Form 4 add maths note

BRIEF NOTES
ADDITIONAL MATHEMATICS
FORM 4
Symbol : f -1
To find the inverse function, change f(x) to
y and find x in tems of y.

CHAPTER 1: FUNCTION
1. f : x  x + 3
x is the object, x + 3 is the image
f : x  x + 3 can be written as
f(x) = x + 3.
To find the image for 2 means
f(2) = 2 + 3 = 5
Image for 2 is 5.
Find the object for 8 means f(x) = 8 what is
the value of x ?
x+3=8; x=5
The object is 5.
If the function is written in the form of
ordered pairs (x, y) , x is the object and y is
the image.
E.g. (2, 4), (2, 6), (3, 7)
Image for 2 is 4 and 6 whereas object for 7
is 3.

Given f : x 

x
, find f -1
3 x

Let f(x) = y
y=

x
3 x

y(3 – x) = x

3y – xy = x
3y = x + xy
= x(1 + y)
x=

4.

3x
3y
, thus f -1(x) =
1 x
1 y

Composite Function
Given f : x  3x – 4 and g : x  2 – 3x,
find
(a) fg(x)
(b) gf(x)
(c) f 2(3)
(d) gf -1(4)
fg(x) = f(2 – 3x) = 3(2 – 3x) - 4
= 6 – 9x – 4 = 2 – 9x
(b) gf(x) = g(3x – 4) = 2 – 3(3x – 4)
= 2 – 9x + 12 = 14 – 9x
(c) f 2(3) = ff(3) = f(9 – 4) = f(5)
= 15 – 4 = 11.
(a)

In the arrow diagram, the set of object is
{1, 2, 3} and the set of image is {4, 5}
2.

For f : x 

5
, x – 3  0, i.e. x  3
x3

(d)

5
because
is undefined.
0
5
Thus, if f : x 
, x  k then k is 3.
x3
3.

Function which maps into itself means f (x)
=x

3
If f : x 
, find the value of x which
x2
is mapped into itself.

3
=x
x2
3 = x(x – 2) = x2 – 2x
Thus, x2 – 2x – 3 = 0
(x – 3)(x + 1) = 0
x = 3 or 1
3.

Inverse Function

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Let y = 3x – 4, x =
Thus f -1(4) =
gf -1(4) = g(

5.

y4
3

8
3

8
8
) = 2 – 3 × = 6
3
3

To find f(x) or g(x) given the composite
function.
Given f(x) = 2x + 8 and fg(x) = 6x + 12,
find g(x).
f(x) = 2x + 8
f[g(x)] = 2g(x) + 8
= 6x + 12
2g(x) = 6x + 12 – 8
= 6x + 4
g(x) = 3x + 2

Given f(x) = 3x – 5 and gf(x) = 9x2 – 30x + 30,
find g(x)

1

gf(x) = 9x2 – 30x + 30
g(3x – 5) = 9x2 – 30x + 30
Let y = 3x – 5, x =

2.

y5
3

Using SOR and POR and the formula x2 –
(SOR)x + POR = 0
Cari persamaan kuadratik dengan
punca

1
dan 3
2

2

y5
 y5
g (y) = 9 
) + 30
  30(
3
 3 

1
+3=
2
1
POR =
×3=
2

= y2 + 10y + 25 – 10y – 50 + 30
= y2 + 5
Thus, g(x) = x2 + 5

SOR =

CHAPATER 2 : QUADRATIC EQUATION
1. Find the roots of quadratic equation
(a) Factorisation

7
2
3
2

Equation is

(b) formula x =
(a)

b  b 2  4ac
2a

Solve 6x2 – 7x – 3 = 0

x2 

2x2 – 7x + 3 = 0

× 2,
3.

7
3
x+
=0
2
2

If ax2 + bx + c = 0 is the general form of
the quadratic equation,
SOR = α + β =
POR = αβ =

(2x – 3)(3x + 1) = 0
2x – 3 = 0, x =

(b)

1
3

The roots are α and 2α

If it cannot be factorised, use the
formula.
Solve 2x2 – 4x – 5 = 0
a = 2, b = 4 and c = 5

(4)  (4)  4  2  (5)
4
4  16  40 4  56
=

4
4
4  56
x=
= 2.871
4
4  56
x=
= 0.8708
4

SOR = α + 2α = 3α =

x=

2. Form equation form roots.
Use the reverse of factorisaton
Find the quadratic equation with roots

1
,
2

2x = 1, (2x – 1) = 0
x = 3, (x – 3) = 0
The equation is
(2x – 1)(x – 3) = 0
2x2 – 7x + 3 = 0
×2,

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m
= m
1

POR = α × 2α = 2α2 = 18

2

x=

c
a

Given that one root is twice the other root
for the quadratic equation x2 + mx + 18 = 0,
find the postive value of m.

