25 de thi thu dai hoc sp hn co dap an chi tiet
TRANSCRIPT
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I HC SPHM H NI THI THI HC CAO NG 2011KHOA TON-TIN MN: TON- KHI A
------------- Thi gian lm bi: 180 pht( khng kthi gian giao )---------------------------------------------------------------------------------------------------------------------------------------------
A. PHN CHUNG CHO MI TH SINHCu I (2 im).
1. Kho st v v th hm s y = x4 4x2 + 32.
Tm m phng trnh
4 2
24 3 log x x m + =
c ng 4 nghim.Cu II (2 im).
1. Gii bt phng trnh: ( ) ( )325 1 5 1 2 0
x x x+
+ +
2. Gii phng trnh: 2 ( 2) 1 2 x x x x + = Cu III (2 im)
1. Tnh gii hn sau: 1 231
tan( 1) 1lim
1
x
x
e x
x
+
2. Cho hnh chp S.ABCD c y l hnh thoi , BAD = . Hai mt bn (SAB) v (SAD) cng vunggc vi mt y, hai mt bn cn li hp vi y mt gc . Cnh SA = a. Tnh din tch xung quanh
v th tch khi chp S.ABCD.
Cu IV (1 im). Cho tam gic ABC vi cc cnh l a, b, c. Chng minh rng:3 3 3 2 2 2 2 2 23 ( ) ( ) ( )a b c abc a b c b c a c a b+ + + + + + + +
B. PHN TCHN:Mi th sinh ch chn cu Va hoc VbCu Va (3 im). Chng trnh cbn
1. Trong mt phng ta Oxy cho ng thng : 2 3 0x y + = v hai im A(1; 0), B(3; - 4).Hy tm trn ng thng mt im M sao cho 3 MA MB+
nh nht.
2. Trong khng gian Oxyz cho hai ng thng: 1 1: 22
x td y t
z t
= =
= +
v 2 : 1 3
1
x td y t
z t
== +
=
.
Lp phng trnh ng thng i qua M(1; 0; 1) v ct c d1 v d2.
3. Tm s phc z tha mn: 2 2 0z z+ = Cu Vb. (3 im).Chng trnh nng cao
1. Trong mt phng ta cho hai ng trn (C1): x2 + y2 = 13 v (C2): (x - 6)2 + y2 = 25 ct nhau tiA(2; 3). Vit phng trnh ng thng i qua A v ct (C1), (C2) theo hai dy cung c di bng nhau.
2. Trong khng gian Oxyz cho hai ng thng: 11
: 2
2
x t
d y t
z t
=
= = +
v 2 : 1 3
1
x t
d y t
z t
=
= + =
.
Lp phng trnh mt cu c ng knh l on vung gc chung ca d1 v d2.3. Trong cc s phc z tha mn iu kin 1 2 1z i+ + = , tm s phc z c modun nh nht.
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I HC SPHM H NI------------------------------------------------------------------------------------------------------------
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1
P N THANG IM THI THI HC, CAO NG NM 2011
MN:TON, Khi A
Cu Ni dung im2
1 1TX D = Gii hn : lim
xy
= +
S bin thin : y = 4x3 - 8x
y = 0 0, 2x x = =
Bng bin thin
x 2 0 2+
y - 0 + 0 - 0 +y + +
3-1 -1
Hm sng bin trn cc khong ( ) ( )2;0 , 2; + v nghch bin trn cc khong
( ) ( ); 2 , 0; 2 Hm st cc i ti x = 0, yCD = 3. Hm st cc tiu ti x = 2 , yCT= -1
th
025
025
025
025
2 1
I
th hm s
4 2
4 3 y x x= +
S nghim ca phng trnh 4 2 24 3 log x x m + = bng s giao im ca th hm s
4 24 3 y x x= + v ng thng y = log2m.
Vy phng trnh c 4 nghim khi v ch khi log2m = 0 hoc 21 log m 3< <
hay m = 1 hoc 2
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I HC SPHM H NI------------------------------------------------------------------------------------------------------------
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21 1
Vit li bt phng trnh di dng5 1 5 1
2 2 02 2
x x
++
t t =5 1
, 0.2
x
t +
>
khi 5 1 1
2
x
t
=
Bt phng trnh c dng
t +1
2 2 0t
2 2 2 1 0t t +
2 1 2 1t +
5 1 5 1
2 2
5 12 1 2 1
2
log ( 2 1) log ( 2 1)
x
x+ +
+ +
+
025
025
025
025
2 1
II
iu kin : 1x
Phng trnh tng ng vi 2 ( 1 1) 2 1 2( 1) 0 x x x x x = (*)
t 1, 0 y x y= . Khi (*) c dng : x2 x(y - 1) 2y 2y2 = 0( 2 )( 1) 0
2 0( 1 0)
x y x y
x y do x y
+ + =
= + +
2
2 1
4 4 02
x x
x x
x
=
+ =
=
025025
05
21 1
1 2 1 23 2 3
31 1
1 23 2 3 23 3
21 1
3 2 3 23 3
1 1
tan( 1) 1 1 tan( 1)lim lim .( 1)
11
1 tan( 1)lim .( 1) lim .( 1)( 1)
1 1
lim( 1) lim( 1)( 1) 9
x x
x x
x
x x
x x
e x e xx x
xx
e x x x x x x
x x
x x x x x
+ + = + +
= + + + + + +
= + + + + + + =
025
05
025
2 1
III
Kng cao SI ca tam gic SBC. Khi AI BC
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3
(nh l 3 ng vung gc) do SIA = S
AI = a.cot , AB = AD =cot
sin
a
, SI =
sin
a
2 2cot. .sin
sinABCDa
S AB AD
= =
A3 2
.
cot
3sinS ABCDa
V
=
Sxq = SSAB + SSAD SSBC + SSCD B I C
=2 cot 1
.(1 )sin sin
a
+
025
025
025
025
1IV
Ta c 3 3 3 2 2 2 2 2 23 ( ) ( ) ( )a b c abc a b c b c a c a b+ + + + + + + + 2 2 2 2 2 2 2 2 2 3
2 2 2 23
cos cos cos2
a b c b c a c a b
ab bc ca
A B C
+ + + + +
+ +
Mt khc
2 2 2 2
cos cos cos (cos cos ).1 (cos cos sin sin )
1 1 3[(cos cos ) 1 ]+ [sin A+sin B]-cos cos
2 2 2
A B C A B A B A B
A B A sB
+ + = +
+ + =
Do 3
cos cos cos2
A B C + +
025
025
05
31 1
Gi I l trung im ca AB, J l trung im ca IB. Khi I(1 ; -2), J(5
; 32
)
Ta c : 3 ( ) 2 2 2 4 MA MB MA MB MB MI MB MJ + = + + = + =
V vy 3 MA MB+
nh nht khi M l hnh chiu vung gc ca J trn ng thng ng thng JM qua J v vung gc vi c phng trnh : 2x y 8 = 0.
Ta im M l nghim ca h
22 3 0 5
2 8 0 19
5
xx y
x yy
=+ =
= =
vy M(19 2
;5 5
)
025
025025
025
Va
2 1
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I HC SPHM H NI------------------------------------------------------------------------------------------------------------
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5
==============================
Do M(2 14 3
; ;5 5 5
), N(
3 14 2; ;
5 5 5).
Mt cu ng knh MN c bn knh R =2
2 2
MN= v tm I(
1 14 1; ;
10 5 10
) c phng trnh
2 2 21 14 1 1
( ) ( ) ( )10 5 10 2 x y z + + + =
025
0253 1
Gi z = x + yi, M(x ; y ) l im biu din s phc z.2 21 2 1 ( 1) ( 2) 1 z i x y+ + = + + + =
ng trn (C) : 2 2( 1) ( 2) 1x y+ + + = c tm (-1;-2)ng thng OI c phng trnh y = 2xS phc z tha mn iu kin v c mdun nh nht khi v ch khi imBiu din n thuc (C) v gn gc ta O nht, chnh l mt trong haigiao im ca ng thng OI v (C)
Khi ta ca n tha
mn h2 2
1 11 1
2 5 5,
2 2( 1) ( 2) 12 2
5 5
x xy x
x yy y
= = + =
+ + + = = = +
Chon z =1 2
1 ( 2 )5 5
i + + +
025
025
025
025
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I HC SPHM H NI THI THI HC CAO NG 2011
KHOA TON-TIN MN: TON- KHI A------------- Thi gian lm bi: 180 pht( khng kthi gian giao )
---------------------------------------------------------------------------------------------------------------------------------------------
PHN CHUNG CHO TT C TH SINH (7 im).Cu I ( 2 im)
Cho hm s 2)2()21( 23 ++++= mxmxmxy (1) m l tham s.
1. Kho st s bin thin v v th (C) ca hm s (1) vi m=2.2. Tm tham s m th ca hm s (1) c tip tuyn to vi ng thng d: 07 =++yx gc ,bit
26
1cos = .
Cu II(2 im)
1. Gii bt phng trnh: 544
2log2
2
1
x
x.
2. Gii phng trnh: ( ) .cos32cos3cos21cos2.2sin3 xxxxx +=++ Cu III(1 im)
Tnh tch phn: I( ) ++
+=
4
02
2111 dx
xx .
Cu IV(1 im)
Cho hnh chp S.ABC c y ABC l tam gic vung cn nh A, AB 2a= . Gi I l trung im ca BC, hnh
chiu vung gc H ca S ln mt y (ABC) tha mn: IHIA 2= , gc gia SC v mt y (ABC) bng 060 .Hy tnh th tch khi chp S.ABC v khong cch t trung im K ca SB ti (SAH).Cu V(1 im)
Cho x, y, z l ba s thc dng thay i v tha mn: xyzzyx ++ 222 . Hy tm gi tr ln nht ca biu thc:
xyz
z
zxy
y
yzx
xP
++
++
+=
222.
PHN TCHN (3 im): Th sinhchchn lm mt trong hai phn ( phn A hoc phn B ).A.Theo chng trnh chun:Cu VI.a(2 im)
1. Trong mt phng Oxy, cho tam gic ABC bit A(3;0), ng cao tnh B c phng trnh 01 =++yx ,trung tuyn tnh C c phng trnh: 2x-y-2=0. Vit phng trnh ng trn ngoi tip tam gic ABC.
2. Trong khng gian vi h trc ta Oxyz, cho cc im A(-1;1;0), B(0;0;-2) v C(1;1;1). Hy vit phng
trnh mt phng (P) qua hai im A v B, ng thi khong cch t C ti mt phng (P) bng 3 .Cu VII.a (1 im)
Cho khai trin: ( ) ( ) 14142
210
2210 ...121 xaxaxaaxxx ++++=+++ . Hy tm gi tr ca 6a .B. Theo chng trnh nng cao:
Cu VI.b (2 im)1. Trong mt phng ta Oxy, cho tam gic ABC bit A(1;-1), B(2;1), din tch bng
11
2v trng tm G
thuc ng thng d: 043 =+yx . Tm ta nh C.
