Exercises in Maths III

download Exercises in Maths III

If you can't read please download the document

  • date post

    19-Feb-2016
  • Category

    Documents

  • view

    215
  • download

    2

Embed Size (px)

description

math exercises with solutions for high school inadvanced level nad in greece

Transcript of Exercises in Maths III

  • . )

    )))

    ' -

    - -

    21 - 30

    , 2014

    '

  • .

    ' % %

    , 2014

    - www.dimoshopoulos.gr

    (

    21(-(30

    - - .

  • , . http://lisari.blogspot.gr/2012/04/25.html

    ' .

    - www.dimoshopoulos.gr

  • 21 f : [,] ! , [,] , f (x) 0 ,

    x [,] , z, Re(z) 0 , Im(z) 0 , | Re(z) | > | Im(z) | .

    z +

    1z

    = f () z 2 +

    1z 2

    = f 2() , :

    ) | z | = 1 . ) f2() < f 2() .

    ) x3 f ()+ f () = 0 , , (1,1) .

    ) z +

    1z

    = f () ! , z +

    1z

    = z +1z

    , '

    z2z + z = z 2z + z z 2z z 2z + z z = 0 zz (z z)(z z) = 0 (z z)(zz 1) = 0

    z z = 0 zz 1 = 02 i Im(z) = 0 | z |2= 1

    Im(z) = 0 (, Im(z) 0 ) | z | = 1 | z | = 1 .

    ) z +

    1z

    = f ()

    z +

    1z

    2

    = f 2() z 2 +1z 2

    + 2 = f 2() f 2()+ 2 = f 2() f 2() f 2() = 2 < 0

    f 2() f 2() < 0 f 2() < f 2() (1)

    ) g(x) = x3 f ()+ f () , x [1,1] ,

    g(x) = 0 , , (1,1) .

    . g [1,1] .

    . g(1) = (1)3 f ()+ f () g(1) = f () f () .

    . g(1) = 13 f ()+ f () g(1) = f ()+ f () .

    g(1) g(1) = [f () f ()][f ()+ f ()] = f2() f 2() < 0 , (1).

    Bolzano, g(x) = 0 , , (1,1) .

    ' .

    - www.dimoshopoulos.gr

  • 22 f : [1,+) ! , f (x) = x

    2 nx .

    ) f.

    ) f (x) = 2012 .

    ) z1 ,z2 ! , | z1 | >1 , | z2 | >1 z1z1 n | z1 | = z2z2 n | z2 | = n16 , :

    . | z1 | = | z2 | = 2 . .

    z1

    + z2

    4 + z1z

    2

    ! .

    ) zz n | z | = 1 , z , > 1.

    ) f = [1,+) -, , ,

    f (x) = (x 2 ) nx + x 2 (nx ) = 2x nx + x 2

    1x

    = 2x nx + x f (x) = x(2nx +1) > 0 ,

    x >1 , nx > 0 , x >1 .

    f [1,+) , .

    ) x 1 , nx 0 , f (x) 0 , f () = [0,+) .

    2012 f () , f (x) = 2012 , , .

    , , f , .

    f , f () = [ f (1) , im

    x+f (x)) .

    f (1) = 12 n1 = 1 0 f (1) = 0 .

    imx+

    f (x) = imx+

    (x 2 nx) = (+) (+) = + .

    f () = [0,+) .

    2012 f () , f (x) = 2012 , , .

    , , f , .

    ) . z1z1 n | z1 | = z2z2 n | z2 | = n16

    | z1 |2 n | z

    1| = | z

    2|2 n | z

    2| = n24 f (| z

    1|) = f (| z

    2|) = 4n2

    f (| z1 |) = f (| z2 |) = 22 n2 f (| z

    1|) = f (| z

    2|) = f (2) .

    f , 1-1, -

    | z

    1| = | z

    2| = 2 .

    ' .

