DE THI DH TOAN 2012-2014.pdf

BO GIAO DUC VA AO TAO E THI TUYE4N SINH AI HOC NA M 2014 Mon: TOAN; Khoi A va Khoi A1

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

Cau 1 (2,0 iem). Cho ham so y = x+ 2x 1 (1).

a) Khao sat s bien thien va ve o th (C) cua ham so (1).

b) Tm toa o iem M thuoc (C) sao cho khoang cach t M en ng thang y = x bang2.

Cau 2 (1,0 iem). Giai phng trnh sin x+ 4cos x = 2 + sin 2x.

Cau 3 (1,0 iem). Tnh dien tch hnh phang gii han bi ng cong y = x2 x+3 va ngthang y = 2x+ 1.

Cau 4 (1,0 iem).

a) Cho so phc z thoa man ieu kien z + (2 + i) z = 3 + 5i. Tm phan thc va phan ao cua z.

b) T mot hop cha 16 the c anh so t 1 en 16, chon ngau nhien 4 the. Tnh xac suate 4 the c chon eu c anh so chan.

Cau 5 (1,0 iem). Trong khong gian vi he toa o Oxyz, cho mat phang (P ) : 2x+y2z1 = 0va ng thang d :

x 21

=y

2 =z + 3

3. Tm toa o giao iem cua d va (P ). Viet phng

trnh mat phang cha d va vuong goc vi (P ).

Cau 6 (1,0 iem). Cho hnh chop S.ABCD co ay ABCD la hnh vuong canh a, SD =3a

2,

hnh chieu vuong goc cua S tren mat phang (ABCD) la trung iem cua canh AB. Tnh theo athe tch khoi chop S.ABCD va khoang cach t A en mat phang (SBD).

Cau 7 (1,0 iem). Trong mat phang vi he toa o Oxy, cho hnh vuong ABCD co iem Mla trung iem cua oan AB va N la iem thuoc oan AC sao cho AN = 3NC . Viet phngtrnh ng thang CD, biet rang M(1; 2) va N(2;1).

Cau 8 (1,0 iem). Giai he phng trnh

{x12 y +

y(12 x2) = 12

x3 8x 1 = 2y 2(x, y R).

Cau 9 (1,0 iem). Cho x, y, z la cac so thc khong am va thoa man ieu kien x2+ y2 + z2 = 2.Tm gia tr ln nhat cua bieu thc

P =x2

x2 + yz + x+ 1+

y + z

x+ y + z + 1 1 + yz

9.

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: . . . . . . . . . . . . . . . . . . .

BO GIAO DUC VA AO TAO AP AN - THANG IE5M E THI TUYE5N SINH AI HOC NA M 2014

E CHNH THC Mon: TOAN; Khoi A va Khoi A1(ap an - Thang iem gom 03 trang)

Cau ap an iem

1 a) (1,0 iem)(2,0) Tap xac nh D = R \ {1}.

S bien thien:- Chieu bien thien: y = 3

(x 1)2 ; y < 0, x D.

Ham so nghch bien tren tng khoang (; 1) va (1; +).0,25

- Gii han va tiem can: limx

y = limx+

y = 1; tiem can ngang: y = 1.limx1

y = ; limx1+

y = +; tiem can ng: x = 1. 0,25

- Bang bien thien:x 1 +y

y1 +

1

!!!!!!!"

!!!!!!!"

0,25

o th:y

xO2

2

1

1

0,25

b) (1,0 iem)

M (C)M(a;a+ 2

a 1), a = 1. 0,25

Khoang cach t M en ng thang y = x la d =

a+ a+ 2a 1

2

. 0,25

d =2 |a2 + 2| = 2|a 1|

[ a2 2a+ 4 = 0a2 + 2a = 0.

0,25

a2 2a+ 4 = 0: phng trnh vo nghiem. a2 + 2a = 0

[ a = 0a = 2. Suy ra toa o iem M can tm la: M(0;2) hoac M(2; 0).

0,25

1

Cau ap an iem2 Phng trnh a cho tng ng vi sinx+ 4 cosx = 2 + 2 sinx cosx 0,25

(1,0) (sinx 2)(2 cosx 1) = 0. 0,25 sinx 2 = 0: phng trnh vo nghiem. 0,25 2 cosx 1 = 0 x =

3+ k2 (k Z).

Nghiem cua phng trnh a cho la: x = 3+ k2 (k Z).

0,25

3(1,0)

Phng trnh hoanh o giao iem cua ng cong y = x2 x+ 3 va ng thangy = 2x+ 1 la x2 x+ 3 = 2x+ 1

[ x = 1x = 2.

0,25

Dien tch hnh phang can tm la S =2

1

|x2 3x+ 2|dx 0,25

=

21

(x2 3x+ 2)dx =

(x33 3x

2

2+ 2x

)21

0,25=

1

6. 0,25

4(1,0)

a) at z = a+ bi (a, b R). T gia thiet suy ra{

3a+ b = 3a b = 5 0,25

a = 2, b = 3. Do o so phc z co phan thc bang 2, phan ao bang 3. 0,25

b) So phan t cua khong gian mau la: C416 = 1820. 0,25

So ket qua thuan li cho bien co 4 the c anh so chan la: C 48 = 70.

Xac suat can tnh la p =70

1820=

1

26.

0,25

5 Goi M la giao iem cua d va (P ), suy ra M(2 + t;2t;3 + 3t). 0,25

(1,0) M (P ) suy ra 2(2 + t) + (2t) 2(3 + 3t) 1 = 0 t = 32. Do o M

(72;3; 3

2

). 0,25

d co vect ch phng u = (1;2; 3), (P ) co vect phap tuyen n = (2; 1;2).Mat phang () can viet phng trnh co vect phap tuyen [u ,n ] = (1; 8; 5).

0,25

Ta co A(2; 0;3) d nen A (). Do o () : (x 2) + 8(y 0) + 5(z + 3) = 0,ngha la () : x + 8y + 5z + 13 = 0.

0,25

6(1,0)

GoiH la trung iem cua AB, suy ra SH (ABCD).Do o SH HD. Ta co SH = SD2 DH2=SD2 (AH2 +AD2) = a.

0,25

Suy ra VS.ABCD =1

3.SH.SABCD =

a3

3. 0,25

Goi K la hnh chieu vuong goc cua H tren BD vaE la hnh chieu vuong goc cua H tren SK . Ta coBD HK va BD SH , nen BD (SHK).Suy ra BD HE . Ma HE SK ,do o HE (SBD).

0,25

Ta co HK = HB. sin KBH =a2

4.

Suy ra HE =HS.HKHS2 +HK2

=a

3. 0,25

A

BC

D

H

S

K

E

Do o d(A, (SBD)) = 2d(H, (SBD)) = 2HE =2a

3.

2

Cau ap an iem7

(1,0)Ta co MN =

10. Goi a la o dai canh cua hnh vuong ABCD,

a > 0. Ta co AM =a

2va AN =

3AC

4=

3a2

4,

nen MN 2 = AM2 + AN 2 2AM.AN. cosMAN = 5a2

8.

Do o5a2

8= 10, ngha la a = 4.

0,25

Goi I(x; y) la trung iem cua CD. Ta co IM = AD = 4A B

CD

M

N

I

va IN =BD

4=2, nen ta co he phng trnh 0,25{

(x 1)2 + (y 2)2 = 16(x 2)2 + (y + 1)2 = 2

[ x = 1; y = 2x =

17

5; y = 6

5.

Vi x = 1; y = 2 ta co I(1;2) va IM = (0; 4).ng thang CD i qua I va co vect phap tuyen la

IM , nen co phng trnh y + 2 = 0. 0,25

Vi x = 175; y = 6

5ta co I

(175; 6

5

)vaIM =

( 12

5;16

5

).

ng thang CD i qua I va co vect phap tuyen laIM , nen co phng trnh 3x4y15 = 0.

0,25

8(1,0)

{x12 y +y(12 x2) = 12 (1)

x3 8x 1 = 2y 2 (2). ieu kien: 23 x 23; 2 y 12.

Ta co x12 y x

2 + 12 y2

vay(12 x2) y + 12 x

2

2

nen x12 y +y(12 x2) 12. Do o (1) { x 0

y = 12 x2.

0,25

Thay vao (2) ta c x3 8x 1 = 210 x2 x3 8x 3 + 2(110 x2) = 0 (x 3)

(x2 + 3x+ 1 +

2(x+ 3)

1 +10 x2

)= 0 (3). 0,25

Do x 0 nen x2 + 3x+ 1 + 2(x+ 3)1 +

10 x2 > 0. 0,25

Do o (3) x = 3. Thay vao he va oi chieu ieu kien ta c nghiem: (x; y) = (3; 3). 0,259

(1,0)Ta co 0 (x y z)2 = x2 + y2 + z2 2xy 2xz + 2yz = 2(1 xy xz + yz),nen x2 + yz + x+ 1 = x(x+ y + z + 1) + (1 xy xz + yz) x(x+ y + z + 1).Suy ra

x2

x2 + yz + x+ 1 xx+ y + z + 1

.

0,25

Mac khac, (x+ y + z)2 = x2 + y2 + z2 + 2x(y + z) + 2yz = 2 + 2yz + 2x(y + z)

2 + 2yz + [x2 + (y + z)2] = 4(1 + yz). Do o P x+ y + zx+ y + z + 1

(x+ y + z)2

36.

0,25

at t = x+ y + z, suy ra t 0 va t2 = (x+ y + z)2 = (x2 + y2 + z2) + 2xy + 2yz + 2zx 2 + (x2 + y2) + (y2 + z2) + (z2 + x2) = 6. Do o 0 t 6.Xet f(t) =

t

t+ 1 t

2

36, vi 0 t 6.

Ta co f (t) =1

(t+ 1)2 t

18= (t 2)(t

2 + 4t+ 9)

18(t+ 1)2, nen f (t) = 0 t = 2.

0,25

Ta co f(0) = 0; f(2) =5

9va f(

6) =

31

306

5, nen f(t) 5

9khi 0 t 6.

Do o P 59. Khi x = y = 1 va z = 0 th P =

5

9. Do o gia tr ln nhat cua P la

5

9.

0,25

Het

3

BO GIAO DUC VA AO TAO E THI TUYE4N SINH AI HOC NA M 2014 Mon: TOAN; Khoi B


Cau 1 (2,0 iem). Cho ham so y = x3 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 = 1.

b) Cho iem A(2; 3). Tm m e o th ham so (1) co hai iem cc tr B va C sao chotam giac ABC can tai A.

Cau 2 (1,0 iem). Giai phng trnh2(sin x 2 cos x) = 2 sin 2x.

Cau 3 (1,0 iem). Tnh tch phan I =2

1

x2 + 3x+ 1

x2 + xdx.