2.Trong khng gian vi h trc Oxyz, cho mt phng (P) 01 =++ zyx ,ng thng d:3
1
1
1
1
2
=
=
zyx
Gi I l giao im ca d v (P). Vit phng trnh ca ng thng nm trong (P), vung gc vi d v cch
I mt khong bng 23 .
Cu VII.b (1 im) Gii phng trnh: .13
=
+
zi
iz
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1
P N THANG IM THI THI HC, CAO NG NM 2011
MN:TON, Khi A
PHN CHUNG CHO TT C TH SINH.
Cu Ni dung im1(1) Kho st hm skhi m = 2Khi m = 2, hm s trthnh: y = x3 3x 2 + 4a) TX: Rb) SBTGii hn: lim ; lim
x xy y
+
= = + 0,25
Chiu bin thin:C y = 3x2 6x; y=0 x =0, x =2
x 0 2 +y + 0 0 +
y
4
0
+
Hm sB trn cc khong ( ; 0) v (2 ; +), nghch bin trn (0 ; 2).
0,25
Hm st cc i ti x = 0, yC = y(0) = 4;Hm st cc tiu ti x = 2, yCT = y(2) = 0.
0,25
c) th:Qua (-1 ;0)Tm i xng:I(1 ; 2)
0,25
2(1) Tm m ...
Gi k l h s gc ca tip tuyn tip tuyn c vctphp )1;(1 = kn
d: c vctphp )1;1(2 =n
Ta c
=
=
=+
+
==
32
2
3
0122612
12
1
26
1.cos
2
12
221
21
k
k
kk
k
k
nn
nn
0,5
I(2)
Yu cu ca bi ton tha mn t nht mt trong hai phng trnh: 1/ ky = (1)
v 2/ ky = (2) c nghim x
=++
=++
3
22)21(23
2
32)21(23
2
2
mxmx
mxmx
0
0
2/
1/
0,25c nghim
1
I
2
2
-1
4
0 x
y
c nghim
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2
034
01282
2
mm
mm
1;4
32
1;
4
1
mm
mm
4
1m hoc
2
1m 0,25
II(2) 1(1) Gii bt phng trnh ...
Bpt
)2(34
2log2
)1(24
2log3
94
2log
044
2log
2
1
2
1
2
2
1
2
2
1
x
x
x
x
x
x
x
x
0,25
. Gii (1): (1)5
16
3
8
04
165
04
83
84
24
x
x
x
x
x
x
x 0,25
. Gii (2): (2) 9
4
17
4
04
49
04
417
4
1
4
2
8
1
x
x
x
x
x
x
x
0,25
Vy bt phng trnh c tp nghim4 4 8 16
; ;17 9 3 5
. 0,25
2(1) Gii PT lng gic
Pt )1cos2()12(cos)cos3(cos)1cos2(2sin3 ++=+ xxxxxx
)1cos2(sin2cossin4)1cos2(2sin3 22 +=+ xxxxxx
0)1sin22sin3)(1cos2( 2 =+++ xxx
0,5
1)6
2sin(22cos2sin301sin22sin3 2 ===++
xxxxx
kx +=6
0,25
)(
23
2
23
2
01cos2 Zk
kx
kx
x
+=
+=
=+
Vy phng trnh c nghim: 23
2 kx += ; 23
2 kx += v kx +=6
0,25
III(1) 1(1) Tnh tch phn.
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3
I( ) ++
+=
4
02
211
1dx
x
x.
t dttdxx
dxdtxt )1(
21211 =
+=++= v
2
22 ttx
=
i cnx 0 4
t 2 4
0,25
Ta c I =
dttt
tdtt
tttdt
t
ttt
+=
+=
+4
22
4
2
4
22
23
2
2 243
2
1243
2
1)1)(22(
2
1
=
++
ttt
t 2ln43
22
1 2
0,5
=4
12ln2 0,25
(1) Tnh thtch v khong cch
Ta c = IHIA 2 H thuc tia i ca tia IA v IA = 2IH
BC = AB 2 a2= ; AI= a ; IH=2
IA=
2
a
AH = AI + IH =2
3a
0,25
Ta c2
545cos.2 0222
aHCAHACAHACHC =+=
V )(ABCSH 060))(;( ==
SCHABCSC
21560tan 0 aHCSH ==
0,25
IV
6
15
2
15)2(
2
1.
3
1.
3
1 32.
aaaSHSV ABCABCS === 0,25
H
K
I
BA
S
C
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4
)(SAHBISHBI
AHBI
Ta c22
1)(;(
2
1))(;(
2
1
))(;(
))(;( aBISAHBdSAHKd
SB
SK
SAHBd
SAHKd=====
0,25
V (1) Tim gi tr ln nht ca P
xyz
z
zxy
y
xyx
xP
++
++
+=
222.
V 0;; >zyx , p dng BT Csi ta c:xyz
z
zxy
y
yzx
xP
222 222++ =
++=
xyzxyz
222
4
1
0,25
++
++=
+++++
xyz
zyx
xyz
xyzxyz
yxxzzy
222
2
1
2
1111111
4
1
2
1
2
1=
xyz
xyz
0,5
Du bng xy ra 3=== zyx . Vy MaxP =2
1
0,25
PHN TCHN:
Cu Ni dung imVIa(2) 1(1) Vit phng trnh ng trn
KH: 022:;01: 21 ==++ yxdyxd
1d c vctphp tuyn )1;1(1 =n v 2d c vctphp tuyn )1;1(2 =n
AC qua im A( 3;0) v c vct ch phng )1;1(1 =n phng trnhAC: 03 =yx .
= 2dACC Ta C l nghim h: )4;1(022
03
=
=C
yx
yx.
0,25
Gi );( BB yxB )2;2
3
(BB yx
M
+
( M l trung im AB)
Ta c B thuc 1d v M thuc 2d nn ta c: )0;1(02
23
01
=+
=++
Byx
yx
B
B
BB
0,25
Gi phng trnh ng trn qua A, B, C c dng:02222 =++++ cbyaxyx .
Thay ta ba im A, B, C vo pt ng trn ta c:
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5
=
=
=
=+
=+
=+
3
2
1
1782
12
96
c
b
a
cba
ca
ca
Pt ng trn qua A, B, C l:
034222 =++ yxyx . Tm I(1;-2) bn knh R = 22
0,5
2(1) Vit phng trnh mt phng (P)
G
i Ocban=
);;( l vct
php tuyn c
a (P)
V (P) qua A(-1 ;1 ;0) pt (P):a(x+1)+b(y-1)+cz=0
M (P) qua B(0;0;-2) a-b-2c=0 b = a-2c
Ta c PT (P):ax+(a-2c)y+cz+2c =0
0,25
d(C;(P)) = 0141623)2(
23 22
222=+=
++
+ caca
ccaa
ca
=
=
ca
ca
7
0,5
TH1: ca = ta chn 1== ca Pt ca (P): x-y+z+2=0
TH2: ca 7= ta chn a =7; c = 1 Pt ca (P):7x+5y+z+2=00,25
VII.a (1 ) Tm h sca khai trin
Ta c 4
3
)12(4
1
1
22++=++
xxx nn
( ) 1012142210 )21(16
9)21(
8
3)21(
16
1)1(21 xxxxxx +++++=+++
0,25
Trong khai trin ( )1421 x+ h s ca 6x l: 61462 C
Trong khai trin ( )1221 x+ h s ca 6x l: 61262 C
Trong khai trin ( )1021 x+ h s ca 6x l: 61062 C
0,5
Vy h s .4174821692
832
161 6106612661466 =++= CCCa 0,25
Tm ta ca im CVI.b(2) 1(1)
Gi ta ca im )3
;3
1();( CCCCyx
GyxC + . V G thuc d
)33;(330433
13 ++==+
+ CCCC
CC xxCxyyx
ng thng AB qua A v c vctch phng )2;1(=AB
0,25
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6
032: = yxptAB
5
11
5
3332
5
11);(
2
11);(.
2
1=
+===
CC
ABC
xxABCdABCdABS
=
=
=
5
17
11165
C
C
C
x
x
x 0,5
TH1: )6;1(1 = CxC
TH2: )5
36;
5
17(
5
17= CxC .
0,25
2(1) Vit phng trnh ca ng thng
(P) c vc tphp tuyn )1;1;1()( =Pn v d c vc tch phng 3;1;1(. =u
)4;2;1()( IPdI =
v dP);( c vc tch phng )2;2;4(;)(
==
unuP
)1;1;2(2 =
0,25
Gi H l hnh chiu ca I trn )(QmpH qua I v vung gc Phng trnh (Q): 0420)4()2()1(2 =+=+ zyxzyx
Gi 11 )()( dQPd = c vcto ch phng
)1;1;0(3)3;3;0(; )()( ==QP nn v 1d qua I
+=
+=
=
tz
ty
x
ptd
4
2
1
:1
Ta c );;0()4;2;1(1 ttIHttHdH =++
=
===
3
323223 2
t
ttIH
0,5
TH1:1
7
1
5
2
1:)7;5;1(3
=
=
=
zyxptHt
TH2:1
1
1
1
2
1:)1;1;1(3
=
+=
=
zyxptHt
0,25
VII.b 1 Gii phng trnh trn tp sphc.K: iz
tzi
izw
+= ta c phng trnh: 0)1)(1(1 23 =++= wwww
0,5
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I HC SPHM H NI----------------------------------------------------------------------------------------------------------------------------
---------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n Tel: 0912.676.613 091.5657.95
7
-----------------------------------------------------
=
+=
=
=++
=
2
31
2
31
1
01
12
iw
iw
w
ww
w
Vi 011 ==
+= zzi
izw
Vi 333)31(2
31
2
31==+
+=
+
+= zizi
i
zi
iziw
Vi 333)31(2
31
2
31==
=
+
= zizi
i
zi
iziw
Vy pt c ba nghim 3;0 == zz v 3=z .
0,5
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I HC SPHM H NIKHOA TON TIN
-----------------
THI THI HC CAO NG 2010MN: TON
Thi gian lm bi: 180 pht (khng kthi gian giao )
-----------------------------------------------------------------------------------------------------------------
I. PHN CHUNG CHO TT C TH SINH
Cu I( 2,0 im): Cho hm s: (C)
1.
Kho st v v th (C) hm s2. Cho im A( 0; a) Tm a t A kc 2 tip tuyn ti th (C) sao cho 2 tip im tng ng nm 2 pha ca trc honh.
Cu II (2,0 im):1. Gii phng trnh lng gic.
2. Gii h phng trnh.
Cu III(1,0 im): Tnh tch phn sau.
