Khoi A-A1 2013

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  • BO GIAO DUC VA AO TAO E THI TUYEN SINH AI HOC NAM 2013 Mon: TOAN; Khoi A va khoi A1

    E CHNH THC Thi gian lam bai: 180 phut, khong ke thi gian phat e

    I. PHAN CHUNG CHO TAT CA TH SINH (7,0 iem)

    Cau 1 (2,0 iem). Cho ham so y = x3 + 3x2 + 3mx 1 (1), vi m la tham so thc.a) Khao sat s bien thien va ve o th cua ham so (1) khi m = 0.

    b) Tm m e ham so (1) nghch bien tren khoang (0; +).

    Cau 2 (1,0 iem). Giai phng trnh 1 + tanx = 22 sin

    (x+

    pi

    4

    ).

    Cau 3 (1,0 iem). Giai he phng trnh

    { x+ 1 + 4

    x 1

    y4 + 2 = y

    x2 + 2x(y 1) + y2 6y + 1 = 0(x, y R).

    Cau 4 (1,0 iem). Tnh tch phan I =2

    1

    x2 1x2

    lnx dx.

    Cau 5 (1,0 iem). Cho hnh chop S.ABC co ay la tam giac vuong tai A, ABC = 30, SBC latam giac eu canh a va mat ben SBC vuong goc vi ay. Tnh theo a the tch cua khoi chopS.ABC va khoang cach t iem C en mat phang (SAB).

    Cau 6 (1,0 iem). Cho cac so thc dng a, b, c thoa man ieu kien (a + c)(b+ c) = 4c2. Tm gia tr

    nho nhat cua bieu thc P =32a3

    (b+ 3c)3+

    32b3

    (a+ 3c)3a2 + b2

    c.

    II. PHAN RIENG (3,0 iem): Th sinh ch c lam mot trong hai phan (phan A hoac phan B)

    A. Theo chng trnh Chuan

    Cau 7.a (1,0 iem). Trong mat phang vi he toa o Oxy, cho hnh ch nhat ABCD co iem C thuocng thang d : 2x+ y + 5 = 0 va A(4; 8). Goi M la iem oi xng cua B qua C , N la hnh chieuvuong goc cua B tren ng thang MD. Tm toa o cac iem B va C , biet rang N(5;4).

    Cau 8.a (1,0 iem). Trong khong gian vi he toa o Oxyz, cho ng thang :x 63 =

    y + 1

    2 =z + 2

    1va iem A(1; 7; 3). Viet phng trnh mat phang (P ) i qua A va vuong goc vi . Tm toa o iemM thuoc sao cho AM = 2

    30.

    Cau 9.a (1,0 iem). Goi S la tap hp tat ca cac so t nhien gom ba ch so phan biet c chon tcac ch so 1; 2; 3; 4; 5; 6; 7. Xac nh so phan t cua S. Chon ngau nhien mot so t S, tnh xac suate so c chon la so chan.

    B. Theo chng trnh Nang cao

    Cau 7.b (1,0 iem). Trong mat phang vi he toa o Oxy, cho ng thang : x y = 0. ngtron (C) co ban knh R =

    10 cat tai hai iem A va B sao cho AB = 4

    2. Tiep tuyen cua (C)

    tai A va B cat nhau tai mot iem thuoc tia Oy. Viet phng trnh ng tron (C).

    Cau 8.b (1,0 iem). Trong khong gian vi he toa o Oxyz, cho mat phang (P ) : 2x+ 3y+ z 11 = 0va mat cau (S) : x2 + y2 + z2 2x+ 4y 2z 8 = 0. Chng minh (P ) tiep xuc vi (S). Tm toa otiep iem cua (P ) va (S).

    Cau 9.b (1,0 iem). Cho so phc z = 1+3 i. Viet dang lng giac cua z. Tm phan thc va phan ao

    cua so phc w = (1 + i)z5.Het

    Th sinh khong c s dung tai lieu. Can bo coi thi khong giai thch g them.

    Ho va ten th sinh: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ; So bao danh: . . . . . . . . . . . . . . . . . . . . .

  • B GIO DC V O TO CHNH THC

    P N THANG IM THI TUYN SINH I HC NM 2013

    Mn: TON; Khi A v khi A1 (p n - thang im gm 04 trang)

    Cu p n ima. (1,0 im) Khi m = 0 ta c 3 23 1y x x .= + Tp xc nh: .D = \ S bin thin:

    - Chiu bin thin: hoc 2' 3 6 ; ' 0y x x y x= + = = 0 2.x = 0,25

    Khong ng bin: (0; 2); cc khong nghch bin: ( ; 0) v (2; ).+ - Cc tr: Hm s t cc tiu ti x = 0, yCT = 1; t cc i ti x = 2, yC = 3. - Gii hn: lim ; lim .

    x xy y += + =

    0,25

    - Bng bin thin:

