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  • 7/28/2019 Chapter 11 Maclaurin Series

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    O xford Fajar Sdn. Bhd. (008974-T) 2012

    CHAPTER 11 MACLAURIN

    SERIES

    Focus on Exam 11

    1 Let f ( x) = e x sin x f(0) = e0

    = 1 f ( x) = ( x cos x + sin x) e x sin x = ( x cos x + sin x) f( x) f (0) = (0 cos 0 + sin 0)(1)

    = 0 f ( x) = ( x cos x + sin x) f ( x) + f ( x)(- x sin x + cos x + cos x)

    = ( x cos x + sin x) f ( x) + f ( x)(- x sin x + 2 cos x) f ( x) = 0 + 1(2)

    = 2Hence, f ( x) = 1 + f (0) x + f (0) x

    2

    2! +

    e x sin x = 1 + 2 x22

    +

    = 1 + x 2 + [Shown]2 y = ln (cos x)

    d yd x

    = sin xcos x

    = - tan x

    d 2 yd x2

    = - sec 2 x = - (1 + tan 2 x)

    d 3 y

    d x3 = - 2 tan x(sec 2 x)

    = - 2(- tan x)(- sec 2 x)

    d yd x

    d 2 yd x2

    = - 2 d yd x

    d 2 yd x2

    [Shown]

    d 4 yd x4

    = - 2 d yd x

    d 3 yd x3

    + d 2 y

    d x2 1- 2d

    2 yd x22

    = - 2 d yd x d 3

    yd x3 - 21d 2

    yd x222

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    ACE AHEAD Mathematics (T) Second Term2

    When x = 0, y = ln (cos 0) = ln 1 = 0

    When x = 0,

    d yd x = - tan 0

    = 0When x = 0,

    d 2 yd x2

    = - (1 + tan 2 0) = - 1

    When x = 0,

    d 3 yd x3

    = - 2 d yd x

    d 2 yd x2

    = - 2(0)( - 1) = 0

    When x = 0,

    d 4 yd x4

    = - 2 d yd x

    d 3 yd x3

    - 21d 2 yd x222

    = - 2(0)(0) - 2(- 1)2

    = - 2

    Hence, ln (cos x) = f (0) + f (0) x + f (0) x2

    2! + f (0) x

    3

    3! + f IV x

    4

    4! +

    = 0 + 0 x - x2

    2 + 0 x3 - 21 x

    4

    242 + = - x

    2

    2 - x

    4

    12 +

    3 y2 = 1 + sin x

    2 y d yd x

    = cos x

    2 yd 2 yd x2

    + d yd x

    2 d yd x

    = - sin x

    2 y d 2

    yd x2

    + 21d yd x22

    = 1 - y 2

    2 yd 2 yd x 2

    + 21d yd x 22

    + y 2 - 1 = 0 [Shown]

    2 yd 3 yd x3

    + d 2 y

    d x2 2 d y

    d x + 41d yd x21

    d 2 yd x22 + 2 y d yd x = 0

    yd 3 yd x3

    + d yd x

    d 2 yd x2

    + 21d yd x21d 2 yd x22 + y d yd x = 0

    yd 3 yd x 3 + 31

    d yd x 21

    d 2 yd x 22 + y

    d yd x = 0

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    Fully Worked Solution 3

    When x = 0, y = 1 + sin 0 = 1

    When x = 0,

    2(1)d yd x = cos 0

    d yd x

    = 12

    When x = 0,

    2(1)d 2 yd x2

    + 211222

    + 12 - 1 = 0

    d 2 yd x2

    = - 14

    When x =

    0,(1)

    d 3 yd x3

    + 311221- 142+ (1) 12 = 0

    d 3 yd x3

    = - 18

    Hence, 1 + sin x = f (0) + f (0) x + f (0) x2

    2! + f (0) x

    3

    3! +

    = 1 + 12

    x - 141 x

    2

    2!2 - 18 1 x3

    3!2 + = 1 + x 2 -

    x 2

    8 - x

    3

    48 +

    4 y = cos x y2 = cos x

    2 y d yd x

    = - sin x

    2 yd 2 yd x2

    + d yd x

    2 d yd x

    = - cos x

    yd 2 yd x2

    + 1d yd x2

    2

    = - 12

    cos x

    y d 2 y

    d x2 + 1d yd x2

    2

    = - 12

    y2

    yd 2 yd x 2

    + 1d yd x 22

    + 12

    y 2 = 0 [Shown]

