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BGIO DC VO TO
CHNH TH C
THI TUYN SINHI HC NM 2011Mn: TON; Khi: D
Th i gian lm bi: 180 pht, khng k th i gian pht
PHN CHUNG CHO TT C TH SINH (7,0 i m)
Cu I (2,0 i m) Cho hm s 2 11
x y
x
+=
+
1. Kho st s bin thin v v th (C) ca hm s cho.2. Tmk ng thng y =
kx + 2k + 1 ct th (C ) ti haiim phn bit A, B sao cho khongcch t A v
B n tr c honh bng nhau.
Cu II (2,0 i m)
1. Gii ph ng trnh sin 2 2cos sin 1 0.tan 3
x x x
x
+ =
+
2. Gii ph ng trnh ( ) ( )22 12
log 8 log 1 1 2 0 ( ). x x x + + + = x
Cu III (1,0 i m) Tnh tch phn4
0
4 1 d .2 1 2
x I x
x
=
+ +
Cu IV (1,0 i m) Cho hnh chpS.ABC c y ABC l tam gic vung ti B, BA
= 3a , BC = 4a;mt phng (SBC ) vung gc v i mt phng ( ABC ). Bit SB =
2 3a v Tnh th tchkhi chpS . ABC v khong cch t im B n mt phng (SAC )
theoa .
30 .SBC =
Cu V (1,0 i m) Tmm hph ng trnh sau c nghim:3 2
2
2 ( 2)( , ).
1 2 x y x xy m
x y x x y m
+ + =
+ =
PHN RING (3,0 i m): Th sinh ch c lm m t trong hai ph n ( ph n A
ho c B )
A. Theo ch ng trnh ChunCu VI.a (2,0 i m) 1. Trong mt phng ta
Oxy, cho tam gic ABC c nh B( 4; 1), tr ng tmG(1; 1) v ng
thng cha phn gic trong ca gc A c ph ng trnh x y 1 = 0. Tm ta
ccnh A v C .
2. Trong khng gian v i h to Oxyz, cho im A(1; 2; 3) v ng thng d
: 1 32 1 2
x y z+ = =
Vit ph ng trnh ng thng i quaim A, vung gc v i ng thng d v ct tr
c Ox.Cu VII.a (1,0 i m) Tm sphc z, bit: z (2+ 3i) z = 1 9i.B. Theo
ch ng trnh Nng caoCu VI.b (2,0 i m) 1. Trong mt phng to Oxy, choim
A(1; 0) v ng trn (C ): x2 + y2 2 x + 4 y 5 = 0. Vit
ph ng trnh ng thng ct (C ) ti haiim M v N sao cho tam gic AMN
vung cn ti A.
2. Trong khng gian v i h ta Oxyz, cho ng thng 1 3:2 4 1
y = = z v mt phng
Vit ph ng trnh mt cu c tm thuc ng thng , bn knh bng 1 vti p xc v
i mt phng ( P ).( ) : 2 2 0. P x y z + =
Cu VII.b (1,0 i m) Tm gi tr nh nht v gi tr l n nht ca hm s 22
3
1 x x
y x
+ +=
+
3 trn
on [0; 2]. ----------- Ht ----------
Th sinh khng c s d ng ti li u. Cn b coi thi khng gi i thch g
thm.Hv tn th sinh:............................................. ;
Sbo danh:................................
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BGIO DC VO TO
CHNH TH C
P N THANGIM THI TUYN SINHI HC NM 2011
Mn: TON; Khi D(p n - thangim gm 04 trang)
P N THANG I M
Cu p n i m
1. (1,0 i m) T p xcnh: { }\ 1 D = . S bin thin:
Chiu bin thin: 21' 0
( 1) y
x=
+,> x D.
Hm s ng bin trn cc khong ( ; 1) v ( 1;+ ).
0,25
Gi i hn v tim cn: lim lim x x
y y +
= = 2; tim cn ngang: y = 2.
= + , = ; tim cn ng: x = 1.( )1
lim x
y
( )1lim
x y
+
0,25
Bng bin thin:
Trang 1/4
0,25
th:
0,25
2. (1,0 i m)
Gi d : y = kx + 2k + 1, suy ra honhgiaoim ca d v (C ) l nghim ph
ng trnh:
kx + 2k + 1 = 2 11 x
+
+ 2 x + 1 = ( x + 1)(kx + 2k + 1) (do x = 1 khng l nghim)
kx2 + (3k 1) x + 2k = 0 (1).
0,25
d ct (C ) ti haiim phn bit A v B, khi v ch khi (1) c hai nghim
phn bit
0
0
k
> 20
6 1 0
k
k k
+ >
0
3 2 2 3 2 2.
k
k k
< > +
(*). 0,25
I(2,0 i m)
Khi: A( x1; kx1 + 2k + 1) v B( x2; kx2 + 2k + 1), x1 v x2 l
nghim ca (1).
x 1 y + +
y
+
+
2
2
2
x
y
1 O
1
0,25d( A, Ox) = d( B, Ox) 1 2 1kx k + + = 2 2 1kx k + +
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Cu p n i m
k ( x1 + x2) + 4k + 2 = 0 (do x1 x2).p dng nh l Viti v i (1),
suy ra: (1 3k ) + 4k + 2 = 0 k = 3, tha mn (*).Vy, gi tr cn tm l: k
= 3.
