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Tuyển Tập Các Đề Thi Đại Học Cao Đẳng Môn Toán ( Có Đáp Án )
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Transcript of Tuyển Tập Các Đề Thi Đại Học Cao Đẳng Môn Toán ( Có Đáp Án )
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7/26/2019 Tuyn Tp Cc Thi i Hc Cao ng Mn Ton ( C p n )
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b gio dc v o to K thi tuyn sinh i hc, cao nG nm 2002------------------------------ Mn thi : ton
chnh thc (Thi gian lm bi: 180 pht)_____________________________________________
Cu I (H : 2,5 im; C : 3,0 im)
Cho hm s : (1) ( l tham s).23223 )1(33 mmxmmxxy +++= m
1. Kho st s bin thin v v th hm s (1) khi .1=m 2. Tm k phng trnh: c ba nghim phn bit.033 2323 =++ kkxx3. Vit phng trnh ng thng i qua hai im cc tr ca th hm s (1).Cu II.(H : 1,5 im; C: 2,0 im)
Cho phng trnh : 0121loglog 232
3 =++ mxx (2) ( l tham s).m
1 Gii phng trnh (2) khi .2=m
2. Tm phng trnh (2) c t nht mt nghim thuc on [m 33;1 ].Cu III. (H : 2,0 im; C : 2,0 im )
1. Tm nghim thuc khong )2;0( ca phng trnh: .32cos
2sin21
3sin3cossin +=
+
++ x
x
xxx5
2. Tnh din tch hnh phng gii hn bi cc ng: .3,|34| 2 +=+= xyxxyCu IV.( H : 2,0 im; C : 3,0 im)1. Cho hnh chp tam gic u nh c di cnh y bng a. GiABCS. ,S v ln ltN
l cc trung im ca cc cnh v Tnh theo din tch tam gic , bit rngSB .SC a AMN mt phng ( vung gc vi mt phng .)AMN )(SBC
2. Trong khng gian vi h to cac vung gc Oxyzcho hai ng thng:
v .
=++
=+
0422
042:1
zyx
zyx
+=
+=
+=
tz
ty
tx
21
2
1
:2
a) Vit phng trnh mt phng cha ng thng)(P 1 v song song vi ng thng .2
b) Cho im . Tm to im)4;1;2(M Hthuc ng thng 2 sao cho on thng H c di nh nht.
Cu V.( H : 2,0 im)1. Trong mt phng vi h to cac vung gc Oxy , xt tam gic vung ti ,ABC A
phng trnh ng thng lBC ,033 =yx cc nh vA B thuc trc honh v
bn knh ng trn ni tip bng 2. Tm ta trng tm ca tam gic .G ABC 2. Cho khai trin nh thc:
nx
n
n
nxx
n
n
xn
x
n
nx
n
nxx
CCCC
+
++
+
=
+
3
1
32
1
13
1
2
1
12
1
032
1
22222222 L
( n l s nguyn dng). Bit rng trong khai trin C v s hng th t13 5 nn C=
bng , tm vn20 n x .----------------------------------------Ht---------------------------------------------
Ghi ch: 1)Th sinhch thi cao ngkhng lm Cu V.
2) Cn b coi thi khng gii thch g thm.
H v tn th sinh:.................................................... S bo danh:.....................
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1
b gio dc v o to Kthi tuyn sinh i hc, cao ng nm 2002------------------------------------- p n v thang im mn ton khi A
Cu Ni dung H CI 1
23 31 xxym +==
Tp xc nh Rx . )2(363' 2 =+= xxxxy ,
=
==
2
00'
2
1
x
xy
10",066" ===+= xyxy
Bng bin thin
+ 210x
'y +0 0
+ 0"y
y + lm U 4
CT 2 C 0 li
=
== 300xxy , 4)1( =y
th:
( Th sinh c th lp 2 bng bin thin)
1 ,0
0,25
0,5
0,25
1 ,5
0,5
0,5
0,5
-1 1 2 3 x0
2
4
y
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2
I 2
Cch I. Ta c 2332323 33033 kkxxkkxx +=+=++ .
t 23 3kka += Da vo th ta thy phng trnh axx =+ 23 3
c 3 nghim phn bit 43040 23 =+= ymm c 2 nghim 21 xx
v 'y i du khi qua 1x v 2x hm s t cc tr ti 1x v 2x .
Ta c 23223 )1(33 mmxmmxxy +++=
( ) .2336333
1 222 mmxmmxxmx ++++
=
T y ta c mmxy += 211 2 v mmxy += 2
22 2 .
Vy phng trnh ng thng i qua 2 im cc tr l mmxy += 2
2 .
1 ,0
0,25
0,25
0,25
0,25
----------
0,25
0,25
0,25 0,25
1 ,0
0,25
0,25
0,25
0,25
-----------
0,25
0,25
0,25 0,25
II 1.
Vi 2=m ta c 051loglog 232
3 =++ xx
iu kin 0>x . t 11log23 += xt ta c
06051 22 =+=+ tttt .2
3
2
1
=
=t
t
5,0
0,25
0,1
0,5
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3
31 =t (loi) ,3
3
2
32 33log3log2 ==== xxxt
33=x tha mn iu kin 0>x .(Th sinh c th gii trc tip hoc t n ph kiu khc)
0,25 0,5
2.
0121loglog 232
3 =++ mxx 2)
iu kin 0>x . t 11log23 += xt ta c
0220121 22 =+=+ mttmtt 3)
.21log13log0]3,1[ 2333 += xtxx
Vy (2) c nghim ]3,1[ 3 khi v ch khi (3) c
nghim [ ]2,1 . t tttf += 2)(
Cch 1.
Hm s )(tf l hm tng trn on ][ 2;1 . Ta c 2)1( =f v 6)2( =f .Phng trnh 22)(222 +=+=+ mtfmtt c nghim [ ]2;1
.20622
222
22)2(
22)1(
+
+
+
+ m
m
m
mf
mf
Cch 2.
TH1. Phng trnh (3) c 2 nghim 21, tt tha mn 21 21
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4
2.
V (0x ; )2 nn ly3
1
=x v
3
52
=x . Ta thy 21,xx tha mn iu
kin2
12sin x . Vy cc nghim cn tm l:
31
=x v
3
52
=x .
(Th sinh c th s dng cc php bin i khc)
Ta thy phng trnh 3|34| 2 +=+ xxx c 2 nghim 01 =x v .52 =x
Mt khc ++ 3|34| 2 xxx [ ]5;0x . Vy
( ) ( ) ( )dxxxxdxxxxdxxxxS ++++++=++=1
0
3
1
22
5
0
2 343343|34|3
( )dxxxx +++5
3
2 343
( ) ( ) ( )dxxxdxxxdxxxS +++++=5
3
2
3
1
2
1
0
2 5635
5
3
233
1
231
0
23
2
5
3
16
2
3
3
1
2
5
3
1
++
++
+= xxxxxxxS
6
109
3
22
3
26
6
13=++=S (.v.d.t)
(Nu th sinh v hnh th khng nht thit phi nu bt ng thc ++ 3|34| 2 xxx [ ]5;0x )
0,25
1 ,0
0,25
0,25
0,25
0,25
0,25
1 ,0
0,25
0,25
0,25
0,25
IV 1. 1 1
x510-1
y
3
32
1
8
-1
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5
S
N
I
M C
A K
B
Gi Kl trung im ca BC v MNSKI =
. T gi thitMNa
BCMN ,22
1== //BC I l trung im ca SKv MN.
Ta c = SACSAB hai trung tuyn tng ng ANAM = AMN cn ti A MNAI .
Mt khc
( ) ( )( ) ( )
( ) ( ) SKAISBCAI
MNAI
AMNAI
MNAMNSBC
AMNSBC
=
.
Suy ra SAK cn ti 2
3a
AKSAA == .
244
3 222222 aaaBKSBSK ===
4
10
84
3
2
222
222 aaaSKSASISAAI ==
== .
Ta c16
10.
2
1 2aAIMNS AMN == (vdt)
ch 1) C th chng minh MNAI nhsau:
( ) ( ) AIMNSAKMNSAKBC .2) C th lm theo phng php ta :Chng hn chn h ta cac vung gc Oxyz sao cho
h
aS
aA
aC
aBK ;
6
3;0,0;
2
3;0,0;0;
2,0;0;
2),0;0;0(
trong h l di ng cao SH ca hnh chp ABCS. .
0,25
0,25
0,25
0,25
0,25
0,25
0,25
0,25
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6
2a)
Cch I. Phng trnh mt phng )(P cha ng thng 1 c dng:
( ) ( ) 042242 =++++ zyxzyx ( 022 + )( ) ( ) ( ) 044222 =+++ zyx
Vy ( ) 2;22; ++=Pnr
.Ta c ( )2;1;12 =ur
// 2 v ( ) 22 1;2;1 M
( )P //( ) ( ) ( )
=
=
PMPM
unP
22
2
2
0
1;2;1
0. rr
Vy ( ) 02: =zxP
Cch II Ta c th chuyn phng trnh 1 sang dng tham s nhsau:
T phng trnh 1 suy ra .02 =zx t
=
=
=
=
'4
2'3
'2
:'2 1
tz
ty
tx
tx
( ) )4;3;2(,0;2;0 111 = uM r // 1 .(Ta c th tm ta im 11 M bng cch cho 020 === zyx
v tnh ( )4;3;221
21;12
11;22
121 =
=u
r).
Ta c ( )2;1;12 =ur
// 2 . T ta c vc t php ca mt phng )(P l :
[ ] ( )1;0;2, 21 == uunPrrr
. Vy phng trnh mt phng )(P i qua ( )0;2;01 Mv ( )1;0;2 =Pn
rl: 02 =zx .
Mt khc ( ) ( ) PM 1;2;12 phng trnh mt phng cn tm l: 02 =zx
5,0
0,25
0,25 -----------
0,25
0,25
0,1
0,5
0,5 -----------
0,5
0,5
2b)
b)Cch I. ( ) MHtttHH +++ 21,2,12 = ( )32;1;1 + ttt
( ) ( ) ( ) 5)1(6111263211 22222 +=+=+++= ttttttMHt gi tr nh nht khi v ch khi ( )3;3;21 Ht =Cch II. ( )tttHH 21;2;12 +++ .MHnh nht ( )4;3;210. 22 HtuMHMH ==
r
5,0 0,25
0,25 -----------
0,25 0,25
0,1 0,5
0,5 -----------
0,5 0,5
V 1.
Ta c ( )0;1BOxBC =I . t axA = ta c );( oaA v
.33 == ayax CC Vy 33; aaC .
T cng thc( )
( )
++=
++=
CBAG
CBAG
yyyy
xxxx
3
13
1
ta c
+
3
)1(3;3
12 aaG .
Cch I.
Ta c :
|1|2|,1|3|,1| === aBCaACaAB . Do
1
0,25
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7
( )212
3.
2
1== aACABS ABC .