Cau 4 (1,0 iem).a) Cho so phc z thoa man ieu kien 2z + 3(1 i) z = 1 9i. Tnh moun cua z.

b) e kiem tra chat lng san pham t mot cong ty sa, ngi ta a gi en bo phankiem nghiem 5 hop sa cam, 4 hop sa dau va 3 hop sa nho. Bo phan kiem nghiemchon ngau nhien 3 hop sa e phan tch mau. Tnh xac suat e 3 hop sa c chonco ca 3 loai.

Cau 5 (1,0 iem). Trong khong gian vi he toa o Oxyz, cho iem A(1; 0;1) va ngthang d :

x 12

=y + 1

2=

z

1 . Viet phng trnh mat phang qua A va vuong goc vi d.Tm toa o hnh chieu vuong goc cua A tren d.Cau 6 (1,0 iem). Cho lang tru ABC.ABC co ay la tam giac eu canh a. Hnh chieuvuong goc cua A tren mat phang (ABC) la trung iem cua canh AB, goc gia ngthang AC va mat ay bang 60. Tnh theo a the tch cua khoi lang tru ABC.A BC vakhoang cach t iem B en mat phang (ACC A).Cau 7 (1,0 iem). Trong mat phang vi he toa o Oxy, cho hnh bnh hanh ABCD. iemM(3; 0) la trung iem cua canh AB, iem H(0;1) la hnh chieu vuong goc cua B trenAD va iem G

(43; 3)

la trong tam cua tam giac BCD. Tm toa o cac iem B va D.

Cau 8 (1,0 iem). Giai he phng trnh{(1 y)x y + x = 2 + (x y 1)y2y2 3x + 6y + 1 = 2x 2y 4x 5y 3

(x, y R).

Cau 9 (1,0 iem). Cho cac so thc a, b, c khong am va thoa man ieu kien (a+ b)c > 0.Tm gia tr nho nhat cua bieu thc

P =

a

b+ c+

b

a+ c+

c

2(a+ b).

HetTh sinh khong c s dung tai lieu. Can bo coi thi khong giai thch g them.Ho va ten th sinh: . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . ; So bao danh: . . . . . . . . . . . . . . . . . .


E CHNH THC Mon: TOAN; Khoi B(ap an - Thang iem gom 03 trang)

Cau ap an iem

1 a) (1,0 iem)(2,0) Vi m = 1, ham so tr thanh: y = x3 3x+ 1.

Tap xac nh: D = R. S bien thien:

- Chieu bien thien: y = 3x2 3; y = 0 x = 1.0,25

Cac khoang ong bien: (;1) va (1; +); khoang nghch bien: (1; 1).- Cc tr: Ham so at cc ai tai x = 1, yC = 3; at cc tieu tai x = 1, yCT = 1.- Gii han tai vo cc: lim

xy = ; lim

x+y = +.

0,25

- Bang bien thien:x 1 1 +y + 0 0 +

y3 +

1 #

#####$

0,25

o th:

x

y

3

11

1

O

1

0,25

b) (1,0 iem)

Ta co y = 3x2 3m.o th ham so (1) co hai iem cc tr phng trnh y = 0 co hai nghiem phan biet m > 0. 0,25

Toa o cac iem cc tr B,C la B(m; 2m3 + 1), C(m;2m3 + 1).Suy ra

BC = (2

m;4m3). 0,25

Goi I la trung iem cua BC, suy ra I(0; 1). Ta co tam giac ABC can tai A AI.BC = 0 0,25

4m+ 8m3 = 0 m = 0 hoac m = 12.

oi chieu ieu kien ton tai cc tr, ta c gia tr m can tm la m =1

2.

0,25

1

Cau ap an iem

2 Phng trnh a cho tng ng vi 2 sinx cosx 22 cos x+

2 sinx 2 = 0. 0,25

(1,0) (sinx 2)(2 cosx+2) = 0. 0,25 sinx2 = 0: phng trnh vo nghiem. 0,25

2 cosx+2 = 0 x = 34

+ k2 (k Z).Nghiem cua phng trnh a cho la: x = 3

4+ k2 (k Z).

0,25

3(1,0)

Ta co I =2

1

x2 + 3x+ 1

x2 + xdx =

21

dx+

21

2x+ 1

x2 + xdx. 0,25

2

1

dx = 1. 0,25

2

1

2x+ 1

x2 + xdx = ln |x2 + x|

21

0,25

= ln 3. Do o I = 1 + ln3. 0,25

4(1,0)

a) at z = a+ bi (a, b R). T gia thiet suy ra{

5a 3b = 13a+ b = 9

0,25

a = 2, b = 3. Do o moun cua z bang 13. 0,25

b) So phan t cua khong gian mau la: C312 = 220. 0,25

So cach chon 3 hop sa co u 3 loai la 5.4.3 = 60. Do o xac suat can tnh la p =60

220=

3

11. 0,25

5 Vect ch phng cua d la u = (2; 2;1). 0,25(1,0) Mat phang (P ) can viet phng trnh la mat phang qua A va nhan u lam vect phap tuyen,

nen (P ) : 2(x 1) + 2(y 0) (z + 1) = 0, ngha la (P ) : 2x+ 2y z 3 = 0. 0,25

Goi H la hnh chieu vuong goc cua A tren d, suy ra H(1 + 2t;1 + 2t;t). 0,25

Ta co H (P ), suy ra 2(1+2t)+2(1+2t)(t)3 = 0 t = 13. Do o H

(53;1

3;1

3

). 0,25

6(1,0)

Goi H la trung iem cua AB, suy ra A H (ABC)va ACH = 60. Do o AH = CH. tan ACH =

3a

2.

0,25

The tch khoi lang tru la VABC.ABC = AH.SABC =33 a3

8. 0,25

Goi I la hnh chieu vuong goc cua H tren AC; K la hnh chieuvuong goc cua H tren AI . Suy ra HK = d(H, (ACC A)).

0,25

Ta co HI = AH. sin IAH =3 a

4,

1

HK2=

1

HI2+

1

HA2=

52

9a2, suy ra HK =

313 a

26. 0,25A

B

A

H

C

BC

I

K

Do o d(B, (ACCA)) = 2d(H, (ACCA)) = 2HK =313 a

13.

2

Cau ap an iem

7(1,0)

Goi E va F lan lt la giao iem cua HM va HGvi BC. Suy ra

HM =

ME va

HG = 2

GF ,

Do o E(6; 1) va F (2; 5).0,25

A

B C

DH

M IG

E F

ng thang BC i qua E va nhanEF lam vect

ch phng, nen BC :x 2y + 8 = 0. ng thangBH i qua H va nhan

EF lam vect phap tuyen, nen

BH : 2x + y + 1 = 0. Toa o iem B thoa man he

phng trnh{x 2y + 8 = 02x+ y + 1 = 0.

Suy ra B(2; 3).

0,25

Do M la trung iem cua AB nen A(4;3).Goi I la giao iem cua AC va BD, suy ra

GA = 4

GI. Do o I

(0;

3

2

).

0,25

Do I la trung iem cua oan BD, nen D(2; 0). 0,25

8(1,0)

{(1 y)x y + x = 2 + (x y 1)y (1)2y2 3x+ 6y + 1 = 2x 2y 4x 5y 3 (2).

ieu kien:

y 0x 2y4x 5y + 3

().

Ta co (1) (1 y)(x y 1) + (x y 1)(1y) = 0

(1 y)(x y 1)( 1

x y + 1 +1

1 +y

)= 0 (3).

0,25

Do1

x y + 1 +1

1 +y> 0 nen (3)

[ y = 1y = x 1.

Vi y = 1, phng trnh (2) tr thanh 9 3x = 0 x = 3.0,25

Vi y = x 1, ieu kien () tr thanh 1 x 2. Phng trnh (2) tr thanh2x2 x 3 = 2 x 2(x2 x 1) + (x 12 x) = 0 (x2 x 1)

[2 +

1

x 1 +2 x]= 0

0,25

x2x1 = 0 x = 15

2. oi chieu ieu kien () va ket hp trng hp tren, ta c

nghiem (x; y) cua he a cho la (3; 1) va(1 +5

2;1 +5

2

).

0,25

9(1,0)

Ta co a+ b+ c 2a(b+ c). Suy ra ab+ c

2aa + b+ c

. 0,25

Tng t,

b

a+ c 2ba+ b+ c

.

Do o P 2(a+ b)a+ b+ c

+c

2(a+ b)=[ 2(a+ b)a+ b+ c

+a+ b+ c

2(a+ b)

] 1

2

0,25

2 12=

3

2. 0,25

Khi a = 0, b = c, b > 0 th P =3

2. Do o gia tr nho nhat cua P la

3

2. 0,25

Het

3

BO GIAO DUC VA AO TAO E THI TUYE4N SINH AI HOC NA M 2014 Mon: TOAN; Khoi D


Cau 1 (2,0 iem). Cho ham so y = x3 3x 2 (1).a) Khao sat s bien thien va ve o th (C) cua ham so (1).

b) Tm toa o iem M thuoc (C) sao cho tiep tuyen cua (C) tai M co he so goc bang 9.

Cau 2 (1,0 iem). Cho so phc z thoa man ieu kien (3z z)(1 + i) 5z = 8i 1.Tnh moun cua z.

Cau 3 (1,0 iem). Tnh tch phan I =

40

(x+ 1) sin 2x dx.

Cau 4 (1,0 iem).

a) Giai phng trnh log 2(x 1) 2 log4(3x 2) + 2 = 0.b) Cho mot a giac eu n nh, n N va n 3. Tm n biet rang a giac a cho co 27

ng cheo.

Cau 5 (1,0 iem). Trong khong gian vi he toa o Oxyz, cho mat phang(P ) : 6x + 3y 2z 1 = 0 va mat cau (S) : x2+y2+z26x4y2z11 = 0. Chngminh mat phang (P ) cat mat cau (S) theo giao tuyen la mot ng tron (C). Tm toao tam cua (C).

Cau 6 (1,0 iem). Cho hnh chop S.ABC co ay ABC la tam giac vuong can tai A, matben SBC la tam giac eu canh a va mat phang (SBC) vuong goc vi mat ay. Tnhtheo a the tch cua khoi chop S.ABC va khoang cach gia hai ng thang SA,BC.

Cau 7 (1,0 iem). Trong mat phang vi he toa o Oxy, cho tam giac ABC co channg phan giac trong cua goc A la iem D(1;1). ng thang AB co phng trnh3x + 2y 9 = 0, tiep tuyen tai A cua ng tron ngoai tiep tam giac ABC co phngtrnh x+ 2y 7 = 0. Viet phng trnh ng thang BC .Cau 8 (1,0 iem). Giai bat phng trnh (x+1)

x + 2+(x+6)

x+ 7 x2+7x+12.