0,25
1. (1,0 i m) iu kin: cos x 0, tan x 3 (*).
Ph ng trnh cho t ng ng v i: sin2 x +
2cos x sin x 1=
0
0,25
2cos x(sin x + 1) (sin x + 1) = 0 (sin x + 1)(2cos x 1)= 0.
0,25
sin x = 1 x = 2
+ k 2 hoc cos x = 12
x = 3
+ k 2 . 0,25
i chiu iu kin (*), suy ra nghim: x = 3
+ k 2 (k Z ). 0,25
2. (1,0im) iu kin: 1 x 1 (*).Khi, ph ng trnh cho t ng ng v i: (
) ( )22 2log 8 log 4 1 1 x = + + x 0,25
8 x2 = 4( 1 1 ) x x+ + (8 x2)2 = 16( )22 2 1+ (1). 0,25t t = 21
x , (1) tr thnh: (7+ t 2)2 = 32(1+ t ) t 4 + 14t 2 32t + 17 = 0
(t 1)2(t 2 + 2t + 17)= 0 t = 1.0,25
II(2,0 i m)
Do, (1) 21 = x 1 x = 0, tha mn (*).Vy, ph ng trnh c nghim: x =
0.
0,25
t t = 2 1+ 4 x = 2(t 2 1), d x = t dt .i cn: x = 0 t = 1; x = 4
t = 3.
0,25
I = 3 3
12 3 d2
t t t t
+ =
32
1102 4 5 2
t t t
+ +
III
dt 0,25
= 33
2
1
2 2 5 10ln 23t
t t t
+ + 0,25
(1,0 i m)
= 34 310ln .3 5
+ 0,25
H SH BC ( H BC ); (SBC ) ( ABC ) SH ( ABC ); SH = SB.sin = SBC
3.a 0,25
Din tch: S ABC = 1
2BA. BC = 6a2.
Th tch: VS.ABC = 13
S ABC .SH = 32 3
IV
.a0,25
H HD AC ( D AC ), HK SD ( K SD) HK (SAC ) HK = d( H , (SAC
)).
BH = SB.cos = 3a BC = 4 HC SBC d( B, (SAC )) = 4.d( H , (SAC
)).
0,25
(1,0 i m)
Ta c AC = 2 2 BA BC + = 5a; HC = BC BH = a HD = BA. HC AC
= 3 .5a
HK = 2 2.SH HDSH HD+ = 3 714a . Vy, d( B, (SAC )) = 4. HK = 6 7
.7a
0,25
V(1,0 i m) H cho t ng ng v i:
2
2
( )(2 )( ) (2 ) 1 2 x x x y m
B
S
A
C D
H K
. x x x y
=
+ = m 0,25
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5/5 Trang 4/4
Cu p n i m
a 3b (3a 3b)i = 1 9i 0,25
3 1
3 3 9a b
a b
=
=0,25
(1,0 i m)
Vy z = 2 i.21.
a
b
=
= 0,25
1. (1,0im)
ng trn (C ) c tm I (1; 2), bn knh bng 10.Ta c: IM = IN v AM = AN
AI MN ; suy ra ph ngtrnh c dng: y = m.
0,25
Honh M , N l nghim ph ng trnh: x2 2 x + m2 + 4m 5 = 0 (1).(1) c
hai nghim phn bit x1 v x2, khi v ch khi:m2 + 4m 6 < 0 (*); khi
ta c: M ( x1; m) v N ( x2; m).
0,25
AM AN = 0 ( x. AM AN 1 1)( x2 1)+ m2 = 0 x1 x2 ( x1 + x2) + m2 +
1 = 0. 0,25
p dng nh l Viti v i (1), suy ra: 2m2 + 4m 6 = 0 m = 1 hoc m = 3,
tha mn (*). Vy, ph ng trnh: y = 1 hoc y = 3.
0,25
2. (1,0im)
Gi I l tm ca mt cu. I , suy ra ta I c dng: I (1 + 2t ; 3 + 4t ;
t ). 0,25 Mt cu ti p xc v i ( P ), khi v ch khi: d( I , ( P )) =
1
2(1 2 ) (3 4 ) 2
3
t t + + + t = 1 0,25
t = 2 hoc t = 1. Suy ra: I (5; 11; 2) hoc I ( 1; 1; 1). 0,25
VI.b(2,0 i m)
Ph ng trnh mt cu:( x 5)2 + ( y 11)2 + ( z 2)2 = 1 hoc ( x + 1)2
+ ( y + 1)2 + ( z + 1)2 = 1.
0,25
2
22 4'( 1) x x
y x
+=
+; 0,25
y' = 0 x = 2 hoc x = 0. 0,25
y(0) = 3, y(2) = 17.3
0,25
VII.b(1,0 i m)
Vy:[ ]0; 2miny = 3, ti x = 0;
[ ]0; 2maxy = 17,
3ti x = 2. 0,25
------------- Ht -------------
A
y
xO
M N
I 2
3
1