    - www.dimoshopoulos.gr

  • . w =

    z1

    + z2

    4 + z1z

    2

    , w = w .

    w =

    z1

    + z2

    4 + z1z

    2

    .

    | z1 | = 2 | z

    1|2 = 4 z

    1z

    1= 4 z

    1=

    4z

    2

    . z

    2=

    4z

    2

    .

    w =

    4z

    1

    +4z

    2

    4 +4z

    1

    4z

    2

    =

    41z

    1

    +1z

    2

    4 1+4

    z1z

    2

    =

    1z

    1

    +1z

    2

    1+4

    z1z

    2

    =

    z2

    + z1

    z1z

    2

    z1z

    2+ 4

    z1z

    2

    =z

    1+ z

    2

    4 + z1z

    2

    = w

    w = w .

    ) zz n | z | = 1 | z |2 n | z | = 1 f (| z |) = 1 (1)

    f (x) > 0 , x >1 , 1 = (1,+) , f (1) = (0,+) .

    1 f () , >1 , f () = 1 .

    f 1 , .

    , (1) f (| z |) = f () f 11

    | z | = , >1 , z

    (0, 0) > 1.

    ' .

    - www.dimoshopoulos.gr

  • 23 z = + i , , ,

    f :! ! ,

    f (x) = | t z + z |dt

    1

    x2+1

    3x + 2 . f - x0 = 1 , :

    ) | z | =

    12

    .

    ) w = 2z i - .) w, .

    ) , f, x'x x A 0 x A 1.

    ) | tz + z | = | z(t +1) | = | z | | t +1 | ,

    f (x) = | z | | t +1 |dt

    1

    x2+1

    3x + 2 f (x) = | z | | t +1 |dt1

    x2+1

    3x + 2 . | t +1 | ! ,

    | t +1 |dt

    1

    x

    ! .

    ,

    | t +1 |dt

    1

    x2+1

    ! x 2 +1

    | t +1 |dt

    1

    x

    , f ! -, 1, .

    , , 1 D

    f= ! , Fermat

    f (1) = 0 (1)

    f (x) = | z | | x2 +1+1 | (x 2 +1 ) 3 = | z | | x 2 + 2 | 2x 3

    f (x) = 2x | z | | x2 + 2 |3 .

    (1) 2 1 | z | | 12 + 2 |3 = 0 6 | z | = 3 | z | =

    36 | z | =

    12

    .

    ' .

    - www.dimoshopoulos.gr

  • ) w = 2z i 2z = w + i z =

    w + i2

    .

    | z | =

    12

    w + i2

    =12

    |w + i |2

    =12 |w + i | = 1 ,

    (0,1) 8 1.

    ) w 2i , - .

    ) ,

    [0,1] , f ' .

    | z | =

    12

    (), f (x) =

    12 | x 2 + 2 | 2x 3 .

    x2 + 2 > 0 , x ! , | x

    2 + 2 | = x 2 + 2 ,

    f (x) = x(x2 + 2)3 = x 3 + 2x 3 f (x) = (x 1)(x 2 + x + 3) .

    12 4 1 3 = 11 < 0 , x

    2 + x + 3 > 0 ,

    x ! , ., f (x) > 0 x 1 > 0 x >1 .

    f x 8 1, -

    f (x) f (1) (2) , x ! .

    f (1) =12

    | t +1 |dt

    1

    12+1

    3 1+ 2 f (1) = 12 | t +1 |dt1

    2

    1 . t +1 > 0 , t [1,2] , | t +1 | = t +1 ,

    | t +1 |dt

    1

    2

    = (t +1)dt1

    2

    = t2

    2+ t

    1

    2

    =22

    2+ 2

    12

    2+1

    =52

    .

    f (1) =

    12521 =

    541 f (1) =

    14

    .

    , (2) f (x)

    14

    , x ! , -

    = f (x)dx

    0

    1

    = 1 f (x)dx0

    1

    = (x ) f (x)dx0

    1

    = [x f (x)]01 x f (x)dx0

    1

    ' .