Cau 9 (1,0 iem). Cho hai so thc x, y thoa man cac ieu kien 1 x 2; 1 y 2.Tm gia tr nho nhat cua bieu thc

P =x+ 2y

x2 + 3y + 5+

y + 2x

y2 + 3x + 5+

1

4(x+ y 1) .

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

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


E CHNH THC Mon: TOAN; Khoi D(ap an - Thang iem gom 03 trang)

Cau ap an iem

1 a) (1,0 iem)(2,0) Tap xac nh D = R.

S bien thien:- Chieu bien thien: y = 3x2 3; y = 0 x = 1.

0,25

Cac khoang ong bien: (;1) va (1; +); khoang nghch bien: (1; 1).- Cc tr: Ham so at cc ai tai x = 1, yC = 0; at cc tieu tai x = 1, yCT = 4.- Gii han tai vo cc: lim

xy = ; lim

x+y = +.

0,25

- Bang bien thien:x 1 1 +y + 0 0 +

y0 +

4 #

#####$

0,25

o th:

x

y

1

4

1

O

20,25

b) (1,0 iem)

M (C)M(a; a3 3a 2). 0,25He so goc cua tiep tuyen tai M bang 9 y (a) = 9 0,25 3a2 3 = 9 a = 2. 0,25Toa o iem M thoa man yeu cau bai toan la M(2; 0) hoac M(2;4). 0,25

2 at z = a+ bi (a, b R). T gia thiet ta c [3(a+ bi) (a bi)](1+ i)5(a+ bi) = 8i1 0,25(1,0)

{

3a+ 4b = 12a b = 8 0,25

{a = 3b = 2. 0,25

Do o moun cua z la

32 + (2)2 = 13. 0,25

1

Cau ap an iem

3(1,0)

I =

40

(x+1) sin 2x dx. at u = x+1 va dv = sin2xdx, suy ra du = dx va v = 12cos 2x. 0,25

Ta co I = 12(x+ 1) cos2x

40+

1

2

40

cos 2xdx 0,25

= 12(x+ 1) cos 2x

40+

1

4sin 2x

40

0,25

=3

4. 0,25

4(1,0)

a) ieu kien: x > 1. Phng trnh a cho tng ng vi log 2x 13x 2 = 2 0,25

x 13x 2 =

1

4 x = 2.

oi chieu ieu kien, ta c nghiem cua phng trnh a cho la x = 2.0,25

b) So ng cheo cua a giac eu n nh la C2n n =n(n 3)

2. 0,25

T gia thiet ta co phng trnhn(n 3)

2= 27

[ n = 9n = 6.

Do n N va n 3 nen ta c gia tr n can tm la n = 9.0,25

5 Mat cau (S) co tam I(3; 2; 1) va ban knh R = 5. 0,25(1,0)

Ta co khoang cach t I en (P ) la d(I, (P )) =|6.3 + 3.2 2.1 1|

62 + 32 + (2)2 = 3 < R.

Do o (P ) cat (S) theo giao tuyen la mot ng tron (C).

0,25

Tam cua (C) la hnh chieu vuong goc H cua I tren (P ). ng thang qua I va vuong goc

vi (P ) co phng trnh lax 36

=y 23

=z 12 . Do H nen H(3+ 6t; 2+ 3t; 1 2t).

0,25

Ta co H (P ), suy ra 6(3+6t)+3(2+3t)2(12t)1 = 0 t = 37. Do o H

(37;5

7;13

7

). 0,25

6(1,0)

Goi H la trung iem cua BC, suy ra AH =BC

2=a

2,

SH (ABC), SH =3 a

2va SABC =

1

2BC.AH =

a2

4.

0,25

The tch khoi chop la VS.ABC =1

3.SH.SABC =

3 a3

24. 0,25

Goi K la hnh chieu vuong goc cua H tren SA, suy raHK SA. Ta co BC (SAH) nen BC HK .Do o HK la ng vuong goc chung cua BC va SA.

0,25AB

C

S

H

K

Ta co1

HK2=

1

SH2+

1

AH2=

16

3a2.

Do o d(BC, SA) = HK =3 a

4.

0,25

2

Cau ap an iem7

(1,0)Toa o iemA thoa man he phng trnh

{3x+ 2y 9 = 0x+ 2y 7 = 0.

Suy ra A(1; 3).0,25

B C

A

DE

Goi la tiep tuyen tai A cua ng tron ngoai tiep tam giacABC va E la giao iem cua vi ng thang BC (do ADkhong vuong goc vi nen E luon ton tai va ta co the gia sEB < EC). Ta co EAB = ACB va BAD = DAC, suy raEAD = EAB + BAD = ACB + DAC = ADE.Do o, tam giac ADE can tai E .

0,25

E la giao iem cua vi ng trung trc cua oan AD, nen

toa o iem E thoa man he phng trnh{x+ 2y 7 = 0y 1 = 0.

Suy ra E(5; 1).

0,25

ng thang BC i qua E va nhanDE = (4; 2) lam vect

ch phng, nen BC : x 2y 3 = 0. 0,25

8(1,0)

ieu kien: x 2. Bat phng trnh a cho tng ng vi(x+ 1)(

x+ 2 2) + (x+ 6)(x+ 7 3) (x2 + 2x 8) 0 0,25

(x 2)( x+ 1

x+ 2+ 2+

x+ 6x+ 7+ 3

x 4) 0 (1). 0,25

Do x 2 nen x+ 2 0 va x+ 6 > 0. Suy rax+ 1x+ 2 + 2

+x+ 6x+ 7 + 3

x 4 =( x+ 2

x+ 2 + 2 x+ 2

2

)+( x+ 6

x+ 7+ 3 x+ 6

2

) 1

x+ 2+ 2< 0.

Do o (1) x 2.

0,25

oi chieu ieu kien, ta c nghiem cua bat phng trnh a cho la: 2 x 2. 0,25

9(1,0)

Do 1 x 2 nen (x 1)(x 2) 0, ngha la x2 + 2 3x. Tng t, y2 + 2 3y.Suy ra P x+ 2y

3x+ 3y + 3+

y + 2x

3y + 3x+ 3+

1

4(x+ y 1) =x+ y

x+ y + 1+

1

4(x+ y 1) .0,25

at t = x+ y, suy ra 2 t 4. Xet f(t) = tt+ 1

+1

4(t 1) , vi 2 t 4.

Ta co f (t) =1

(t+ 1)2 1

4(t 1)2 . Suy ra f(t) = 0 t = 3.

0,25

Ma f(2) =11

12; f(3) =

7

8; f(4) =

53

60nen f(t) f(3) = 7

8. Do o P 7

8. 0,25

Khi x = 1, y = 2 th P =7

8. Vay gia tr nho nhat cua P la

7

8. 0,25

Het

3

BO GIAO DUC VA AO TAO E THI TUYE4N SINH AI HOC NA M 2013 Mon: TOAN; Khoi A va khoi A1


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 = 2

2 sin

(x+

4

).


{ x+ 1 + 4

x 1

y4 + 2 = y

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


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 . x Tp xc nh: .D \x 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)f v (2; ). f - 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

of of f f

0,25

- Bng bin thin:

Trang 1/4

0,25

x th:

0,25

b. (1,0 im)

Ta c 2' 3 6 3y x x .mHm s (1) nghch bin trn khong (0; )f khi v ch khi ' 0, 0y xd !

0,25

2 2 , 0.m x x x d ! 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

f + f 0 2 0 0 +

+ f

f 1

3

2 O

y

x

3

1

x

( )f x

0 + f1 0

0 +

1

+ f'( )f x

d 0,25

/4

Cu p n im

iu kin: Phng trnh cho tng ng vi cos 0.x z sin1 2(sin cocos

x s )x xx 0,25

(sin cos )(2cos 1) 0.x x x 0,25 sin cos 0 ( )4

x x x k kx ] . 0,25

2 (1,0 im)

2cos 1 0 2 ( )3

x x k kx r ] .

i chiu iu kin ta c nghim: 4

x k hoc 2 ( )3

x k k r ] . 0,25

44

2 2

1 1 2

2 ( 1) 6 1 0 (2)

x x y yx x y y y

(1)

,

iu kin: T (2) ta c suy ra 1.x t 24 ( 1)y x y 0.y t0,25

3 (1,0 im)

t 4 1,u x suy ra u Phng trnh (1) tr thnh: 0.t 4 42 2 (3).u u y y

Xt 4( ) 2 ,f t t t vi Ta c 0.t t3

4

2'( ) 1 0, 0.2

tf t tt

! t

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 t

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 1x 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 A BC. M (SBC) vung gc vi (ABC) theo giao tuyn BC, nn SH A (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 A (ABC), suy ra SA = SB = a. Gi I l trung im ca AB, suy ra SI A 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 SAB S SI AB'

3

0,25

S

A B

C

I

H

/4

Cu p n im

t ,ax yc 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 t 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

t .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 t

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 yt

0,25

t t x Suy ra t v .y ! 0 3 2( 1) 2 6.P t t tt

Ta c 2 2( )3 ( )

4 4x y tx y xy x y t d . nn ( 2)( 6) 0t t t Do 2.t t

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

1'( ) 3( 1) .2 6

tf t tt t

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

ttt t

d

3 2 , nn

3 2'( ) 3 0.2

f t t ! Suy ra ( ) (2) 1 2.f t ft Do 1 2P t .

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 A DM, suy ra BN A 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 uJG

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

/4

Cu p n im

S 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 t

2 2.2

ABAH 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 y Hx 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).IVy 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 ) 11 0 1.t t t t (3;1;2).M 0,25 1 31 3 22 2

z i i

0,25 9.b

(1,0 im)

2 cos sin .3 3

i 0,25

5 5 5 52 cos sin 16(1 3 ).3 3

z i iSuy ra 0,25

16( 3 1) 16(1 3) .w i Do 0,25

Vy w c phn thc l 16( v phn o l 3 1) 16 (1 3).

------------- Ht -------------

BO GIAO DUC VA AO TAO E THI TUYE4N SINH AI HOC NA M 2013 Mon: TOAN; Khoi B


I. PHAN CHUNG CHO TAT CA TH SINH (7,0 iem)Cau 1 (2,0 iem). Cho ham so y = 2x3 3(m+ 1)x2 + 6mx (1), vi m la tham so thc.a) Khao sat s bien thien va ve o th cua ham so (1) khi m = 1.b) Tm m e o th ham so (1) co hai iem cc tr A va B sao cho ng thang AB vuong goc ving thang y = x+ 2.

Cau 2 (1,0 iem). Giai phng trnh sin 5x + 2cos2 x = 1.


{2x2 + y2 3xy + 3x 2y + 1 = 04x2 y2 + x+ 4 = 2x+ y +x+ 4y (x, y R).


0

x2 x2 dx.