Ta c( )
|1|3|1|3
132 2
+
=
++=
aa
a
BCACAB
Sr = .2
13
|1|=
+
a
Vy .232|1| +=a
TH1.
+++=
3
326;
3
347332 11 Ga
TH2
=
3
326;
3
134132 22 Ga .
Cch II.
y C
I
O B A x
GiI l tm ng trn ni tip ABC . V 22 == Iyr .
Phng trnh ( ) 3213
11.30: 0 =
== Ixx
xtgyBI .
TH1 Nu A v O khc pha i vi .321+= IxB T 2),( =ACId
.3232 +=+= Ixa
++
3
326;
3
3471G
TH 2.Nu A v O cng pha i vi .321= IxB Tng t
ta c .3212 == Ixa
3
326;
3
1342G
0,25
0,25
0,25 -----------
0,25
0,25
0,25
2.
T 13 5 nn CC = ta c 3n v1
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8
( ) ( ) 02835
6
)2)(1(
!1
!5!3!3
! 2 ==
=
nnnnnn
n
n
n
n
41 =n (loi) hoc .72 =n
Vi 7=n ta c
.4421402.2.3514022 2223
34
21
3
7 ====
xC xxxxx
0,25
0,25
0,5
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b gio dc v o to k thi tuyn sinh i hc, cao ng nm 2002 chnh thc Mn thi : ton, Khi B.
(Thi gian lm bi : 180 pht)_____________________________________________
Cu I (H : 2,0 im; C : 2,5 im) Cho hm s : ( ) 109 224 ++= xmmxy (1) (m l tham s).1. Kho st s bin thin v v th ca hm s (1) khi 1=m .2. Tm m hm s (1) c ba im cc tr.
Cu II (H : 3,0 im; C : 3,0 im)
1. Gii phng trnh: xxxx 6cos5sin4cos3sin 2222 = .
2. Gii bt phng trnh: ( ) 1)729(loglog 3 xx .
3. Gii h phng trnh:
++=+
=
.2
3
yxyx
yxyx
Cu III ( H : 1,0 im; C : 1,5 im) Tnh din tch ca hnh phng gii hn bi cc ng :
4
42x
y = v24
2xy= .
Cu IV (H : 3,0 im ; C : 3,0 im)1. Trong mt phng vi h ta cac vung gc Oxy cho hnh ch nht ABCD c tm
0;
2
1I , phng trnh ng thng AB l 022 =+ yx v ADAB 2= . Tm ta cc nh
DCBA ,,, bit rng nh A c honh m.2. Cho hnh lp phng 1111 DCBABCDA c cnh bng a .
a) Tnh theo a khong cch gia hai ng thng BA1 v DB1 .
b) Gi PNM ,, ln lt l cc trung im ca cc cnh CDBB ,1 , 11DA . Tnh gc gia
hai ng thng MPv NC1 .
Cu V (H : 1,0 im) Cho a gic u nAAA 221 L ,2( n n nguyn ) ni tip ng trn ( )O . Bit rng s tam gic c cc nh l 3 trong n2 im nAAA 221 ,,, L nhiu gp 20 ln s hnh ch nht
c cc nh l 4 trong n2 imn
AAA221
,,, L , tm n .
--------------------------------------Ht-------------------------------------------Ghi ch : 1)Th sinhch thicao ngkhng lm Cu IV 2. b) v Cu V.
2) Cn b coi thi khng gii thch g thm.
H v tn th sinh:................................................................... S bo danh:...............................
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1
B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2002------------------------- p n v thang im thi chnh thc
Mn ton, khi b
Cu Ni dung H CI 1 Vi 1=m ta c 108 24 += xxy l hm chn th i xng qua Oy .
Tp xc nh Rx , ( )44164' 23 == xxxxy , 0'=y
=
=
2
0
x
x
,3
4121612" 22
== xxy
3
20" == xy .
Bng bin thin:
+
23
20
3
22x
'y 0 + 0 0 +
"y + 0 0 + + 10 +
y lm U C U lmCT li CT6 6
Hai im cc tiu : ( )6;21 A v ( )6;22 A .Mt im cc i: ( )10;0B .
Hai im un:
9
10;
3
21U v
9
10;
3
22U .
Giao im ca th vi trc tung l ( )10;0B . th ct trc honh ti 4 im c honh :
64 +=x v 64 =x .
(Th sinh c th lp 2 bng bin thin)
0,1
0,25
0,5
0,25
5,1
0,5
0,5
0,5
x0
10
y
-6
-2 2
A2A1
B
U1 U2
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2
I 2 ( ) ( )922924' 2223 +=+= mmxxxmmxy ,
=+
==
092
00'
22 mmx
xy
Hm s c ba im cc tr phng trnh 0'=y c 3 nghim
phn bit (khi 'y i du khi qua cc nghim) ph
ng trnh092 22 =+mmx c 2 nghim phn bit khc 0.
092 22 =+mmx
=
m
mx
m
2
90
22 . Phng trnh 092
22 =+mmx
c 2 nghim khc 0
>
>
>
x
xxx
x
x (2).
Do 173log9 >>x nn ( ) xx 729log)1( 3 ( ) 072333729 2 xxxx (3).
t xt 3= th (3) tr thnh
2938980722 xttt x .Kt hp vi iu kin (2) ta c nghim ca bt phng trnh l:
273log9 < x .
0,1
0,25
0,25
0,25
0,25
0,1
0,25
0,25
0,25
0,25
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3
3
++=+
=
).2(2
)1(3
yxyx
yxyx iu kin: )3(
.0
0
+
yx
yx
( )
+=
==
.101)1( 63
yx
yxyxyx
Thay y= vo (2), gii ra ta c .1== yx
Thay 1+= yx vo (2), gii ra ta c:2
1,
2
3== yx .
Kt hp vi iu kin (3) h phng trnh c 2 nghim:
1,1 == yx v2
1,
2
3== yx
Ch :Th sinh c th nng hai v ca (1) ln lu tha bc 6 di n kt qu:
+=
=
.1yx
yx
0,1 0,25
0,25
0,25
0,25
0,1 0,25
0,25
0,25
0,25
III
Tm giao im ca hai ng cong4
42x
y = v24
2xy= :
44
2x =
24
2x8804
432
224
===+ xxxx
.
Trn 8;8 ta c24
2x
44
2x v do hnh i xng qua trc tung
nn dxxxS
=
8
0
22
24442 21
8
0
2
8
0
2
22116 SSdxxdxx == .
tnh 1S ta dng php i bin tx sin4= , khi4
0
t th 80 x .
tdtdx cos4= v
>
4;00cos
tt . Do
0,1
0,25
0,25
5,1
0,5
0,25
x0-4 4
2
y
-2 2 2 2
2 A2A1
4
x4y
2
=24
xy
2
=
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4
( ) 422cos18cos16164
0
4
0
2
8
0
2
1 +=+===
dtttdtdxxS .
3
8
26
1
22
1 8
0
3
8
0
2
2 === xdxxS . Vy 34
221 +== SSS .
Ch : Th sinh c th tnh din tch dxxx
S
=8
8
22
2444 .
0,25
0,25
0,5
0,25
IV 1
Khong cch t I n ng thngAB bng2
55= AD v
2
5
== IBIA .
Do BA, l cc giao im ca ng thng AB vi ng trn tm I v bn
knh2
5=R . Vy ta BA, l nghim ca h :
=+
=+2
2
2
2
5
2
1
022
yx
yx
Gii h ta c ( ) ( )2;2,0;2 BA (v 0
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5
IV 2a) Tm khong cch gia BA1 v DB1 .
Cch I. Chn h ta cac vung gc Oxyzsao cho
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )aaDaaaCaaBaaCaAaDaBA ;;0,;;;;0;;0;;;0;0,0;;0,0;0;,0;0;0 1111
( ) ( ) ( )0;0;,;;,;0; 1111 aBAaaaDBaaBA === v ( )22211 ;2;, aaaDBBA = .
Vy ( )[ ] 66,
.,,
2
3
11
1111
11
a
a
a
DBBA
BADBBADBBAd === .
Cch II. ( ) DBBADCABBAADBA
ABBA11111
1
11
.
Tng t DBCA 111 ( )111 BCADB .
Gi ( )111 BCADBG = . Do aCBBBAB === 11111 nn
GGCGBGA == 11 l tm tam gic u 11BCA c cnh bng 2a .
GiI l trung im ca BA1 th IG l ng vung gc chung ca BA1 v
DB1 , nn ( )62
3
3
1
3
1, 1111
aBAICIGDBBAd ==== .
Ch :
Th sinh c th vit phng trnh mt phng ( )P cha BA1 v song song vi
DB1 l: 02 =++ azyx v tnh khong cch t 1B (hoc t D ) ti ( )P ,
hoc vit phng trnh mt phng ( )Q cha DB1 v song song vi BA1 l:
022 =++ azyx v tnh khong cch t 1A (hoc t B) ti ( )Q .
0,1
0,25
0,25
0,25
0,25
0,25
0,25
0,25
5,1
0,25
0,5
0,25
0,5
0,25
0,5
0,5
x
D
D
CB1
A1
z
y
x
A
CB
I
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6
2b)Cch I.
T Cch Ica 2a) ta tm c
a
aPa
aN
aaM ;
2;0,0;;
2,
2;0;
0.;0;2
,2
;2
; 11 =
=
= NCMPa
aNC
aaaMP .
Vy NCMP 1 .
Cch II.
Gi El trung im ca 1CC th ( ) 11CCDDME hnh chiu vung gc caPtrn ( )11CCDD l 1ED . Ta c
NCEDNCDNCCEDCECDCNC 11110
111111 90 === . T y
theo nh l ba ng vung gc ta c NCMP 1 .
0,1
0,25
0,5
0,25
0,25
0,25
0,25 0,25
V
S tam gic c cc nh l 3 trong n2 im nAAA 221 ,,, L l3
2nC .
Gi ng cho ca a gic u nAAA 221 L i qua tm ng trn ( )O lng cho ln th a gic cho c n ng cho ln.
Mi hnh ch nht c cc nh l 4 trong n2 im nAAA 221 ,,, L c cc ng
cho l hai ng cho ln. Ngc li, vi mi cp ng cho ln ta c cc umt ca chng l 4 nh ca mt hnh ch nht. Vy s hnh ch nht ni trnbng s cp ng cho ln ca a gic nAAA 221 L tc
2
nC .
Theo gi thit th:
0,1
0,25
0,25
D1A
B1 C1
C
B
A
ME
N
P
y
x
z
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7
( )( ) ( )
( )( ) ( )2
120
6
2212.2
!2!2
!20
!32!3
!220 232
=
=
=
nnnnn
n
n
n
nCC nn
81512 == nn .
Ch :
Th sinh c th tm s hnh ch nht bng cc cch khc. Nu l lun ng i
n kt qu s hnh ch nht l2
)1( nnth cho im ti a phn ny.