    yd 3 yd x3

    + d 2 y

    d x2

    d yd x

    + 2 d yd x

    d 2 y

    d x2 + 1

    2 2 y d y

    d x = 0

    yd 3 yd x3

    + 3d yd x

    d 2 yd x2

    + y d yd x

    = 0

    yd 4 yd x4 +

    d 3 yd x3

    d yd x + 3

    d yd x

    d 3 yd x3 +

    d 2 yd x2 3

    d 2 yd x2 + y

    d 2 yd x2 +

    d yd x

    d yd x = 0

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    ACE AHEAD Mathematics (T) Second Term4

    y d 4 y

    d x4 + 4 d y

    d x d

    3 yd x3

    + 31d 2 yd x222

    + y d 2 y

    d x2 +1d yd x2

    2

    = 0

    When x = 0,y = cos 0 = 1

    When x = 0,

    2(1) d yd x

    = - sin 0

    d yd x

    = 0

    When x = 0,

    (1)d 2 yd x2

    + 02 + 12

    (1) 2 = 0

    d 2 yd x2

    = - 12

    When x = 0,

    (1)d 3 yd x3

    + 3(0) 1- 122+ (1)(0) = 0

    d 3 yd x3

    = 0

    When x = 0,

    (1)d 4 yd x4

    + 4(0)(0) + 31- 1222

    +(1)1- 122 + 02 = 0

    d 4 yd x4

    = - 14

    Hence, cos x = f (0) + f (0) x + f (0) x2

    2! + f (0) x

    3

    3! + f IV(0) x

    4

    4! +

    = 1 - 12

    1 x22 2 - 14 1 x4

    242 + = 1 - x

    2

    4 - x

    4

    96 +

    5 y = sin - 1 xsin y = x

    (cos y)

    1d y

    d x2 = 1

    1 - sin 2 y 1d yd x2 = 11 - x2 1d yd x2 = 1

    (1 - x2)1d yd x22

    = 1

    (1 - x2) 21d yd x2d 2 y

    d x2 + 1d yd x2

    2

    (- 2 x) = 0

    (1 - x 2) d2

    yd x 2 - x

    d yd x = 0 [Shown]

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    Fully Worked Solution 5

    (1 - x2) d 3 y

    d x3 + d

    2 yd x2

    (- 2 x) - x d 2 y

    d x2 - d y

    d x = 0

    When x = 0, y = sin -1 0 = 0

    When x = 0,1 - 021d yd x2 = 1

    d yd x

    = 1

    When x = 0,

    (1 - 02) d 2 y

    d x2 - (0)(1) = 0

    d 2 yd x2 = 0

    When x = 0,

    (1 - 02) d 3 y

    d x3 + (0)[ -2(0)] - (0)(0) - 1 = 0

    d 3 yd x3

    = 1

    Hence, sin - 1 x = f (0) + f (0) x + f (0) x2

    2! + f (0) x

    3

    3! +

    = x - x 3

    6 +

    6 0.4

    0.1e2 cos x d x =

    0.4

    0.1 11 + x + x2

    2! + x

    3

    3!211 - x2

    2!2d x =

    0.4

    0.1 11 - x2

    2 + x - x

    3

    2 + x

    2

    2 + x

    3

    6 2d x =

    0.4

    0.1 11 + x - x3

    3 2 d x

    = 3 x + x2

    2 - x4

    1240.4

    0.1

    = 0.4 + 0.42

    2 - 0.4

    4

    12 - 10.1 + 0.1 22 - 0.1

    4

    12 2 = 0.373

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    ACE AHEAD Mathematics (T) Second Term6

    7 0.5

    0 sin x1 - x d x

    = 0.5

    0(sin x)(1 - x)- 1 d x

    = 0.5

    0 1 x - x3

    3 2[1 + - 1C 1(- x) + - 1C 2(- x)2 + - 1C 3(- x)3] d x

    = 0.5

    0 1 x - x3

    3 2(1 + x + x2

    + x3

    ) d x

    = 0.5

    0 1 x + x2 + x3 - x3

    3 2d x =

    0.5

    0 1 x + x2 + 2 x3

    3 2d x = 3 x22 + x

    3

    3 + x

    4

    6 400.5

    = 0.52

    2 + 0.5

    3

    3 + 0.5

    4

    6 - 0

    = 0.177

    8 e2 x ln 1 - 2 x = e2 x312 ln (1 - 2 x)4 = 1

    2 31 + 2 x + (2 x)

    2

    2! + (2 x)

    3

    3! + (2 x)

    4

    4! + 4 3-2 x - (- 2 x)

    2

    2 + (- 2 x)

    3

    3 - (- 2 x)