Cau 5 (1,0 iem). Cho hnh chop S.ABCD co ay la hnh vuong canh a, mat ben SAB la tam giaceu va nam trong mat phang vuong goc vi mat phang ay. Tnh theo a the tch cua khoi chopS.ABCD va khoang cach t iem A en mat phang (SCD).

Cau 6 (1,0 iem). Cho a, b, c la cac so thc dng. Tm gia tr ln nhat cua bieu thc

P =4

a2 + b2 + c2 + 4 9

(a + b)

(a + 2c)(b+ 2c).

II. PHAN RIENG (3,0 iem): Th sinh ch c lam mot trong hai phan (phan A hoac phan B)A. Theo chng trnh ChuanCau 7.a (1,0 iem). Trong mat phang vi he toa o Oxy, cho hnh thang can ABCD co hai ngcheo vuong goc vi nhau va AD = 3BC. ng thang BD co phng trnh x+ 2y 6 = 0 va tamgiac ABD co trc tam la H(3; 2). Tm toa o cac nh C va D.

Cau 8.a (1,0 iem). Trong khong gian vi he toa o Oxyz, cho iem A(3; 5; 0) va mat phang(P ) : 2x + 3y z 7 = 0. Viet phng trnh ng thang i qua A va vuong goc vi (P ). Tm toao iem oi xng cua A qua (P ).

Cau 9.a (1,0 iem). Co hai chiec hop cha bi. Hop th nhat cha 4 vien bi o va 3 vien bi trang,hop th hai cha 2 vien bi o va 4 vien bi trang. Lay ngau nhien t moi hop ra 1 vien bi, tnh xacsuat e 2 vien bi c lay ra co cung mau.

B. Theo chng trnh Nang caoCau 7.b (1,0 iem). Trong mat phang vi he toa o Oxy, cho tam giac ABC co chan ng cao hat nh A la H

(175;1

5

), chan ng phan giac trong cua goc A la D(5; 3) va trung iem cua canh

AB la M(0; 1). Tm toa o nh C .Cau 8.b (1,0 iem). Trong khong gian vi he toa o Oxyz, cho cac iem A(1;1; 1), B(1; 2; 3) vang thang :

x+ 1

2 =y 21

=z 33

. Viet phng trnh ng thang i qua A, vuong goc vi

hai ng thang AB va .

Cau 9.b (1,0 iem). Giai he phng trnh

{x2 + 2y = 4x 12 log3(x 1) log3(y + 1) = 0.

HetTh sinh khong c s dung tai lieu. Can bo coi thi khong giai thch g them.Ho va ten th sinh: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ; So bao danh: . . . . . . . . . . . . . . . . . . . . .



Mn: TON; Khi B (p n - thang im gm 04 trang)


Khi m = 1 ta c 32 6y x x .x Tp xc nh: .D \x S bin thin:

- Chiu bin thin: 2' 6 6; ' 0 1.y x y x r

0,25

Cc khong ng bin: v ( ; 1)f (1; ); f khong nghch bin: (1; 1). - Cc tr: Hm s t cc tiu ti x = 1, yCT = 4; t cc i ti x = 1, yC = 4. - Gii hn: lim ; lim .

x xy y

of of f f

0,25

- Bng bin thin:

Trang 1/4

0,25

x th:

0,25

b. (1,0 im)

Ta c hoc 2' 6 6( 1) 6 ; ' 0 1y x m x m y x .x m 0,25 iu kin th hm s c hai im cc tr l 1.mz 0,25 Ta c 3 2(1;3 1), ( ; 3 ).A m B m m m H s gc ca ng thng AB l 2( 1)k m .ng thng AB vung gc vi ng thng 2y x khi v ch khi 1k 0,25

1 (2,0 im)

0m hoc 2.m Vy gi tr m cn tm l hoc 0m 2.m 0,25

x 'y

y

f + f 1 1 0 0 + +

+ f

f 4

4

1 O

y

x

4

1

4

/4

Cu p n imPhng trnh cho tng ng vi sin 5 cos 2 0x x 0,25

cos 5 cos22

x x 0,25

5 2 2 ( )2

x x k k r ] 0,25

2 (1,0 im)

26 3 ( ) 2

14 7

x kk

x k

] . 0,25

2 2

2 2

2 3 3 2 1 0

4 4 2 4

x y xy x yx y x x y x y

(1)

(2)

0x y x y t t

iu kin: . T (1) ta c 2 0, 4 1y x hoc 2 1y x0,25

.

x Vi thay vo (2) ta c 1,y x 23 3 3 1 5x x x x 4 23( ) ( 1 3 1) ( 2 5 4) 0x x x x x x

2 1 1( ) 31 3 1 2 5 4

x xx x x x

0,25

0

2 0 0x x x hoc Khi ta c nghim ( ;1.x )x y l v (0;1) (1;2). 0,25

3 (1,0 im)

x Vi thay vo (2) ta c 2 1y x , 3 3 4 1 9 4x x x 3 ( 4 1 1) ( 9 4 2) 0x x x

4 934 1 1 9 4 2

xx x

0 0.x Khi ta c nghim ( ; )x y l (0 ; 1).

i chiu iu kin ta c nghim ( ; )x y ca h cho l v (0;1) (1;2).

0,25

t 22 d d .t t x x t x Khi 0x th 2,t khi 1x th 1.t 0,25

Suy ra 2

2

1dI t

4

t 0,25

23

13t 0,25

(1,0 im)

2 2 1.3 0,25

Gi H l trung im ca AB, suy ra SH A AB v 3 .2

aSH M (SAB) vung gc vi (ABCD) theo giao tuyn AB, nn SH A (ABCD).

0,25

Do 3

.1 3. .3 6S ABCD ABCD

aV S H S 0,25

Do AB || CD v HAB nn ( , ( )) ( , ( )).d A SCD d H SCD Gi K l trung im ca CD v I l hnh chiu vung gc ca H trn SK. Ta c HKACD. M SHACD CDA(SHK) CD A HI. Do HI A(SCD).

0,25

5 (1,0 im)

Suy ra 2 2

. 2( ,( )) .7

SH HK ad A SCD HISH HK

S

I A

1 0,25

B C H

D

K

/4

Cu p n im

Ta c: 2 2 2 2 24 2 4 4( ) ( 2 )( 2 ) ( ) 2(2 2

a b c a b ab ac bca b a c b c a b a b c d d ). 0,25

t 2 2 2 4,t a b c suy ra v 2t ! 24 9 .

2( 4)P t td

Xt 24 9( ) ,

2( 4)f t t t

vi Ta c 2.t !3 2

2 2 2 2 2 24 9 ( 4)(4 7 4 16'( ) .

( 4) ( 4)t t t t tf t

t t t t

)

.Vi t > 2 ta c 3 2 34 7 4 16 4( 4) (7 4) 0t t t t t t ! Do '( ) 0 4.f t t

0,25

Bng bin thin:

T bng bin thin ta c 5 .8

Pd

0,25

6 (1,0 im)

Khi ta c 2a b c 5 .8

P Vy gi tr ln nht ca P l 5 .8

0,25

Gi I l giao im ca AC v BD .IB ICM IB ICA nn 'IBC vung cn ti I n o45 .ICB BH A AD BH A BC 'HBC vung cn ti B I l trung im ca on thng HC.

0,25

Do CH A BD v trung im I ca CH thuc BD nn ta

im C tha mn h 2( 3) ( 2) 0

3 22 62 2

x yx y

0.

Do ( 1;6).C

0,25

Ta c 1 33

IC IB BC ID ICID ID AD 2 2 1010 5 2.

2CHCD IC ID IC 0,25

7.a (1,0 im)

Ta c (6 2 ; )D t t v 5 2CD suy ra 2 2 1(7 2 ) ( 6) 507.

tt t t

Do hoc (4;1)D ( 8;7).D 0,25

(P) c vct php tuyn (2;3; 1).n JG

0,25

ng thng ' qua A v vung gc vi (P) nhn nJG

lm vct ch phng, nn c phng trnh 3 5 .

2 3 1x y z

0,25

Gi B l im i xng ca A qua (P), suy ra B thuc '. Do (3 2 ;5 3 ; ).B t t t 0,25

8.a (1,0 im)

Trung im ca on thng AB thuc (P) nn 10 32(3 ) 3 7 0 2.2 2

t tt t

Do ( 1; 1; 2).B 0,25

S cch chn 2 vin bi, mi vin t mt hp l: 7.6 42. 0,25 S cch chn 2 vin bi , mi vin t mt hp l: 4.2 8. 0,25 S cch chn 2 vin bi trng, mi vin t mt hp l: 3.4 12. 0,25

9.a (1,0 im)

Xc sut 2 vin bi c ly ra c cng mu l: 8 12 10 .42 21

p 0,25

A D

B C

H

I

t

( )

2 + f 4 0 +

f t

f

58

0

f '( )t

/4

Cu p n imTa c H AH v AH HDA nn AH c phng trnh:

2 3 0x y . Do (3 2 ; ).A a a 0,25 Do M l trung im ca AB nn MA = MH. Suy ra 2 2(3 2 ) ( 1) 13 3a a a hoc 1 .

5a

Do A khc H nn ( 3;3).A 0,25

Phng trnh ng thng AD l 3 0.y Gi N l im i xng ca M qua AD. Suy ra N AC v ta im N tha mn h

1 3 02

1. 0.( 1) 0

y

x y

(0;5).N0,25

7.b

ng thng AC c phng trnh: 2 3 15 0x y

(1,0 im)

. ng thng BC c phng trnh: 2 7x y 0.

Suy ra ta im C tha mn h: 2 7 02 3 15 0.

x yx y

Do C (9;11).

0,25

Ta c vect ch phng ca ' l 2;3;2 ,AB JJJG

( 2;1;3).u JG

0,25

ng thng vung gc vi AB v ', c vect ch phng l , .v AB u JG JJJG JG

0,25

Suy ra v 7; 2; 4 . JG

0,25

8.b (1,0 im)

ng thng i qua A, vung gc vi AB v ' c phng trnh l: 1 1 .7 2 4

x y z 1 0,25

iu kin: H cho tng ng vi 1; 1.x y! !2

3 3

2 4 1log ( 1) log ( 1)x y x

x y

0,25

2 2 3 02

x xy x

0,25

1, 33, 1.

x yx y

0,25

9.b (1,0 im)

i chiu iu kin ta c nghim ( ; )x y ca h cho l (3 ;1). 0,25

------------- Ht -------------

D B C H

M N

A

BO GIAO DUC VA AO TAO E THI TUYE4N SINH AI HOC NA M 2013 Mon: TOAN; Khoi D


I. PHAN CHUNG CHO TAT CA TH SINH (7,0 iem)Cau 1 (2,0 iem). Cho ham so y = 2x3 3mx2 + (m 1)x+ 1 (1), vi m la tham so thc.a) Khao sat s bien thien va ve o th cua ham so (1) khi m = 1.b) Tm m e ng thang y = x+ 1 cat o th ham so (1) tai ba iem phan biet.

