De Cuong Toan 2014-2015

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Đề Cương Toán 2014-2015 ôn thi đại học

Transcript of De Cuong Toan 2014-2015

  • UBND TNH BC NINH5 GIO DC V O TO

    CNGN THI THPT QUC GIA MN TON

    0m hc 2014 - 2015

    %c Ninh, thng 11 nm 2014

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  • GD&0Ninh0ngnthiTHPTqugia2014-2015

    CHUYN 1. KHO ST HM S V CC BI TON LIN QUANBin son v su tm: Ng Vn Khnh GV trng THPT Nguyn Vn C

    1. Ch 1: Bi ton v tip tuyn1.1. Dng 1: Tip tuyn ca th hm s ti mt Lm

    0 0M( , ) ( ) : ( )x y C y f x =

    * Tnh ' '( )y f x= ; tnh '0

    ( )k f x= (h s gc ca tip tuyn)

    * Tip tuyn ca th hm s ( )y f x= ti Km ( )0 0;M x y c phng trnh( )'0 0 0( )y y f x x x- = - vi 0 0( )y f x=

    V d 1: Cho hm s 3 3 5y x x= - + (C). Vit phng trnh tip tuyn ca th (C):a) 6i Km A (-1; 7).b) 6i Km c honh x = 2.c) 6i Km c tung y =5.

    Gii:a) Phng trnh tip tuyn ca (C) ti Km

    0 0 0( ; )M x y c dng:

    0 0 0'( )( )y y f x x x- = -

    Ta c 2' 3 3y x= - '( 1) 0y - = .

    Do phng trnh tip tuyn ca (C) ti Km A(-1; 7) l: 7 0y - = hay y = 7.b) T 2 7x y= = .

    y(2) = 9. Do phng trnh tip tuyn ca (C) ti Km c honh x = 2 l:7 9( 2) 7 9 18 9 11y x y x y x- = - - = - = -

    c) Ta c: 3 30

    5 3 5 5 3 0 3

    3

    x

    y x x x x x

    x

    =

    = - + = - = = - =

    +) Phng trnh tip tuyn ti ca (C) ti Km (0; 5).Ta c y(0) = -3.Do phng trnh tip tuyn l: 5 3( 0)y x- = - - hay y = -3x +5.

    +) Phng trnh tip tuyn ti ca (C) ti Km ( 3;5)- .2'( 3) 3( 3) 3 6y - = - - =

    Do phng trnh tip tuyn l: 5 6( 3)y x- = + hay 6 6 3 5y x= + + .

    +) Tng t phng trnh tip tuyn ca (C) ti ( 3;5)- l: 6 6 3 5y x= - + .

    V d 2: Cho th (C) ca hm s 3 22 2 4y x x x= - + - .a) Vit phng trnh tip tuyn vi (C) ti giao Km ca (C) vi trc honh.

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  • GD&0Ninh0ngnthiTHPTqugia2014-2015

    b) Vit phng trnh tip tuyn vi (C) ti giao Km ca (C) vi trc tung. c) Vit phng trnh tip tuyn vi (C) ti Km x0 tha mn y(x0) = 0.Gii:

    Ta c 2' 3 4 2y x x= - + . Gi ( )0 0;M x y l tip Km th tip tuyn c phng trnh:

    0 0 0 0 0 0'( )( ) '( )( ) (1)y y y x x x y y x x x y- = - = - +

    a) Khi ( )M C Ox= th y0 = 0 v x0 l nghim phng trnh:3 22 2 4 0 2x x x x- + - = = ; y(2) = 6, thay cc gi tr bit vo (1) ta c phng

    trnh tip tuyn: 6( 2)y x= -

    b) Khi ( )M C Oy= th x0 = 0 0 (0) 4y y = = - v 0'( ) '(0) 2y x y= = , thay ccgi trbit vo (1) ta c phng trnh tip tuyn: 2 4y x= - .

    c) Khi x0 l nghim phng trnh y= 0. Ta c: y = 6x 4.

    y = 00 0

    2 2 886 4 0

    3 3 27x x x y y

    - = = = = = - ;

    0

    2 2'( ) '

    3 3y x y

    = =

    Thay cc gi tr bit vo (1) ta c phng trnh tip tuyn: 2 1003 27

    y x= -

    V d 3: Cho hm s 3 3 1y x x= - + (C) a) Vit phng trnh tip tuyn d vi (C) tai Km c honh x=2. b)Tip tuyn d ct li th (C) ti Km N, tm ta ca Km N.Gii

    a) Tip tuyn d ti Km M ca th (C) c honh 0 0

    2 3x y= =

    Ta c 20

    '( ) 3 3 '( ) '(2) 9y x x y x y= - = =

    Phng trnh tip tuyn d ti Km M ca th (C) l

    0 0 0'( )( ) 9( 2) 3 9 15y y x x x y y x y x= - + = - + = -

    8y phng trnh tip tuyn d ti Km M ca th (C) l 9 15y x= -b) Gi s tip tuyn d ct (C) ti N