=3
4
42 cos.sin
xx
dxI
Cu IV(1,0 im): Cho ba s thc tha mn ,Chng minh rng:
Cu V(1,0 im): Cho t din ABCD c AC = AD = , BC = BD = a, khong cch t B n mt phng
(ACD) bng . Tnh gc gia hai mt phng (ACD) v (BCD). Bit th ca khi t din ABCD bng .
II. PHN RING (Th sinh chc lm 1 trong 2 phn A hoc B)A. Theo chng trnh chun.
Cu VIa(2,0 im):1. Trong khng gian vi h ta Oxyz cho 4 im : A(1;2; 2) B(-1;2;-1) C(1;6;-1) D(-1;6;2). Tm ta
hnh chiu vung gc ca im A trn mt phng (BCD)2. Trong mp vi h ta Oxy cho ng trn : x2 +y2 -2x +6y -15=0 (C ).
Vit PT ng thng () vung gc vi ng thng : 4x-3y+2 =0 v ct ng trn (C) ti A; Bsao cho AB = 6
Cu VIIa(1,0 im): Xc nh h s ca x5 trong khai trin (2+x +3x2 )15B. Theo chng trnh nng cao.
Cu VIb(2,0 im):1. Trong khng gian vi h ta Oxyz cho 4 im : A(1;2; 2) B(-1;2;-1) C(1;6;-1) D(-1;6;2). Tm ta
hnh chiu vung gc ca im A trn mt phng (BCD)2. Trong mp vi h ta Oxy cho ng trn : x2 +y2 -2x +6y -15=0 (C ).
Vit PT ng thng ( ) vung gc vi ng thng : 4x-3y+2 =0 v ct ng trn (C) ti A; Bsao cho AB = 6
Cu VIIb(1,0 im):Gii phng trnh:
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P N
CU NI DUNG I
I 2,0
1 1,0
TX: D= R\{1}
y=Hm s lung nghch bin trn D v khng c cc tr
0,25
Gii hn:
PT ng TC: x=1; PT ng TCN: y=1
0,25
Bng bin thin:t - 1 +
f(t) - +f(t)
1 +
- 1
0,25
th: 0,25
2 1,0
Gi k l h s gc ca t i qua A(0;a). PT t d c dng y= kx+a (d) 0,25
x
y f x( ) =x+2
x-1
1
4
-2
-2 O 1 2 3
5/2
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H cho trthnh
Vy h d cho c mt nghim
0,25
III 1,0
=3
4
42 cos.sin
xx
dxI =
3
4
22 cos.2sin.4
xx
dx
t : t = tanx
i cn: x =
x =
0,5
Khi 3
438)
32
1()2
1(
)1( 3
1
33
1
22
3
12
22
=++=++=+
= t
tt
dtttt
dttI
0,5
IV 1,0
BT cn chng minh tng ng vi
Nhn xt: Do nn l cc s thc dng
0,25
Xt : A = vi x,y > 0 Chia t v mu cho v t t = ta c A = vi t > 0
Xt hm s f(t) = trn (0;+ ) Ta c : f(t) = Bng bin thin:
t 0 1 +
f(t) - 0 +f(t)
1 1
0,5
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Da vao bng bin thin ta c f(t) vi mi t > 0 T A = vi x,y > 0; du bng xy ra khi t = 1 nn x = y.
Do vai tr l nh nhau nn BT cn chng minh tng ng
p dng BT c si ta c Thay vo ta suy BT c chng minh, du ng thc xy ra khi a = b = c =
0,25
V 1,0
Gi E l trung im ca CD, k BH AE
Ta c ACD cn ti A nn CD AE
Tng t BCD cn ti B nn CD BE
Suy ra CD (ABE) CD BH
M BH AE suy ra BH (ACD)
Do BH = v gc gia hai mt phng
(ACD) v (BCD) l
0,25
Th tch ca khi t din ABCD l
M
Khi : l 2 nghim ca pt: x2 - x + = 0
trng hp v DE
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VIa 2,0
1 1,0
Ta c ;[ , ] = (12; -6;8)
Mp (BCD) i qua B v c VTPT =(6;-3;4) nn c PT: 6x-3y+4z+16=0Gi d l t i qua A v vung gc vi mp(BCD) th d c PT:
0,5
Hnh chiu vung gc H ca A ln mp(BCD) l giao im ca d vi mp(BCD)Ta ca H l nghim ca h :
Vy H( -2; -4; -4)
0,5
2 1,0
ng trn ( C) c tm I(1;-3); bn knh R=5Gi H l trung im AB th AH=3 v IH AB suy ra IH =4Mt khc IH= d( I; )V || d: 4x-3y+2=0 nn PT ca c dng3x+4y+c=0
0,5
d(I; )=
vy c 2 t tha mn bi ton: 3x+4y+29=0 v 3x+4y-11=0
0,5
VIIa 1,0
Ta c (2+x+3x2 )15 =M =Vy (2+x+3x2 )15 =
0,5
Theo gt vi x5 ta c cc cp s : (k=3; i=2) ( k=4; i=1) (k=5; i=0)Vy h s ca x5 trong khai trin trn l :a=
0,5
VIb 1,0
K: x > 1 Vi K trn phng trnh cho tng ng
0,25
0,5
I
A H B
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Vy phng trnh cho c mt nghim :
0,25http://www.VNMATH.com
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I HC SPHM H NI
KHOA TON TIN
-----------------
THI THI HC -- CAO NG 2010
MN: TON
Thi gian lm bi: 180 pht (khng kthi gian giao )
------------------------------------------------------------------------------------------------------------------------------
Cu I: (2 im) Cho hm s: ( )3 23 1 9 2y x m x x m= + + + (1) c th l (Cm)
1) Kho st v v th hm s (1) vi m=1.2) Xc nh m (Cm) c cc i, cc tiu v hai im cc i cc tiu i xng vi nhau qua
ng thng1
2y x= .
Cu II: (2,5 im)1) Gii phng trnh:
( ) ( )3sin 2 cos 3 2 3 os 3 3 os2 8 3 cos sinx 3 3 0 x x c x c x x+ + = .2) Gii bt phng trnh :
( )22 12
1 1log 4 5 log
2 7x x
x
+ >
+ .
3) Tnh din tch hnh ph
ng gi
i h
n b
i cc
ng: y = x.sin2x, y = 2x, x = 2
.Cu III: (2 im)
1) Cho hnh lng tr ABC.ABC c y ABC l tam gic u cnh a, cnh bn hp vi y mtgc l 45
0. Gi P l trung im BC, chn ng vung gc h t A xung (ABC) l H sao cho
1
2 AP AH =
. gi K l trung im AA, ( ) l mt phng cha HK v song song vi BC ct BB v CC
ti M, N. Tnh t s th tch' ' '
ABCKMN
A B C KMN
V
V.
2) Gii h phng trnh sau trong tp s phc:( )
2
2
2 2 2 2
65
6 0
a a
a aa b ab b a a
+ =
+ + + + =
Cu IV: (2,5 im)1) Cho m bng hng trng v n bng hng nhung khc nhau. Tnh xc sut lyc 5 bng hng trong c t nht 3 bng hng nhung? Bit m, n l nghim ca h sau:
2 2 1
3
1
9 19
2 2
720
m
m n m
n
C C A
P
+
+ +
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P N THI THI HC
Bi1
1
Khi m = 1 ta c hm s: 3 26 9 1 y x x x= +
BBT:x - 1 3 +
y/
+ 0 - 0 +
3 + y
- 1
1
2 9)1(63' 2 ++= xmxy
hm s c cc i, cc tiu:
09.3)1(9' 2 >+= m );31()31;( ++ m
Ta c ( ) 14)22(29)1(633
1
3
1 22+++++
+= mxmmxmx
mxy
Vy ng thng i qua hai im cc i v cc tiu l
14)22(2 2 +++= mxmmy
V hai im cc i v cc tiu i xng qua t xy2
1= ta c iu kin cn l
[ ] 12
1.)22(2 2 =+ mm
=
==+
3
10322
m
mmm
Khi m = 1 ptt i qua hai im C v CT l:y = - 2x + 5. Ta trung im C v
CT l:
=++
=+
==+
12
10)(2
2
2
2
4
22121
21
xxyy
xx
Ta trung im C v CT l (2; 1) thuc ng thng xy2
1= 1= m tm .
Khi m = -3 ptt i qua hai im C v CT l: y = -2x 11.3= m khng tha mn.
Vy m = 1 tha mn iu kin bi.
1
Bi2
1 phng trnh a v:
=
=
=
=+
=
=+
)(4cos
1cos
3tan
04cos3cos
0sincos3
0)8cos6cos2)(sincos3(
2
2
loaix
x
x
xx
xx
xxxx
=
+= k
kx
kx,
2
3
1
2
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k:
>
+
>+
>+
7
);1()5;(
07
0542
x
x
x
xx)1()5;7( + x
T pt7
1log2)54(log 2
2
2+
>+x
xx 2 22 2
27log ( 4 5) log ( 7)
5 x x x x
+ > + <
Kt hp iu kin: Vy BPT c nghim: )5
27;7(
x
0.75
3 Ta c: x.sin2x = 2x x.sin2x 2x = 0 x(sin2x 2) =0 x = 0Din tch hnh phng l:
==2
0
2
0)22(sin)22sin.(
dxxxdxxxxS
t
=
=
=
=
xx
v
dxdu
dxxdv
xu
22
2cos)22(sin 44424
222
=+= S (vdt)
0.75
Bi3
1 Gi Q, I, J ln lt l trung im BC, BB, CC
ta c:
2
3aAP = 3aAH =
V ''AHA vung cn ti H.
Vy 3' aHA = Ta c
4
3
2
3.
2
12
aaaSABC == (vdt)
4
3
4
3.3
32
'''
aaaV CBABCA == (
vtt) (1)V ''AHA vung cn
( )CCBBHKAAHK ''' G i E = MN KH BM = PE= CN (2)
m AA = 22' AHHA + = 633 22 aaa =+
4
6
2
6 aCNPEBM
aAK ====
Ta c th tch K.MNJI l:
1.
31 1 6
'2 4 4
MNJIV S KE
aKE KH AA
=
= = =
26 6
. . ( )4 4
MNJI
a aS MN MI a dvdt = = =
2 31 6 6
( )3 4 4 8
KMNJI
a a aV dvtt = =
3 3
2 3
' ' '
318 8
3 2
8 8
ABCKMN
A B C KMN
a a
V
a aV
= =
+
1
45
E
K
J
I
A
B
C
C'
B'
A'
P
H
Q
N
M
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2 K: 02 + aa
T (1) 06)(5)( 222 =++ aaaa
=+
=+
6
1
2
2
aa
aa
Khi 12 =+ aa thay vo (2)
2
1 23.
26 0
1 23.
