THI TUYN SINH I HC NM 2011 Mn : TON ; Khi: D PHN CHUNG CHO TT C TH SINH (7,0 im)2x 1 x 1 1. Kho st s bin thin v v th (C) ca hm s cho 2. Tm k ng thng y = kx + 2k +1 ct th (C) ti hai im phn bit A, B sao cho khong cch t A v B n trc honh bng nhau. Cu II (2,0 im) sin2x 2cos x sin x 1 1. Gii phng trnh 0 tan x 3 2. Gii phng trnh log 2 (8 x 2 ) log 1 ( 1 x 1 x ) 2 0 (x )

Cu I (2,0 im) Cho hm s y

2

4x 1 dx 2x 1 2 0 Cu IV (1,0 im) Cho hnh chp S.ABC c y ABC l tam gic vung ti B, BA = 3a, BC = 4a; mt phng (SBC) vung gc vi mt phng (ABC). Bit SB = 2a 3 v SBC = 300 . Tnh th tch khi chp S.ABC v khong cch t im B n mt phng (SAC) theo a. 2 x3 ( y 2) x 2 xy m ( x, y ) Cu V (1,0 im) Tm m h phng trnh sau c nghim 2 x x y 1 2m PHN RING (3,0 im) : Th sinh ch c lm mt trong hai phn (phn A hoc B) Cu VI.a (2,0 im) 1. Trong mt phng ta Oxy, cho tam gic ABC c nh B(-4; 1), trng tm G(1; 1) v ng thng cha phn gic trong ca gc A c phng trnh x y 1 = 0. Tm ta cc nh A v C. x 1 y z 3 2. Trong khng gian vi h ta Oxyz, cho im A(1; 2; 3) v ng thng d : . 2 1 2 Vit phng trnh ng thng i qua im A, vung gc vi ng thng d v ct trc Ox. Cu VII.a (1,0 im) Tm s phc z, bit : z (2 3i) z 1 9i B. Theo chng trnh Nng cao Cu VI.b (2,0 im) 1. Trong mt phng ta Oxy, cho im A(1; 0) v ng trn (C) : x2 + y2 2x + 4y 5 = 0. Vit phng trnh ng thng ct (C) ti im M v N sao cho tam gic AMN vung cn ti A. x 1 y 3 z 2. Trong khng gian vi h ta Oxyz, cho ng thng : v mt phng 2 4 1 (P) : 2x y + 2z = 0. Vit phng trnh mt cu c tm thuc ng thng , bn knh bng 1 v tip xc vi mt phng (P). 2 x 2 3x 3 Cu VII.b (1,0 im) Tm gi tr nh nht v gi tr ln nht ca hm s y trn on x 1 [0;2]. ----- Ht -----

4

Cu III (1,0 im) Tnh tch phn I

BI GII GI Cu I : 1. Kho st v v th (C) D = R \ {-1} 1 y/ = > 0 vi mi x ( x 1)2 v lim y lim yx 1 x 1

D x = -1 l TC

x

lim y

2

y = 2 l TCN x y/ y 2 + + -1 + 2 +

BBT :

Hm s ng bin trn tng khong xc nh, khng c cc tr. th hm s :y

2 -1 O

x

2. Pt honh giao im : 2x 1 kx 2k 1 x 1 kx2 + (3k - 1)x + 2k = 0 (x = -1 khng l nghim) Ycbt : k 0 v = k2 - 6k + 1 > 0 k < 3 2 2 k 3 2 2 v k 0 (*) Khong cch t A v B n Ox bng nhau kxA kxB (loai ) 1 3k kxA 2k 1 kxB 2k 1 yA = yB k( ) 4k 2 0 k ( xA xB ) 4k 2 0 k k = 3 (tha k (*) ). Vy YCBT k=3 Cu II : sin 2 x 2cos x sin x 1 3 ; cosx 0 1) 0 k : tg x tan x 3 Pt sin2x + 2cosx sinx 1 = 0 2sinxcosx + 2cosx (sinx + 1) = 0 2cosx (sinx + 1) (sinx + 1)= 0 (2cosx 1)(sinx + 1) = 0

cos x sin x

2)

k2 2 log 2 (8 x 2 ) log 2 ( 1 x 1 x)log 2 (8 x )2

1 2 1

x x

3

k2

so k ta c nghim ca pt : x

3

k2

(k

)