0,5
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B gio dc v o to K thi Tuyn sinh i hc ,cao ng nm 2002 chnh thc Mn thi : Ton Khi D
(Thi gian lm bi : 180 pht) _________________________________________
CuI ( H : 3 im ; C : 4 im).
Cho hm s :( )
1x
mx1m2y
2
= (1) ( m l tham s ).
1. Kho st s bin thin v v th (C) ca hm s (1) ng vi m = -1.
2. Tnh din tch hnh phng gii hn bi ng cong (C) v hai trc ta .3. Tm m thca hm s (1) tip xc vi ng thng xy= .
Cu II ( H : 2 im ; C : 3 im).
1. Gii bt phng trnh : ( )x3x2
. 02x3x22
.
2. Gii h phng trnh :
=+
+
=+
.y22
24
y4y52
x
1xx
2x3
Cu III
( H : 1 im ; C : 1 im ). Tm x thuc on [0 ; 14 ] nghim ng phng trnh : 04xcos3x2cos4x3cos =+ .
Cu IV ( H : 2 im ; C : 2 im).1. Cho hnh t din ABCD c cnh AD vung gc vi mt phng (ABC); AC = AD = 4 cm ;
AB = 3 cm ; BC = 5 cm . Tnh khong cch t im A ti mt phng (BCD).2. Trong khng gian vi h ta cac vung gc Oxyz, cho mt phng (P) : 02yx2 =+
v ng thng md :( ) ( )
( )
=++++
=+++
02m4z1m2mx
01mym1x1m2( m l tham s ).
Xc nh m ng thng md song song vi mt phng (P).
Cu V (H : 2 im ).
1. Tm s nguyn dng n sao cho 243C2....C4C2C nnn2
n1n
0n =++++ .
2. Trong mt phng vi h ta cac vung gc Oxy , cho elip (E) c phng trnh
19
y
16
x 22=+ . Xt im M chuyn ng trn tia Ox v im N chuyn ng trn tia Oy sao cho
ng thng MN lun tip xc vi (E). Xc nh ta ca M , N on MN c di nhnht . Tnh gi tr nh nht .
-------------------------Ht-------------------------
Ch :
1. Th sinh ch thi cao ng khng lm cu V 2. Cn b coi thi khng gii thch g thm.
H v tn th sinh : ................................................................ S bo danh.............................
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1
B gio dc v o to K thi tuyn sinh i hc , cao ng nm 2002Mn Ton khi D
p n v thang im thi chnh thc
Cu Ni dung im
H CI
3 41. 1 1,5
Khi m = -1 ,ta c1x
43
1x
1x3y
=
=
-TX : 1x
- CBT :( )
>
= 1x,01x
4y2
, hm s khng c cc tr.1/4 1/4
3ylimx
=
; =+=+ 1x1x
ylim;ylim .
- BBT :
x - 1 +
y/ + ++
y -3 -3
- 1/4 1/4- TC: x=1 l tim cn ng v =
ylim
1x.
y=-3 l tim cn ngang v 3ylimx
=
1/4 1/4
- Giao vi cc trc : x = 0 y = 1; y = 0 x = - 1/3. 1/4- th :
x
y
1/4 1/2
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2
2. 1
1,5
Din tch cn tnh l :
dx1x
1x3S
0
3/1
=
1/4 1/2
=
0
3/1
0
3/1 1x
dx
4dx3 1/4 1/4
3/1
01xln4
3
1.3
=
1/4 1/2
3
4ln41 += ( vdt).
1/4 1/43. 1 1
K hiu( )
1x
mx1m2)x(f
2
= . Yu cu bi ton tng ng vi tm
m h phng trnh sau c nghim:
(H)( )
=
=
.x)x(f
x)x(f//
1/4 1/4
Ta c (H)
( )
( )
=
=
01x
mx
01x
mx
/2
2
1/4 1/4
( )
( )( ) ( )( )
=
+
=
0
1x
mx1xmx2
0
1x
mx
2
2
2
1/4 1/4Ta thy vi 1m ; x = m lun tho mn h ( H ) . V vy 1m , (H)lun c nghim , ng thi khi m = 1 th h ( H ) v nghim. Do th hm s (1) tip xc vi ng thng y = x khi v ch khi 1m .
S : 1m . 1/4 1/4II
2 31. 1 1,5
Bt phng trnh
>
=
0x3x
02x3x2
02x3x2
2
2
2
1/4 1/2
TH 1: .2
1x2x02x3x202x3x2 22 ====
1/4 1/4
TH 2:
>
>
0x3x
02x3x2
0x3x
02x3x22
2
2
2
>
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3
3x2
1x <
1/4 1/4
T hai trng hp trn suy ra S: 3x2x2
1x =
1/4 1/4
2.1 1,5
H phng trnh
=
=
y2
y4y52x
2x3
1/4 1/2
=+
>=
0y4y5y
0y223
x
1/4 1/4
===
>=
4y1y0y
0y2x
1/4 1/4
=
=
=
=
4y
2x
1y
0x
1/4 1/2
III1 1
Phng trnh ( ) ( ) 01x2cos4xcos3x3cos =++ 0xcos8xcos4 23 =
( ) 02xcosxcos4 2 = 0xcos = 1/4 1/2
+
= k2
x .1/4 1/4
[ ] 3k2k1k0k14;0x ==== 1/4
S : ;2
x
= 2
3x
= ;
2
5x
= ;
2
7x
= .
1/4 1/4IV
2 21. 1 1Cch 1T gi thit suy ra tam gic ABC vung ti A , do .ACAB 1/4 1/4Li c ( )ABCmpAD ABAD v ACAD , nn AB, AC, AD imt vung gc vi nhau. 1/4 1/4Do c th chn h to cac vung gc, gc A sao cho B(3;0;0) ,
C(0;4;0), D( 0;0;4). Mt phng (BCD) c phng trnh :
014
z
4
y
3
x=++ .
1/4 1/4
Khong cch cn tnh l :17
346
16
1
16
1
9
1
1=
++
(cm).
1/4 1/4
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4
Cch 2T gi thit suy ra tam gic ABC vung ti A , do .ACAB 1/4 1/4Li c ( )ABCmpAD ABAD v ACAD , nn AB, AC, AD imt vung gc vi nhau. 1/4 1/4
D
H C
A E
BGi AE l ng cao ca tam gic ABC; AH l ng cao ca tam gicADE th AH chnh l khong cch cn tnh.
D dng chng minh c h thc:2222 AC
1
AB
1
AD
1
AH
1++= .
1/4 1/4Thay AC=AD=4 cm; AB = 3 cm vo h thc trn ta tnh c:
cm17
346AH=
1/4 1/4Cch 3:T gi thit suy ra tam gic ABC vung ti A , do .ACAB 1/4 1/4
Li c ( )ABCmpAD ABAD v ACAD , nn AB, AC, AD imt vung gc vi nhau. 1/4 1/4
Gi V l th tch t din ABCD, ta c V= 8ADACAB6
1= .
p dng cng thc)BCD(dt
V3AH
= vi V = 8 v dt( BCD) =2 34
ta tnh c cm17
346AH= .
1/2 1/22 1 1
Cch 1:Mt phng (P) c vect php tuyn ( )0;1;2n
. ng thng md c vec
t ch phng ( )( ) ( ) ( )( )m1m;1m2;1m2m1u 2 ++
. 1/4 1/4
Suy ra
u .
n =3(2m+1).
md song song vi (P)
)P(d
nu
m
1/4 1/4
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5
( )
=
PA,dA
0n.u
m
Ta c : iu kin 0n.u =
2
1m =
1/4 1/4
Mt khc khi m = - 1/2 th md c phng trnh :
=
=
0x
01y, mi im
A( 0;1;a) ca ng thng ny u khng nm trong (P), nn iu kin( )PA,dA m c tho mn. S : m = - 1/2 1/4 1/4
Cch 2:Vit phng trnh dmdi dng tham s ta c
=
+=
+=
m)t.m(12z
t1)(2m1y
1)tm)(2m(1x2
1/4 1/4
md // (P) h phng trnh n t sau
=+
=
+=
+=
02yx2
t)m1(m2z
t)1m2(1y
t)1m2)(m1(x2
v nghim
1/4 1/4 phng trnh n t sau 3(2m+1)t+1 = 0 v nghim 1/4 1/4 m=-1/2 1/4 1/4Cch 3:
md // (P) h phng trnh n x, y, z sau
(H) ( ) ( )
=++++
=+++=+
02m4z)1m2(mx
01myx1x1m2
02yx2
v nghim 1/4 1/4
T 2 phng trnh u ca h phng trnh trn suy ra
+=
=
3
4m2y
3
1mx
1/4 1/4
Th x , y tm c vo phng trnh th ba ta c :
)6m11m(3
1z)1m2( 2 ++=+
1/4 1/4
H (H) v nghim2
1m =
1/4 1/4V
2
1. 1
Ta c : ( ) =
=+n
0k
kkn
nxC1x ,
1/4
Cho x = 2 ta c ==n
0k
kknn 2C31/4
5n32433 5n === . 1/2
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6
2. 1Cch 1Gi s M(m;0) v N(0;n) vi m > 0 , n > 0 l hai im chuyn ng trnhai tia Ox v Oy.
ng thng MN c phng trnh : 01n
y
m
x=+
1/4ng thng ny tip xc vi (E) khi v ch khi :
1n
19
m
116
22
=
+
.
1/4Theo BT Csi ta c :
( )2
2
2
2
22
22222
n
m9
m
n1625
n
9
m
16nmnmMN ++=
++=+=
499.16225 =+ 7MN 1/4
ng thc xy ra
>>
=+=
0n,0m
49nm
n
m9
m
n16
22
2
2
2
2
21n,72m == .
KL: Vi 21;0N,0;72M th MN t GTNN v GTNN (MN) = 7. 1/4Cch 2Gi s M(m;0) v N(0;n) vi m > 0 , n > 0 l hai im chuyn ng trnhai tia Ox v Oy.
ng thng MN c ph
ng trnh : 01n
y
m
x=+
1/4ng thng ny tip xc vi (E) khi v ch khi :
1n
19
m
116
22
=
+
.
1/4Theo bt ng thc Bunhiacpski ta c
( ) 49n
3.n
m
4.m
n
9
m
16nmnmMN
2
22
22222 =
+
++=+= .
7MN 1/4
- ng thc xy ra
>>
=+
=
0n,0m
7nmn
3
:nm
4
:m22 21n,72m == .
KL: Vi 21;0N,0;72M th MN t GTNN v GTNN (MN) = 7. 1/4Cch 3:
Phng trnh tip tuyn ti im (x0; y0) thuc (E) : 19
yy
16
xx 00 =+
1/4
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7
Suy ra to ca M v N l
0;
x
16M
0
v
0y
9;0N
+
+=+=
20
2
20
220
20
20
2
20
22
y
9
x
16
9
y
16
x
y
9
x
16MN
1/4S dng bt ng thc Csi hoc Bunhiacpski (nhcch 1 hoc cch 2)ta c : 22 7MN
1/4
- ng thc xy ra7
213y;
7
78x 00 == .
