Boxes in Additional Mathematics
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8/9/2019 Boxes in Additional Mathematics
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BOXES IN ADDITIONAL MATHEMATICSBY: CG. SHAIFUR AZURA BIN SUHAIMI
Box #1: Inverse Function
2
5-x=f-1(x)
5+2x=f(x)
Box # 2 : Inverse Function
2-
5x 5+
3x
terms in thebox remainunchanged
5-3x
=f-1(x)
5+2x=f(x)
Box # 3 : Form the quadratic equation given the roots
(a) 3 and -4 are the roots
Roots
= 0Eqn: +x -12x2
= 04 )+( xx=- 4
= 0)3-x(x=3
(b) 2 and3
1are the roots
Roots
= 0Eqn: -7x +23x2
= 01 )-(3xx=13
= 0)2-x(x=2
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Box # 4 : Form the quadratic equation using given roots
Example: Given that and are the roots of quadratic equation 2x2 + 7x -15 = 0.Form an equation with the roots of 2 and 2.
Form
1
xx
eqn:
1
cbxax 2
x2 -30+7x-304()-72(+)
()(+)
-15
2-7
2
2x2+7x-15=0: a=2, b=7, c=-15
+c
a-b
a+
a
a
Box # 5: Index Numbers
Alternative Method: Using Index Box
r o w fo r p ri c e in
i n d e x n u m b
r o w f o r Q u a n
P r i c e e ta d a p e r k a ta a n '
y e a r ', in d e x n u
o r p r ic e i n d e x m
1 0 0
I1 0 0
Q1Q0
S p e c i f iY e a r
B a s eY e a r
orI
QQ10
100= 10 100QxIQ =
Example:
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Q1: The mean electric bill of Ali increases from RM420 a month in 2006 to RM630 a
month in 2009. Calculate the index number of the electric bill of Ali in 2009 based
on 2006.
I
630420
100
20092006
630100420 xI =
420
63000=I
150=I
Box # 6 : RELATED RATES OF CHANGES
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dd
THE RATES BOX
column
for time
column for he ight,
radius & x
column for Area,
Volume & y
t
h, r, x
h, r, x
A, v, y
Briefing about Rates Box:
s y m b o l X = t i
r e s p e c t tod iv id e o r
d i f f e re n t i a
dd
T H E R A T E S B
t
h , r , x
h , r , x
A , v , y
d dA
r
r
t
column for Are a
column for radius column
for time
THE RATES BOX
1)dt
drX
dr
dA
dt
dA=
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Refer to general
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dd
THE RATES BOX
column
for timecolumn for radius
column for
volume
t
r
r
v
2)dt
drX
dr
dv
dt
dv=
d dhv
h t
column forvolume
column for height columnfor time
THE RATES BOX
3)dt
dhX
dh
dv
dt
dv=
d dcolumnfor time
THE RATES BOX
column for height
column forvolume
tx
v x
4)dt
dxX
dx
dv
dt
dv=
AFS Technic: Answering For Success Technic
Refer to s here
Refer to cylinder
Refer to cube
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dd
h,r,x
h,r,x t
S,A, V, Y
2: Find the value
[ ] [ ]xrhXyvAsdt
yvAsd,,,,,
),,,(=
ddh,r,x
h,r,x t
S,A, V, Y
3: Find the value
[ ][ ]xrh
yvAs
dt
xrhd
,,
,,,),,(=
Example: SPM 2005, Q20
hhv 83
1 3+= 8)3(
3
1 2+= h
dh
dv
10=dt
dv= h2 + 8
= (2)2 + 8
?=dt
dh, when h = 2
dh
dv= 12
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12
10dd
h
h t
V
3: Find the value
12
10=
dt
dh
= 0.833cms-1
Example 2: Sasbadi Nexus, pg 152
12 = cms
dt
dr, spherical balloon
r = 7 cm. Find ?=dt
dv
Solution:
V sphere =3
3
4r
2)3(3
4r
dr
dv=
= 24 r
= 4 (7)2
= 196
2196dd
r
r t
V
2196 Xdt
dv=
= 392
SMALL CHANGES IN QUANTITIES
Using Small Changes Box
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x
yit
dx
dy
x
0lim.1
= , when x is very small
dx
dy
x
y
xdx
dyy =
2. Using Small Changes Box
divide or by
y = dydx
x
dx is a reference,
dx = 1
= times
dx
dy
x
y
3. Small changes in X
x increase from, 2 to 2.01
To find x:
x = xend - xstart = 2.01 - 2
= 0.01
endstart
x:22.01
4. ynew = yinitial +y
A. Example 1:
y = x3 +2 , find the approximate increase
in y or y when x increases from 2 to
2.01
solution:
x: 2 2.01201.2 =dx
= 0.01
23x
dx
dy=
= 3(2)
2
= 12Using the small changes box:
Depend on question:May be y, A, etc
Depend on question:
May be dy, dA, etc
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C. Example 3:
Given y = 416
x, find the value of
dx
dy
when x = 2. Hence, find the
approximate value of 4)04.2(16 .
Solution:
y =4
16
x=16x -4 = 4)2(
16,
yinitial = 1)2(16
4=
5)4(16
= xdx
dy
= 16(-4) (2)-5
= -2
x : 2 04.2
x = 2.04 -2
= 0.04
ynew = ?
y -2
0.04 1
y = 0.04 x -2= -0.08
ynew = yinitial + y= 1 + (-0.08)
= 0.92
Check: using calculator
press < >
< = >
Answer shown on a screen as0.9233845426
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