Cau 2 (1,0 iem). Giai phng trnh sin 3x+ cos 2x sinx = 0.

Cau 3 (1,0 iem). Giai phng trnh 2 log2 x+ log 12

(1x

)=

1

2log

2

(x 2x+ 2

).


0

(x+ 1)2

x2 + 1dx.

Cau 5 (1,0 iem). Cho hnh chop S.ABCD co ay la hnh thoi canh a, canh ben SA vuong gocvi ay, BAD = 120, M la trung iem cua canh BC va SMA = 45. Tnh theo a the tch cuakhoi chop S.ABCD va khoang cach t iem D en mat phang (SBC).

Cau 6 (1,0 iem). Cho x, y la cac so thc dng thoa man ieu kien xy y 1. Tm gia tr lnnhat cua bieu thc P =

x+ yx2 xy + 3y2

x 2y6(x+ y)

.

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 tam giac ABC co iem M( 9

2;3

2

)la trung iem cua canh AB, iem H(2; 4) va iem I(1; 1) lan lt la chan ng cao ke t Bva tam ng tron ngoai tiep tam giac ABC . Tm toa o iem C .

Cau 8.a (1,0 iem). Trong khong gian vi he toa o Oxyz, cho cac iem A(1;1;2), B(0; 1; 1)va mat phang (P ) : x+y+z1 = 0. Tm toa o hnh chieu vuong goc cua A tren (P ). Viet phngtrnh mat phang i qua A,B va vuong goc vi (P ).

Cau 9.a (1,0 iem). Cho so phc z thoa man ieu kien (1 + i)(z i) + 2z = 2i. Tnh moun cuaso phc w =

z 2z + 1z2

.

B. Theo chng trnh Nang caoCau 7.b (1,0 iem). Trong mat phang vi he toa o Oxy, cho ng tron (C) : (x1)2+(y1)2 = 4va ng thang : y 3 = 0. Tam giac MNP co trc tam trung vi tam cua (C), cac nh Nva P thuoc , nh M va trung iem cua canh MN thuoc (C). Tm toa o iem P .

Cau 8.b (1,0 iem). Trong khong gian vi he toa o Oxyz, cho iem A(1; 3;2) va mat phang(P ) : x 2y 2z + 5 = 0. Tnh khoang cach t A en (P ). Viet phng trnh mat phang i quaA va song song vi (P ).

Cau 9.b (1,0 iem). Tm gia tr ln nhat va gia tr nho nhat cua ham so f(x) =2x2 3x+ 3

x+ 1tren oan [0; 2].

HetTh sinh khong c s dung tai lieu. Can bo coi thi khong giai thch g them.Ho va ten th sinh: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ; So bao danh: . . . . . . . . . . . . . . . . . . . .



Mn: TON; Khi D (p n - thang im gm 04 trang)


Khi m = 1 ta c 3 22 3 1y x x . x Tp xc nh: .D \x S bin thin:

- Chiu bin thin: hoc 2' 6 6 ; ' 0 0y x x y x 1.x

0,25

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

x xy y

of of f f

0,25

- Bng bin thin:

Trang 1/4

0,25

x th:

0,25

b. (1,0 im) Phng trnh honh giao im ca th hm s (1) vi ng thng l 1y x

3 22 3 ( 1) 1 1x mx m x x 0,25

2

0

2 3 0 (*

xx mx m ).

Yu cu ca bi ton phng trnh (*) c hai nghim phn bit khc 0 0,25

29 80

m mm

! z0 0,25

1 (2,0 im)

x 'y

y

f + f 0 1 0 0 + +

+ f

f 0

1

1 O

y

x

1

0m hoc 8 .9

m! 0,25

/4

Cu p n im

Phng trnh cho tng ng vi 2cos 2 sin cos 2 0x x x 0,25 cos 2 (2sin 1) 0.x x 0,25

cos2 0 ( ).4 2

x x k kx ] 0,25

2 (1,0 im)

262sin 1 0 ( ).

7 2 6

x kx k

x k

x

]

Vy nghim ca phng trnh cho l 4 2

x k , 2,6

x k 7 2 ( )6

x k k ] .

0,25

iu kin: Phng trnh cho tng ng vi 0 x 1.2

2 21

x x xx

0,25

2

2 2 1 2(1 ) 1 1 1x x x x

x x x x

0 0,25

2 01

xx

(do 01

xx!

) 0,25

3 (1,0 im)

4 2 3.x i chiu vi iu kin ta c nghim ca phng trnh cho l 4 2 3.x

0,25

Ta c 1 1 1

2 20 0 0

2 21 d d1 1

x xI x xx x

d .x 0,25 1 1

00

d 1x xx . 0,25 1

122 0

0

2 d ln( 1) ln 21

x x xx

x . 0,25

4 (1,0 im)

Do . 1 ln 2I 0,25 n no o120 60BAD ABC ABC ' u

32

aAM 2 3 .2ABCD

aS 0,25

SAM' vung ti A c n o45SMA SAM'

vung cn ti A 3 .2

aSA AM

Do 3

.1 . .3 4S ABCD ABCD

aV S A S

0,25

Do AD||BC nn ( ,( )) ( ,( )).d D SBC d A SBC Gi H l hnh chiu vung gc ca A trn SM. Ta c AM BCA v SA BCA ( )BC SAMA

( ) ( ,( )) .BC AH AH SBC d A SBC AH A A

0,25

5 (1,0 im)

Ta c 2 6 ,2 4

AM aAH

S

H

suy ra 6( ,( )) .4

ad D SBC 0,25

A

B C

M

D

/4

Cu p n im

Do 0, 0, 1x y xy y! ! d nn 2

2 21 1 1 1 1 1 10 .

4 2x yy y yy y

d d 4 0,25

t ,xt suy ra y 10 .4

t d Khi 2

1 2 .6( 1)3

t tP tt t

Xt 2

1 2( ) ,6( 1)3

t tf t tt t

vi 10 .

4t d Ta c 22 3

7 3 1'( ) .2( 1)2 ( 3)

tf ttt t

Vi 104

t d ta c v 2 3 ( 1) 3 3; 7 3 6t t t t t ! 1 1.t !

Do 2 3

7 3 7 3 16 3 32 ( 3)

t tt t ! !

v 21 .

22( 1)t1 !

Suy ra 1 1'( ) 0.

23f t ! !

0,25

Do 1 5 7( ) .4 3 30

P f t f d 0,25

6 (1,0 im)

Khi 12

x v ta c 2,y 5 7 .3 30

P Vy gi tr ln nht ca P l 5 7 .3 30 0,25

7 1;2 2

IM .JJJG Ta c M AB v AB IMA nn ng

thng AB c phng trnh 7 33x y 0. 0,25

( ;7 33).A AB A a a Do M l trung im ca AB nn ( 9; 7 30).B a a Ta c . 0HA HB HA HBA

JJJG JJJG

2 9 20 0 4a a a hoc 5.a 0,25

x Vi a 4 ( 4;5), ( 5; 2).A B Ta c BH ACA nn ng thng AC c phng trnh 2 6 0x y . Do

(6 2 ; ).C c c T IC = IA suy ra (7 Do c

2 22 ) ( 1) 25.c c 1 hoc 5.c Do C khc A, suy ra (4;1).C

0,25

7.a (1,0 im)

A

M

B

C H

I

x Vi a 5 ( 5; 2), ( 4;5).A B Ta c BH ACA nn ng thng AC c phng trnh 2 8x y 0. Do

( ;2 8).C t t T IC = IA suy ra Do 2 2( 1) (2 7) 25.t t 1t hoc 5.t Do C khc A, suy ra ( 1;6).C

0,25

Gi H l hnh chiu vung gc ca A trn (P). Suy ra ( 1 ; 1 ; 2 ).H t t t 0,25

5( ) ( 1 ) ( 1 ) ( 2 ) 1 0 .3

H P t t t t Do 2 2 1; ; .3 3 3

H 0,25

Gi (Q) l mt phng cn vit phng trnh. Ta c (1;2;3)AB JJJG

v vect php tuyn ca (P) l Do (Q) c vect php tuyn l (1;1;1).n

JG' ( 1;2; 1).n JG

0,25

8.a (1,0 im)

Phng trnh ca mt phng (Q) l: 2 1x y z 0. 0,25

iu kin ca bi ton tng ng vi (3 ) 1 3i z i 0,25 .z i 0,25

Suy ra 1 3 .w i 0,25

9.a (1,0 im)

0,25 Do mun ca w l 10.

/4

Cu p n imTa c tm ca (C) l ng thng IM vung gc vi ' nn c phng trnh

(1;1).I1.x Do (1; ).M a 0,25

Do ( )M C nn .2( 1) 4a Suy ra 1a hoc 3.a M M ' nn ta c (1; 1).M 0,25

( ;3).N N b' Trung im ca MN thuc (C)

2

21 1 1 1 4 5b 2

b hoc b 3.

Do hoc (5;3)N ( 3;3).N

0,25

7.b (1,0 im)

( ;3).P P c' - Khi t (5;3),N MP INA

JJJG JJG suy ra 1.c Do ( 1;3).P

I

M

- Khi ( 3;3),N t MP INAJJJG JJG

suy ra 3.c Do (3;3).P0,25

2 2 2

| ( 1) 2.3 2( 2) 5|( ,( ))1 ( 2) ( 2)

d A P

0,25

2 .3

0,25

Vect php tuyn ca (P) l (1; 2; 2).n JG

0,25

8.b

P N

(1,0 im)

Phng trnh mt phng cn tm l 2 2 3 0x y z . 0,25

Ta c ( )f x xc nh v lin tc trn on [0 ; ;2]2

22 4 6'( ) .

( 1)x xf x

x

0,25

Vi ta c [0; 2]x '( ) 0 1.f x x 0,25

9.b (1,0 im)

Ta c 5(0) 3; (1) 1; (2) .3

f f f 0,25 Gi tr nh nht ca f(x) trn on [0; 2] l 1; gi tr ln nht ca f(x) trn on [0; 2] l 3. 0,25

------------- Ht -------------

B GIO DC V O TO

CHNH THC

THI TUYN SINH I HC NM 2012 Mn: TON; Khi A v khi A1

Thi gian lm bi: 180 pht, khng k thi gian pht

I. PHN CHUNG CHO TT C TH SINH (7,0 im) Cu 1 (2,0 im). Cho hm s vi m l tham s thc. 4 2 22( 1) (1),y x m x m

a) Kho st s bin thin v v th ca hm s (1) khi 0.m b) Tm m th ca hm s (1) c ba im cc tr to thnh ba nh ca mt tam gic vung.

