de mau toan2015.pdf
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B GIO DC V O TO
THI MINH HA - K THI THPT QUC GIA NM 2015 Mn: TON
Thi gian lm bi: 180 pht.
Cu 1.(2,0 im) Cho hm s 2 1.1
xyx
=
+
a) Kho st s bin thin v v th (C) ca hm s cho. b) Vit phng trnh tip tuyn ca th (C), bit tip im c honh 1.x =
Cu 2.(1,0 im) a) Cho gc tha mn:
2< < v 3sin .
5= Tnh 2
tan .
1 tan A =
+
b) Cho s phc z tha mn h thc: (1 ) (3 ) 2 6 .i z i z i+ + = Tnh mun ca z. Cu 3.(0,5 im) Gii phng trnh: 3 3log ( 2) 1 log .x x+ =
Cu 4.(1,0 im) Gii bt phng trnh: 2 22 3( 2 2).x x x x x+ +
Cu 5.(1,0 im) Tnh tch phn: 2
3
1
(2 ln )d .I x x x= +
Cu 6.(1,0 im) Cho hnh chp S.ABC c y ABC l tam gic vung ti B, AC = 2a, o30 ,ACB = Hnh chiu vung gc H ca nh S trn mt y l trung im ca cnh AC v 2 .SH a= Tnh theo a th tch khi chp S.ABC v khong cch t im C n mt phng (SAB). Cu 7.(1,0 im) Trong mt phng vi h ta Oxy , cho tam gic OAB c cc nh A v B thuc ng thng : 4 3 12 0x y + = v im (6; 6)K l tm ng trn bng tip gc O. Gi C l im nm trn sao cho AC AO= v cc im C, B nm khc pha nhau so vi im A. Bit im C c
honh bng 24 ,5
tm ta ca cc nh A, B.
Cu 8.(1,0 im) Trong khng gian vi h ta Oxyz, cho hai im (2; 0; 0)A v (1; 1; 1).B Vit phng trnh mt phng trung trc (P) ca on thng AB v phng trnh mt cu tm O, tip xc vi (P). Cu 9.(0,5 im) Hai th sinh A v B tham gia mt bui thi vn p. Cn b hi thi a cho mi th sinh mt b cu hi thi gm 10 cu hi khc nhau, c ng trong 10 phong b dn kn, c hnh thc ging ht nhau, mi phong b ng 1 cu hi; th sinh chn 3 phong b trong s xc nh cu hi thi ca mnh. Bit rng b 10 cu hi thi dnh cho cc th sinh l nh nhau, tnh xc sut 3 cu hi A chn v 3 cu hi B chn l ging nhau.
Cu 10.(1,0 im) Xt s thc x. Tm gi tr nh nht ca biu thc sau: 2
2 2
3 2 2 1 1 13 2 3 3 3 2 3 3 3
+ += + +
+ + + + +
( ).
( ) ( )
x xP
x x x x
----------- HT -----------
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B GIO DC V O TO P N - THANG IM THI MINH HA - K THI THPT QUC GIA NM 2015
Mn: TON
CU P N IM Cu 1
(2,0 im)
a) (1,0 im) Tp xc nh: { }\ 1 .D = Gii hn v tim cn:
( 1)lim
xy
+ = ,
( 1)lim
xy
= + ; lim lim 2.
x xy y
+= =
Suy ra, th hm s c mt tim cn ng l ng thng 1x = v mt tim cn ngang l ng thng 2.y =
0,25
S bin thin:
- Chiu bin thin: y' = 23
( 1)x + > 0 x D.
Suy ra, hm s ng bin trn mi khong ( ); 1 v ( )1; + . - Cc tr: Hm s cho khng c cc tr.
0,25
Lu : Cho php th sinh khng nu kt lun v cc tr ca hm s.
- Bng bin thin:
x 1 +
y' + +
y + 2 2
0,25
th (C):
0,25
O x
y
1 1
2
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b) (1,0 im) Tung 0y ca tip im l: 0
1(1) .2
y y= = 0,25
Suy ra h s gc k ca tip tuyn l: 3'(1) .4
k y= = 0,25
Do , phng trnh ca tip tuyn l: 3 1( 1) ;4 2
y x= + 0,25
hay 3 1 .4 4
y x= 0,25
Cu 2 (1,0 im)
a) (0,5 im) Ta c: 22
tan 3tan .cos sin .cos cos.
1 tan 5A = = = =
+ (1) 0,25
22 2 3 16cos 1 sin 1 .