    Xt phng trnh

    ( )( )3 3 2 23 1 9 15 12 16 0 2 2 8 0 4x

    x x x x x x x xx

    =- + = - - + = - + - = = -8y ( )4; 51N - - l Km cn tmV d 4: Cho hm s 3 3 1 ( )y x x C= - + v Km 0 0( , )A x y (C), tip tuyn ca th

    (C) ti Km A ct (C) ti Km B khc Km A. tm honh Km B theo0

    x

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  • GD&0Ninh0ngnthiTHPTqugia2014-2015

    /i gii: V Km

    0 0( , )A x y (C) 3

    0 0 03 1y x x = - + , ' 2 ' 2

    0 03 3 ( ) 3 3y x y x x= - = -

    Tip tuyn ca th hm c dng:' 2 3

    0 0 0 0 0 0 02 30 0 0

    ( )( ) (3 3)( ) 3 1

    (3 3)( ) 2 1 ( )

    y y x x x y y x x x x x

    y x x x x d

    = - + = - - + - +

    = - - - +

    Phng trnh honh giao Km ca (d) v (C):3 2 3 3 2 3 2

    0 0 0 0 0 0 0

    200

    000

    3 1 (3 3)( ) 2 1 3 2 0 ( ) ( 2 ) 0

    ( ) 0( 0)

    22 0

    x x x x x x x x x x x x x x

    x xx xx

    x xx x

    - + = - - - + - + = - + = =- = = -+ =

    8y Km B c honh 0

    2B

    x x= -

    V d 5: Cho hm s 3 21 2 33

    y x x x= - + (C). Vit phng trnh tip tuyn d ca th

    (C) ti i m c honh 0

    x tha mn ''0

    ( ) 0y x = v chng minh d l tip tuyn ca (C) cJ s gc nh nht.Gii

    Ta c ' 2 ''4 3 2 4y x x y x= - + = -

    0 0 0

    2''( ) 0 2 4 0 2 (2; )

    3y x x x M= - = =

    Khi tip tuyn ti M c h s gc0

    k = ' '0

    ( ) (2) 1y x y= = -

    8y tip tuyn d ca th (C) ti Km 22;3

    M

    c phng trnh ( )'0 0 0( )y y f x x x- = -

    suy ra ( )2 1 23

    y x- = - - hay 83

    y x= - +

    Tip tuyn d c h s gc0

    k = -1/t khc tip tuyn ca thi (C) ti Km by k trn (C) c h s gc

    ( )2' 2 0( ) 4 3 2 1 1k y x x x x k= = - + = - - - =

    Du = xy ra 1x = nn ta tip im trng vi 22;3

    M

    8y tip tuyn d ca (C) ti Km 22;3

    M

    c h s gc nh nht.

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  • GD&0Ninh0ngnthiTHPTqugia2014-2015

    V d 6: Vit phng trnh tip tuyn vi th (C): 21

    xy

    x+

    =-

    ti cc giao Km ca (C)

    Xi ng thng (d): 3 2y x= - .+ Phng trnh honh giao Km ca (d) v (C):

    23 2 2 (3 2)( 1)

    1x

    x x x xx

    += - + = - -

    - (x = 1 khng phi l nghim phng trnh)

    23 6 0 0 ( 2) 2 ( 4)x x x y x y - = = = - = =

    8y c hai giao Km l: M1(0; -2) v M2(2; 4)

    + Ta c:2

    3'

    ( 1)y

    x

    -=

    -.

    + Ti tip Km M1(0; -2) th y(0) = -3 nn tip tuyn c phng trnh: 3 2y x= - -+ Ti tip Km M2(2; 4) th y(2) = -3 nn tip tuyn c phng trnh: 3 10y x= - +Tm li c hai tip tuyn tha mn yu cu bi ton l: 3 2y x= - - v 3 10y x= - + .

    V d 7: Cho hm s 3 21 13 2 3

    my x x= - + (Cm).)i M l Km thuc th (Cm) c honh

    bng -1. Tm m tip tuyn vi (Cm) ti M song song vi ng thng d: 5x-y=0Gii

    Ta c ' 2y x mx= -ng thng d: 5x-y=0 c h s gc bng 5, nn tip tuyn ti M song song vi ngthng d trc ht ta cn c '( 1) 5 1 5 4y m m- = + = =

    Khi 4m = ta c hm s 3 21 123 3

    y x x= - + ta c0

    1x = - th0

    2y = -

    Phng trnh tip tuyn c dng '0 0 0

    ( )( ) 5( 1) 2 5 3y y x x x y y x y x= - + = + - = +

    R rng tip tuyn song song vi ng thng d8y 4m = l gi tr cn tm.V d 8: Cho hm s 3 23y x x m= - + (1).Tm m tip tuyn ca th (1) ti Km c honh bng 1 ct cc trc Ox, Oy ln

    Nt ti cc Km A v B sao cho din tch tam gic OAB bng 32

    .

