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=

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    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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