3
2

3x + 1 = 0, x = 

b
a

4.

α2 = 9
α = 9  3
When α = 3, 3α = 9 = m, m = 9 (not
accepted)
When α = 3, 3α = 9 = m, thus m = 9
Types of roots
(a) 2 real and distinct roots.
b2 – 4ac > 0
(b) 2 real and equal roots
b2 – 4ac = 0
(c)

1
and 3
2

No real root
b2 – 4ac < 0

(d)

Real root (distinct or same)
b2 – 4ac ≥ 0

Find the range of values of k in which the
equation 2x2 – 3x + k = 0 has two real and
distinct roots.
For two real and distinct roots
b2 – 4ac > 0

2

(3)2 – 4(2)k > 0
9 – 8k > 0
8k < 9

3.

9
k<
8

CHAPTER 3: QUADRATIC FUNCTIONS
1. To find the maximum/minimum value by
completing the square.

Find the range of value of x for which x2 –
7x – 8 < 0
x2 – 7x – 8 < 0
Note: If the
(x – 8)(x + 1) < 0
coefficient of x2
x = 8, x = 1
is negative, the
Sketch the graph
shape of the
graph is‘n’

Given f(x) = 2x2 – 6x + 8, find the
maximum or minimum value and state the
corresponding value of x.
f(x) = 2x2 – 6x + 8
= 2[x2 – 3x] + 8
2

Quadratic Inequality
(a) Factorise
(b) Find the roots
(c) Sketch the graph and determine the
range of x from the graph.

2

3 3
= 2[x – 3x +      ] + 8
2 2
3
9
=2[(x  )2  ] + 8
2
4
3
9
= 2 (x  )2 
+8
2
2
3
7
= 2(x  )2 +
2
2
2

The minimum value (the coefficient of x2
is positive and the graph is ‘u’ shaped) is

7
3
3
when x 
= 0, or x = .
2
2
2
2.

To sketch quadratic function
(a) Determine the y-intercept and the xintercept (if available)
(b) Determine the maximum or minimum
value.
(c) Determine the third point opposite to
the y-intercept.
Sketch the graph f(x) = x2 – 8x + 6
(a)
(b)

(c)

Y-intercept = 6
f(x) = x2 – 8x + 42 – 42 + 6
= (x – 4)2 – 16 + 6
= (x – 4)2 – 10
Min value = 10 when x – 4 = 0, x =
4. Min point (4, 10)
when x = 8, f(8) = 82 – 8(8) + 6 = 6

From the sketch, (x  8)(x + 1) < 0
1 < x < 8
4.

Types of Roots
(a) If the graph intersects the x-axis at
two different points 2 real and
distinct roots  b2 – 4ac > 0
(b) If the graph touches the x-axis,  2
equal roots b2 – 4ac = 0
(c) If the graph does not intersect the xaxis,(or the graph is always positiv or
always negative.)  no real root  b2
– 4ac < 0
The graph y = nx2 + 4x + n  3 does not
intersect the x-axis for n < a and n > b, find
the value of a and b.
y = nx2 + 4x + n – 3 does not intersect the
x-axis  no real root  b2 – 4ac < 0
42 – 4n(n – 3) < 0
16 – 4n2 + 12n < 0
0 < 4n2 – 12n – 16
4
n2 – 3n – 4 > 0
(n – 4)(n + 1) > 0
n = 4, n = 1

From the graph, for (n – 4)(n + 1) > 0, n <
1 and n > 4
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3

 a = 1 and b = 4

3.