- Khi 21;0N,0;72M v GTNN (MN) = 7 1/4
-----------------------Ht----------------------
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8
B gio dc v o to K thi tuyn sinh i hc ,cao ng nm 2002------------------------ ---------------------------------------------
H
ng dn chm thi mn ton khi D
Cu I:
1. -Nu TS lm sai bc no th k t tr i s khng c im.-Nu TS xc nh ng hm s v ch tm ng 2 tim cn th c 1/4 im.
2. Nu TS lm sai bc no th k t tr i s khng c im.3. -Nu TS dng iu kin nghim kp th khng c im.
-Nu TS khng loi gi tr m = 1 th b tr 1/4 im.
Cu II:
1. -Nu TS lm sai bc no th k t tr i s khng c im.-Nu TS kt lun nghim sai b tr 1/4 im .
-Nu TS s dng iu kin sai:
> l trung im cnh CC .'
a) Tnh th tch khi t din 'BDA M theo a v b .
b) Xc nh t sa
b hai mt phng v( ' )A BD ( )BD vung gc vi nhau.
Cu 4 ( 2 im).
1) Tm h s ca s hng chax8trong khai trin nh thc Niutn ca
n
xx
+ 5
3
1 , bit rng
)3(73
14
+= +++ nCC
nn
nn
( n l s nguyn dng,x> 0, l s t hp chp k ca n phn t).knC
2) Tnh tch phn
+=
32
52 4xx
dxI .
Cu 5 (1 im).Cho x,y, zl ba s dng v x+y+ z1. Chng minh rng
.821
1
1
2
2
2
2
2
2 +++++z
zy
yx
x
HT
Ghi ch: Cn b coi thi khng gii thch g thm.
H v tn th sinh: .. . S bo danh: .
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1
B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003
p n thang im thi chnh thc Mn thi : ton Khi A
Ni dung imCu 1. 2im
1)
Khi2 1 1
1 .1 1
x xm y x
x x
+ = = =
+ Tp xc nh: \{ 1 }.R
+2
2 2
01 2' 1 . ' 0
2.( 1) ( 1)
xx xy y
xx x
= += + = = =
+ [ ] =
=
01
1lim)(lim
xxy
xxtim cn xin ca th l: y = .
=
yx 1lim tim cn ng ca th l: 1=x .
Bng bin thin:
th khng ct trc honh. th ct trc tung ti im (0; 1).
1 im
0,25
0,5
0, 25
x 0 1 2 + y 0 + + 0
+ + 3y CT C
1
y
xO 1 2
3
1
1
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2
2)
th hm s1
2
++=
x
mxmxy ct trc honh ti 2 im phn bit c honh
dng phng trnh 2( ) 0f x mx x m= + + = c 2 nghim dng phn bit khc 1
2
0
1 4 0
(1) 2 1 0
10, 0
m
m
f m
mS P
m m
= >
= + = > = >
0
1
12 01 2
2
0
m
m
m
m
m
x x x x x .
Cch 2. t 43
1( ) 2 ( ) min ( ) 0
4
= + + = >
xf x x x f x f x f
R
.
Trng hp ny h v nghim.Vy nghim ca h phng trnh l:
1 5 1 5 1 5 1 5( ; ) (1;1), ; , ;
2 2 2 2x y
+ + =
.
0, 25
Cu 3. 3im
1)
Cch 1.t AB a= . Gi Hl hnh chiu vung gc ca B trn AC, suy ra BH AC,m BD (AAC) BD AC, do AC(BHD) ACDH.Vy gc
phng nh din [ ], ' ,B A C D l gc BHD .Xt 'A DC vung ti D c DH l ng cao, ta c . ' . 'DH A C CD A D=
. '
'
CD A DDH
A C =
. 2 2
3 3
a a a
a= = . Tng t, 'A BC vung tiB cBHl ng
cao v2
3
aBH= .
Mt khc:
2 2 2
2 2 2 2 2 2 22 2 . cos 2. cos3 3 3a a aa BD BH DH BH DH BHD BHD= = + = + ,
do 1
cos2
BHD= o120BHD = .
Cch 2. Ta cBDACBD AC (nh l ba ng vung gc).Tng t,BCAC(BCD) AC. Gi H l giao im ca 'A Cv ( ' )BC D
BHD l gc phng ca [ ]; ' ;B A C D .
Cc tam gic vungHAB,HAD, HAC bng nhau HB=HC=HD
H l tm BCD u o120BHD = .
1 im
0, 25
0, 25
0, 25
0, 25
hoc
0, 25
0,25 0,5
A
A
B C
D
D
CB
H
I
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4
2)a) T gi thit ta c
)2
;;();;('0);;;( b
aaMbaaCaaC .
Vy ( ; ; 0), (0; ; )2
bBD a a BM a= =
2, ; ;
2 2
ab abBD BM a
=
.
( )23
' ; 0; , . ' .2
a bBA a b BD BM BA
= =
Do 2
'1
, . '6 4
BDA Ma b
V BD BM BA = =
.
b) Mt phng ( )BDM c vct php tuyn l 21 , ; ;2 2
ab abn BD BM a
= =
,
mt phng ( ' )BD c vct php tuyn l 22 , ' ( ; ; )n BD BA ab ab a = =
.
Do 2 2 2 2
41 2( ) ( ' ) . 0 02 2
a b a bBDM A BD n n a a b = + = =
1ab
= .
2 im
0, 25
0, 25
0, 25
0, 25
0, 5
0, 5
Cu 4. 2im
1)
Ta c ( )1 14 3 3 3 37( 3) 7( 3)n n n n nn n n n nC C n C C C n+ ++ + + + + = + + = + ( 2)( 3)
7( 3) 2 7.2! 14 12.2!
n nn n n
+ + = + + = = =
S hng tng qut ca khai trin l ( )12
5 60 113 2 2
12 12.
kk
kk kC x x C x
=
.
Ta c
60 1182
60 118 4.
2
= = =
kk
x x k
Do h s ca s hng cha 8x l .495)!412(!4
!12412 =
=C
2) Tnh tch phn2 3
2 25 4
xdxI
x x=
+ .
t 22
4
4
dxt x dt
x
= + =+
v 2 2 4.x t=
Vi 5x= th 3t= , vi 2 3x= th 4t= .
Khi 2 3 4 4
22 23 35
1 1 1
4 2 244
xdx dtI dt
t ttx x
= = = + +
4
3
1 2 1 5ln ln .
4 2 4 3
t
t
= = +
1 im
0, 5
0, 25
0, 25
1 im
0, 25
0, 25
0,25
0, 25
A
A
BC
D
D
CB
y
x
z
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5
Cu 5. 1im
Vi mi ,u v
ta c | | | | | | (*)u v u v+ +
(v ( )22 22 2 2| | 2 . | | | | 2 | | . | | | | | |u v u v u v u v u v u v+ = + + + + = +
)
t ,1
;
=
xxa
=
yyb
1; ,
=
zzc
1; .
p dng bt ng thc (*) ta c | | | | | | | | | | | | .a b c a b c a b c+ + + + + +
Vy
22 2 2 2
2 2 2
1 1 1 1 1 1( )P x y z x y z
y zx y z
= + + + + + + + + + +
.
Cch 1. Ta c
( )22
22 3 31 1 1 1 9
( ) 3 3 9P x y z xyz tx y z xyz t
+ + + + + + = +
, vi
( )
22
3 1
0 3 9
x y z
t xyz t
+ + = <
.
t2
9 9 1( ) 9 '( ) 9 0, 0; ( )
9Q t t Q t t Q t
t t
= + = <
gim trn1
0;9
1( ) 82.
9Q t Q
=
Vy ( ) 82.P Q t
(Du = xy ra khi 13
x y z= = = ).Cch 2.
Ta c2 2
2 2 21 1 1 1 1 1( ) 81( ) 80( )x y z x y z x y zx y z x y z
+ + + + + = + + + + + + +
21 1 118( ) 80( ) 162 80 82.x y z x y zx y z
+ + + + + + =
Vy 82.P
(Du = xy ra khi 13
x y z= = = ).Ghi ch:Cu ny cn c nhiu c ch gii khc.
0, 25
0, 25
0, 25
0, 25
hoc
0,25
0,5
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B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003----------------------- Mn thi : ton khi B
chnh thc Thi gian lm bi: 180 pht_______________________________________________
Cu 1(2 im). Cho hm s ( l tham s).3 23 (1)y x x m= + m 1) Tm th hm s (1) c hai im phn bit i xng vi nhau qua gc ta .m 2) Kho st s bin thin v v th hm s (1) khi m =2.Cu 2(2 im).
1) Gii phng trnh2
otg tg 4sin 2sin2
x x xcx
+ = .
2) Gii h phng trnh
2
2
2
2
23
23 .
yy
x
xx
y
+=
+=
Cu 3(3 im).1) Trong mt phng vi h ta cac vung gc Ox cho tam gic cy ABC
0, 90 .AB AC BAC= = Bit (1; 1)M l trung im cnh BCv2
; 03
G l trng
tm tam gic . Tm ta cc nh .
ABC , ,A B C
2) Cho hnh lng tr ng c y l mt hnh thoi cnh ,
gc
. ' ' ' 'ABCD A B C D ABCD a
060BAD= . Gi l trung im cnh v l trung im cnh ' .Chng minh rng bn im
' NAA CC
', , ,B M D N
'
cng thuc mt mt phng. Hy tnh
di cnh ' theo a t gicAA B MDN l hnh vung.
3) Trong khng gian vi h ta cac vung gc Ox cho hai im
v im sao cho . Tnh khong cch t
trung im
yz
0)(2; 0; 0), (0; 0; 8)A B C (0; 6;AC
=
I ca BCn ng thng OA .Cu 4 (2 im).
1) Tm gi tr ln nht v nh nht ca hm s 24 .y x x= +
2) Tnh tch phn
4 2
0
1 2sin
1 sin 2I dxx
= + .Cu 5 (1 im). Cho l s nguyn dng. Tnh tngn
2 3 10 1 22 1 2 1 2 1
2 3 1
nn
n n nC C Cn
+ + + + +
+ nC
( C l s t hp chp kca phn t).kn n
----------------------------------Ht---------------------------------
Ghi ch: Cn b coi thi khng gii thch g thm.
H v tn th sinh.. S bo danh
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1
B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003
p n thang im thi chnh thc Mn thi : ton Khi B
Ni dung imCu 1. 2im
1) th hm s (1) c hai im phn bit i xng nhau qua gc ta tn ti 0 0x sao cho 0 0( ) ( )y x y x=
tn ti 0 0x sao cho3 2 3 2
0 0 0 03 ( ) 3( )x x m x x m + = +
tn ti 0 0x sao cho2
03x m=
0m > .2) Kho st s bin thin v v th hm s khi m = 2.