2
ib
b bi
b
=
= +
=
;
+=
=
=++
2
31
2
31
012
ia
ia
aa
Khi 62 =+ aa
=
=
2
3
a
aThay vo (2) 2
1 5
26 6 6 0
1 5
2
b
b b
b
+=
+ =
=
Vy h pt c nghim (a, b) l:
+
2
31;
2
231,
2
31;
2
231 iiii
+
+
2
31;
2
231,
2
31;
2
231 iiii;
+
+
2
51;2,
2
51;2,
2
51;3,
2
51;3
Bi4 1)
=
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=
2255
6;0 aAB ; 2 2 2
10 100 100 12525 25 25
3 9 9 9a a a = = = =
3
55= a Vy phng trnh ng thng:
3
55,
3
55=
= xx
3)ng thng d2 c PTTS l:
+=
+=
+=
'51
'2
'21
tz
ty
tx
vectCP ca d1 v d2 l:1 2
(1;1; 1), (2;1;5)d d
u u= =
VTPT ca mp() l1 2. (6; 7; 1)
d dn u u
= =
pt mp() c dng 6x 7y z + D = 0ng thng d1 v d2 ln lt i qua 2 M(2; 2; 3) v N(1; 2; 1)
( , ( )) ( , ( ))
|12 14 3 | | 6 14 1 |
| 5 | | 9 | 7
d M d N
D D
D D D
=
+ = +
+ = + =
Vy PT mp() l: 3x y 4z + 7 0=
Bi 5
Ta c: P + 3 = 22
32
2
32
2
3
111a
a
cc
c
bb
b
a+
+
++
+
++
+
24
1
121224
6 2
2
2
2
3b
b
a
b
aP
++
+
+
+
=+ 24
1
1212
2
2
2
2
3c
c
b
c
b ++
+
+
+
+
24
1
1212
2
2
2
2
3 a
a
c
a
c ++
+
+
+
+ 3
6
3
6
3
6
2163
2163
2163
cba++
6
222
3 82
9)(222
3
22
3=+++ cbaP
2
3
22
3
22
9
22
3
22
96 3
== P
PMin khi a = b = c = 1
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1
I HC SPHM H NIKHOA TON TIN
-----------------
THI THI HC CAO NG 2010MN: TON
Thi gian lm bi: 180 pht (khng kthi gian giao )
---------------------------------------------------------------------------------------------------------
( thi gm 2 trang )
I. PHN CHUNG DNH CHO TT C TH SINH (7,0 im)
Cu I: (2,0 im) Cho hm s 4 2 2 42 2y x m x m m= + + (1), vi m l tham s.1. Kho st s bin thin v v th ca hm s (1)khi 1m = .2. Chng minh th hm s (1) lun ct trc Ox ti t nht hai im phn bit,
vi mi 0m < .
Cu II: (2,0 im)1. Gii phng trnh :
2sin 2 4sin 16
x x
+ + =
.
2. Tm cc gi tr ca tham sm sao cho h phng trnh2
1
y x m
y xy
=
+ =
c nghim
duy nht.Cu III: (2,0 im)
1. Tm nguyn hm ca hm s ( )( )
( )
2
4
1
2 1
xf x
x
=
+.
2. Vi mi s thc dng ; ;x y z tha iu kin 1x y z+ + . Tm gi tr nh nht
ca biu thc: 1 1 12P x y zx y z
= + + + + +
.
Cu IV: (1,0 im) Cho khi t din ABCD. Trn cc cnhBC,BD,ACln lt lycc im M, N, P sao cho 4 , 2BC BM BD BN = = v 3AC AP= . Mt phng (MNP) chiakhi t dinABCD lm hai phn. Tnh t s th tch gia hai phn .
II. PHN RING (3,0 im)
Tt c th sinh chc lm mt trong hai phn: A hoc B.
A. Theo chng trnh ChunCu Va: (1,0 im )Trong mt phng ta (Oxy), chong thng ( ) : 2 4 0d x y = .
Lp phng trnh ng trn tip xc vi cc trc ta v c tm trn ng thng (d).Cu VIa: (2,0 im)
1. Gii phng trnh : log log4 22 8x xx = .
2. Vit phng trnh cc ng thng ct th hm s 12
xy
x
=
ti hai im
phn bit sao cho honh v tung ca mi im l cc s nguyn.
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2
B. Theo chng trnh Nng caoCu Vb: (1,0 im) Trong khng gian Oxyz , cho cc im
( ) ( ) ( )1;3;5 , 4;3;2 , 0;2;1A B C . Tm ta tm ng trn ngoi tip tam gicABC.
Cu VIb: (2,0 im)1. Gii bt phng trnh :
( )2 4 82 1 log log log 0x x x+ + < .2. Tm m th hm s ( )3 25 5y x m x mx= + c im un trn th
hm s 3y x= .
........HT.......................................
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3
I HC SPHM H NIKHOA TON - TIN
P N THI THI HC CAO NG 2010
Mn thi: TON
CU NI DUNG IM
Khi 4 21 2 3m y x x= = + .Tp xc nh D=R .
0,25
Gii hn: lim ; limx x
y y +
= + = + .
( )3 2' 4 4 4 1y x x x x= = . ' 0 0, 1y x x= = = .0,25
Bng bin thin:
Hm sng bin trn khong ( ) ( )1; 0 , 1; + v nghch bin
trn khong ( ) ( ); 1 , 0;1 .
Hm st C ti 0, 3CDx y= = v t CT ti 1, 2CTx y = .
0,25
1(1,0)
th ct Oy ti (0;3). thi xng qua Oy. 0,25 Phng trnh HG ca th (1) v Ox:
4 2 2 42 2 0x m x m m + + = ().0,25
t ( )2 0t x t= , ta c : 2 2 42 2 0t m t m m + + = (). 0,25
Ta c : ' 2 0m = > v 22 0S m= > vi mi 0m > .Nn PT () c nghim dng.
0,25
Cu I(2,0)
2(1,0)
KL: PT () c t nht 2 nghim phn bit (pcm). 0,25
PT 3 sin 2 cos 2 4sin 1 0x x x + + = 22 3 sin cos 2sin 4sin 0x x x x + = .
0,25
( )2 3 cos sin 2 sin 0x x x + = . 0,25
Khi : 5sin 3 cos 2 sin 1 23 6
x x x x k
= = = +
. 0,25
1(1,0)
Khi: sin 0x x k = = .
KL: nghim PT l 5, 26
x k x k
= = + . 0,25
Ta c : 2x y m= , nn : 22 1y my y = . 0,25
PT1
12
y
m yy
= +
( v y = 0 PTVN). 0,25
Cu II(2,0)
2(1,0)
Xt ( ) ( ) 21 1
2 ' 1 0f y y f yy y
= + = + > 0,25
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Khi: 2t= th 2log 2 4( )x x th= = . KL: Nghim PT2, 4x x= = .
0,25
Ta c: 112
yx
= +
0,25
Suy ra: ; 2 1 3, 1x y Z x x x = = = 0,25
Ta cc im trn th c honh v tung lnhng s
nguyn l ( ) ( )1;0 , 3;2A B
0,25
2(1,0)
KL: PT ng thng cn tm l 1 0x y = . 0,25
Ta c: ( )3;0; 3 3 2AB AB= =
. 0,25 Tng t: 3 2BC CA= = . 0,25 Do : ABC u, suy ra tm I ng trn ngoi tipABC ltrng tm ca n.
0,25
Cu Vb(1,0)
KL:5 8 8
; ;3 3 3I
. 0,25
K : 0x > . t 2logt x= , ta c : ( )1 03t
t t+ + < 0,25
BPT 2 43 4 0 03
t t t + < < < . 0,25
1(1,0)
KL: 2 34 1
log 0 13 2 2
x x < < < < . 0,50
Ta c: ( )2' 3 2 5 5 ; " 6 2 10y x m x m y x m= + = + . 0,25
5" 03
my x = = ; yi du qua 53
mx = .
Suy ra: ( ) ( )3
2 5 5 55;
3 27 3
m m mmU
+
l im un0,50
Cu VIb(2,0)
2(1,0)
KL: 5m = . 0,25
..HT...
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I HC SPHM H NI
KHOA TON TIN
-----------------
THI THI HC CAO NG 2010
MN: TON
Thi gian lm bi: 180 pht (khng kthi gian giao )
------------------------------------------------------------------------------------------------------------------
I. Phn chung cho tt c th sinh (7 im)
Bi 1: Cho hm s 3 21 5
3
3 3
y x x x= + +
1. Kho st v v th hm s.2. Gi A v B l giao im ca (C) v trc Ox. Chng minh rng trn th (C) tn ti hai im cngnhn on AB di mt gc vung.Bi 2 : Gii phng trnh sau
1. 2 2 2sin cos 02 4 2
x xtg x
=
2. 32222 2
=+ xxxx
Bi 3 : 1. Tm gi tr ln nht v gi tr nh nht ca hm s y =1
12
+
+
x
xtrn on [1; 2].
2. Tnh tch phn : I = dxxx 2
0
2
Bi 4: Cho hai mp(P) v (Q) vung vi nhau, c giao tuyn l ng thng (). Trn () ly hai im Av B vi AB = a. Trong mp(P) ly im C, trong mp(Q) ly im D sao cho AC, BD cng vung gc vi() v AC = BD = AB.Tnh bn knh mt cu ngoi tip t din ABCD v tnh khong cch t A n mp (BCD) theo a.II) Phn ring (3 im)- Th sinh chc lm 1 trong 2 phn.
A. PHN IBi 5a1. Trong mp Oxy cho ng trn (C) :
(x 1)2 +(y 2)2 = 4 v ng thng (d) : x y 1 = 0.
Vit phng trnh ng trn (C) i xng vi ng trn (C) qua (d). Tm ta cc giao im ca(C) v (C).2. Trong khng gian Oxyz cho ng thng (dk): l giao tuyn ca hai mt phng(P): x +3ky z +2 = 0; (Q): kx y +z +1 = 0Tm k ng thng (dk) vung vi(R): x y 2z +5 = 0Bi 6a. T mt t gm 6 bn nam v 5 bn n, chn ngu nhin 5 bn xp vo bn u theo nhng tht khc nhau. Tnh xc sut sao cho trong cch xp trn c ng 3 bn nam.B. PHN IIBi 5b. Cho hai ng thng
( )
x 2 t
d : y 1 t
z 2t
= + = =
v ( )
x 0 2t '
d' : y 3
z 1 t '
= = = +
1) Chng minh (d) v (d) cho nhau. Hy vit pt ng vung gc chung ca (d) v (d).2) Vit pt mp song song cch u (d) v (d)
Bi 6b: Cho hm s2 2 4
2
x xy
x
+=
c th (C), chng minh (C) c tm i xng
________________HT ________________
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P N
Ni dung Li gii chi titBi 1: Cho hm s
3 21 533 3
y x x x= + +
1) Kho st v v thhm s.