Cu 2 (1,0 im). Gii phng trnh 3 sin 2 cos 2 2cos 1.x x x

Cu 3 (1,0 im). Gii h phng trnh 3 2 3 2

2 2

3 9 22 3 9( , ).1

2

x x x y y yx y

x y x y

\

Cu 4 (1,0 im). Tnh tch phn 3

21

1 ln( 1) d .xI xx

Cu 5 (1,0 im). Cho hnh chp c y l tam gic u cnh a. Hnh chiu vung gc ca trn mt phng (ABC) l im H thuc cnh AB sao cho

.S ABC S2 .HA HB Gc gia ng thng SC v mt

phng (ABC) bng Tnh th tch ca khi chp S.ABC v tnh khong cch gia hai ng thng SA v BC theo a.

o60 .

Cu 6 (1,0 im). Cho cc s thc , ,x y z tha mn iu kin 0.x y z Tm gi tr nh nht ca biu thc | | | | | | 2 2 23 3 3 6 6 6x y y z z xP x .y z

.ND

II. PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn ring (phn A hoc phn B) A. Theo chng trnh Chun Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hnh vung ABCD. Gi M l trung im

ca cnh BC, N l im trn cnh CD sao cho CN 2 Gi s 11 1;2 2M v ng thng AN c phng trnh Tm ta im A. 2 3x y 0.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho ng thng 1 2:1 2 1

x y zd v im Vit phng trnh mt cu (S) c tm I v ct d ti hai im A, B sao cho tam gic IAB vung ti I.

(0;0;3).I

Cu 9.a (1,0 im). Cho n l s nguyn dng tha mn 15 nnC 3

nC . Tm s hng cha 5x trong khai

trin nh thc Niu-tn ca 2 1 , 0.14n

nx xx

zB. Theo chng trnh Nng cao Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn Vit phng trnh chnh tc ca elip (E), bit rng (E) c di trc ln bng 8 v (E) ct (C) ti bn im to thnh bn nh ca mt hnh vung.

2 2( ): 8.C x y

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho ng thng 1 2: ,2 1 1

x y zd mt

phng v im( ): 2 5 0P x y z (1; 1;2).A Vit phng trnh ng thng ' ct d v (P) ln lt ti M v N sao cho A l trung im ca on thng MN.

Cu 9.b (1,0 im). Cho s phc z tha mn 5( ) 21

z i iz

.

Tnh mun ca s phc 21 .w z z

---------- HT ---------- Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm. H v tn th sinh:....................................................................; S bo danh: ..............................................



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

Cu p n im

a) (1,0 im)

Khi ta c: 0,m 4 22 .y x x x Tp xc nh: .D \x S bin thin: Chiu bin thin: 3' 4 4 ;y x x ' 0y 0x hoc 1.x r

0,25

Cc khong nghch bin: cc khong ng bin: (( ; 1)f v (0; 1); 1; 0) v ( 1; ).f Cc tr: Hm s t cc tiu ti 1,x r yCT 1; t cc i ti 0,x yC 0. Gii hn: lim lim .

x xy y

of of f

0,25

Bng bin thin:

0,25

x th:

0,25

Trang 1/4

b) (1,0 im)

Ta c 3 2' 4 4( 1) 4 ( 1).y x m x x x m th hm s c 3 im cc tr khi v ch khi 1 0m ! (*). 1m !

0,25

Cc im cc tr ca th l 2(0; ),A m ( 1; 2B m m 1) v ( 1; 2 1).m C m Suy ra: 2( 1; ( 1)AB m m

JJJG) v 2( 1; ( 1)AC m m ).

JJJG

0,25

Ta c nn tam gic ABC vung khi v ch khi AB AC . 0AB AC JJJG JJJG

0,25

1 (2,0 im)

. Kt hp (*), ta c gi tr m cn tm l 4( 1) ( 1) 0m m 0.m 0,25

fy

'y 0 + 0 0 + x 1 0 1 f f

1

0

1

f

O

2

1

1

1

2

8

x

y

Cu p n im

Phng trnh cho tng ng vi ( 3 sin cos 1)cos 0.x x x 0,25 cos 0 ( )2

x x k kx ] . 0,25

3 sin cos 1 0x xx cos cos3 3x 0,25

2 (1,0 im)

2x k hoc 2 2 ( )3

x k k ] .

Vy nghim ca phng trnh cho l ,2

x k 2x k v 2 2 ( ). 3

x k k ]0,25

H cho tng ng vi: 3 3

2 2

( 1) 12( 1) ( 1) 12( 1) (1)

1 1 1. (2)2 2

x x y y

x y

0,25

T (2), suy ra 11 12

x d d v 11 12

y d d 3 112 2

x d d v 1 31 .2 2

y d d

Xt hm s 3( ) 12f t t t trn 3 3;2 2

, ta c

2'( ) 3( 4) 0f t t , suy ra f(t) nghch bin. 0,25

Do (1) x 1 y 1 y x 2 (3).

Thay vo (2), ta c 2 21 3 12 2x x 24 8 3x x 0 12x hoc 3 .2x 0,25

3 (1,0 im)

Thay vo (3), ta c nghim ca h l 1 3( ; ) ;2 2x y hoc 3 1( ; ) ; .2 2x y 0,25 t u v 1 ln( 1)x 2

dd , suy ra dd1

xux

v

1 .vxvx

x

0,25

33

1 1

1 ln( 1)( 1)

x dxIx x x

0,25

31

2 ln 2 1 13 1

dxx x

3

1

2 ln 2 ln3 1

xx

0,25

4 (1,0 im)

2 2ln3 ln 2.3 3

0,25

Ta c nSCH l gc gia SC v (ABC), suy ra n o60 .SCH Gi D l trung im ca cnh AB. Ta c: ,

6aHD 3 ,

2aCD

2 2 7 ,3

aHC HD CD o 21.tan60 .3

aSH HC

0,25

2 3

.1 1 21 3. . . . 73 3 3 4 12S ABC ABC

a a aV SH S' . 0,25

K Ax//BC. Gi N v K ln lt l hnh chiu vung gc

ca H trn Ax v SN. Ta c BC//(SAN) v 32

BA H A nn

3( , ) ( ,( )) ( ,( )).2

d SA BC d B SAN d H SAN Ta cng c ( )Ax SHNA nn .Ax HKA Do

(HK SAN ).A Suy ra d H( ,( )) .

Trang 2/4

SAN HK

0,25

5 (1,0 im)

o2 2

2 3 . 4212

, sin 60 , .3 3a a SH HN aAH HN AH HK

SH HN

Vy

S

B

C

Hx

N

K

D A

42( , ) .8

ad SA BC 0,25

Cu p n im

Ta chng minh 3 1 (*). ,t t tt t 0Xt hm ( ) 3 1tf t t , c '( ) 3 ln 3 1 0, 0tf t t ! t (0) 0f v , suy ra (*) ng. p dng (*), ta c | | | | | |3 3 3 3 | | | | |x y y z z x |.x y y z z x t

0,25

p dng bt ng thc | , ta c: | | | | |a b a b t 2 2 2 2(| | | | | |) | | | | | | | |(| | | |) | |(| | | |)x y y z z x x y y z z x x y y z z x y z z x x y

2 2 2| |(| | | |) 2 | | | | | | .z x x y y z x y y z z x t 0,25

Do 22 2 2 2 2 2| | | | | | 2 | | | | | | 6 6 6 2 .x y y z z x x y y z z x x y z x y z t M suy ra 0,x y z 2 2 2| | | | | | 6 6 6 .x y y z z x x y z t

0,25

6 (1,0 im)

Suy ra | | | | | | 2 2 23 3 3 6 6 6x y y z z xP x t3.y z Khi x y z 0 th du bng xy ra. Vy gi tr nh nht ca P bng 3.

0,25

Gi H l giao im ca AN v BD. K ng thng qua H v song song vi AB, ct AD v BC ln lt ti P v Q. t HP x. Suy ra PD x, AP 3x v HQ 3x. Ta c QC x, nn MQ x. Do 'AHP = 'HMQ, suy ra

.AH HMA

0,25

Trang 3/4

Hn na, ta cng c .AH HM

Do AM 2 2 ( ,( ))MH d M AN 3 10 .2

0,25

AAN, suy ra A(t; 2t 3). 3 10

2MA 2 211 7 4522 2t t 2 0,25

7.a (1,0 im)

t t2 5 4 0

A B

C D N

M

H P Q

t 1 hoc t 4. Vy: (1; 1)A hoc (4;5).A 0,25

Vc t ch phng ca d l Gi H l trung im ca AB, suy ra IH A AB. (1; 2; 1).a JJG

Ta c nn ta H c dng H d ( 1; 2 ; 2) ( 1; 2 ; 1).H t t t IH t t t JJJG

0,25

IH A AB . 0 IH a JJJG JJG

1 4 1 0t t t 13

t 2 2 2; ; .3 3 3IH JJJG

0,25

Tam gic IAH vung cn ti H, suy ra bn knh mt cu (S) l 2 62 .3

R IA IH 0,25

8.a (1,0 im)

Do phng trnh mt cu cn tm l 2 2 2 8( ): ( 3) .3

S x y z 0,25

15 nn nC C 3 ( 1)( 2)5

6n n nn 0,25

(v n nguyn dng). 7n 0,25

Khi 7 77 72 2 2

14 377 7

0 0

( 1)1 1 1 .14 2 2 2

n k k kkk k

kk k

Cnx x xC xx x x

0,25

9.a (1,0 im)

S hng cha 5x tng ng vi 14 3 5k k 3 .

Do s hng cn tm l 3 3

5 574

( 1) . 35 . 0,25

162C x x

Cu p n im

Phng trnh chnh tc ca (E) c dng: 2 2

2 2 1,x ya b

vi v 2 80a b! ! .a Suy ra a 4. 0,25

Do (E) v (C) cng nhn Ox v Oy lm trc i xng v cc giao im l cc nh ca mt hnh vung nn (E) v (C) c mt giao im vi ta dng ( ; ), 0.A t t t !

0,25

A(C) t t2 2 8,

Trang 4/4

suy ra t 2. 0,25

7.b (1,0 im)

(2;2) ( )A E 24 4 1

16 b 2 16.b

3

Phng trnh chnh tc ca (E) l 2 2

1.16163

x y 0,25

M thuc d, suy ra ta ca M c dng M(2t 1; t; t 2). 0,25

MN nhn A l trung im, suy ra N(3 2t; 2 t; 2 t). 0,25

N(P) t3 2 2 2(2 ) 5 0t t t 2, suy ra M(3; 2; 4). 0,25

8.b (1,0 im)

ng thng ' i qua A v M c phng trnh 1 1:2 3 2

x y z 2 ' . 0,25

t ( , ), 1.z a bi a b z z \

Ta c 5( ) 2 (3 2) ( 7 6)1

z i i a b a b iz

0 0,25

3 27 6

a ba b

00

11.

ab

0,25

Do Suy ra 1 .z i 3 .i2 21 1 1 (1 ) 2w z z i i 0,25

9.b (1,0 im)

Vy 2 3 13.w i 0,25

x2

2

O

y

A

------------- HT -------------

B GIO DC V O TO

CHNH THC

THI TUYN SINH I HC NM 2012 Mn: TON; Khi B


I. PHN CHUNG CHO TT C TH SINH (7,0 im)

Cu 1 (2,0 im). Cho hm s m l tham s thc. 3 2 33 3 (y x mx m 1),a) Kho st s bin thin v v th ca hm s (1) khi 1.m b) Tm m th hm s (1) c hai im cc tr A v B sao cho tam gic OAB c din tch bng 48.