5 25
= = =
(2)
V ;2pi
pi
nn cos 0.< Do , t (2) suy ra 4cos .5
= (3)
Th (3) vo (1), ta c 12 .25
A =
0,25
b) (0,5 im) t z = a + bi, ( ,a b ); khi z a bi= . Do , k hiu () l h thc cho trong bi, ta c: () (1 )( ) (3 )( ) 2 6i a bi i a bi i+ + + = (4 2 2) (6 2 ) 0a b b i + =
0,25
{4 2 2 06 2 0a bb = = { 23.ab == Do 2 2| | 2 3 13.z = + =
0,25
Cu 3 (0,5 im)
iu kin xc nh: 0.x > (1) Vi iu kin , k hiu (2) l phng trnh cho, ta c: (2) 3 3log ( 2) log 1x x+ + = 3 3log ( ( 2)) log 3x x + =
0,25
2 2 3 0x x+ = 1x = (do (1)). 0,25
Cu 4 (1,0 im)
iu kin xc nh: 1 3.x + (1) Vi iu kin , k hiu (2) l bt phng trnh cho, ta c: (2) 2 22 2 2 ( 1)( 2) 3( 2 2)x x x x x x x+ + +
0,25
( 2)( 1) ( 2) 2( 1)x x x x x x + + ( )( )( 2) 2 ( 1) ( 2) ( 1) 0.x x x x x x + + + (3) Do vi mi x tha mn (1), ta c ( 2) ( 1) 0x x x + + > nn (3) ( 2) 2 ( 1)x x x +
0,50
2 6 4 0x x 3 13 3 13.x + (4) Kt hp (1) v (4), ta c tp nghim ca bt phng trnh cho l:
1 3 ; 3 13 . + +
0,25
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Cu 5 (1,0 im) Ta c:
2 23
1 1
2 d ln d .I x x x x= + (1) 0,25
t 2
31
1
2 dI x x= v 2
21
ln d .I x x= Ta c:
24
11
1 15.
2 2I x= =
0,25
2 22 2
2 1 11 1
.ln d(ln ) 2ln 2 d 2ln 2 2ln 2 1.I x x x x x x= = = =
Vy 1 213 2 ln 2.2
I I I= + = + 0,50
Cu 6 (1,0 im)
Theo gi thit, 12
HA HC AC a= = = v SH mp(ABC).
Xt v. ABC, ta c: o.cos 2 .cos 30 3 .BC AC ACB a a= = = 0,25
Do o 21 1 3. .sin .2 . 3 .sin 30 .2 2 2ABC
S AC BC ACB a a a= = =
Vy 3
2.
1 1 3 6. . 2 . .
3 3 2 6S ABC ABCaV SH S a a= = =
0,25
V CA = 2HA nn d(C, (SAB)) = 2d(H, (SAB)). (1) Gi N l trung im ca AB, ta c HN l ng trung bnh ca ABC. Do HN // BC. Suy ra AB HN. Li c AB SH nn AB mp(SHN). Do mp(SAB) mp(SHN). M SN l giao tuyn ca hai mt phng va nu, nn trong mp(SHN), h HK SN, ta c HK mp(SAB). V vy d(H, (SAB)) = HK. Kt hp vi (1), suy ra d(C, (SAB)) = 2HK. (2)
0,25
V SH mp(ABC) nn SH HN. Xt v. SHN, ta c:
2 2 2 2 21 1 1 1 1
.
2HK SH HN a HN= + = +
V HN l ng trung bnh ca ABC nn 1 3 .2 2
aHN BC= =
Do 2 2 2 21 1 4 11
.
2 3 6HK a a a= + = Suy ra 66 .
11aHK = (3)
Th (3) vo (2), ta c ( ) 2 66, ( ) .11
ad C SAB =
0,25
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Cu 7 (1,0 im)
Trn , ly im D sao cho BD = BO v D, A nm khc pha nhau so vi B. Gi E l giao im ca cc ng thng KA v OC; gi F l giao im ca cc ng thng KB v OD. V K l tm ng trn bng tip gc O ca OAB nn KE l phn gic ca gc
.OAC M OAC l tam gic cn ti A (do AO = AC, theo gt) nn suy ra KE cng l ng trung trc ca OC. Do E l trung im ca OC v KC = KO. Xt tng t i vi KF, ta cng c F l trung im ca OD v KD = KO. Suy ra CKD cn ti K. Do , h KH , ta c H l trung im ca CD. Nh vy: + A l giao ca v ng trung trc 1d ca on thng OC; (1) + B l giao ca v ng trung trc 2d ca on thng OD, vi D l im i xng ca C qua H v H l hnh chiu vung gc ca K trn . (2)
0,50
V C v c honh 0245
x = (gt) nn gi 0y l tung ca C, ta c:
0244. 3 12 0.5
y+ = Suy ra 012
.