3 21 533 3
y x x x= + + , D = R
y = x2 +2x 3 , y = 0 1
3
x
x
=
=
lim , limx x
y y +
= = +
y tng trn (- ; -3) v (1 ; + )y gim trn (-3; 1),C(-3; -32/2) v CT(1; 0)
+
-
+
0
32
3
BBT_
f(x)
f'(x)
x - +1-3
00 +
2) Gi A v B l giao imca (C) v trc Ox. Chng
minh rng trn th (C)tn ti hai im cng nhnon AB di mt gcvung.
Phng trnh honh 3 211 5
3 053 3
x x x x
x
=+ + =
=
=> A(-5; 0) v B(1; 0)
Gi M thuc (C) => M 3 21 5
; 33 3
a a a a
+ +
khc A v B
3 2 3 21 5 1 55 ; 3 ; 1 ; 33 3 3 3
AM a a a a BM a a a a
+ + + + +
Theo gi thit AM BM . 0AM BM =
(a +5)(a -1) +[1
3(a -1)2(a +5)]2 = 0
Do M khc A v B nn a khc -5 v a khc 1 nn pt trn tng ng
1 +1
9(a -1)3(a +5) = 0 hay a4 +2a3 -12a2 +14a +4 = 0 (*)
t y = a4 +2a3 -12a2 +14a +4 c tp xc nh D = R
y = 4a3 +6a2 -12a +14 ; y = 0 c 1 nghim thc0 0
7 2043a y
2 16
=>
++
9
+
y0
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S x = 2k+ v x =
k+4
(k Z)
2) 32222 2
=+ xxxx
t t =xx 22 , vi t > 0 th phng trnh thnh: 34 =
tt t2 3t 4 = 0
t = 1 (loi) hoc t = 4
xx
2
2 = 22 x2 x 2 = 0S x = 1 v x = 2
Bi 3:1) Tm gi tr ln nht vgi tr nh nht ca hm s
y =1
12
+
+
x
xtrn on
[1; 2].
y =1
12
+
+
x
xc min xc nh D = [1; 2]
y =)1(1
122
++
xx
x= 0 khi x = 1
So snh cc gi tr y(1) = 2 ; y(-1) = 0 v y(2) =5
53ta c
max y = y(1) = 2 ; min y = y(-1) = 0
2) Tnh tch phn
I = dxxx 2
0
2 I = ( ) ( )dxxxdxxxdxxx
+=
2
1
21
0
22
0
2
= 12323
2
1
231
0
23
=
+
xxxx
Bi 4: Cho hai mp(P) v(Q) vung vi nhau, cgiao tuyn l ng thng(). Trn () ly hai imA v B vi AB = a. Trongmp(P) ly im C, trongmp(Q) ly im D sao cho
AC, BD cng vung gcvi () v AC = BD = AB.Tnh bn knh mt cungoi tip t din ABCDv tnh khong cch t An mt phng (BCD) theoa.
Cch 1: Cc gc B, C u nhn DC dimt gc vung nn mt cu ngoi tipc tm I l trung im ca BC.V AF// BD => F l trung im ca BC
V ABC vung cn nn AF BC (1)V DB (ABC)=> DB AF (2)(1) v (2) th AF (DBC) Nn d(A;
(BCD)) =AF =2
2a
Cch 2: Chn h trc Oxyz sao cho A(0; 0; 0), B(0; a; 0), C(0; 0; a), I(x; y;
z). Theo gi thit ta c: IA = IB = IC = ID =21
CD = R
x = y = z =2
a R = IA =
2
3a=> n BCD = (0; a
2; a2)
Pt (BCD): y +z a = 0 => d(A; (BCD)) =2
2a
PHN RING
A. Phn 1Bi 5a :1) Trong mp Oxy chong trn (C) :(x 1)2 +(y 2)2 = 4 vng thng(d) : x y 1 = 0.Vit phng trnh ngtrn (C) i xng ving trn (C) qua (d).
(C) c tm I(1 ; 2) v bn knh R = 2.() l ng thng qua I v () (d) c pt : (x 1) +(y 2) = 0
Giao im H ca () vi (d):1 0
3 0
x y
x y
=
+ ==> H(2; 1)
Gi I l im i xng ca I qua (d) th H l trung im ca II.p dng cng thc trung im => I(3; 0)V R = R = 2 nn phng trnh (C): (x 3)2 +y2 = 4.
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Tm ta cc giao imca (C) v (C). Gii h
( ) ( )
( )
( ) ( )2 2 2 2
2 2
( ) : 1 2 4 1 2 4
1 0( ') : 3 4
C x y x y
x yC x y
+ = + =
= + =
Ta c giao im A(1; 0) v B(3; 2)2) Trong khng gian Oxyzcho ng thng (dk): lgiao tuyn ca hai mtphng(P): x +3ky z +2 = 0,
(Q): kx y +z +1 = 0Tm k ng thng (dk)vung vi(R): x y 2z +5 = 0
(P) c vtpt ( )1;3 ; 1n k=
, (Q) c vtpt ( )' ; 1;1n k=
=> ng thng dk c vtcp ( )2, ' 3 1; 1; 3 1u n n k k k = =
Ga thit dk vung (R) nn ta c , 0Ru n =
2
2
3 2 1 0
3 6 3 0 1
2 2 0
k k
k k k
k
+ + =
+ = =
+ =
. Vy k = 1 l p s
Chn ngu nhin 5 bn v sp th t (ch ngi)
=> khng gian mu gm 511A (phn t)
K hiu A l bin c: Trong cch xp trn c ng 3 bn namChn 5 bn, trong
- Chn 3 nam t 6 nam, c 36C cch.
- Chn 2 n t 5 n, c 25C cch.- Xp 5 bn chn vo 5 v tr khc nhau th c 5! Cch.-T theo quy tc nhn ta c bin c A chn 5 ngi vo 5 v tr (trong
c ng 3 nam) l: n(A) = 36C .25C .5!
Bi 6a:T mt t gm 6 bn namv 5 bn n, chn ngunhin 5 bn xp vo bnu theo nhng th t khcnhau. Tnh xc sut sao
cho trong cch xp trn cng 3 bn nam.
Vy:3 26 5
511
. .5!( ) 0,433
C CP A
A=
B. Phn 2
Bi 5b. Cho hai ngthng
( )x 2 t
d : y 1 t
2t
= + = =
v
( )
x 0 2t '
d' : y 3
z 1 t '
= = = +
1) Chng minh (d) v (d)cho nhau. Hy vitphng trnh ng vunggc chung ca (d) v (d).
d qua im M(2; 1; 0) c vtcp ( )1; 1; 2u =
d qua im M(0; 3; 1) c vtcp ( )' 2;0;1u=
( )' 2; 2; 1M M
v ( ), ' 1; 5; 2u u =
, ' . ' 2 10 2 10 0u u M M = + + =
nn d v d cho nhau.
ng vung gc chung (D) ct (d) ti I => I(2 +t; 1 t; 2t)(d)v (D) ct (d) ti J =>J(2s; 3; 1 +s) (d)
=> (2 2 ; 2 ; 1 2 ) JI t s t t s= + + +
Theo gi thit ng vung gc chung nn. 0
. ' 0
JI u
JI u
=
=
1
(2 2 ) (2 ) 2 4 2 0 3( 4 2 4 ) 1 2 0
1t s t t s t
t s t ss
+ + + + + =
= + = =
Lc 5 4 2
; ;3 3 3
I
, J(2; 3; 0) v1 5 2
; ;3 3 3
JI
=
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1
THI THI HC, CAO NG NM 2010Mn thi: TON
THI THAM KHO Thi gian: 180 pht ( khng kthi gian giao )--------------------------------------------------------------------------------------------------------------------
Cu 1:( 2,0 im)Cho hm s
2 1
1
xy
x
=
c th (C).
1) Kho st s bin thin v v th (C) ca hm s.2) Tm m ( )m ng thng y x m= + ct (C) ti hai im A, B sao cho 4AB
Cu 2: (2,0 im)1) Gii phng trnh ( )3 cos 2 2cos sin 1 0x x x+ = 2) Gii phng trnh 2 2
2 1 4
2
2log 2log log ( )x x x
x =
Cu 3:(1,0 im)
Tnh tch phn1
0
2I
1
xdx
x=
+
Cu 4:(1,0 im)Cho lng tr ABC.ABC c cnh bn bng a, y ABC l tam gic u, hnh chiu ca A trn
(ABC) trng vi trng tm G ca ABC. Cnh bn to vi y gc 060 . Tnh th tch lng trABC.ABC theo a.
Cu 5:(1,0 im)Trong h to Oxy, cho hai ng thng 1 2: 3 4 20 0, : 1 0d x y d x y+ = + + =
Vit phng trnh ng trn (C) bit rng (C) c bn knh R=5, tip xc vi 1d v c tm nm trn
Cu 6:( 1,0 im)Trong khng gian Oxyz, cho mt cu (S) v mt phng (P) c phng trnh(S): 2 2 2 4 4 2 16 0x y z x y z+ + + = ( ) : 2 2 1 0P x y z+ + =
Vit phng trnh mt phng (Q) song song vi (P) v khong cch t tm mt cu (S) n mt phngbng 3.
Cu 7:( 1,0 im).Cho s phc z tho mn ( )1 3 4i z i+ = . Tnh 2010z .
Cu 8:(1,0 im)
Cho cc s thc khng m x, y, z tho mn 2 2 24
3
x y z+ + = .
Tm gi tr ln nht ca biu thc ( )4
3P x y zx y z
= + + ++ +
.HtTh sinh khng c sdng ti liu. Cn b coi thi khng gii thch g thm.H v tn th sinh:; S bo danh:Ch k gim th:
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Hng dn chm TONCu Ni dung iu1
2,0))1,0
1)Kho st s bin thin v v th hm s2 1
1
xy
x
=
1. Tp xc nh: \ {1}D = 2. S bin thin ca hm s* Gii hn ti v cc, gii hn v cc ca hm s. Tim cn ca th hm s.
12
2 1lim lim lim 2
111
x xx
x xyx
x
= = =
=> th hm s nhn ng thng y=2 lm tim cn ngang
1 11 1
2 1 2 1lim lim ;lim lim
1 1x xx x
x xy y
x x+ +
= = + = =
=> th hm s nhn ng thng x=1 lm tim cn ng
0,
* Lp bng bin thin
2
1' 0
( 1)y x D
x
= <
, y khng xc nh x=1
Hm s nghch bin trn tng khong xc nh ca n. Hm s khng c cc tr.
0,
bng bin thin
x - 1 + y - || -
y 2 +
- 2
0.
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3
3. th- Giao ca th hm s v Ox: y=0=>x=1/2
- Giao ca th hm s v Oy: x=0=>y=1- th hm s nhn im I(1;2) lm tm i xng.