CHAPTER 4: SIMULTANEOUS
EQUATIONS
To solve between one linear and one non-linear
equation.
Method : Substitution
Example : Solve
x + 2y = 4 --------(1)

2x 2 y

 5 -----(2)
y
x
from (2), × xy
2x2 + 2y2 = 5xy ------------(3)
from (1), x = 4 – 2y
substitute in (3)
2(4 – 2y)2 + 2y2 = 5(4 – 2y)y
2(16 – 16y + 4y2) + 2y2 = 20y – 10y2
8y2 + 10y2 + 2y2– 32y – 20y + 32 = 0
20y2 – 52y + 32 = 0
4
5y2 – 13y + 8 = 0
(5y – 8)(y – 1) = 0

loga xn = nloga x

4.

loga b =

5.

loga a = 1

6.

loga 1 = 0

log c a
log c b

Example: Find the value of

5
log4 8 – 2log4 3 +
3

log4 18

5
log4 8 – 2log4 3 + log4 18
3
5

8 3  18
32
32  18
= log4
= log4 64 = log4 43
9
= log4

= 3log4 4 = 3 × 1 = 3

y=

8
or 1
5
8
8
16 4
y = , x = 4 – 2( ) = 4 
=
5
5
5
5

To solve index equations, change to the same
base if possible. If not possible to change to the
same base take logarithm on both sides of the
equation.

y = 1, x = 4 – 2 = 2

Example: Solve 3.27x-1 = 93x

4
8
Thus, x = 2, y = 1 and x = , y = .
5
5
!Note Be careful not to make the
mistake
(4 – 2y)2 =16 + 4y2 wrong
If the equations are joined, you have to
separate them.
2

2

Solve x + y = x + 2y = 3
x2 + y2 = 3
and
x + 2y = 3
CHAPTER 5: INDEX AND LOGARTHM
Index form:
b = ax
Logarithm form
loga b = x

3.27x-1 = 93x
3 × 33(x-1) = 32(3x)
31 + 3x – 3 = 36x
1 + 3x – 3 = 6x
2 = 3x
x= 

2
3

Example: Solve 5x+3 – 7 = 0
5x+3 – 7 = 0
5x+3 = 7
log 5x+3 = log 7
(x + 3)log 5 = log 7
x +3=

log 7
= 1.209
log 5

x = 1.209 – 3 = 1.791
Example: Solve

Logarithm Law :
1. loga x + logay = loga xy
2.

loga x – loga y = loga

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x
y

log a 384  log a 144  log a 6 = 4
log

a

384  6
=4
144

4

log a 16 = 4

 a = a
4

16 =

x y
 1
a b
b
y  int ercept
Graident = m = 
= 
a
x  int ercept
Intercept form:

2

a=  4

General form: ax + by + c = 0

CHAPTER 6: COORDINATE GEOMETRY
1. Distance between A(x1, y1) and
B(x2, y2)
AB =

The equation of straight line given the
gradient, m, and passes through the point
(x1, y1) :
y – y1 = m(x – x1)

( x2  x1 )2  ( y2  y1 )2

Example: If M(2k, k) and N(2k + 1, k – 3) are
two points equidistant from the origin O. Find
the value of k.

Equation of a straight line passing throug
two points (x1, y1) and (x2, y2) is

y  y1 y2  y1

x  x1 x2  x1

MO = ON

(2k )2  k 2  (2k  1)2  (k  3)2
Square,
4k2 + k2 = 4k2 + 4k + 1 + k2 – 6k + 9
0 = 2k + 9
2k = 9

k=

9
2

Example: Find the equatioon of the straight line
(a) with gradient 3 and passes through
(1, 2)
(b) passes through (2, 5) and (4, 8)
(a)

2.

Point which divides a line segment in
the ratio m : n

 nx1  mx2 ny1  my2 
 nm , nm 



Equation of straight line
y (2) = 3(x – 1)
y + 2 = 3x – 3
y = 3x – 5

(b) Equation of straight line

y 5

x2
y 5

x2

Example: Given Q(2, k) divides the line which
joins P(1, 1) and R(5, 9) in the ratio m : n. Find
(a) the ratio m : n
(b) the value of k

85
42
3
2

2(y – 5) = 3(x – 2)
2y – 10 = 3x – 6
2y = 3x + 4
(a)

n  5m
=2
nm

3.

n + 5m = 2n + 2m
5m – 2m = 2n – n
3m = n

(b)

2.