    Gii8i

    0 01 2x y m= = - M(1 ; m 2)

    - Tip tuyn ti M l d: 20 0 0

    (3 6 )( ) 2y x x x x m= - - + -

    d: y = -3x + m + 2.

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  • GD&0Ninh0ngnthiTHPTqugia2014-2015

    - d ct trc Ox ti A: 2 20 3 2 ; 03 3A A

    m mx m x A

    + + = - + + =

    - d ct trc Oy ti B: 2 (0 ; 2)B

    y m B m= + +

    - 23 1 3 2| || | | || | 3 2 3 ( 2) 92 2 2 3OAB

    mS OA OB OA OB m m

    += = = + = + =

    2 3 1

    2 3 5

    m m

    m m

    + = = + = - = - 8y m = 1 v m = - 5

    1.2. Dng 2: Vit tip tuyn ca thi hm s ( )y f x= (C) khi bit trc h s gc ca n

    + Gi0 0

    ( , )M x y l tip Km, gii phng trnh '0 0

    ( )f x k x x= = ,0 0

    ( )y f x=

    + n y tr v Fng 1,ta d dng lp c tip tuyn ca th:

    0 0( )y k x x y= - +

    Cc dng biu din h s gc k:

    *) Cho trc tip: 35; 1; 3; ...7

    k k k k= = = =

    *) Tip tuyn to vi chiu dng ca trc Ox mt gc a , vi 0 0 0 215 ;30 ;45 ; ; .... .3 3p p

    a

    Khi h s gc k = tan a .*) Tip tuyn song song vi ng thng (d): y = ax + b. Khi h s gc k = a.

    *) Tip tuyn vung gc vi ng thng (d): y = ax + b 11ka ka

    - = - = .

    *) Tip tuyn to vi ng thng (d): y = ax + b mt gc a . Khi : tan1k a

    kaa

    -=

    +.

    V d 9: Cho hm s 3 23y x x= - (C). Vit phng trnh tip tuyn ca th (C) bit hU gc ca tip tuyn k = -3.Gii:

    Ta c: 2' 3 6y x x= -

    )i0 0

    ( ; )M x y l tip im Tip tuyn ti M c h s gc ' 20 0 0

    ( ) 3 6k f x x x= = -

    Theo gi thit, h s gc ca tip tuyn k = - 3 nn:2 20 0 0 0 0

    3 6 3 2 1 0 1x x x x x- = - - + = =

    V0 0

    1 2 (1; 2)x y M= = - - .

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  • GD&0Ninh0ngnthiTHPTqugia2014-2015

    Phng trinh tip tuyn cn tm l 3( 1) 2 3 1y x y x= - - - = - +

    V d 10: Vit phng trnh tip tuyn ca th hm s 3 23 1y x x= - + (C). Bit tiptuyn song song vi ng thng y = 9x + 6.Gii:

    Ta c: 2' 3 6y x x= -

    )i0 0

    ( ; )M x y l tip Km Tip tuyn ti M c h s gc ' 20 0 0

    ( ) 3 6k f x x x= = -

    Theo gi thit, tip tuyn song song vi ng thng y = 9x + +6 tip tuyn c h

    U gc k = 9 02 20 0 0 0

    0

    1 ( 1; 3)3 6 9 2 3 0

    3 (3;1)

    x Mx x x x

    x M

    = - - -- = - - = = Phng trinh tip tuyn ca (C) ti M(-1;-3) l: 9( 1) 3 9 6y x y x= + - = + (loi)Phng trinh tip tuyn ca (C) ti M(3;1) l: 9( 3) 1 9 26y x y x= - + = -

    V d 11: Cho hm s 3 3 2y x x= - + (C). Vit phng trnh tip tuyn ca (C) bit tip

    tuyn vung gc vi ng thng 19

    y x-

    = .

    Gii:Ta c 2' 3 3y x= - . Do tip tuyn ca (C) bit tip tuyn vung gc vi ng

    thng 19

    y x-

    = nn h s gc ca tip tuyn k = 9.

    Do 2 2' 3 3 9 4 2.y k x x x= - = = = +) Vi x = 2 4y = . Pttt ti Km c honh x = 2 l:

    9( 2) 4 9 14.y x y x= - + = -

    +) Vi 2 0x y= - = . Pttt ti Km c honh x = - 2 l:9( 2) 0 9 18y x y x= + + = + .

    Vy c hai tip tuyn c (C) vung gc vi ng thng 19

    y x-

    = l:

    y =9x - 14 v y = 9x + 18.

    V d 12: Lp phng trnh tip tuyn vi th (C) ca hm s: 4 21 24

    y x x= + ,

    bit tip tuyn vung gc vi ng thng (d): 5 2010 0x y+ - = .

    Gii:

    (d) c phng trnh: 1 4025

    y x= - + nn (d) c h s gc l - 15

    .

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  • GD&0Ninh0ngnthiTHPTqugia2014-2015

    )i D l tip tuyn cn tm c h s gc k th 1 . 1 5 ( ( ))5