    Trang 1/4

    0,25

    th:

    0,25

    b. (1,0 im)

    Ta c 2' 3 6 3y x x= + + .mHm s (1) nghch bin trn khong (0; )+ khi v ch khi ' 0, 0y x > 0,25

    2 2 , 0.m x x x > Xt 2( ) 2f x x x= vi Ta c 0.x > '( ) 2 2; '( ) 0 1.f x x f x x= = =

    0,25

    Bng bin thin:

    0,25

    1 (2,0 im)

    Da vo bng bin thin ta c gi tr m tha mn yu cu ca bi ton l m 1.

    x 'y

    y

    + 0 2 0 0 +

    +

    1

    3

    2 O

    y

    x

    3

    1

    x

    ( )f x

    0 + 1 0

    0 +

    1

    + '( )f x

    0,25

  • Trang 2/4

    Cu p n imiu kin: Phng trnh cho tng ng vi cos 0.x sin1 2(sin co

    cosx s )x xx

    + = + 0,25 (sin cos )(2cos 1) 0.x x x + = 0,25

    sin cos 0 ( )4

    x x x k k + = = + ] . 0,25

    2 (1,0 im)

    2cos 1 0 2 ( )3

    x x k k = = + ] .

    i chiu iu kin ta c nghim: 4

    x k= + hoc 2 ( )3

    x k k= + ] . 0,25

    44

    2 2

    1 1 2

    2 ( 1) 6 1 0 (2)

    x x y y

    x x y y y

    + + + = + + + =(1)

    ,

    iu kin: T (2) ta c suy ra 1.x 24 ( 1)y x y= + 0.y 0,25

    3 (1,0 im)

    t 4 1,u x= suy ra u Phng trnh (1) tr thnh: 0. 4 42 2 (3).u u y y+ + = + + Xt 4( ) 2 ,f t t= + + t vi Ta c 0.t

    3

    4

    2'( ) 1 0, 0.2

    tf t tt

    = + > +

    Do phng trnh (3) tng ng vi ,y u= ngha l 4 1.x y= +

    0,25

    Thay vo phng trnh (2) ta c 7 4( 2 4) 0 (4).y y y y+ + =Hm c 7 4( ) 2 4g y y y y= + + 6 3'( ) 7 8 1 0g y y y= + + > vi mi 0.y 0,25

    M nn (4) c hai nghim khng m l (1) 0,g = 0y = v 1.y = Vi ta c nghim ( ; vi 0y = ) (1; 0);x y = 1y = ta c nghim ( ; ) (2; 1).x y = Vy nghim ( ; )x y ca h cho l v (1; 0) (2; 1).

    0,25

    t 2

    21 dln , d d d , .x xu x v x u v x 1

    x xx= = = = + 0,25

    Ta c 22

    1 1

    1 1ln dI x x x 1 xx x x

    = + + 0,25

    2 2

    1 1

    1 1lnx x xx x

    = + 0,25

    4 (1,0 im)

    5 3ln 2 .2 2

    = 0,25 Gi H l trung im ca BC, suy ra SH BC. M (SBC) vung gc vi (ABC) theo giao tuyn BC, nn SH (ABC). 0,25

    Ta c BC = a, suy ra 3 ;2

    aSH = osin30 ;2aAC BC= =

    o 3cos30 .2

    aAB BC= =

    Do 3

    .1 . .6 1S ABC

    a .6

    H AB AC= =V S

    0,25

    Tam gic ABC vung ti A v H l trung im ca BC nn HA = HB. M SH (ABC), suy ra SA = SB = a. Gi I l trung im ca AB, suy ra SI AB.

    0,25

    5 (1,0 im)

    Do 2

    2 13 .4 4

    AB aSI SB= =

    Suy ra . .3 6 39( ,( )) .

    . 1S ABC S ABC

    SAB

    V V ad C SABS SI AB

    = = =3

    0,25

    S

    A B

    C

    I

    H

  • Trang 3/4

    Cu p n imt ,ax y

    c c= = .b Ta c iu kin ca bi ton tr thnh 0, 0.x y> > 3.xy x y+ + =

    Khi 33 2 2

    3 33232 .