    4

    4 + 4

    = 12

    11 + 2 x + 4 x22 + 8 x3

    6 + 16 x

    4

    24 + 21-2 x - 4 x22 - 8 x

    3

    3 - 16 x

    4

    4 + 2

    = 12

    11 + 2 x + 2 x2 + 43 x3 + 23 x4 + 21- 2 x - 2 x2 - 83 x3 - 4 x4 + 2 = 1

    2 1- 2 x - 2 x2 - 83 x3 - 4 x4 - 4 x2 - 4 x3 - 163 x4 - 4 x3 - 4 x4 - 83 x4 + 2

    = 12

    1- 2 x - 6 x2 - 323 x3 - 16 x4 - 2 = - x - 3 x 2 - 16

    3 x 3 - 8 x 4 -

    When e 2 x ln 1 - 2 x = e15 ln 4

    5,

    1 - 2 x = 45

    x = 110

    Hence, e15 ln 4

    5 = - 1

    10 - 3

    100- 16

    3000 - 8

    10 000

    = - 1021

    7500

    [Shown]

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    Fully Worked Solution 7

    9 f ( x) = cosec 12 + x2f ( x) = - cosec 12 + x2cot 12 + x2f ( x) = - cosec 1

    2

    + x23- cosec 2 12

    + x24+ cot 12

    + x23cosec 12

    + x2cot 12

    + x24 = cosec 3 12 + x2 + cot 2 12 + x2 cosec 12 + x2 = cosec 3 12 + x2+3cosec 2 12 + x2- 14 cosec 12 + x2 = cosec 3 12 + x2 + cosec 3 12 + x2 - cosec 12 + x2 = 2 cosec 3 12 + x2 - cosec 12 + x2 f ( x ) = 2[f ( x )]3 - f ( x ) [Shown]

    f ( x) = 6[f ( x)]2 f ( x) - f ( x)f IV ( x) = 6[f ( x)]2 f ( x) + f ( x) 12 f ( x) f ( x) - f ( x)

    = 6[ f ( x)]2 f ( x) + 12 f ( x)[f ( x)]2 - f ( x)

    When x = 0,

    f (0) = cosec 12 + 02 = 1 f (0) = - cosec 12 + 02cot 12 + 02 = 0

    f (0) = 2(1) 3 - 1 = 1

    f (0) = 6(1) 2(0) - 0 = 0

    f IV

    (0) = 6(1)2

    (1) + 12(1)(0)2

    - 1 = 5

    f ( x) = f (0) + f (0) x + f (0) x2

    2! + f (0) x

    3

    3! + f IV(0) x

    4

    4! +

    cosec 12 + x2 = 1 + x 2

    2 + 5 x

    4

    24+

    Hence, cosec 12 + 1122 1 + 12 111222

    + 524

    111224

    1.00348

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    ACE AHEAD Mathematics (T) Second Term8

    10 1 - cos x x sin x

    =1 - 11 x22! + x

    4

    4! + 2

    x1 x x33! + x5

    5! + 2

    =

    x2

    2 - x

    4

    24 +

    x2 - x4

    6 + x

    6

    120 +

    =12

    - x4

    24 +

    1 - x2

    6 + x

    4

    120 +

    Dividing each term by x2

    Hence, lim x0

    1 - cos x x sin x

    =12

    - 0 +

    1 - 0 + 0 +

    = 12

    11 Let f ( x) = ln (1 + sin x)

    f ( x) = cos x1 + sin x

    f ( x) = (1 + sin x)(- sin x) - cos x (cos x)(1 + sin x)2

    = - sin x - sin2 x - cos 2 x

    (1 + sin x)2

    = - sin x - (sin2 x + cos 2 x)

    (1 + sin x)2

    = - sin x - 1(1 + sin x)2

    = - (sin x + 1)(1 + sin x)2

    = - 11 + sin x

    = -(1 + sin x)- 1f ( x) = (1 + sin x)- 2 cos x

    = - cos x(1 + sin x)2

    When x = 0,f (0) = ln (1 + sin 0)

    = 0f (0) = cos 01 + sin 0

    = 1

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    Fully Worked Solution 9

    f (0) = - 11 + sin 0

    = - 1

    f (0) = cos 0(1 + sin 0) 2

    = 1f ( x)= f (0) + f (0) x + f (0) x

    2

    2!+ f (0) x

    3

    3!+

    ln (1 + sin x) = x - x 2

    2 + x

    3

    6 +

    lim x0

    ln (1+ sin x) - x

    x2= lim

    x0 1 x x22 + x

    3

    6 + 2 x

    x2

    = lim x0

    - x2

    2 + x

    3

    6 +

    x2

    = lim x0 1- 12 + x6 + 2

    = - 12