5y =
T , trung im E ca OC c ta l 12 6;5 5
v ng thng OC c
phng trnh: 2 0.x y+ = Suy ra phng trnh ca 1d l: 2 6 0.x y = Do , theo (1), ta ca A l nghim ca h phng trnh:
{4 3 12 02 6 0.x yx y+ = = Gii h trn, ta c A = (3; 0).
0,25
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Gi d l ng thng i qua K(6; 6) v vung gc vi , ta c phng trnh ca d l: 3 4 6 0.x y + = T y, do H l giao im ca v d nn ta ca H l nghim ca h phng trnh:
{4 3 12 03 4 6 0.x yx y+ = + = Gii h trn, ta c 6 12; .
5 5H =
Suy ra 12 36; .
5 5D =
Do , trung im F ca OD c ta l 6 18;5 5
v ng thng OD c
phng trnh: 3 0.x y+ = Suy ra phng trnh ca 2d l: 3 12 0.x y + = Do , theo (2), ta ca B l nghim ca h phng trnh:
{4 3 12 03 12 0.x yx y+ = + = Gii h trn, ta c B = (0; 4).
0,25
Cu 8 (1,0 im) Gi M l trung im ca AB, ta c
3 1 1; ; .
2 2 2M =
V (P) l mt phng trung trc ca AB nn (P) i qua M v ( 1; 1; 1)AB =
l mt vect php tuyn ca (P).
0,25
Suy ra, phng trnh ca (P) l: 3 1 1( 1) ( 1) 02 2 2
x y z + + + =
hay: 2 2 2 1 0.x y z + = 0,25
Ta c 2 2 2
| 1| 1( , ( )) .2 32 ( 2) 2
d O P = =+ +
0,25
Do , phng trnh mt cu tm O, tip xc vi (P) l: 2 2 2 112
x y z+ + =
hay 2 2 212 12 12 1 0.x y z+ + = 0,25
Cu 9 (0,5 im)
Khng gian mu l tp hp gm tt c cc cp hai b 3 cu hi, m v tr th nht ca cp l b 3 cu hi th sinh A chn v v tr th hai ca cp l b 3 cu hi th sinh B chn. V A cng nh B u c 310C cch chn 3 cu hi t 10 cu hi thi nn theo quy
tc nhn, ta c ( )2310( ) C .n = 0,25
K hiu X l bin c b 3 cu hi A chn v b 3 cu hi B chn l ging nhau. V vi mi cch chn 3 cu hi ca A, B ch c duy nht cch chn 3 cu hi ging nh A nn ( ) 3 310 10C .1 C .Xn = = V vy ( ) ( )
310
2 331010
C 1 1( ) .( ) C 120CXnP X
n
= = = =
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Cu 10 (1,0 im)
Trong mt phng vi h ta Oxy, vi mi s thc x, xt cc im ( ; 1)A x x + , 3 1
;2 2
B
v 3 1; .2 2
C
Khi , ta c ,OA OB OCPa b c
= + + trong a = BC, b = CA v c = AB.
0,25
Gi G l trng tm ABC, ta c: . . . 3 . . .
. . . 2 . . .a b cOA GA OB GB OC GC OA GA OB GB OC GCPa GA b GB c GC a m b m c m
= + + = + +
,
trong ,a bm m v cm tng ng l di ng trung tuyn xut pht t A, B, C ca ABC.
0,25
Theo bt ng thc C si cho hai s thc khng m, ta c
( )( )
2 2 2 2
2 2 2 2 2 2 2
1. . 3 2 2
2 33 2 21
. .
22 3 2 3
aa m a b c a
a b c a a b c
= +
+ + + + =
Bng cch tng t, ta cng c: 2 2 2
.
2 3ba b cb m + + v
2 2 2
. .
2 3ca b c
c m+ +
Suy ra ( )2 2 23 3 . . . .P OAGA OB GB OC GCa b c + ++ + (1)
0,25
Ta c: . . . . . . .OAGA OB GB OC GC OA GA OB GB OC GC+ + + +
(2)
( ) ( ) ( )( )
( )2 2 2
2 2 22 2 2
. . .
. . .
.
4. (3)
9 3a b c
OAGA OB GB OC GCOG GA GA OG GB GB OG GC GC
OG GA GB GC GA GB GCa b c
m m m
+ +
= + + + + +
= + + + + +
+ += + + =
T (1), (2) v (3), suy ra 3.P Hn na, bng kim tra trc tip ta thy 3P = khi x = 0. Vy min 3.P =
0,25
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