    ( 3) ( 3)yxP x

    y x= + ++ + y

    v> >

    Vi mi u ta c 0, 03

    3 3 3 3 3 ( )3( ) 3 ( ) ( ) ( )4 4u v .v u v uv u v u v u v ++ = + + + + =u

    Do 333 23

    3 332 ( ) 2 3 332 8 83 3 3 3 9( 3) ( 3)

    y y x y xy xx xy x xy x yy x

    + + + + + = + + + + + + + .y

    0,25

    Thay 3xy x= y vo biu thc trn ta c 333 3

    3 332 ( 1)( 6)32 8 (2( 6)( 3) ( 3)

    y x y x yx x yx yy x+ + + + = + + + + + 1) . Do

    3 2 2 3 2 3 2( 1) ( 1) ( ) 2 ( 1) ( ) 2( ) 6.P x y x y x y x y xy x y x y x y + + = + + = + + + +

    0,25

    t t x Suy ra t v .y= + > 0 3 2( 1) 2 6.P t t t + Ta c

    2 2( )3 ( )4 4

    x y tx y xy x y t+= + + + + = + . nn ( 2)( 6) 0t t + Do 2.t

    Xt 3 2( ) ( 1) 2 6,f t t t t= + vi t Ta c 2. 22

    1'( ) 3( 1) .2 6

    tf t tt t

    += +

    Vi mi t ta c v 2 23( 1) 3t 221 7 71 1 2 2( 1) 72 6

    ttt t

    + = + + =+ + 3 2 , nn

    3 2'( ) 3 0.2

    f t > Suy ra ( ) (2) 1 2.f t f = Do 1 2P .

    0,25

    6 (1,0 im)

    Khi a th b c= = 1 2P = . Do gi tr nh nht ca P l 1 2 . 0,25 Do C d nn ( ; 2 5).C t t Gi I l tm ca hnh ch nht ABCD, suy ra I l trung im ca AC.

    Do ( )4 2 3; .2 2t tI + 0,25 Tam gic BDN vung ti N nn IN = IB. Suy ra IN = IA. Do ta c phng trnh

    ( ) ( )2 22 24 22 3 45 4 4 82 2 2t tt t + + = +

    7.a (1,0 im)

    32+

    1.t = Suy ra C(1; 7).

    0,25

    Do M i xng vi B qua C nn CM = CB. M CB = AD v CM||AD nn t gic ACMD l hnh bnh hnh. Suy ra AC||DM. Theo gi thit, BN DM, suy ra BN AC v CB = CN. Vy B l im i xng ca N qua AC.

    0,25

    ng thng AC c phng trnh: 3 4 0.

    .

    x y+ + =ng thng BN qua N v vung gc vi AC nn c phng trnh 3 17 0x y = Do (3 17; ).B a a+ Trung im ca BN thuc AC nn

    3 17 5 43 4 0 7.2 2

    a a a+ + + + = = ( 4; 7).B Vy 0,25

    c vct ch phng l ( 3; 2;1).u = JG 0,25 (P) qua A v nhn u

    JG lm vct php tuyn, nn (P) c phng trnh

    3( 1) 2( 7) ( 3) 0 3 2 14 0.x y z x y z + = + = 0,25 M thuc nn (6 3 ; 1 2 ; 2 ).M t t + t 0,25

    8.a (1,0 im)

    2 2 2 22 30 (6 3 1) ( 1 2 7) ( 2 3) 120 7 4 3 0AM t t t t t= + + + = = 1t = hoc 3 .7t Suy ra M= (3; 3; 1) hoc ( )51 1 17; ;7 7 7M . 0,25

    A D

    B C M

    N I

  • Trang 4/4

    Cu p n imS phn t ca S l 37A 0,25 = 210. 0,25 S cch chn mt s chn t S l 3.6.5 90= (cch). 0,25

    9.a (1,0 im)

    Xc sut cn tnh bng 90 3 .210 7

    = 0,25 Gi M l giao im ca tip tuyn ti A v B ca (C), H l giao im ca AB v IM. Khi (0; ),M t vi H l trung im

    ca AB. Suy ra

    0;t 2 2.

    2ABAH = =

    0,25

    2 21 1 1 ,

    AH AM AI= + 2 suy ra 2 10.AM =

    Do 2 2 4 2.MH AM AH= = M | |( , ) ,

    2tMH d M= = nn 8.t = Do (0; 8).M

    0,25

    ng thng IM qua M v vung gc vi nn c phng trnh 8 0.x y+ = Do ta im H tha mn h

    .0

    (4;4)8 0

    x yH

    x y = + =

    0,25

    7.b (1,0 im)

    A

    I

    B

    H

    M

    Ta c 2 2 12 ,4

    IH IA AH HM= = = nn 1 .4

    IH HM=JJJG JJJJG Do (5;3).I

    Vy ng trn (C) c phng trnh 2 2( 5) ( 3) 10x y + = .0,25

    (S) c tm v bn knh (1; 2;1)I 14.R = 0,25

    2 2 2

    | 2.1 3( 2) 1.1 11| 14( ,( )) .142 3 1

    d I P R+ + =+ +

    = = Do (P) tip xc vi (S). 0,25

    8.b (1,0 im)

    Gi M l tip im ca (P) v (S). Suy ra M thuc ng thng qua I v vung gc vi (P). 0,25 (1 2 ; 2 3 ;1 ).M t t+ + + t Do

    Do M thuc (P) nn Vy 2(1 2 ) 3( 2 3 ) (1 )