    - www.dimoshopoulos.gr

    1 1 +

    f (x) +

    f (x) 2 1

  • = 1 f (1)0 f (0) x(x 3 + 2x 3)dx

    0

    1

    = 14 (x 4 + 2x 2 3x)dx0

    1

    =

    14

    x 5

    5+

    2x 3

    3

    3x 2

    2

    0

    1

    .

    g(x) =

    x 5

    5+

    2x 3

    3

    3x 2

    2,

    =

    14[g(x)]

    01 =

    14[g(1)g(0)] =

    14g(1)+ g(0) (3)

    g(1) =

    15

    5+

    2 13

    3

    3 12

    2=

    15

    +23

    32

    =6 + 20 45

    30 g(1) =

    1930

    .

    g(0) =

    05

    5+

    2 03

    3

    3 02

    2 g(0) = 0 .

    (3) =

    141930

    + 0 =14

    +1930

    =15 + 38

    60 =

    5360

    .. .

    24 z = + i (, !) f (x) = | z xi | , x ! , (1,2) .

    ) 1 | z | 3 . z ;

    ) f (x) = 2 | z | x , , (1,1) .

    ) C

    f (1,2) , f (1) = 2 , '

    | z 1 i | = 2 | z i | = 2 ,

    (0,1) D 2.

    ,

    () () () 1 | z | 3 .

    | z | = 1 , ,

    z =i .

    , | z | = 3 , ,

    z = 3i .

    ) f (x)2 | z | x = 0 .

    g(x) = f (x)2 | z | x , x [1,1] , g(x) = 0

    , , (1,1) .

    ' .

    - www.dimoshopoulos.gr

  • . f (x) = | z xi | = |+ ixi | = |+ (x)i | f (x) = 2 + (x)2 .

    f ! , g [1,1] .

    . g(1) = f (1)2 | z | (1) = | z (1) i | + 2 | z | g(1) = | z + i | + 2 | z | .

    | | z | | i | | | z + i | | z | + | i | | | z |1 | | z + i | | z | +1 .

    () | z |1 | z |1 0 , | | z |1 | = | z |1 .

    | z |1 | z + i | 2 | z | + | z |1 | z + i | + 2 | z | 3 | z |1 g(1) .

    | z |1

    3 | z | 3 3 | z |1 31 3 | z |1 2 > 0 g(1) > 0 .

    III. g(1) = f (1)2 | z | 1 = 22 | z | =2(| z |1) g(1) 0 , | z |1 0 .

    g(1) g(1) 0 , :

    i) g(1) g(1) < 0 , Bolzano g(x) B 0

    , , (1,1) .

    ii) g(1) g(1) = 0 , g(1) = 0 ( g(1) > 0 ), 1

    g(x) B 0.

    (i) (ii) , , g(x) = 0 , -

    , [1,1) .

    ' .

    - www.dimoshopoulos.gr

  • 25 f : [,] ! z = + i , z1 = + f () i

    z2 = + f () i . 3(z2 z 2) 4izz = 4 i Re(z

    1z

    2) ,

    C

    f ,

    , x'x. f (x) = 0 , , (,) .

    z2 z 2 = (z z )(z + z ) = 2 i Im(z) 2Re(z) = 2i 2 = 4 i .

    , z1z2 = [+ f () i ][ f () i ] = f () i + f () i f () f () i2

    z1z2 = [ + f () f ()]+ [ f () f ()]i .

    ,

    3 4 i 4i | z |2 = 4i [ + f () f ()] 3(2 + 2) = + f () f ()

    f () f () = 22 2 f () f () =(2 + 2 2) f () f () =()2 .

    [,] < < 0 ()2 > 0 ,

    f () f () < 0 .

    f [,] , Bolzano,

    f (x) = 0 , , (,) .

    ' .

    - www.dimoshopoulos.gr

  • 26) z. z (1+ i)z + (1 i)z + 4 = 0 , z, , ().

    ) () () -

    f, f (x) = x + +

    x 2

    ex + , , ! .

    ) g(x) =

    f (x)x

    , x 0 . - g.

    )

    z + iz + z i z + 4 = 0 (z + z )+ i(z z )+ 4 = 0 2Re(z)+ i 2 i Im(z)+ 4