSE
HOW TO DETERMINE 1-1

SKETCH F(X) AND INVERSE
Domain and Range f(x) and f-1(x)

18

TOPIC: 8 LIMITS Modul I Lo

LIMITS

SUBSTITUTE

0 number   number
0 0 

factorize X conjugate  

lim 6 1 lim 4  x
x1 x  1 x2 x  2

lim x2  5    2  
x2  4 0 0
x2

 lim  x  x  2  2  lim x 1
x2 2 x x1 x 1


 lim  x 1 2   lim x 1  x 1
 x1 x 1 x 1
x2

 1 x 1 x 1
4
 lim
x1 x 1 x2 

 lim x 1
x1

2

By: Madam Nuraini Abdullah

Kolej Matrikulasi Pahang

ove Maths

CONTINUITY

0  asymptote f (x)is said to be continuous
 at x  a if it satisfy the
following condition
i)f (a)is defined
ii) lim f (x)exists

xa

iii) lim f (x)  f (a)
xa

Factorize the highest power

lim 3x2 1 lim 3x 1 HV
x2 1 x x2  1 oe
x rr
it
x2  3  1  x  3  1  zi
x2  x2   x  oc
lim 1  lim na
1  x2  x  1  tl
x x2  1 x2  a
 l
30 
1 0
  3 x  3  1 
 x 
 lim
x    1 
x  1 x2 
 x, x  0 
 x  x, x  0 

  lim 3  0  3
x 1 1  0

19

Modul I Love Maths

TOPIC:9 DIFFERENTIATIONS

PROBLEMS: HOW TO DIFFERENTIATE?

METHOD:

GENERAL POWER RULE DIFFERENTIATE LAWN

y  2x  65 y  ln  x2  5

METHOD METHOD

Sing: y  ln  x2  5
Bring down power
Tips!!
Reduce power 1.Use law of logarithm
Differentiate bracket 2.Put a slash / on n

Example 1 (the letter l seems like number 1)
easy to remember 1 / (…)
y  2x  65
3.Then write number 1 and 2 on the top
dy  52x  64 2 4.Proceed to step 2, differentiate bracket
dx Example 1

12

y  ln  x2  5

Bring down power  Put a slash / on n and it become 1/ x 2  5
Reduce power
Differentiate  dy l (2x)
x2  5
bracket dx

Move to 2nd step: Differentiate bracket

Example 2 Example 2 12

y  4 cos 3 y  4 ln 3x5  2

dy  4 3  cos  2  sin 
dx
 Put a slash / on n and it become 1/ x 2  5
Bring down power
Reduce power 4
Differentiate 3x5  2
bracket  dy (15x4 )

dx

Move to 2nd step: Differentiate bracket

By: Madam Nuraini Abdullah 20

Kolej Matrikulasi Pahang

Modul I Love Maths

DIFFERENTIATE TRIGO DIFFERENTIATE EXPONENTIAL

y  sin 3x  4 y  e2x2 5

METHOD METHOD

Tips!! Use ACRONIM DI-CO-DI Tips!! Use ACRONIM CO-DI-PO-LN
1.Copy the exponent
1. Write number 1 and 2 on the top 2.DIfferentiate POwer
3.LAWN base
2. Differentiate nom 1,

3. Copy angle

4. Differentiate nom 2

Example 1 12 Example 1 y  e2x2 5

y  sin 3x  4 dy  e2x2 5 (4x) ln e
dx
dy
dx  cos 3x  4.3 1.Copy the exponent
2.DIfferentiate POwer
3.LAWN base
Differentiate nom 1

Copy angle

Differentiate

Nom 2

Example 2 12 Example 2 y  32x2

y  cos x2  3  dy 32x2 2xln 3

dy   sin  x2  32x dx 
dx
1.Copy the exponent
2.DIfferentiate POwer
3.LAWN base
Differentiate nom 1
Copy angle
Differentiate
Nom 2

By: Madam Nuraini Abdullah 21

Kolej Matrikulasi Pahang

Modul I Love Maths
IMPLICIT DIFFERENTIATION

1ST DERIVATIVE

 x  52  y3  7x  6y

2x  51  3y2 dy  7  6 dy S1:Differentiate each terms with respect to
dx dx
x
dy dy dy
3y2 dx 6 dx  7 2x  51 S2: Collect dx on the left

 dy dy
dx
dx
3y2 6  7  2x 10 S3: Factorize

dy  17  2x S4:Make dy as a subject
dx 3y2 6 dx

2nd DERIVATIVE

dy  17  2x u
dx 3y2 6 v

u  17  2x v  3y2 6

u '  2 v '  6 y dy
dx

d2y  vu ' uv '
dx2 v2

 d 2 y3y2 6  2  17  2x  6 y dy  dy
 dx2   dx  dx
S1:Differentiate
3y2  6 2

6 y2 12  102 y dy  12 xy dy 
dx dx 
 
3y2  6 2

6 y2 12  102 y  17  2x   12 xy  17  2x  S2:Substitute dy
 3y 2 6   3y2  6  dx
 
3y2 6 2

18 y3  36 y  36 y2  72  1743 y  408xy  24 x 2 y d2y
  3y2 6 2 S3: Simplify dx2

By: Madam Nuraini Abdullah 22

Kolej Matrikulasi Pahang

Modul I Love Maths
PARAMETRIC DIFFERENTIATION

1ST DERIVATIVE

x  4  t3 y  t2  2t  2 S1:Differentiate x and y with respect to t

dx  3t 2 dy  2t 2
dt dt

dy  dy  dt S2: Form a Chain Rule dy  dy  dt
dx dt dx dx dt dx

dy   2t  2 1 S3: Substitute into Chain Rule formulae
dx 3t 2

dy  2t  2  2 t 2  2 t 1 S4: Simplify
dx 3t 2 3 3 2nd DERIVATIVE

d2y  d  dy   dt S1: Form a Chain Rule d2y  d  dy   dt
dx2 dt  dx  dx dx2 dt  dx  dx

d2y  d  2 t 2  2 t 1   1 S2: Differentiate dy and multiply with dt
dx2 dt  3 3 3t 2 dx dx

d2y    4 t 3  2 t 2   1
dx2  3 3  3t 2

   4  2   1
 3t 3 3t 2  3t 2

42
 9t5  9t 4

 4  2t S3: Simplify
9t 5

By: Madam Nuraini Abdullah 23

Kolej Matrikulasi Pahang