0,
)1,0 2)Honh giao im ca ng thng y=x+m (d) v th (C) l nghim ca phng trnh
( ) ( )
2 1
1
2 1 1 (*)
xx m
x
x x x m
= +
= +
( x=1 khng phi l nghim ca (*))2 ( 3) 1 0x m x m + + = (1)
0,
2 2( 3) 4(1 ) 2 5 0m m m m m = = + >
Do (d) lun ct (C) ti hai im phn bit1 1 2 2
( ; ), ( ; )A x y B x y vi1 2,x x l hai nghim ca (1)
0,
Theo vit1 2 1 2
3 ; 1x x m x x m+ = = . V , ( )A B d nn 1 1 2 2;y x m y x m= + = +
( )22 2 2
1 2 1 2 1 22( ) 2 4 2( 2 5)AB x x x x x x m m = = + = +
0,
2 2 21
4 16 2( 2 5) 16 2 3 03
mAB AB m m m m
m
= = = + = =
=
0,
Cu 2:2,0)
1)Gii phng trnh ( )3 cos 2 2cos sin 1 0x x x+ =
3 cos 2 sin 2 2cosx x x + =
3 1
cos 2 sin 2 cos2 2x x x + =
0,
cos 2 cos sin 2 sin cos6 6
x x x
+ =
cos(2 ) cos6
x x
=
0,
I(1;2)
2
1 y
x
O
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4
2 26
( )
2 26
x x k
k
x x k
= +
= +
0,
26( )
2
18 3
x kk
kx
= +
= +
KL
0,
)1,02)Gii phng trnh 2 2
2 1 4
2
2log 2log log ( )x x x
x = (1)
KX:x>0
( ) 22 2 22
1 log 2log logx x
x
+ =
0,
2
2 2log 3log 2 0(*)x x + = 0,
t t=log2xThay vo (*) ta c
2 3 2 0
1
2
t t
t
t
+ =
=
=
0,
t=1 ta c log2x=1 x=2
t=2 ta c log2x=2 x=4
kt hp vi KX phng trnh cho c 2 nghim l x=2 v x=4
0,
Cu 3:1,0) Tnh tch phn
1
0
2I
1
xdx
x=
+
t 2 2t x x t dx tdt = = =
32 4
11
xdx t dt
tx=
++
Nu0 0
1 1
x t
x t
= =
= =
0,
1 132
0 0
4 14 ( 1 )
1 1
tI dt t t dt
t t= = +
+ +
0,
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3 2 1
0
1 14( ln 1 ) )
3 2t t t t = + +
0,
104ln2
3=
0,
Cu 4:
1,0)
A'
G
M'
C'
B'
C
B
A
Hnh chiu ca AA trn (ABC) l AG nn gc to bi AAv (ABC) l 0' 60AA G = gi Ml trung im BC A,G, M thng hng
0,
t x=AB
ABC u cnh x c AM l ng cao 3 2 3
' ' , ' ' '2 3 3
x xA M A G A M = = =
Trong AAG vung c AG=AAsin600=
3
2
a; 0
3 3' ' os60
2 3 2
a x aA G AA c x= = = =
0,
din tch ABC l2 2
0 21 3 3 3 3 3. .sin 60 ( )
2 4 4 2 16ABC
x a aS AB AC
= = = =
0,
th tch khi lng tr l2 3
. ' ' '
3 3 3 9.
2 16 32ABC A B C ABC
a a aV AG S
= = =
0,
Cu 5:1,0)
Gi s l 2( ; 1 )I t t d tm ca ng trn (C)
V (C) tip xc vi1
d nn
12 2
3 4( 1 ) 20( , ) 5
3 4
t td I d R
+ = =
+
0,
24 25 124 25
24 25 49
t tt
t t
+ = = + =
+ = =
0,
Vi1
1 (1; 2)t I= ta c phng trnh ng trn
( ) ( ) ( )2 2
11 2 25C x y + + =
0,
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Vi1
49 ( 49;48)t I= ta c phng trnh ng trn
( ) ( ) ( )2 2
249 48 25C x y+ + =
0,
Cu 6:1,0)
(S): 2 2 2 4 4 2 16 0x y z x y z+ + + =
(S) c tm I(2;2;-1)
phng trnh mt phng (Q) c dng: 2 2 0x y z D+ + =
iu kin 1(*)D
0,
( , ( )) 3d I P =2 2 2
| 2.2 1.2 2( 1) |3
2 1 ( 2)
D+ + =
+ +
0,
1| 8 | 9
17
DD
D
= + =
=
Kt hp vi iu kin (*) ta c D = -17
0,
Vy phng trnh ca (Q) 2 2 17 0x y z+ = 0,Cu 7:1,0)
( )
( )2 2
1 3 4
4 1 343
1 3 1 ( 3)
i z i
i iiz i
i
+ =
= = = ++ +
0,
3 12( ) 2 cos sin
2 2 6 6i i
= + = +
0,
Theo cng thc Moa-vr
2010 2010 2010 20102 cos sin6 6
z i
= +
0,
( )2010 20102 1 2= = 0,
Cu 8:1,0)
t t=x+y+zTa c
( )22 2 2 2 2 2 24 2 33( ) 4 2
3 3
43
x y z x y z x y z t t
A tt
+ + + + + +
= +
0,
Xt hm s
4( ) 3f t t
t= +
trn
2 3;2
3
2
2 2
4 3 4 2 3'( ) 3 0
3
tf t t
t t
= =
2 3'( ) 0
3f t t = =
Hm s f(t) ng bin trn2 3
;23
do ( ) (2) 8f t f =
0
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7
Du ng thc xy ra khi t=2
Do 8A Du = xy ra khi v ch khi ( )2 2 2 2 23( )
32
x y z x y zx y z
x y z
+ + = + + = = =
+ + =
Vy gi tr ln nht ca A l 8
0,
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I HC SPHM H NI================================================================================
THI THI HC - CAO NG NM 2010Mn Thi: TON
THI THAM KHO Thi gian: 180 pht ( khng kthi gian giao )---------------------------------------------------------------------------------------------------------------------
I.Phn chung cho tt c th sinh(7 im)
Cu I(2 im). Cho hm s212
+
+=x
xy c th l (C)
1.Kho st s bin thin v v th ca hm s2.Chng minh ng thng: mxy += lun lun ct th (C) ti hai im phn bit A, B. Tm m
on AB c di nh nht.Cu II(2 im)
1.Gii phng trnh 9sinx + 6cosx - 3sin2x + cos2x = 8
2.Gii bt phng trnh )3(log53loglog 242
222 > xxx
Cu III(1 im). Tm nguyn hm =xx
dxI
53
cos.sin
Cu IV(1 im). Cho lng tr tam gic ABC.A1B1C1 c tt c cc cnh bng a, gc to bi cnh bn v mt phy bng 300. Hnh chiu H ca im A trn mt phng (A1B1C1) thuc ng thng B1C1. Tnh khong cch hai ng thng AA1 v B1C1 theo a.Cu V(1 im).Cho cc s dng x, y, z tha mn 2 2 2x y z 1+ + = . Tm gi tr nh nht ca biu thc
2 2 2 2 2 2
x y zP
y z z x x y= + +
+ + +
II.Phn ring(3 im)1.Theo chng trnh chunCu Via:
1.Trong mt phng vi h ta Oxy cho ng trn (C) c phng trnh (x-1) 2 + (y+2)2 = 9 v ng th
d: x + y + m = 0. Tm m trn ng thng d c duy nht mt im A m t k c hai tip tuyn AB, ACng trn (C) (B, C l hai tip im) sao cho tam gic ABC vung.
2.Cho im A(10; 2; -1) v ng thng d c phng trnh
+=
=
+=
tz
ty
tx
31
21
. Lp phng trnh mt phng (P
qua A, song song vi d v khong cch t d ti (P) l ln nht.Cu VIIa:1). C bao nhiu s t nhin c 4 ch s khc nhau v khc 0 m trong mi s lun lun c mt hai s chn v hai ch s l.
2) Gii phng trnh: )(,14
Cziz
iz=
+
2.Theo chng trnh nng cao (3 im)Cu VIb(2 im)
1.Trong mt phng vi h ta Oxy cho ng trn (C): x2 + y2 - 2x + 4y - 4 = 0 v ng thng dphng trnh x + y + m = 0. Tm m trn ng thng d c duy nht mt im A m t k c hai tip tuAB, AC ti ng trn (C) (B, C l hai tip im) sao cho tam gic ABC vung.
2.Cho im A(10; 2; -1) v ng thng d c phng trnh3
1
12
1 ==
zyx. Lp phng trnh mt ph
(P) i qua A, song song vi d v khong cch t d ti (P) l ln nht.Cu VIIb(1 im)C bao nhiu s t nhin c 5 ch s khc nhau m trong mi s lun lun c mt hai chchn v ba ch s l.
-------------------------------------------------------------
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I HC SPHM H NI================================================================================
P N
I.Phn dnh cho tt c cc th snhCuI p n im
2. (0,75 im)Honh giao im ca th (C ) v ng thng d l nghim ca phng
trnh
=++
+=
+
+
)1(021)4(2
212
2mxmx
xmx
x
x
Do (1) c mmmvam =++>+= 0321)2).(4()2(01 22 nn ngthng d lun lun ct th (C ) ti hai im phn bit A, B
0,25
Ta c yA = m xA; yB = m xB nn AB2 = (xA xB)
2 + (yA yB)2 = 2(m2 + 12)
suy ra AB ngn nht AB2 nh nht m = 0. Khi 24=AB
0,5
1. (1 im)Phng trnh cho tng ng vi9sinx + 6cosx 6sinx.cosx + 1 2sin2x = 8 6cosx(1 sinx) (2sin2x 9sinx + 7) = 0 6cosx(1 sinx) (sinx 1)(2sinx 7) = 0
0,5
(1-sinx)(6cosx + 2sinx 7) = 0
=+
=
)(07sin2cos6
0sin1
VNxx
x
0,25
22
kx += 0,25
2. (1 im)
K:
>
03loglog
02
222 xx
x
Bt phng trnh cho tng ng vi)1()3(log53loglog 2
22
22 > xxx
t t = log2x,
BPT (1) )3(5)1)(3()3(5322 >+> tttttt
0,5
4log3
1log
43
1
)3(5)3)(1(
3
1
2
2
2x
x
t
t
ttt
t
t
0,25
II(2im)
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I HC SPHM H NI================================================================================
Cx
xxxdttt
tt
dtt
ttt
+++=+++=
+++=
22433
3
246
tan2
1tanln3tan
2
3tan
4
1)
33(
133
0,5
Do )( 111 CBAAH nn gc 1AA H l gc gia AA1 v (A1B1C1), theo gi thit
th gc HAA1 bng 300. Xt tam gic vung AHA1 c AA1 = a, gc
HAA1 =300
2
31
aHA = . Do tam gic A1B1C1 l tam gic u cnh a, H
thuc B1C1 v2
31
aHA = nn A1H vung gc vi B1C1. Mt khc 11CBAH
nn )( 111 HAACB 0,5
K ng cao HK ca tam gic AA1H th HK chnh l khong cch gia AA1v B1C10,25
Cu IV
1 im
Ta c AA1.HK = A1H.AH4
3.