Parallel and Perpendicular Line
Parallel lines,
m1 = m2
Perpendicular lines,
m1 × m2 = 1

m 1
 thus, m : n = 1 : 3
n 3
3  1  1 9
=k
1 3
12
3 =k
4

Example: Find the equation of the straight line
which is parallel to the line 2y = 3x – 5 and
passes through (1, 4)

Equation of a straight line
Gradient form: y = mx + c

m=

2y = 3x – 5 , y =

3
5
x2
2

3
, passes through (1, 4)
2

Persamaan garis lurus ialah

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5

y–4=

3
(x – 1)
2

Standard deviation = variance
Example: For the data3, 5, 5, 6, 7, 8 find the
(a) mean
(b) variance
(c) standard deviation

2y – 8 = 3x – 3
2y = 3x + 5
Example: Find the equation of the straight line

x

x

(a)

passes through (2, 3)

(b) variance,  2 =

N

= 5. 667

9  25  25  36  49  64  34 
 
6
 6 

x y
(4) 4
=
  1 , m1 =
3 4
3
3
4
× m2 = 1
3
3
m2 =  , passes through (2, 3)
4

2

2

=

208  34 
= 2.556

6  6 
 

(c)

The equation of the straight line is
y–3= 

=

355 6 7 8

6

x y
which is perpendicular to the line   1 and
3 4

3
(x – 2)
4

standard deviation =  =

2.

Grouped Data
Mean, x 

4y – 12 = 3x + 6
4y + 3x = 18

 fx
f

i

2.556 = 1.599

xi = mid-point

f = frequency

4. Equation of Locus
Example: Find the equation of the locus for P
which moves such that its distance from Q(1, 2)
and R(2, 3) is in the ratio 1 : 2

Median,
1
2

M=L+

N  Fcu
c
fm

L = lower boundary of the median class
N = total frequency
Fcu = cumulative frequency before the
median class
fm = frequency of median class
c = class interval size

Let P(x, y), Q(1, 2), R(2, 3)
PQ : PR = 1 : 2

PQ 1

PR 2
PR = 2PQ

( x  2)2  ( y  3)2  2 ( x  1)2  ( y  2)2

Mode is obtained from a histogram
frequency

Square,
x2 + 4x + 4 + y2 – 6y + 9 =
4(x2 – 2x + 1 + y2 – 4y + 4) =
4x2 + 4y2 – 8x – 16y + 20
0 = 4x2 – x2 + 4y2 – y2 – 12x – 10y + 7
3x2 + 3y2 – 12x – 10y + 7 = 0
CHAPTER 7: STATISTICS
1. Ungrouped Data
Mean, x 

x
N

Variance,  2 =
=

Mode

 ( x  x)


2

N
x2

N

Standard deviation,  =

 fx
f

i



 x

2

class

2

 ( x) 2

Example:

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6

The table shows the marks obtained in a test.
Marks
Frequency
10 – 14
2
15 – 19
5
20 – 24
8
25 – 29
12
30 – 34
10
35 – 39
7
40 – 44
6
Find,
(a) mean mark
(b) median
(c) mode
(d) standard devition
Mark
10 – 14
15 – 19
20 – 24
25 – 29
30 – 34
35 – 39
40 – 44

(a)

(b)

f
2
5
8
12
10
7
6

Mean = x 

xi
12
17
22
27
32
37
42

fxi
24
85
176
324
320
259
252

 fx
f

i

=

fxi2
288
1445
3872
8748
10240
9583
10584

(d)

 =

i

2

 ( x) 2

44760
 28.82
50

=

=

65.76

= 8.109
CHAPTER 8: DIFFERENTIATION

dy
represents the gradient of a curve at a point.
dx
dy
= f (x) = first derivative
dx
= gradient function.

C.F.
2
7
15
27
37
44
50

d
(ax n )  anx n 1
dx
Differentiation of Polynomials
1. Differentiate with respect to x:
(a) y = 3x4 + 2x3 – 5x – 2

x
2
(c) y = 2
x
(b) y =

1440
= 28.8
50

(a)

1
1
N   50 = 25
2
2

y = 3x4 + 2x3 – 5x – 2

dy
= 12x3 + 6x2 – 5
dx
1
(b) y = x = x 2
1
dy 1 1 1 1  1
 x2  x 2 =
dx 2
2
2 x
2
(c) y = 2 = 2x-2
x
4
dy
= 4x-3 = 3
dx
x

Median class = 25 – 29
M = 24.5 +

 fx
f

25  15
 5 = 28.67
12

(c)
Frequency

Differentiation of Product

d
dv
du
(uv)  u  v
dx
dx
dx
2.