Khi 2m= hm s tr thnh 3 23 2.y x x= +
Tp xc nh : .
2 0' 3 6 , ' 02.
xy x x yx
== = =
" 6 6. '' 0 1.y x y x= = =
"y trit tiu v i du qua 1 (1;0)x= l im un.
Bng bin thin:
th ct trc honh ti cc im (1; 0), (1 3; 0) v ct trc tung ti im (0;2) .
1 im0, 25
0, 25
0,25
0,25 1 im
0,25
0,25
0,25
0,25
x 0 2 + y + 0 0 +
2 + C CT
y 2
x
y
O
2
21
2
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2
Cu 2. 2im
1) Gii phng trnh:2
cotg tg 4sin 2 (1).sin2
x x xx
+ =
iu kin:sin 0
(*).cos 0
x
x
Khi (1)cos sin 2
4sin2sin cos sin 2
x xx
x x + =
2 2cos sin 24sin2
sin cos sin 2
x xx
x x x
+ =
22cos 2 4sin 2 2x x + = 22cos 2 cos 2 1 0x x =
cos 2 1
1cos2
32
x kx
kx
== = +=
( )kZ .
Kt hp vi iu kin (*) ta c nghim ca (1) l
( ).3
x k k= + Z
2) Gii h phng trnh
2
2
2
2
23 (1)
2
3 (2).
yy
x
x
x y
+=
+=
iu kin 0, 0x y .
Khi h cho tng ng vi2 2
2 22 2
( )(3 ) 03 2
3 2.3 2
x y xy x yx y y
xy xxy x
+ + == +
= += +
TH1:2 2
1
1.3 2
x y x
yxy x
= =
== +
TH2:2 2
3 0
3 2
xy x y
xy x
+ + =
= + v nghim, v t (1) v (2) ta c , 0x y> .
Vy nghim ca h phng trnh l: 1.x y= =
1 im
0,25
0,25
0,25
0,25
1 im
0,25
0,5
0,25
Cu 3. 3im1)V G l trng tm BC v l trung im BCnn
3 ( 1;3)MA MG= =
(0;2)A .
Phng trnh BC i qua (1; 1)M v vung gc vi
( 1, 3)MA=
l: 1( 1) 3( 1) 0 3 4 0 (1).x y x y + + = + + =
Ta thy 10MB MC MA= = = ta ,B C tha mn
phng trnh: 2 2( 1) ( 1) 10 (2).x y + + =
Gii h (1),(2) ta
c ta ca ,B Cl (4; 0), ( 2; 2).
2)Ta c ' // 'A M NC A MCN= l hnh bnh hnh,do 'A C v MN ct nhau ti trung im I cami ng. Mt khcADCBl hnh bnh hnh nntrung imI caACcng chnh l trung im ca
BD.Vy MN v BD ct nhau ti trung im Icami ng nnBMDNl hnh bnh hnh. Do B,
M, D, Ncng thuc mt mt phng.Mt khc DM2 = DA2+ AM2= DC2 + CN2= DN2,
hay DM = DN. Vy hnh bnh hnh BMDN l hnh thoi. Do BMDN l hnh
1 im
0,25
0,25
0,250,251 im
0,5
GA
B
C
M.
D
A
D C
B N
M
A B
C
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3
vung MN = BD AC = BD AC2= BD2= BB2+BD23a2 = BB2+ a2
BB= 2a AA= 2a .3)
T (0;6;0)AC=
vA(2; 0; 0) suy ra C(2; 6; 0), do I(1; 3; 4).
Phng trnh mt phng () quaIv vung gc vi OA l : 1 0.x = ta giao im ca () vi OA lK(1; 0; 0).
khong cch tIn OAl 2 2 2(1 1) (0 3) (0 4) 5.IK= + + =
0,5
1 im0,25
0,25
0,25
0,25
Cu 4. 2im
1) Tm gi tr ln nht v nh nht ca hm s 24 .y x x= +
Tp xc nh: [ ]2; 2 .
2' 1
4
xy =
,
22 2
0' 0 4 2
4
xy x x x
x x
= = =
=.
Ta c ( 2) 2, ( 2) 2 2, (2) 2y y y = = = ,
Vy[ 2;2]max ( 2) 2 2y y
= = v[ 2;2]min ( 2) 2y y
= = .
2) Tnh tch phn
4 2
0
1 2sin.
1 sin 2I dx
x
=
+
Ta c
4 42
0 0
1 2sin cos 2
1 sin 2 1 sin 2
x xI dx dx
x x
= =
+ + .
t 1 sin 2 2cos 2t x dt xdx= + = .
Vi 0x= th 1,t= vi
4x= th 2t= .
Khi 2
1
21 1 1ln | | ln 2.
12 2 2
dtI t
t= = =
1 im
0,25
0,25
0,25
0,25
1 im
0,25
0,25
0,25
0,25
Cu 5. 1im
Ta c 0 1 2 2(1 ) ...n n nn n n nx C C x C x C x+ = + + + + .
Suy ra ( )2 2
0 1 2 2
1 1
(1 ) ...n n nn n n nx dx C C x C x C x dx+ = + + + +
22 2 3 1
1 0 1 2
11
1 (1 ) ...1 2 3 1
n
n nn n n nx x xx C x C C Cn n
+
+
+ = + + + + + +
2 3 1 1 10 1 22 1 2 1 2 1 3 2
2 3 1 1
n n nn
n n n nC C C C n n
+ + + + + + + =
+ + .
0,5
0,5
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B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003---------------------- Mn thi: ton Khi D chnh thc Thi gian lm bi: 180 pht
_______________________________________________
Cu 1 (2 im).1) Kho st s bin thin v v th ca hm s
2 2 4 (1)
2
x xy
x
+=
.
2) Tm ng thng d ym : 2 2m mx m= + ct th ca hm s (1) ti hai im
phn bit.Cu 2 (2 im).
1) Gii phng trnh 2 2 2
sin tg cos 02 4 2
x xx
=
.
2) Gii phng trnh .2 222 2x x x x + = 3
Cu 3 (3 im).1) Trong mt phng vi h ta cac vung gc cho ng trnOxy
4)2()1(:)( 22 =+ yxC v ng thng : 1 0d x y = .
Vit phng trnh ng trn ( i xng vi ng trn qua ng thng
Tm ta cc giao im ca v .
')C
(C
( )C .d
) ( ')C
2) Trong khng gian vi h ta cac vung gc Oxyz cho ng thng
3 2:
1 0.k
x ky zd
kx y z
0+ + =
+ + =
Tm ng thng vung gc vi mt phngk kd ( ) : 2 5 0P x y z + = .
3) Cho hai mt phng v vung gc vi nhau, c giao tuyn l ng thng( )P ( )Q .Trn ly hai im vi ,A B AB a= . Trong mt phng ly im , trong
mt phng ( ly im sao cho ,
( )P C
)Q D AC BD cng vung gc vi v
. Tnh bn knh mt cu ngoi tip t din v tnh khongcch t n mt phngAC BD
A
AB== ABCD( )BCD theo .a
Cu 4 ( 2 im).
1) Tm gi tr ln nht v gi tr nh nht ca hm s2
1
1
xy
x
+=
+ trn on [ ]1; 2 .
2) Tnh tch phn2
2
0
I x x d= x .
Cu 5 (1 im).
Vi l s nguyn dng, gin 3 3na l h s ca3 3nx trong khai trin thnh a
thc ca ( 1 . Tm n 2 ) ( 2)nx x+ + n 3 3 26na n= .
------------------------------------------------ Ht ------------------------------------------------Ghi ch: Cn b coi thi khng gii thch g thm.
H v tn th sinh:.. . S bo danh:
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B gio dc v o to k thi tuyn sinh i hc, cao ng nm 2003
p n thang im thi chnh thc Mn thi : ton Khi D
Ni dung im
Cu 1. 2im
1) Kho st s bin thin v v th ca hm s2 2 4
2
x xy
x
+=
. 1 im
Tp xc nh :R \{ 2 }.
Ta c2 2 4 4
.2 2
x xy x
x x
+= = +
2
2 2
04 4' 1 . ' 0
4.( 2) ( 2)
xx xy y
xx x
== = = =
[ ] 4
lim lim 02x xy x x = = tim cn xin ca th l: y x= ,
tim cn ng ca th l:2
limx
y
= 2x= .
Bng bin thin:
th khng ct trc honh. th ct trc tung ti im (0; 2).
0,25
0,5
0,25
2) 1 im
ng thng ct th hm s (1) ti 2 im phn bitmd
phng trnh4
2 22
x mx mx
+ = +
c hai nghim phn bit khc 2
2( 1)( 2) 4m x = c hai nghim phn bit khc 2 1 0m > 1.m >
Vy gi tr cn tm lm 1.m>
0,5
0,5
x
2
6
2
2 4O
y
x 0 2 4 + y + 0 0 +
2 + + y C CT
6
1
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Cu 2. 2im
1) Gii phng trnh 2 2 2
tg cos 02 4 2
x xx
sin (1)= 1 im
iu kin: (*). Khi cos 0x
( )2
2
1 sin 1(1) 1 cos 1 cos
2 2 2cos
xx x
x
= +
( ) ( )2 21 sin sin 1 cos cosx x = + x
( ) ( )1 sin (1 cos )(1 cos ) 1 cos (1 sin )(1 sin )x x x x + = + + x
( )1 sin (1 cos )(sin cos ) 0x x x x + + =
2sin 1 2
cos 1 2
tg 1
4
x kx
x x k
xk
= +=
= = + = = +
( )kZ .
Kt hp iu kin (*) ta c nghim ca phng trnh l:
2
4
k
k
= + = +
( ) .kZ
0,5
0,25
0,25
2) Gii phng trnh (1).2 222 2x x x x + 3= 1 im
t .2
2 0x xt t= >
Khi (1) tr thnh 24
3 3 4 0 ( 1)( 4) 0t t t t t t
= = + = =4t (v t )0>
Vy2 22 4x x x x = =2
1
2.
= =
x
x
Do nghim ca phng trnh l1
2.
= =
x
x
0,5
0,5
Cu 3. 3im
1) 1 im
T ( ) suy ra c tm v bn knh2 2: ( 1) ( 2) 4 + =C x y ( )C (1;2)I 2.R=
ng thng c vct php tuyn l nd (1; 1).= uur
Do ng thng i qua
v vung gc vi d c phng trnh:(1;2)I 1 2
1 1
x yx y 3 0
= +
= .
Ta giao im ca v l nghim ca h phng trnh:H d
1 0 2 (2;1).3 0 1
x y x Hx y y
= = + = =
Gi l im i xng vi qua . Khi J (1;2)I d
2 3(3;0)
2 0J H I
J H I
x x xJ
y x x
= =
= =.
V i xng vi ( qua nn c tm l v bn knh
Do c phng trnh l:
( ')C
(C
)C d ( ')C
2 2
(3;0)J 2.R=
') ( 3) 4 +x y = .