1
1 a
AA
AHHAHK ==
0,25
Cu V1 im
T gi thit 2 2 2x y z 1+ + = suy ra 0 x,y,z 1< <
p dng bt ng thc C-si cho ba s dng 2 2 22x ,1 x ,1 x ta c
( ) ( )( ) ( )
( )
2 2 22 2
2 2 2 23 3
2
2
2
2x 1 x 1 x 22x 1 x 2x 1 x
3 3
2x 1 x3 3
x 3 3x
1 x 2
+ +
2
2 2
x 3 3x (1)
y z 2
+
0,5
A1
A B
C
C1
B1
K
H
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I HC SPHM H NI================================================================================
Tng t ta c
2
2 2
2
2 2
y 3 3y (2)
z x 2z 3 3
z (3)x y 2
+
+
Cng tng v (1), (2), (3) ta c
( )2 2 22 2 2 2 2 2x y z 3 3 3 3
x y z (*)y z z x x y 2 2
+ + + + =+ + +
Du bng (*) xy ra khi3
x y z3
= = =
Vy 3 3minP2
=
0,5
1.T phng trnh chnh tc ca ng trn ta c tm I(1;-2), R = 3, t A kc 2 tip tuyn AB, AC ti ng trn v ACAB => t gic ABIC l hnh
vung cnh bng 3 23=IA
0,5
=
===
7
56123
2
1
m
mm
m
0,52. (1 im)
Gi H l hnh chiu ca A trn d, mt phng (P) i qua A v (P)//d, khi khong cch gia d v (P) l khong cch t H n (P).G.s im I l hnh chiu ca H ln (P), ta c HIAH => HI ln nht khi IA
Vy (P) cn tm l mt phng i qua A v nhn AH lm vc t php tuyn.
0,5
CuVIa2im
)31;;21( tttHdH ++ v H l hnh chiu ca A trn d nn
)3;1;2((0. == uuAHdAH l vc t ch phng ca d)
)5;1;7()4;1;3( AHH Vy (P): 7(x 10) + (y 2) 5(z + 1) = 0 7x + y -5z -77 = 0
0,5
1) T gi thit bi ton ta thy c 624 =C cch chn 2 ch s chn (v khng c
s 0)v 1025 =C cch chn 2 ch s l => c25C .
25C = 60 b 4 s tha mn bi
ton
Mi b 4 s nh th c 4! s c thnh lp. Vy c tt c 24C .25C .4! = 1440 s
0,5Cu
VIIa1im
2) pt z(z2 1) = 0 z = 0 ; x = 1; z = -1 0,5
2.Ban nng cao.1.( 1 im)Cu
VIa2im
T phng trnh chnh tc ca ng trn ta c tm I(1;-2), R = 3, t A k c 2tip tuyn AB, AC ti ng trn v ACAB => t gic ABIC l hnh vung cnh
bng 3 23=IA
0,5
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I HC SPHM H NI================================================================================
=
===
7
56123
2
1
m
mm
m
0,52.Gi H l hnh chiu ca A trn d, mt phng (P) i qua A v (P)//d, khi khong cch gia d v (P) l khong cch t H n (P).Gi s im I l hnh chiu ca H ln (P), ta c HIAH => HI ln nht khi IA
Vy (P) cn tm l mt phng i qua A v nhn AH lm vc t php tuyn.
0,5
)31;;21( tttHdH ++ v H l hnh chiu ca A trn d nn
)3;1;2((0. == uuAHdAH l vc t ch phng ca d)
)5;1;7()4;1;3( AHH Vy (P): 7(x 10) + (y 2) 5(z + 1) = 0 7x + y -5z -77 = 0
0,5
T gi thit bi ton ta thy c 1025 =C cch chn 2 ch s chn (k c s c ch s
0 ng u) v 35C =10 cch chn 2 ch s l => c25C .
35C = 100 b 5 s c chn.
0,5CuVIIa1im Mi b 5 s nh th c 5! s c thnh lp => c tt c 25C .
35C .5! = 12000 s.
Mt khc s cc s c lp nh trn m c ch s 0 ng u l 960!4..
3
5
1
4=
CC .Vy c tt c 12000 960 = 11040 s tha mn bi ton
0,5
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1
THI THI HC - CAO NG NM 2010Mn thi: TON
THI THAM KHO Thi gian: 180 pht ( khng kthi gian giao )--------------------------------------------------------------------------------------------------------------------
Cu 1:( 2,0 im)Cho hm s
2 1
1
xy
x
=
c th (C).
1) Kho st s bin thin v v th (C) ca hm s.2) Tm m ( )m ng thng y x m= + ct (C) ti hai im A, B sao cho 4AB
Cu 2: (2,0 im)1) Gii phng trnh ( )3 cos 2 2cos sin 1 0x x x+ = 2) Gii phng trnh 2 2
2 1 4
2
2log 2log log ( )x x x
x =
Cu 3:(1,0 im)
Tnh tch phn1
0
2I
1
xdx
x=
+
Cu 4:(1,0 im)Cho lng tr ABC.ABC c cnh bn bng a, y ABC l tam gic u, hnh chiu ca A trn
(ABC) trng vi trng tm G ca ABC. Cnh bn to vi y gc 060 . Tnh th tch lng trABC.ABC theo a.
Cu 5:(1,0 im)Trong h to Oxy, cho hai ng thng 1 2: 3 4 20 0, : 1 0d x y d x y+ = + + =
Vit phng trnh ng trn (C) bit rng (C) c bn knh R=5, tip xc vi 1d v c tm nm trn
Cu 6:( 1,0 im)Trong khng gian Oxyz, cho mt cu (S) v mt phng (P) c phng trnh(S): 2 2 2 4 4 2 16 0x y z x y z+ + + = ( ) : 2 2 1 0P x y z+ + =
Vit phng trnh mt phng (Q) song song vi (P) v khong cch t tm mt cu (S) n mt phngbng 3.
Cu 7:( 1,0 im).Cho s phc z tho mn ( )1 3 4i z i+ = . Tnh 2010z .
Cu 8:(1,0 im)
Cho cc s thc khng m x, y, z tho mn 2 2 24
3
x y z+ + = .
Tm gi tr ln nht ca biu thc ( )4
3P x y zx y z
= + + ++ +
.HtTh sinh khng c sdng ti liu. Cn b coi thi khng gii thch g thm.H v tn th sinh:; S bo danh:Ch k gim th:
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2
Hng dn chm TONCu Ni dung iu12,0))1,0
1)Kho st s bin thin v v th hm s2 1
1
xy
x
=
1. Tp xc nh: \ {1}D = 2. S bin thin ca hm s* Gii hn ti v cc, gii hn v cc ca hm s. Tim cn ca th hm s.
12
2 1lim lim lim 2
111
x xx
x xyx
x
= = =
=> th hm s nhn ng thng y=2 lm tim cn ngang
1 11 1
2 1 2 1
lim lim ;lim lim1 1x xx x
x x
y yx x+ +
= = + = =
=> th hm s nhn ng thng x=1 lm tim cn ng
0,
* Lp bng bin thin
2
1' 0
( 1)y x D
x
= <
, y khng xc nh x=1
Hm s nghch bin trn tng khong xc nh ca n. Hm s khng c cc tr.
0,
bng bin thinx - 1 +
y - || -
y 2 +
- 2
0.
3. th- Giao ca th hm s v Ox: y=0=>x=1/2
- Giao ca th hm s v Oy: x=0=>y=1- th hm s nhn im I(1;2) lm tm i xng.
0,
I(1;2)
2
1 y
x
O
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3
)1,0 2)Honh giao im ca ng thng y=x+m (d) v th (C) l nghim ca phng trnh
( ) ( )
2 1
1
2 1 1 (*)
xx m
x
x x x m
= +
= +
( x=1 khng phi l nghim ca (*))2 ( 3) 1 0x m x m + + = (1)
0,
2 2( 3) 4(1 ) 2 5 0m m m m m = = + >
Do (d) lun ct (C) ti hai im phn bit1 1 2 2
( ; ), ( ; )A x y B x y vi1 2,x x l hai nghim ca (1)
0,
Theo vit1 2 1 2
3 ; 1x x m x x m+ = = . V , ( )A B d nn 1 1 2 2;y x m y x m= + = +
( )22 2 2
1 2 1 2 1 22( ) 2 4 2( 2 5)AB x x x x x x m m = = + = +
0,
2 2 21
4 16 2( 2 5) 16 2 3 03
mAB AB m m m m
m
= = = + = =
=
0,
Cu 2:2,0) 1)Gii phng trnh ( )3 cos 2 2cos sin 1 0x x x+ = 3 cos 2 sin 2 2cosx x x + =
3 1cos 2 sin 2 cos
2 2x x x + =
0,
cos 2 cos sin 2 sin cos6 6
x x x
+ =
cos(2 ) cos6x x
=
0,
2 26
( )
2 26
x x k
k
x x k
= +
= +
0,
26
( )2
18 3
x k
kk
x
= +
= +
KL
0,
)1,02)Gii phng trnh 2 2
2 1 4
2
2log 2log log ( )x x x
x = (1)
KX:x>0
( ) 22 2 22
1 log 2log logx xx
+ =
0,
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-------------------------------------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n 0912.
4
2
2 2log 3log 2 0(*)x x + = 0,
t t=log2xThay vo (*) ta c
2 3 2 0
1
2
t t
t
t
+ =
=
=
0,
t=1 ta c log2x=1 x=2
t=2 ta c log2x=2 x=4
kt hp vi KX phng trnh cho c 2 nghim l x=2 v x=4
0,
Cu 3:1,0) Tnh tch phn
1
0
2I
1
xdx
x=
+
t 2 2t x x t dx tdt = = =
32 4
11
xdx t dt
tx=
++
Nu0 0
1 1
x t
x t
= =
= =
0,
1 132
0 0
4 14 ( 1 )
1 1
tI dt t t dt
t t= = +
+ +
0,
3 2 1
01 14( ln 1 ) )3 2
t t t t = + + 0,
104ln2
3=
0,
Cu 4:1,0)
A'
GM'
C'
B'
C
B
A
0,
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I HC SPHM H NI====================================================================
-------------------------------------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n 0912.