Cu 2 (1,0 im). Gii phng trnh 2(cos 3 sin ) cos cos 3 sin 1.x x x x x

Cu 3 (1,0 im). Gii bt phng trnh 21 4 1 3 .x x x t x

Cu 4 (1,0 im). Tnh tch phn 1 3

4 20

d .3 2xI x

x x

Cu 5 (1,0 im). Cho hnh chp tam gic u S.ABC vi 2 , .SA a AB a Gi H l hnh chiu vung gc ca A trn cnh SC. Chng minh SC vung gc vi mt phng (ABH). Tnh th tch ca khi chp S.ABH theo a.

Cu 6 (1,0 im). Cho cc s thc x, y, z tha mn cc iu kin 0x y z v Tm gi tr ln nht ca biu thc

2 2 2 1.x y z 5 5 5.P x y z

II. PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn ring (phn A hoc phn B) A. Theo chng trnh Chun

Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho cc ng trn 2 21( ): 4,C x y v ng thng 2 22( ): 12 18 0C x y x : 4d x y 0. Vit phng trnh ng trn c tm

thuc tip xc vi d v ct ti hai im phn bit A v B sao cho AB vung gc vi d. 2( )C , 1( )C

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho ng thng 1:2 1 2

x y zd

v hai

im Vit phng trnh mt cu i qua A, B v c tm thuc ng thng d. (2;1;0),A ( 2;3;2).B Cu 9.a (1,0 im). Trong mt lp hc gm c 15 hc sinh nam v 10 hc sinh n. Gio vin gi ngu nhin 4 hc sinh ln bng gii bi tp. Tnh xc sut 4 hc sinh c gi c c nam v n. B. Theo chng trnh Nng cao Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho hnh thoi ABCD c v ng trn tip xc vi cc cnh ca hnh thoi c phng trnh

2AC BD 2 2 4.x y Vit phng trnh chnh

tc ca elip (E) i qua cc nh A, B, C, D ca hnh thoi. Bit A thuc Ox. Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho Vit phng trnh mt phng (P) qua A v ct cc trc Ox, Oy ln lt ti B, C sao cho tam gic ABC c trng tm thuc ng thng AM.

(0;0;3), (1;2;0).A M

Cu 9.b (1,0 im). Gi z1 v z2 l hai nghim phc ca phng trnh 2 2 3 4 0.z i z Vit dng lng gic ca z1 v z2.

---------- HT ---------- Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm. H v tn th sinh: ................................................................... ; S bo danh:............................................. .



Mn: TON; Khi B (p n - thang im gm 04 trang)

Cu p n im

a) (1,0 im)

Khi ta c: . 1,m 3 23 3y x x x Tp xc nh: .D \x S bin thin:

Chiu bin thin: ' 02' 3 6 ;y x x y 0x hoc 2.x

0,25

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

xy

of f lim .

xy

o f f

0,25

Bng bin thin:

0,25

x th:

0,25

b) (1,0 im) 2' 3 6 ;y x mx ' 0 hoc y 0x 2 .x m

th hm s c 2 im cc tr khi v ch khi 0m z (*). 0,25

Cc im cc tr ca th l 3(0; 3 )A m v 3(2 ; ).B m m Suy ra v 33 | |OA m ( , ( )) 2 | | .d B OA m

0,25

48OABS' 3 4 4 8m 0,25

1 (2,0 im)

tha mn (*). 2,m r 0,25

O2

3

1 x

y

f

1

3

fy

'y + 0 0 +

x 0 2 f f

Trang 1/4

Phng trnh cho tng ng vi: cos2 3 sin 2 cos 3 sinx x x x 0,25

co s 2 cos3 3x x 0,25 2 2 ( ). 3 3x x k k r ] 0,25

2 (1,0 im)

2 23

x k hoc 2 ( )3

x k k ] . 0,25

iu kin: 0 2 hoc 3xd d 2x t 3 (*). Nhn xt: l nghim ca bt phng trnh cho. 0x

Vi bt phng trnh cho tng ng vi: 0,x! 1 1 4 3x xxx

t (1). 0,25

t 1 (2),t xx

bt phng trnh (1) tr thnh 2 6 3t t t 2 2

3 03 0

6 (3 )

tt

t t

t t

0,25

5.2

t t Thay vo (2) ta c 1 5 22

x xx

t t hoc 12

x d 0,25

3 (1,0 im)

104

x d hoc . Kt hp (*) v nghim 4x t 0,x ta c tp nghim ca bt phng

trnh cho l: 10; [4; ).4

f 0,25

t t x suy ra Vi 2 , .2dt xdx 0x th 0;t vi 1x th 1.t 0,25

Khi 1 12

2 20 0

1 .2 d 1 d2 2 (( 1)( 2)

x x x t tIt tx x

1)( 2) 0,25

1 100

1 2 1 1d ln| 2| ln| 1|2 2 1 2

t t tt t

0,25

4 (1,0 im)

3ln3 ln2.2

0,25

Gi D l trung im ca cnh AB v O l tm ca 'ABC. Ta c AB CDA v AB SOA nn (AB SCD),A do .AB SCA 0,25

Mt khc ,SC AHA suy ra S ( ).C ABHA 0,25

Ta c: 3 3,2 3

a aCD OC nn 2 2 33.3

aSO SC OC

Do . 114

SO CD aDHSC

. Suy ra 21 1. .

2 8ABHaS AB DH'

10,25

5 (1,0 im)

Ta c 2 2 7 .4aSH SC HC SC CD DH

Do 3

.1 7. .3 9S ABH ABH

a116

H S' V S 0,25

O D

B

A

H

C

S

Trang 2/4

Vi v ta c: 0x y z 2 2 2 1,x y z 2 2 2 2 20 ( ) 2 ( ) 2 1 2 2 ,x y z x y z x y z yz x yz nn 2 1.

2yz x

Mt khc 2 2 21 ,

2 2y z xyz d suy ra:

22 1 1 ,

2 2xx d do 6 6

3 3x d d (*).

0,25

Khi : P 5 2 2 3 3 2 2( )( ) ( )x y z y z y z y z

25 2 2 2 2 1(1 ) ( )( ) ( ) 2x x y z y z yz y z x x 25 2 2 2 21 1(1 ) (1 ) 2 2x x x x x x x x 35 2 .4 x x

0,25

Xt hm 3( ) 2f x x x trn 6 6;3 3

,

suy ra 2'( ) 6 1;f x x 6'( ) 0 .6

f x x r

Ta c 6 6 69

,3 6

f f

6 6 .3 6

f f

69

Do 6( ) .9

f x d

Suy ra 5 6 .36

P d

0,25

6 (1,0 im)

Khi 6 ,3 6

x y z 6 th du bng xy ra. Vy gi tr ln nht ca P l 5 6 .36

0,25

(C1) c tm l gc ta O. Gi I l tm ca ng trn (C) cn vit phng trnh, ta c .A

Trang 3/4

B OIA M AB dA v O d nn OI//d, do OI c phng trnh y x.

0,25

Mt khc 2( )I C , nn ta ca I tha mn h:

2 23

(3;3).312 18 0

y x xI

yx y x

0,25

Do (C) tip xc vi d nn (C) c bn knh ( , ) 2 2.R d I d 0,25

7.a (1,0 im)

Vy phng trnh ca (C) l 2 2( 3) ( 3) 8x y . 0,25

Gi (S) l mt cu cn vit phng trnh v I l tm ca (S). Do nn ta ca im I c dng I d (1 2 ; ; 2 ).I t t t 0,25

Do nn , ( )A B S ,AI BI suy ra .2 2 2 2 2 2(2 1) ( 1) 4 (2 3) ( 3) (2 2) 1t t t t t t t 0,25

Do v bn knh mt cu l ( 1; 1; 2)I 17.IA 0,25

8.a (1,0 im)

Vy, phng trnh mt cu (S) cn tm l 2 2 2( 1) ( 1) ( 2) 17x y z . 0,25

S cch chn 4 hc sinh trong lp l C 425 12650. 0,25

S cch chn 4 hc sinh c c nam v n l 1 3 2 2 3 115 10 15 10 15 10. . .C C C C C C 0,25

11075. 0,25

9.a (1,0 im)

Xc sut cn tnh l 11075 443.12650 506

P 0,25

B

A I d

(C2)

(C)

(C1)

/4

Gi s 2 2

2 2( ): 1( 0).x yE aa b

b ! ! Hnh thoi ABCD c

2AC BD v A, B, C, D thuc (E) suy ra OA 2 .OB 0,25

Khng mt tnh tng qut, ta c th xem v ( ;0)A a

0; .2aB Gi H l hnh chiu vung gc ca O trn AB, suy ra OH l bn knh ca ng trn ( ) 2 2: 4.C x y

0,25

Ta c: 2 2 2 21 1 1 1 1 4 .4 OH OA OB a a 2 0,25

7.b (1,0 im)

Suy ra do b Vy phng trnh chnh tc ca (E) l 2 20,a 2 5. 2 2

1.20 5x y 0,25

Do ,B Ox C Oy nn ta ca B v C c dng: B b v C c ( ; 0; 0) (0; ; 0). 0,25

Gi G l trng tm ca tam gic ABC, suy ra: ; ; 1 .G 3 3b c 0,25 Ta c nn ng thng AM c phng trnh (1;2; 3)AM

JJJJG 3.1 2 3x y z

Do G thuc ng thng AM nn 2 .3 6 3b c Suy ra 2b v 4.c

0,25

8.b (1,0 im)

Do phng trnh ca mt phng (P) l 1,2 4 3x y z ngha l ( ) : 6 3 4 12 0.P x y z 0,25

Phng trnh bc hai 2 2 3 4 0z i z c bit thc 4.' 0,25

Suy ra phng trnh c hai nghim: 1 1 3z i v 2 1 3z i . 0,25

x Dng lng gic ca l 1z 1 2 cos sin .3 3

z i 0,25

9.b (1,0 im)

x Dng lng gic ca l 2z 22 22 cos sin .3 3

z i 0,25

O

H

x

y

D

A

B

C

---------- HT ----------

B GIO DC V O TO

CHNH THC

THI TUYN SINH I HC NM 2012 Mn: TON; Khi D


I. PHN CHUNG CHO TT C TH SINH (7,0 im)

Cu 1 (2,0 im). Cho hm s 3 2 22 22(3 1) (1),3 3

y x mx m x m l tham s thc. a) Kho st s bin thin v v th ca hm s (1) khi 1.m b) Tm m hm s (1) c hai im cc tr 1x v 2x sao cho 1 2 1 22( ) 1.x x x x

Cu 2 (1,0 im). Gii phng trnh sin 3 cos3 sin cos 2 cos 2 .x x x x x

Cu 3 (1,0 im). Gii h phng trnh 3 2 2 22 0

( , ).2 2 0

xy xx y

x x y x y xy y

\

Cu 4 (1,0 im). Tnh tch phn

4

0

(1 sin 2 )d .I x x x

')

Cu 5 (1,0 im). Cho hnh hp ng c y l hnh vung, tam gic vung cn, . Tnh th tch ca khi t din v khong cch t im A n mt phng(

. ' ' ' 'ABCD A B C D 'A AC'AC a ' 'ABBC BCD theo a.