Ta cc giao im ca ( v l nghim ca h phng trnh:)C ( ')C
2 2
2 2 22 2
1 0 1( 1) ( 2) 4 1, 0
3, 2.( 3) 4 2 8 6 0( 3) 4
x y y xx y x y
x yx y x xx y
= = + = = = = = + = + = + =
Vy ta giao im ca v ( l v( )C ')C (1;0)A (3;2).B
0,5
0,25
0,25
2
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2) 1 im
Ta c cp vect php tuyn ca hai mt phng xc nh lkd 1 (1;3 ; 1)= uur
n k
v . Vect php tuyn ca l2 ( ; 1;1)= uur
n k ( )P (1; 1; 2)= r
n .
ng thng c vect ch phng l:kd
21 2, (3 1; 1; 1 3 ) 0 k k k
r
Nn21 1 3
1.1 1 2
k k kk = = =
Vy gi tr cn tm l
0,5
0,5
3) 1 imTa c (P) (Q) v = (P) (Q), mAC AC (Q) AC AD, hay
. Tng t, ta cBDnn
BD(P), do CBD . VyAvB
A,Bnm trn mt cu ng knh CD.
090=CAD 090=
V bn knh ca mt cu l:
2 21
2 2
CDR BC BD= = +
2 2 21 3
2 2
aAB AC BD= + + = .
GiHl trung im caBCAHBC. DoBD(P) nn BD AHAH(BCD).
VyAHl khong cch t A n mt phng (BCD)v1 2
.2 2
aAH BC= =
0,25
0,25
0,5
Cu 4. 2im
1) Tm gi tr ln nht v gi tr nh nht ca hm s2
1
1
xy
x
+=
+trn on [ ]1; 2 . 1 im
2 3
1' .
( 1)
xy
x
=
+
' 0 1y x= = .
Ta c 3
( 1) 0, 2, (2) .5
y(1)y y = = =
Vy [ ]1;2 (1) 2max y y = = v [ ]1;2min ( 1) 0.y y = =
0,5
0,5
2) Tnh tch phn2
2
0
I x x d= x . 1 im
Ta c 2 0 0 1x x , suy ra1 2
2 2
0 1
( ) ( )= + I x x dx x x dx
1 22 3 3 2
0 1
1.2 3 3 2
= + =
x x x x
0,5
0,5
u n n k = =
r uur uur
3 1( ) || kd P u n
r r
k 1.=k
.
A B
C
D
P
Q
H
3
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Cu 5. 1im
Cch 1:Ta c ( 2 0 2 1 2 2 2 2 41) ...n n n nn n nnnC x C x C x C
+ = + + + + ,
0 1 1 2 2 2 3 3 3( 2) 2 2 2 ... 2n n n n n nn n n nnnx C x C x C x C x C
+ = + + + + + .
D dng kim tra 1, 2= =n n khng tha mn iu kin bi ton.
Vi th3n 3 3 2 3 2 2 1.n n n n nx x x x x = =
Do h s ca 3 3nx trong khai trin thnh a thc ca l2( 1) ( 2+ +n nx x )
nC3 0 3 1 1
3 3 2 . . 2. .n n n na C C C = + .
Vy2
3 3
52 (2 3 4)
26 26 73
2
= + = =
=
n
nn n n
a n nn
Vy l gi tr cn tm (v nguyn dng).5=n nCch 2:
Ta c
2 3 2
3 3 2
20 0 0 0
1 2( 1) ( 2) 1 1
1 22 .
n n
n n n
i kn n n nn i k n i i k k k
n n n ni k i k
x x x xx
x C C x C x C xxx
= = = =
+ + = + +
= =
Trong khai trin trn, lu tha ca l 3 3n khi 2 3i k = 3k
, hayTa ch c hai trng hp tha iu kin ny l
2 3i k+ = .0,i= = hoc i 1, 1k= = .
Nn h s ca 3 3nx l .0 3 3 1 13 3 . .2 . .2n n n n na C C C C = +
Do 2
3 3
52 (2 3 4)26 26 7
32
= + = =
=
n
nn n na n n
n
Vy l gi tr cn tm (v nguyn dng).5=n n
0,75
0,25
hoc
0,75
0,25
4
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B gio dc v o to thi tuyn sinh i hc, cao ng nm 2004------------------------------ Mn thi : Ton , Khi A
chnh thc Thi gian lm bi : 180 pht, khng k thi gian pht --------------------------------------------------------------
Cu I (2 im)
Cho hm s2x 3x 3
y2(x 1)
+ =
(1).
1) Kho st hm s (1).2) Tm m ng thng y = m ct th hm s (1) ti hai im A, B sao cho AB = 1.
Cu II (2 im)
1) Gii bt phng trnh22(x 16) 7 x
x 3 >x 3 x 3
+
.
2)
Gii h phng trnh 1 442 2
1
log (y x) log 1y
x y 25.
=
+ =
Cu III (3 im)
1) Trong mt phng vi h ta Oxy cho hai im ( )A 0; 2 v ( )B 3; 1 . Tm ta trctm v ta tm ng trn ngoi tip ca tam gic OAB.
2) Trong khng gian vi h ta Oxyz cho hnh chp S.ABCD c y ABCD l hnh thoi,
AC ct BD ti gc ta O. Bit A(2; 0; 0), B(0; 1; 0), S(0; 0; 2 2 ). Gi M l trung im
ca cnh SC.a) Tnh gc v khong cch gia hai ng thng SA, BM.
b) Gi s mt phng (ABM) ct ng thng SD ti im N. Tnh th tch khi chp S.ABMN.
Cu IV (2 im)
1) Tnh tch phn I =2
1
xdx
1 x 1+ .
2) Tm h s ca x8trong khai trin thnh a thc ca8
21 x (1 x) + .
Cu V (1 im)
Cho tam gic ABC khng t, tha mn iu kin cos2A + 2 2 cosB + 2 2 cosC = 3.Tnh ba gc ca tam gic ABC.
------------------------------------------------------------------------------------------------------------------------
Cn b coi thi khng gii thch g thm.
H v tn th sinh............................................................................S bo danh.................................................
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1
B gio dc v o to p n - Thang im..................... thi tuyn sinh i hc, cao ng nm 2004
........................................... chnh thc Mn:Ton, Khi A
(p n - thang im c 4 trang)
Cu Ni dung imI 2,0
I.1 (1,0 im)
( )12332
+=
x
xxy =
( )
1 1x 1
2 2 x 1 +
.
a) Tp xc nh: { }R \ 1 .
b) S bin thin:
2
x(2 x)y '
2(x 1)
=
; y ' 0 x 0, x 2= = = . 0,25
yC= y(2) =1
2 , yCT= y(0) =
3
2.
ng thng x = 1 l tim cn ng.
ng thng 1
y x 12
= + l tim cn xin. 0,25
Bng bin thin:x 0 1 2 +
y'
0 + + 0
y + + 1
2
3
2
0,25
c) th:
0,25
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2
I.2 (1,0 im)Phng trnh honh giao im ca th hm s vi ng thng y = ml :
( ) m
x
xx=
+
12
332 ( ) 023322 =++ mxmx (*). 0,25
Phng trnh (*) c hai nghim phn bit khi v ch khi:
0> 24m 4m 3 0 > 3m2
> hoc 1m2
< (**). 0,25
Vi iu kin (**), ng thng y = m ct th hm s ti hai im A, B c honh x1, x2 l nghim ca phng trnh (*).
AB = 1 121 = xx 2
1 2x x 1 = ( )1 22
1 2x x 4x x 1+ = 0,25
( ) ( ) 123432 2 = mm
1 5m
2
= (tho mn (**)) 0,25
II 2,0
II.1 (1,0 im)
iu kin : x 4 . 0,25
Bt phng trnh cho tng ng vi bt phng trnh:2 2
2(x 16) x 3 7 x 2(x 16) 10 2x + > > 0,25
+ Nu x > 5 th bt phng trnh c tho mn, v v tri dng, v phi m. 0,25
+ Nu 4 x 5 th hai v ca bt phng trnh khng m. Bnh phng hai v ta
c: ( ) ( )22 22 x 16 10 2x x 20x 66 0 > + < 10 34 x 10 34 < < + .
Kt hp vi iu kin 4 x 5 ta c: 10 34 x 5 < . p s: x 10 34> 0,25
II.2 (1,0 im)
iu kin: y > x v y > 0.
( ) 11
loglog 44
1 =y
xy ( ) 11
loglog 44 =y
xy 0,25
4y x
log 1y
=
4
3yx = . 0,25
Th vo phng trnh x2+ y2= 25 ta c:
2
23y y 25 y 4.4
+ = =
0,25
So snh vi iu kin , ta c y = 4, suy ra x= 3 (tha mn y > x).Vy nghim ca h phng trnh l (3; 4). 0,25
III 3,0 III.1 (1,0 im)
+ ng thng qua O, vung gc vi BA( 3 ; 3)
c phng trnh 3x 3y 0+ = .
ng thng qua B, vung gc vi OA(0; 2)
c phng trnh y = 1
( ng thng qua A, vung gc vi BO( 3 ; 1)
c phng trnh 3x y 2 0+ = )0,25
Gii h hai (trong ba) phng trnh trn ta c trc tm H( 3; 1) 0,25
+ ng trung trc cnh OA c phng trnh y = 1.ng trung trc cnh OB c phng trnh 3x y 2 0+ + = .
( ng trung trc cnh AB c phng trnh 3x 3y 0+ = ).0,25
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3
Gii h hai (trong ba) phng trnh trn ta c tm ng trn ngoi tip tam gicOAB l ( )I 3 ; 1 . 0,25
III.2.a (1,0 im)
+ Ta c: ( )C 2; 0; 0 , ( )D 0; 1; 0 , ( )2;0;1M ,
( )22;0;2 =SA , ( )BM 1; 1; 2=
. 0,25 Gi l gc gia SA v BM.
Ta c: ( ) SA.BM 3
cos cos SA, BM2SA . BM
= = =
30 = .0,25
+ Ta c: ( )SA, BM 2 2; 0; 2 =
, ( )AB 2; 1; 0=
. 0,25
Vy:
( )SA, BM AB 2 6
d SA,BM
3SA,BM
= =
0,25
III.2.b (1,0 im)
Ta c MN // AB // CD N l trung im SD
2;
2
1;0N .
0,25
( )SA 2; 0; 2 2=
, ( )2;0;1 =SM , ( )22;1;0 =SB , 1SN 0; ; 22
=
( )SA, SM 0; 4 2; 0 =
. 0,25
S.ABM 1 2 2V SA,SM SB6 3
= =
0,25
S.AMN
1 2V SA,SM SN
6 3 = =
S.ABMN S.ABM S.AMNV V V 2= + = 0,25
IV 2,0 IV.1 (1,0 im)
2
1
xI dx
1 x 1=
+ . t: 1= xt 1
2+=tx tdtdx 2= .