2 (x1 x)

[-1;1]) 8 x2 = 4( 1 x

2 log 2 ( 1 x

1 x ) (*)

1 x t t = 1 x 2 2 (*) thnh (t 2) (t + 4t + 8) = 0 1 x 1 x =2 t=2 Cu III : 4 4x 1 dx I= 2x 1 2 0

x = 0 (nhn)

2 x 1 2 => (t - 2)dt = dx 5 (2t 2 8t 5)(t 2) 10 34 3 dt (2t 2 12t 21 )dt 10 ln => I = t t 3 5 3 3 Cu IV : Gi H l hnh chiu ca S xung BC. V (SBC) (ABC) nn SH (ABC) Ta c SH = a 3 1 1 1 Th tch khi (SABC) = S ABC .SH ( 3a.4a).a 3 2a 3 3 3 3 2 Ta c : Tam gic SAC vung ti S v SA = a 21 ; SC = 2a; AC = 5a. Din tch (SAC) = a 2 21 3VSABC 3.2a 3 3 6a d(B,(SAC)) = = 2 S SAC a 21 7 Cu V : 1 ( x 2 x)(2 x y ) m u x 2 x (dk u ) H t 4 2 ( x x) (2 x y ) 1 2m v 2 x y (v )5

t t =

S B J A H I

C

H thnh :

u v 1 2m uv m

v (1 2m) u u2 u m(2u 1)

v 1 2m u u2 u 2u 1 m (1)

t f(u) =

u2 u ,u 2u 1u f/(u) f(u)

1 / ; f (u) = 41 4+1

2u 2 2u 1 / ;f (u)=0 (2u 1) 2

u

1 2

3

(loi) hay u

1 2

3

3

2 0 2 3 2

+

5 8

Vy h c nghim Cu VIa :

(1) c nghim thuc

1 ; 4

m

2 2

3

3 7 BG M ;1 2 2 Gi N l im i xng ca B qua phn gic trong ca gc A v H l giao im ca ng thng BN.

1. Gi M l trung im ca AC, ta c BM

vi

ng thng BN c phng trnh : x + y + 3 = 0 x y 3 0 => Ta H l nghim ca h phng trnh : x y 1 0 H l trung im ca BN

H ( 1; 2)

N (2; 5) y N 2 yH yB 5 ng thng AC qua 2 im M, N nn c pt : 4x y 13 = 0 A l giao im ca ng thng v ng thng AC nn ta A l nghim 4 x y 13 0 ca h : A(4;3) x y 1 0M l trung im ca AC 2.

xN

2 xH

xB

2

xC yC

2 xM 2 yM

xA yA

3 1

C(3; 1)

Gi M l giao im ca ng thng vi Ox M (m; 0; 0) AM = (m 1; -2; -3) AM d m = -1 AM = (-2; -2; -3) AM . ad = 0 x 1 y 2 z 3 Vy pt l 2 2 3 Cu VII.a : Gi z = a + bi (a, b R). Khi z (2 + 3i) z = 1 9i a + bi (2 + 3i)(a bi) = 1 9i a 3b 1 a 2 (a + 3b) + (3b 3a)i = 1 9i 3b 3a 9 b 1 Vy z = 2 iCu VI.b : 1. ng trn (C) c tm I (1; -2), R = 10 AI (0; 2) . V I v A cch u M, N nn MN MN = 2 d A/ MN 2 bd I / MN b 2

AI, vy pt MN c dng : y = b

d

2 I / MN

MN 2

2

R21

b22

2b 3 0

b 1v b

3

Vy Pt : 2.

:y=1;

:y=

3x 1 2t y 3 4t z t

Phng trnh tham s ng thng I ( ) I (1 + 2t; 3 + 4t; t) 2(1 2t ) (3 4t ) 2t

d (I, P) =

3

=1

t = 2 hay t = -1

I1 (5; 11; 2) Pt mt cu (S) : (x 5)2 + (y 11)2 + (z 2)2 = 1 I2 (-1; -1; -1) Pt mt cu (S) : (x + 1)2 + (y + 1)2 + (z + 1)2 = 1 Cu VII.b : 2x2 4x / Ta c : y = ( x 1) 2 y/ = 0 x = 0 v x = 2 (loi) 17 m y(0) = 3 v y(2) = 3 17 Vy GTLN l v GTNN l 3 3