Differentiate with respect to x:
y = (3x + 2)(4 – 5x)

dy
= (3x + 2) × 5 + (4 – 5x)× 3
dx
From the graph, mode = 28 mark

= 15x – 10 + 12 – 15x
= 2 – 30x

Differentiation of Quotient

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7

dv
d  u  v du  u dx
 dx 2
 
dx  v 
v
3x  4
3. Differentiate
with respect to x
2x  5
3x  4
y=
2x  5
dy (2 x  5)3  (3x  4)2

dx
(2 x  5) 2
6 x  15  6 x  8
23
=
= 
2
(2 x  5)
(2 x  5) 2

Differentiation of Composite Function

d
(ax  b)n = n(ax + b)n-1 × a
dx
4.

Differentiate with respect to x :
(a) (3x + 5)8
(b) (2x – 1)4(3x + 2)5
(a)

Note:
you must
differentiate
the function in
the brackets.

y = (3x + 5)8

dy
= 8(3x + 5)7 × 3
dx
y = (2x – 1)4(3x + 2)5

dy
= (2x – 1)45(3x + 2)4 × 3 + (3x +
dx
2)54(2x – 1)3 × 2
= 15(2x – 1)4(3x + 2)4 +
8(2x – 1)3(3x + 2)5
= (2x – 1)3(3x + 2)4[15(2x – 1) + 8(3x
+ 2)]
= (2x – 1)3(3x + 2)4[30x – 15 + 24x +
16]
= (2x – 1)3(3x + 2)4(54x + 1)

dy
= 4x  8
dx
For turning point

dy
=0
dx

4x – 8 = 0
x=2
x = 2, y = 2(4) – 16 + 3 = 5

d2y
= 4 > 0, thus the point (2, 5) is a
dx 2
minimum point.
Rate of Change of Related Quantities
Example: The radius of a circle increases which
a rate of 0.2 cm s-1, find the rate of change of the
area of the circle when the radius is 5 cm.

dA
= 2r
dr
dr
= 0.2 cm s-1
dt
dA dA dr


dt dr dt
= 2r × 0.2
= 0.4 r
When r = 5 cm,

dA
= 0.4 × 5
dt
= 2 cm2 s-1

Equation of Tangent and Normal
Gradient of tangent = gradient of curve =

Maximum and Minimum Value
Given y = 2x2 – 8x + 3. Find the coordinates of
the turning point. Hence, determine if the turning
point is maximum or minimum.
y = 2x2 – 8x + 3

A = r2

= 24(3x + 5)7

(b)

y – 0 = 1( x – 1)
y = x – 1.

dy
dx

Example: Find the equation of the tangent to the
curve y = 3x2 – 5x + 2 at the point x = 1.

Small Changes and Approximation

y

dy
 x
dx

y = 3x2 – 5x + 2

Example: Given y = 2x2 – 5x + 3, find the small
change in y when x increases from 2 to 2.01

dy
= 6x – 5
dx

y = 2x2 – 5x + 3

x = 1, y = 3 – 5 + 2 = 0

dy
=6–5=1
dx
Equation of tangent :

zefry@sas.edu.my

dy
= 4x – 5
dx
 x = 2.02 – 2 = 0.01

8

y =

dy
 x
dx

(c)

= (4x – 5) × 0.01
Substitute the original value, x = 2,
 y = (8 – 5) × 0.01
= 0.03
Thus the small increment in y is 0.03.

p1
 100 = 125
60
60
p1 = 125 ×
= RM75
100

CHAPTER 9: INDEX NUMBER
1.

Price Index, I =

p1
× 100
p0

p1 = price at a certain time
p0 = price in the base year

2.

Composite index I 

 Iw
w

I = price index
w = weightage
Example:
Item
Book
Beg
Shirt
Shoes

Price index
100
x
125
140

Weightage
6
2
y
3

The table above shows the price indices and the
weightage for four items in the year 2004 based
in the year 2000 as base year.
If the price of a beg in the year 2000 and 2004
are RM40 and RM44 respectively. The
composite index for 2004 is 116. Find
(a) the value of x
(b) the value of y
(c) the price of a shirt in 2004, if the price in
2000 was RM60.
(a)
(b)

44
× 100 = 110
40
6  100  2 110  125 y  3 140
= 116
62 y3
600  220  125 y  420
= 116
11  y

x=

1240 + 125y = 116(11 + y)
1240 + 125y = 1276 + 116y
125y – 116y = 1276 – 1240
9y = 36
y=4

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9

Form 4 add maths note

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