01 == tx , 12 == tx . 0,25
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4
Ta c:
1 1 12 32
0 0 0
t 1 t t 2I 2t dt 2 dt 2 t t 2 dt
1 t 1 t t 1
+ + = = = +
+ + +
0,25
I
1
3 2
0
1 12 t t 2t 2 ln t 1
3 2
= + +
0,25
1 1 11I 2 2 2 ln 2 4ln 23 2 3 = + =
. 0,25
IV.2 (1, 0 im)
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
8 2 3 42 0 1 2 2 4 3 6 4 8
8 8 8 8 8
5 6 7 85 10 6 12 7 14 8 16
8 8 8 8
1 x 1 x C C x 1 x C x 1 x C x 1 x C x 1 x
C x 1 x C x 1 x C x 1 x C x 1 x
+ = + + + +
+ + + +
0,25
Bc ca x trong 3 s hng u nh hn 8, bc ca x trong 4 s hng cui ln hn 8. 0,25
Vy x8 ch c trong cc s hng th t, th nm,vi h s tng ng l:3 2 4 0
8 3 8 4C .C , C .C 0,25
Suy ra a8 168 70 238= + = . 0,25V 1,0
Gi 3cos22cos222cos ++= CBAM
32
cos2
cos2221cos2 2
+
+= CBCB
A . 0,25
Do 0
2sin >
A, 1
2cos
CBnn 2
AM 2cos A 4 2 sin 4
2 + . 0,25
Mt khc tam gic ABC khng t nn 0cos A , AA coscos2 . Suy ra:
42sin24cos2 + A
AM 42sin242sin212 2 +
=
AA
22
sin242
sin4 2 += AA
012
sin222
=
A. Vy 0M . 0,25
Theo gi thit: M = 0
=
=
=
2
1
2sin
12
cos
coscos2
A
CB
AA
A 90
B C 45
=
= =
0,25
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B gio dc v o to------------------------
chnh thc
thi tuyn sinh i hc, cao ng nm 2004Mn: Ton, Khi B
Thi gian lm bi: 180 pht, khng k thi gian pht -------------------------------------------
Cu I(2 im)
Cho hm s y = xxx 3231 23 + (1) c th (C).
1) Kho st hm s (1).2)Vit phng trnh tip tuyn ca (C) ti im un v chng minh rng l tip tuyn ca (C)
c h s gc nh nht.
Cu II(2 im)
1)
Gii phng trnh xtgxx 2)sin1(32sin5 = .
2) Tm gi tr ln nht v gi tr nh nht ca hm s
x
xy
2ln= trn on [1; 3e ].
Cu III(3 im)1) Trong mt phng vi h ta Oxy cho hai im A(1; 1), B(4; 3 ). Tm im C thuc ng
thng 012 = yx sao cho khong cch t C n ng thng AB bng 6.
2) Cho hnh chp t gic u S.ABCD c cnh y bng a, gc gia cnh bn v mt y bng
( o0 < < o90 ). Tnh tang ca gc gia hai mt phng (SAB) v (ABCD) theo . Tnh thtch khi chp S.ABCD theo a v .
3) Trong khng gian vi h ta Oxyz cho im A )4;2;4( v ng thng d:
+=
=
+=
.41
1
23
tz
ty
tx
Vit phng trnh ng thng i qua im A, ct v vung gc vi ng thng d.
Cu IV(2 im)
1) Tnh tch phn I = dxx
xxe
+
1
lnln31.
2) Trong mt mn hc, thy gio c 30 cu hi khc nhau gm 5 cu hi kh, 10 cu hi trungbnh, 15 cu hi d. T 30 cu hi c th lp c bao nhiu kim tra, mi gm 5 cu
hi khc nhau, sao cho trong mi nht thit phi c 3 loi cu hi (kh, trung bnh, d) vs cu hi d khng t hn 2 ?
Cu V(1 im)Xc nh m phng trnh sau c nghim
22422 1112211 xxxxxm ++=
++ .
------------------------------------------------------------------------------------------------------------------------
Cn b coi thi khng gii thch g thm.
H v tn th sinh.................................................................................................S bo danh...........................
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1
B gio dc v o to p n - Thang im..................... thi tuyn sinh i hc, cao ng nm 2004
........................................... chnh thc Mn:Ton,Khi B
(p n - thang im c 4 trang)
Cu Ni dung imI 2,0
1 Kho st hm s (1,0 im)
3 21y x 2x 3x3
= + (1).
a) Tp xc nh: R.b) S bin thin:y' = x2 4x + 3; 3,10' === xxy . 0,25
yC= y(1) = 43 , yCT= y(3) = 0; y" = 2x 4, y'' = 0 ( ) 2x 2, y 2 3 = = . th
hm s li trn khong ( ; 2), lm trn khong ( 2; + ) v c im un l
2U 2;
3
.
0,25
Bng bin thin:x 1 3 +
y' + 0 0 +
y 43
+
0
0,25
c) th:Giao im ca th vi cc trcOx, Oy l cc im ( ) ( )0;0 , 3;0 .
0,25
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2
2 Vit phng trnh tip tuyn ca (C) ti im un, ...(1,0 im)
Ti im un U 22;
3
, tip tuyn ca (C) c h s gc 1)2(' =y . 0,25
Tip tuyn ti im un ca th (C) c phng trnh:
2 8y 1.(x 2) y x
3 3
= + = + . 0,25
H s gc tip tuyn ca th (C) ti im c honh x bng:
y'(x) = x2 34 + x = 1)2( 2 x 1 y' (x) y' (2), x.0,25
Du " =" xy ra khi v ch khi x = 2 ( l honh im un).Do tip tuyn ca th (C) ti im un c h s gc nh nht.
0,25
II 2,01 Gii phng trnh (1,0 im)
5sinx 2 = 3 tg2x ( 1 sinx ) (1) .
iu kin: cosx 0 x k , k Z2
+ (*). 0,25
Khi (1) 2
2
3sin x5sin x 2 (1 sin x)
1 sin x =
02sin3sin2 2 =+ xx . 0,25
2
1sin = x hoc 2sin =x (v nghim).
0,25
+
== 262
1sin kxx hoc +
= 2
6
5kx , Zk ( tho mn (*)).
0,25
2 Tm gi tr ln nht v gi tr nh nht ca hm s (1,0 im)
y =2ln x
x
2
ln x(2 ln x )y '
x
= 0,25
y'= 03
2 3
ln x 0 x 1 [1; e ]
ln x 2 x e [1; e ].
= = = =
0.25
Khi : y(1) = 0, 2 32 3
4 9y(e ) , y(e )
e e= =
0,25
So snh 3 gi tr trn, ta c:33
2
2[1; e ][1; e ]
4max y khi x e , min y 0 khi x 1
e= = = = .
0,25
III 3,01 Tm im C (1,0 im)
Phng trnh ng thng AB:4
1
3
1
=
yx 4x + 3y 7 = 0. 0,25
Gi s );( yxC . Theo gi thit ta c: 012 = yx (1).
d(C, (AB)) = 62 2
4x 3y 37 0 (2a)4x 3y 76
4x 3y 23 0 (2b).4 3
+ =+ = + + =+
0,25
Gii h (1), (2a) ta c: C1( 7 ; 3). 0,25
Gii h (1), (2b) ta c: 2 43 27C ;11 11 . 0,25
2 Tnh gc v th tch (1,0 im)
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3
Gi giao im ca AC v BD lO th SO (ABCD) , suy ra
SAO= .
Gi trung im ca AB l M thOM AB v ABSM Gcgia hai mt phng (SAB) v(ABCD) lSMO .
0,25
Tam gic OAB vung cn ti O, nn === tg
aSO
aOA
aOM
2
2
2
2,
2.
Do : SO
tgSMO 2 tgOM= = . 0,25
2 3
S.ABCD ABCD
1 1 a 2 2V S .SO a tg a tg .
3 3 2 6= = = 0,50
3 Vit phng trnh ng thng (1,0 im)
ng thng d c vect ch phng )4;1;2( =v . 0,25 B d )41;1;23( tttB ++ (vi mt s thc t no ).
( )AB 1 2t;3 t; 5 4t = + +
. 0,25
AB d 0. =vAB 2(1 2t) (3 t) 4( 5 4t) 0 + + + = t = 1. 0,25
AB (3; 2; 1) =
Phng trnh ca1
4
2
2
3
4:
=
+=
+
zyx. 0,25
IV 2,01 Tnh tch phn (1,0 im)
dxx
xxI
e
+
=1
lnln31.
t: 2 dx
t 1 3ln x t 1 3ln x 2tdt 3x
= + = + = .
x 1 t 1= = , x e t 2= = . 0,25Ta c: ( )
2 222 4 2
1 1
2 t 1 2I t dt t t dt
3 3 9
= = .
0,25
25 3
1
2 1 1I t t
9 5 3
=
.
0,25
I =
135
116.
0,25
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4
2 Xc nh s kim tra lp c ... (1,0 im)Mi kim tra phi c s cu d l 2 hoc 3, nn c cc trng hp sau: c 2 cu d, 2 cu trung bnh, 1 cu kh, th s cch chn l:
23625.. 15210
215 =CCC . 0,25
c 2 cu d, 1 cu trung bnh, 2 cu kh, th s cch chn l:
10500..25
110
215 =CCC .
0,25 c 3 cu d, 1 cu trung bnh, 1 cu kh, th s cch chn l:
22750.. 15110
315 =CCC . 0,25
V cc cch chn trn i mt khc nhau, nn s kim tra c th lp c l:56875227501050023625 =++ . 0,25
V Xc nh m phng trnh c nghim 1,0 iu kin: 1 x 1.t t 2 21 x 1 x= + .
Ta c: 2 21 x 1 x t 0+ , t = 0 khi x = 0.2 4t 2 2 1 x 2 t 2= , t = 2 khi x = 1.
Tp gi tr ca t l [0; 2 ] ( t lin tc trn on [ 1; 1]). 0,25
Phng trnh cho tr thnh: m ( ) 2t 2 t t 2+ = + +
2t t 2m
t 2
+ + =
+(*)
Xt f(t) =2t t 2
t 2
+ +
+vi 0 t 2 . Ta c f(t) lin tc trn on [0; 2 ].
Phng trnh cho c nghim x Phng trnh (*) c nghim t [0; 2 ]
]2;0[]2;0[
)(max)(min tfmtf .0,25
Ta c: f '(t) = ( )
2
2
t 4t0, t 0; 2
t 2
+ f(t) nghch bin trn [0;2 ].
0,25
Suy ra:[0; 2 ] [0; 2 ]
min f (t) f ( 2) 2 1 ; max f (t) f (0) 1= = = = .