5
Hnh chiu ca AA trn (ABC) l AG nn gc to bi AAv (ABC) l 0' 60AA G = gi Ml trung im BC A,G, M thng hngt x=AB
ABC u cnh x c AM l ng cao 3 2 3
' ' , ' ' '2 3 3
x xA M A G A M = = =
Trong AAG vung c AG=AAsin600=
3
2
a; 0
3 3' ' os60
2 3 2
a x aA G AA c x= = = =
0,
din tch ABC l2 2
0 21 3 3 3 3 3. .sin 60 ( )
2 4 4 2 16ABC
x a aS AB AC
= = = =
0,
th tch khi lng tr l2 3
. ' ' '
3 3 3 9.
2 16 32ABC A B C ABC
a a aV AG S
= = =
0,
Cu 5:
1,0)
Gi s l 2( ; 1 )I t t d tm ca ng trn (C)
V (C) tip xc vi 1d nn
12 2
3 4( 1 ) 20( , ) 5
3 4
t td I d R
+ = =
+
0,
24 25 124 25
24 25 49
t tt
t t
+ = = + =
+ = =
0,
Vi 11 (1; 2)t I= ta c phng trnh ng trn
( ) ( ) ( )2 2
1 1 2 25C x y + + =
0,
Vi1
49 ( 49;48)t I= ta c phng trnh ng trn
( ) ( ) ( )2 2
2 49 48 25C x y+ + =
0,
Cu 6:1,0)
(S): 2 2 2 4 4 2 16 0x y z x y z+ + + =
(S) c tm I(2;2;-1)
phng trnh mt phng (Q) c dng: 2 2 0x y z D+ + = iu kin 1(*)D
0,
( , ( )) 3d I P =2 2 2
| 2.2 1.2 2( 1) |3
2 1 ( 2)
D+ + =
+ +
0,
1| 8 | 9
17
DD
D
= + =
=
Kt hp vi iu kin (*) ta c D = -17
0,
Vy phng trnh ca (Q) 2 2 17 0x y z+ = 0,Cu 7:1,0)
( )
( )2 2
1 3 4
4 1 343
1 3 1 ( 3)
i z i
i iiz i
i
+ =
= = = ++ +
0,
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-------------------------------------------------------------------------------------------------------------------------------------------------Gv: Trn Quang Thu n 0912.
6
3 12( ) 2 cos sin
2 2 6 6i i
= + = +
0,
Theo cng thc Moa-vr
2010 2010 2010 2010
2 cos sin6 6z i
= +
0,
( )2010 20102 1 2= = 0,
Cu 8:1,0)
t t=x+y+zTa c
( )22 2 2 2 2 2 24 2 3
3( ) 4 23 3
43
x y z x y z x y z t t
A t
t
+ + + + + +
= +
0,
Xt hm s4
( ) 3f t t t
= +
trn
2 3;2
3
2
2 2
4 3 4 2 3'( ) 3 0
3
tf t t
t t
= =
2 3'( ) 0
3f t t = =
Hm s f(t) ng bin trn2 3
;23
do ( ) (2) 8f t f =
Du ng thc xy ra khi t=2
0
Do 8A Du = xy ra khi v ch khi ( )2 2 2 2 23( )
32
x y z x y zx y z
x y z
+ + = + + = = =
+ + =
Vy gi tr ln nht ca A l 8
0,
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-----------------------------------------------------------------------------------------------------------------------------------
Gv: Trn Quang Thu n 0912.676.613 09
I HC SPHM H NIKHOA TON TIN
P N THI THmn : Ton
hng dn chm v biu im
Ni dung imCu I : (3,0 im)
Cho hm s y = x3
- 3x2
+2 c th (C) trong h ta Oxy1. Kho st v v th hm s2. Gi E l tm i xng ca th (C).Vit phng trnh ng thng qua E v ct (C) ti
ba im E, A, B phn bit sao cho din tch tam gic OAB bng 2
a) Tp xc nh : R 0,25b) S bin thin
* Gii hnx -
, limyx
Limy+
= + = 0,25
1.
(2,0)
* Bng bin thin
y = 3x
2
-6x , y= 0
0
2
x
x
=
=
x - 0 2 +y + 0 - 0 +
y2 +
- -2
Hm sng bin trn cc khong (- ;0) v ( 2 ; +)
Nghch bi
n trn (0; 2)Hm st cc i ti x = 0, yc = 2
t cc tiu ti x =2, yct = -2
0,25
0,5
0,25
c. th+ im cc i, cc tiu :(0;2), (2;-2)+ Giao vi Oy : (0;2)+ Giao vi Ox :
NX :0,5
1
2E
O x
y
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Gv: Trn Quang Thu n 0912.676.613 09
+E (1;0)0,25
2
(1,0)
+ PT ng thng qua E, tha mn yu cu bi ton phi c dng y = k(x-1)( Do trng hp x =1 khng tha mn)
Hong giao im ca (C ) v l nghim ca PT: (x-1)(x2-2x-2-k)=0
+ ct (C ) ti 3 im phn bit th PT x2
-2x-2-k = 0 phi c hai nghim phnbit khc 1 k>-3 0,25
+ Tnh c dtOAB =1
( , ).2
d O AB = 3k k+
0,25
+ T gi thit suy ra k c 3 gi tr -1; -1 3 .
KL : C 3 ng thng tha mn yu cu l y = -x +1 ; y = ( ) ( )1 3 1x 0,25
Cu II : (2,0 im)1. Tm gi tr ln nht v nh nht ca hm s y = 2cosx + sin2x trn [0; 2]
2. Tnh tch phn
( ) ( )
3 2
2 2
0 1 1 2 1
x I dx
x x
=
+ + + +
+ H m s lin tc trn [0;2]
+ Tnh y = 2cos2x - 2sinx, [ ]0;2x
y= 0 5 3
; ;6 6 2
x
0,5
1.
(1,0)+) y(0)=2, 3 3 5 3 3 3( ) ; ( ) ; ( ) 0; (2 ) 2
6 2 6 2 2 y y y y = = = =
0,25
Suy ra[ ] [ ]0;20;2
3 3 3 3ax , min
2 2m y y
= = 0,25
+ t 2 1 x t+ + = x =(t-2)2 -1, dx = 2(t-2)dt ; x =0 t =3, x = 3 t = 4 0,25
2.
(1,0) + a v4
2
3
42 362 16 I t dt
t t
= +
0,25
+ Tnh ra c I = -12+ 42ln4
3 0,5
Cu III : (1,0 im)Cho hnh chp t gic u S.ABCD, bit khong cch gia AB v mt phng(SCD) bng 2. Gc gia mt bn v mt y bng 600 .Tnh th tch hnh chpS.ABCD
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+ Goij I, J ln lt l trung im caAB v CD, H l hnh chiu ca I trnSJ. Chng tc IH = 2 v
gc 060SJI =
+ Gi O l tm y, chng minh c
SO = 2,4
IJ=3
+ Tnh c VS.ABCD =32
9( vtt)
0,5
0,25
0,25
Cu IV : (1,0 im)Tm cc cp s thc (x ; y) tha mn phng trnh sau:
4 3 2 2 3 21 1 4 2 2 2 2 x x y x y x y x xye e x x y xy x + + ++ = + + +
+ t 4 3 2 2 3 21 , x 1x x y x y u y x xy v + = + + =
PT trthnh 2u ve e u v+ = + + (2)
+ Xt f(t)=et - t - 1. Chng tc( ) 0,
( ) 0 0
f t t
f t t
= =
T PT (2) u = v = 0
0,25
0,25
+ Gii h4 3 2 2
3 2
1 0
1 0
x x y x y
x y x xy
+ =
+ + =
( )2
2 3
2 3
1
1
x xy x y
x xy x y
=
= +.
t2
3
x xy a
x y b
=
= , gii ra ta c
1
0
a
b
=
=hoc
2
3
a
b
=
=
+ Thay trli tm c hai cp (x;y) l (1;0) v (-1;0) . Kt lun
0,25
0,25
Cu V : (2,0 im)Trong khng gian vi h ta Oxyz, cho im M( 1; -1; 1) v hai ng
thng 11
:1 2 3
x y zd
+= =
v 2
1 4:
1 2 5
x y zd
= =
1. Chng minh rng im M v cc ng thng d1 v d2 cng nm trn mt mtphng. Vit phng trnh mt phng
2. Gi A, B, C ln lt l hnh chiu ca im M trn Ox, Oy, Oz . Vit phngtrnh ng thng nm trong mt phng (ABC) sao cho ct ng thng
(d2) ng thi vung gc vi (d1)
d1qua M1(0;-1;0), vc tch phng 1 (1; 2; 3)u
d2 qua M2(0;1;-4), 2 (1;2;5)u
0,25
1.
(1,0)
+ Chng t d1 v d2ng phng v vit c PT mp(d1,d2) : - x - 2y + z -2 = 0
+ Chng t Mmp(d1,d2). Kt lun
0,5
0,25
S
A
BC
D
IJ
60
0
O
H
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2.
(1,0)
+ A(1;0;0), B(0; -1;0), C(0;0;1); mp(ABC): x - y + z -1 = 0
+ d2ct (ABC) ti H(1 3
;0;2 2
+ ng thng cn tm c vc tch phng ( )1, ABCu u n =
=(-5;-4;1) , ng
thi i qua H
Suy ra PT :
15
2
4
3
2
x t
y t
z t
=
=
= +
0,25
0,25
0,25
0,25
Cu VI : (1,0 im)Gii phng trnh sau trn tp cc s phc bit n c mt nghim thc:
3 2(5 ) 4( 1) 12 12 0 z i z i z i + + + =
+ Gi nghim thc l a thay vo pt suy ra h3 2
2
5 4 12 06
4 12 0
a a aa
a a
= =
+ + = 0,25
+ Khi PT cho tng ng vi
( )( )2
2
6 (1 ) 2 2 0
6
(1 ) 2 2 0
z z i z i
z
z i z i
+ + =
=
+ + =
0,25
+ Gii ra c cc nghim l 6, 2i v -1-i . Kt lun 0,5
- Trn y ch l hng dn lm bi; phi l lun hp l mi cho im- Nhng cch gii khc ng vn c im ti a- im ton bi c lm trn n 0,5
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2
B. Theo chng trnh Nng caoCu Vb:1. Trong mpOxy, cho ng trn (C): x2 + y2 6x + 5 = 0. Tm M thuc trc tung sao cho qua
M kc hai tip tuyn ca (C) m gc gia hai tip tuyn bng 600.2.Trong khng gian vi h ta Oxyz, cho im M(2 ; 1 ; 0) v ng thng d vi
d :x 1 y 1 z
2 1 1
+= =
.Vit phng trnh chnh tc ca ng thng i qua im M,
ct v vung gc vi ng thng d v tm to ca im M i xng vi M qua d
Cu VIb: Gii h phng trnh
3 3log log 2
2 2
4 4 4
4 2 ( )
log ( ) 1 log 2 log ( 3 )
xyxy
x y x x y
= +
+ + = + +
....HT.
(Cn b coi thi khng gii thch g thm)
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