Μαθηματικά Κατεύθυνσης Απαντήσεις Θέματων...

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Μαθηματικά Κατεύθυνσης Απαντήσεις Θέματων Πανελληνίων 2010

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  • 2010

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    | | 2010 | |

    |

    1. 2. 3. 4. / / / /

    |

    0,22 =+ zz

    z

    1.

    0222222 22 =+=+=+ zzzzz

    z iz +=11 , iz =12

    2.

    ( ) iiz 21 221 =+= ( ) iiz 2122

    2 ==

    pi ( ) ( ) ( ) ( ) ( ) ( ) 02222 10051005100510051005221005212010220101 ==+=+=+ iiiizzzz

    3.

    2|)34(||20||)34(||11||34||||34| 21 =+=++=+=+ iwiiwiiiwzziw

    )3,4( K 2=R

    4.

    5)(2534)( 222 ==+= OKOK

    325)(|| min === ROKw

    725)(|| max =+=+= ROKw

    7||3 w

    4

    3 K

    0

  • 2010

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    |

    Rxxxxf ++= ),1ln(2)( 2 1.

    01

    )1(2)( 22

    >+

    ++=

    x

    xxxf Rx f R

    :

    To 12 ++ xx < 0 Rxxx >++ ,012

    2.

    +

    +=+

    11)23(ln)23(2 4

    22

    x

    xxx ( ) ( )++= 1ln1)23(ln)23(22 422 xxxx

    ( ) ( )++=++ 1)23(ln)23(21ln2 242 xxxx 112 )23()( = fxfxf = 232 xx

    2,10232 ===+ xxxx

    3.

    ( )22

    2

    1

    )1(2)(+

    =

    x

    xxf

    1010)( 2 === xxxf

    pi )2ln2,1( + : 2ln1)1(1)2ln2()1()1()1( +=+=++= xyxyxffy (1)

    pi )2ln2,1( + : 2ln13)1(3)2ln2()1()1()1( +==+= xyxyxffy (2)

    () (1) (2) pipi + )2ln1,0(

    4.

    340

    322)1ln(2))1ln(2())1ln(2()(

    1

    1

    21

    1

    21

    1

    221

    1

    21

    1

    =+=++=++=++==

    dxxxdxxdxxxxdxxxxdxxxfI

    +=1

    1

    21 )1ln( dxxxI ( ) pi, :

    )( xf

    )( xf

    ( )( )2ln2,1

    )1(,1..

    +

    f

    1x

    +

    1

    ( )( )2ln2,1

    )1(,1..

    +

    f

  • 2010

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    A- )1ln()( 2 += xxxg pi pi ( )()( xgxg = ) [-1,1] pi pi

    - =+

    =+=

    1

    1

    221

    1

    21 )1ln(2)1ln( dxx

    xdxxxI

    =

    +=

    +=

    +

    +=

    1

    12

    1

    12

    31

    12

    21

    1

    22

    110

    12

    2)1ln(

    2dx

    x

    xxdx

    x

    xdxx

    xxx

    x

    02

    )1ln(01

    1

    1

    21

    12

    1

    1

    =

    ++=

    ++=

    xdxx

    xxdx

    - 12 += xu , xdxdu 2= 0ln21)1ln(

    2

    2

    1

    1

    21 ==+=

    ududxxxI

    |

    1.

    ++=x

    dtttf

    txxf

    0 )(3)( 3)0( =f Rxxxf ,)(

    Rxxxf

    xfxxf

    xxxfxxf

    xxf

    =

    +

    =

    += ,)()(

    )()(

    )(1)(

    2.

    ( ) ==+== )(2)(2)()(2))()((2)()(2)(2)()( 2 xfxfxxfxfxfxxfxfxfxxfxfxg

    0)(2)(2)(2))(()()(2)(2))(()(2

    1==

    ==

    xfxfxfxxfxxf

    xfxfxxfxf cxg =)(

    cg == 9)0( 9)( =xg , Rx

    3. 09)(2)(9)(2)(9)( 22 === xxfxfxxfxfxg pi )(xf

    9)( 2 += xxxf pipi 9)( 2 += xxxf pipi pi 3)0( =f pi

    Rxxxxf ++= ,9)( 2

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    4.

    +

    =1

    )()(x

    x

    dttfxh , Rx pi )1()( +< xhxh

    ++

    +=+=1

    00

    1

    0

    0

    )()()()()(xxx

    x

    dttfdttfdttfdttfxh

    0)()1()1()()( >+=++= xfxfxfxfxh h R

    pi

    +

    +

    ++=

    =

    x

    xx

    xxfxf

    xf f R

    0)()1()1()(1 >+++++