Cu 6 (1,0 im). Cho cc s thc ,x y tha mn Tm gi tr nh nht ca biu thc

2 2( 4) ( 4) 2 32.x y xy d3 3 3( 1)( 2).A x y xy x y

II. PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn ring (phn A hoc phn B) A. Theo chng trnh Chun Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hnh ch nht ABCD. Cc ng thng AC

v AD ln lt c phng trnh l v 3x y 0 4 0;x y ng thng BD i qua im 1;1 .3M Tm ta cc nh ca hnh ch nht ABCD. Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho mt phng v im Vit phng trnh mt cu tm I v ct (P) theo mt ng trn c bn knh bng 4.

( ): 2 2 10 0P x y z (2;1;3).I

Cu 9.a (1,0 im). Cho s phc z tha mn 2(1 2 )(2 ) 7 8 .1

ii z ii

Tm mun ca s phc 1 .w z i

B. Theo chng trnh Nng cao Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho ng thng Vit phng trnh ng trn c tm thuc d, ct trc Ox ti A v B, ct trc Oy ti C v D sao cho

: 2 3 0.d x y 2.AB CD

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho ng thng 1 1:2 1 1

x yd

z v hai

im Xc nh ta im M thuc d sao cho tam gic AMB vung ti M. (1; 1;2),A (2; 1;0).B

Cu 9.b (1,0 im). Gii phng trnh 2 3(1 ) 5 0z i z i trn tp hp cc s phc. ---------- HT ----------

Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm. H v tn th sinh:....................................................................; S bo danh: ..............................................



Mn: TON; Khi D (p n - thang im gm 04 trang)

Cu p n im a) (1,0 im)

Khi hm s tr thnh 1,m 3 22 24 .3 3

y x x x

x Tp xc nh: .D \x S bin thin:

- Chiu bin thin: hoc 22 2 4; 0y x x y xc c 1 2.x

0,25

Cc khong ng bin: v ( ; 1f ) (2; );f khong nghch bin . ( 1;2)- Cc tr: Hm s t cc i ti 1,x yC 3, t cc tiu ti 2,x yCT 6. - Gii hn: lim , lim ,

x xy y

of of f f

0,25

- Bng bin thin:

0,25

x th:

0,25

b) (1,0 im)

Ta c . 2 22 2 2(3 1)y x mx mc 0,25 th hm s c hai im cc tr khi v ch khi phng trnh 0yc c hai nghim phn bit

213 4 0m ! 2 1313

m ! hoc 2 13 .13

m 0,25

Ta c: 1 2x x m v 21 2 1 3 ,x x m do 21 2 1 22( ) 1 1 3 2 1x x x x m m 0,25

1 (2,0 im)

0m hoc 2 .3

m Kim tra iu kin ta c 2 .3

m 0,25

f

f 3

6

y

'y + 0 0 + x f 1 2 f

x 1 O 2

6

3

y

Trang 1/4

Cu p n im Phng trnh cho tng ng vi: (2sin 2cos 2)cos2 0.x x x 0,25

cos2 0 ( ).4 2

kx x kx ] 0,25

2sin 2cos 2 0x xx 1cos 4 2x 0,25

2 (1,0 im)

7 212

x k hoc 2 ( )12

x k k ] . Vy cc nghim ca phng trnh cho l:

,4 2

kx 7 2,12

x k 2 ( )12

x k k ] . 0,25

H cho tng ng vi: 22 0 (1)

(2)(2 1)( ) 0

xy x

x y x y

0,25

2 1 0 2x y y xx 1. Thay vo (1) ta c 2 1 51 0 .2

x x x r

Do ta c cc nghim 1 5( ; ) ; 52

x y

v 1 5( ; ) ; 5 .2

x y

0,25

2 0 2.x y yx x Thay vo (1) ta c 3 22 0 ( 1)( 2) 0x x x x x 0,25

3 (1,0 im)

1.x Do ta c nghim ( ; ) (1; 1).x y Vy h phng trnh cho c cc nghim l:

( ; ) (1; 1),x y 1 5( ; ) ; 52

x y

, 1 5( ; ) ; 5 .2

x y

0,25

4 4 4 42 24

0 0 0 00

d sin 2 d sin 2 d sin 22 32xI x x x x x x x x x x x d . 0,25

t suy ra ;d sin 2 d ,u x v x x 1d d ; cos22

u x v x . 0,25

Khi

4 4

4

00 0

1 1 1sin 2 d cos2 cos2 d cos2 d2 2 2

4

0

x x x x x x x x x 0,25

4 (1,0 im)

4

0

1 1sin 2 .4 4

x Do 2 1 .

32 4I 0,25

Tam gic A ACc vung cn ti A v A C ac nn A A ACc .

2a Do .

2aAB B Cc c 0,25

3'

1 1' '. ' '. . ' .3 6ABB C ABB

aV B C S B C AB BBc c ' 2

48 0,25

Gi H l chn ng cao k t A ca .A ABc' Ta c 'AH A BA v AH BCA nn ( ' ),AH A BCA

ngha l (AH BCD ').A Do ( ,( ')).AH d A BCD 0,25

5 (1,0 im)

Ta c 2 2 21 1 1 6 .

' 2AH AB AA a

Do 6( ,( ')) .6

ad A BCD AH 0,25

A B

C D

'A

'D 'C

'B

H

Trang 2/4

Cu p n im

Ta c 2 2( 4) ( 4) 2 32 2( ) 8( ) 0 0x y x y x yx y xy d 8. d d d 0,25

3( ) 3( ) 6 6A x y x y xy 3 23( ) ( ) 3( )2

x y x y x yt 6.

Xt hm s: 3 23( ) 3 62

f t t t t trn on [0 ; 8].

Ta c 2( ) 3 3 3,f t t tc 1 5( ) 02

f t t c hoc 1 52

t (loi).

0,25

Ta c 1 5 17 5 5(0) 6, , (8) 398.2 4

f f f

Suy ra 17 5 5 .4

A t 0,25

6 (1,0 im)

Khi 1 54

x y th du bng xy ra. Vy gi tr nh nht ca A l 17 5 5 . 4 0,25

Ta im A tha mn h 3 0

4 0x yx y

( 3;1). A 0,25

Gi N l im thuc AC sao cho MN//AD. Suy ra MN c

phng trnh l 4 0.3

x y V N thuc AC, nn ta

ca im N tha mn h 4 0 11; .3

33 0

x yN

x y

0,25

ng trung trc ' ca MN i qua trung im ca MN v vung gc vi AD, nn c phng trnh l 0.x y Gi I v K ln lt l giao im ca ' vi AC v AD.

Suy ra ta ca im I tha mn h 0

3 0x yx y

,

v ta ca im K tha mn h 04 0.

x yx y

Do I(0; 0) v K(2;2).

0,25

7.a (1,0 im)

2 (3; 1);AC AI C JJJG JJG

2 ( 1;3);AD AK D JJJG JJJG

(1; 3).BC AD B JJJG JJJG

0,25

Gi H l hnh chiu vung gc ca I trn (P). Suy ra H l tm ca ng trn giao tuyn ca mt phng (P) v mt cu (S) cn vit phng trnh. 0,25

Ta c ( ;( )) 3.IH d I P 0,25

Bn knh ca mt cu (S) l: 2 23 4 5R . 0,25

8.a (1,0 im)

Phng trnh ca mt cu (S) l: 2 2 2( 2) ( 1) ( 3) 25x y z . 0,25

Ta c: 2(1 2 )(2 ) 7 8 (2 ) 4 71

ii z i i z ii

0,25

3 2 .z i 0,25

Do 4 3 .w i 0,25

9.a (1,0 im)

Mun ca w l 2 24 3 5 . 0,25

I

N

M D C

B A

K

Trang 3/4

Cu p n im Gi I l tm ca ng trn (C) cn vit phng trnh. Do nn ta ca I c dng I d ( ;2 3).I t t 0,25

( , ) ( , )AB CD d I Ox d I Oy | | | 2 3 | 1t t t hoc 3.t 0,25 x Vi ta c nn 1t ( 1;1),I ( ; ) 1.d I Ox Suy ra, bn knh ca (C) l 2 21 1 2. Do 2 2( ): ( 1) ( 1) 2.C x y

0,25

7.b (1,0 im)

x Vi ta c nn 3t ( 3; 3),I ( ; ) 3.d I Ox Suy ra, bn knh ca (C) l 2 23 1 10. Do 2 2( ): ( 3) ( 3) 10.C x y

0,25

Do M d nn ta ca im M c dng (1 2 ; 1 ; ).M t t t 0,25 Ta c (2 ; ; 2), ( 1 2 ; ; ).AM t t t BM t t t

JJJJG JJJJG

Tam gic AMB vung ti M . 0AM BM JJJJG JJJJG

0,25

2 22 ( 1 2 ) ( 2) 0 6 4 0t t t t t t t 0,25

8.b (1,0 im)

0t hoc 2 .3

t Do 1; 1;0M hoc 7 5 2; ;3 3 3

M . 0,25

Phng trnh bc hai c bit thc 2 3(1 ) 5 0z i z i 2 .i' 0,25 2(1 ) .i 0,25

Do nghim ca phng trnh l 3(1 ) (1 ) 1 22

i iz i 0,25

9.b (1,0 im)

hoc 3(1 ) (1 ) 2 .2

i iz i 0,25

------------- HT-------------

Trang 4/4

A2014A2014-DAB2014B2014-DAD2014D2014-DAA2013A2013-DAB2013B2013-DAD2013D2013-DAA2012A2012-DAB2012B2012-DAD2012D2012-DA

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