Vy gi tr ca m cn tm l 2 1 m 1 . 0,25
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B gio dc v o to thi tuyn sinh i hc, cao ng nm 2004------------------------ Mn: Ton, Khi D chnh thc Thi gian lm bi: 180 pht, khng k thi gian pht
-------------------------------------------Cu I (2 im)
Cho hm s 3 2y x 3mx 9x 1= + + (1) vi m l tham s.1) Kho st hm s (1) khi m = 2.2) Tm m im un ca th hm s (1) thuc ng thng y = x + 1.
Cu II(2 im)1) Gii phng trnh .sin2sin)cossin2()1cos2( xxxxx =+
2) Tm m h phng trnh sau c nghim
=+
=+
.31
1
myyxx
yx
Cu III(3 im)1) Trong mt phng vi h ta Oxy cho tam gic ABC c cc nh );0();0;4();0;1( mCBA
vi 0m . Tm ta trng tm G ca tam gic ABC theo m. Xc nh m tam gic GABvung ti G.
2) Trong khng gian vi h ta Oxyz cho hnh lng tr ng 111. CBAABC . Bit ),0;0;(aA
0,0),;0;(),0;1;0(),0;0;( 1 >> babaBCaB .
a) Tnh khong cch gia hai ng thng CB1 v 1AC theo .,ba
b) Cho ba, thay i, nhng lun tha mn 4=+ba . Tm ba, khong cch gia hai ng
thng CB1 v 1AC ln nht.
3) Trong khng gian vi h ta Oxyz cho ba im )1;1;1(),0;0;1(),1;0;2( CBA v mtphng (P): 02 =++ zyx . Vit phng trnh mt cu i qua ba im A, B, C v c tmthuc mt phng (P).
Cu IV(2 im)
1) Tnh tch phn I = 3
2
2 )ln( dxxx .
2) Tm cc s hng khng cha x trong khai trin nh thc Niutn ca
7
43 1
+
x
x vi x > 0.
Cu V(1 im)Chng minh rng phng trnh sau c ng mt nghim
01225 = xxx .
---------------------------------------------------------------------------------------------------------------------Cn b coi thi khng gii thch g thm.
H v tn th sinh.............................................................S bo danh........................................
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1
B gio dc v o to p n - Thang im..................... thi tuyn sinh i hc, cao ng nm 2004
........................................... chnh thc Mn:Ton, Khi D
(p n - thang im c 4 trang)Cu
Ni dung im
I 2,0
1 Kho st hm s (1,0 im)
1962 23 ++== xxxym .a) Tp xc nh: R .b) S bin thin:
2 2y ' 3x 12x 9 3(x 4x 3)= + = + ; y ' 0 x 1, x 3= = = . 0,25yC= y(1) = 5 , yCT= y(3) =1. y'' = 6x 12 = 0 x = 2 y = 3. th hms li trn khong ( ; 2), lm trn khong );2( + v c im un l
)3;2(U . 0,25Bng bin thin:
x 1 3 +
y' + 0 0 +
y 5 +
10,25
c) th:
th hm s ct trc Oy ti im (0; 1).
0,252 Tm m im un ca th hm s ...(1,0 im)
y = x3 3mx2+ 9x + 1 (1); y' = 3x2 6mx + 9; y'' = 6x 6m .y"= 0 x = m y = 2m3 + 9m + 1. 0,25y" i du t m sang dng khi i qua x = m, nn im un ca th hm s(1) l I( m; 2m3+ 9m +1). 0,25
I thuc
ng thng y = x + 1 2m3
+ 9m + 1 = m + 1 0,252m(4 m2) = 0 m = 0 hoc 2=m . 0,25
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2
II 2,01 Gii phng trnh (1,0 im)
( 2cosx 1) (2sinx + cosx) = sin2x sinx ( 2cosx 1) (sinx + cosx) = 0. 0,25
2cosx 1= 0 cosx =1
x k2 , k
2 3
= + Z .
0,25 sinx + cosx = 0 tgx = 1 x k , k
4
= + Z .
0,25
Vy phng trnh c nghim l: x k23
= + v x k , k
4
= + Z .
0,252 Tm m h phng trnh c nghim (1,0 im)
t: u = x , v y , u 0, v 0.= H cho tr thnh:3 3
u v 1
u v 1 3m
+ =
+ = (*)
0,25u v 1
uv m
+ =
=
u, v l hai nghim ca phng trnh: t2 t + m = 0 (**).
0,25H cho c nghim (x; y) H (*) c nghim u 0, v 0 Phng trnh(**) c hai nghim t khng m. 0,25
1 4m 0
1S 1 0 0 m .
4P m 0
=
=
=
0,25
III 3,01 Tnh to trng tm G ca tam gic ABC v tm m... (1,0 im)
Trng tm G ca tam gic ABC c ta :
A B C A B CG G
x x x y y y mx 1; y3 3 3
+ + + += = = = . Vy G(1; m3
).0,25
Tam gic ABC vung gc ti G GA.GB 0=
. 0,25m m
GA( 2; ), GB(3; )3 3
.0,25
GA.GB 0=
2m6 0
9 + = m 3 6 = .
0,252 Tnh khong cch gia B1C v AC1,... (1,0 im)
a) T gi thit suy ra:
1 1C (0; 1; b), B C (a; 1; b)=
1 1AC ( a; 1; b), AB ( 2a; 0; b)= =
0,25
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3
( )
1 1 1
1 12 2
1 1
B C, AC AB abd B C, AC
a bB C, AC
= = +
.
0,25
b) p dng bt ng thc Csi, ta c:
1 1
2 2
ab ab 1 1 a bd(B C; AC ) ab 2
22ab 2 2a b
+= = =
+
.
0,25 Du "=" xy ra khi v ch khi a = b = 2.
Vy khong cch gia B1C v AC1ln nht bng 2 khi a = b = 2. 0,25
3 Vit phng trnh mt cu (1,0 im)I(x; y; z) l tm mt cu cn tm I (P) v IA = IB = IC .Ta c: IA2= (x 2)2 + y2+ ( z 1)2 ;
IB2= (x 1)2+ y2+ z2;IC2= (x 1)2 + (y 1)2+ ( z 1)2. 0,25
Suy ra h phng trnh:
=
=
=++
22
22
02
ICIB
IBIA
zyx
=+
=+
=++
12
2
zyzx
zyx
0,25
.0;1 === yzx 0,25 == 1IAR Phng trnh mt cu l ( x 1)2+ y2+ ( z 1)2=1. 0,25IV 2,0
1 Tnh tch phn (1,0 im)
I =3
2
2
ln(x x) dx . t2
2
2x 1du dxu ln(x x)
x xdv dx
v x
==
= =
.
0,25 3 332
22 2
2x 1 1I x ln(x x) dx 3ln 6 2 ln 2 2 dx
x 1 x 1
= = +
0,25
( )3
23ln 6 2 ln 2 2x ln x 1= + .
0,25
I = 3ln6 2ln2 2 ln2 = 3ln3 2. 0,25 2 Tm s hng khng cha x... (1, 0 im)
Ta c: ( )7 k7 7 k
k3 374 4
k 0
1 1x C x
x x
=
+ =
0,25
7 k k 28 7k 7 7k k3 4 127 7
k 0 k 0
C x x C x
= =
= = .0,25
S hng khng cha x l s hng tng ng vi k (k Z, 0 k 7) tho mn:
4012
728==
k
k.
0,25
S hng khng cha x cn tm l 47C 35= . 0,25
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4
V Chng minh phng trnh c nghim duy nht 1,0x5 x2 2x 1 = 0 (1) .
(1) x5= ( x + 1)20 x 0 (x + 1)21 x51 x 1. 0,25
Vi x 1: Xt hm s 5 2f (x) x x 2x 1= . Khi f(x) l hm s lin tc
vi mi x 1.Ta c:
f(1) = 3 < 0, f(2) = 23 > 0. Suy ra f(x) = 0 c nghim thuc ( 1; 2). (2) 0,25
f '( x) = 4 4 4 45x 2x 2 (2x 2x) (2x 2) x = + + .3 4 42x(x 1) 2(x 1) x 0, x 1= + + > . 0,25
Suy ra f(x) ng bin trn [ 1; +) (3).T (1), (2), (3) suy ra phng trnh cho c ng mt nghim. 0,25
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BGIO DC V O TO-----------------------
CHNH THC
THI TUYN SINH I HC, CAO NG NM 2005Mn: TON, khi A
Thi gian lm bi: 180 pht, khng kthi gian pht ----------------------------------------
Cu I (2 im)
Gi m(C ) l thca hm s1
y m xx
= + (*) ( m l tham s).
1)
Kho st sbin thin v vthca hm s(*) khi 1m .4
=
2)
Tm m hm s (*) c cc trv khong cch tim cc tiu ca m(C ) n tim
cn xin ca m(C ) bng1
.2
Cu II (2 im)
1) Gii bt phng trnh 5x 1 x 1 2x 4. >
2) Gii phng trnh 2 2cos 3x cos 2x cos x 0. =
Cu III (3 im)1) Trong mt phng vi hta Oxy cho hai ng thng
1d : x y 0 = v 2d : 2x y 1 0.+ =
Tm ta cc nh hnh vung ABCD bit rng nh A thuc 1d , nh C thuc 2d
v cc nh B, D thuc trc honh.
2) Trong khng gian vi hta Oxyz cho ng thngx 1 y 3 z 3
d :1 2 1
+ = =
v mt
phng (P) : 2x y 2z 9 0.+ + =
a)
Tm ta im I thuc d sao cho khong cch t I n mt phng (P) bng 2. b)Tm ta giao im A ca ng thng d v mt phng (P). Vit phng trnh
tham sca ng thng nm trong mt phng (P), bit i qua A v vung
gc vi d.
Cu IV (2 im)
1) Tnh tch phn2
0
sin 2x sin xI dx.
1 3cos x
+=
+
2) Tm snguyn dng n sao cho1 2 2 3 3 4 2n 2n 1
2n 1 2n 1 2n 1 2n 1 2n 1C 2.2C 3.2 C 4.2 C (2n 1).2 C 2005+
+ + + + + + + + + =L
( knC l sthp chp k ca n phn t).
Cu V (1 im)
Cho x, y, z l cc sdng tha mn1 1 1
4.x y z
+ + = Chng minh rng
1 1 11.
2x y z x 2y z x y 2z+ +
+ + + + + +
------------------------------ Ht -----------------------------Cn bcoi thi khng gii thch g thm.
Hv tn th sinh................................................. sbo danh........................................
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1
BGIO DC V O TO---------------------CHNH THC
P N THANG IMTHI TUYN SINH I HC, CAO NG NM 2005
----------------------------------------Mn: TON, Khi A
(p n thang im gm 4 trang)
Cu Ni dung imI 2,0
I.1 1,01 1 1
m y x4 4 x
= = + .
a) TX: \{0}.
b) Sbin thin:2
2 2
1 1 x 4y '
4 x 4x
= = , y ' 0 x 2,x 2.= = =
0,25
yC ( ) ( )CTy 2 1, y y 2 1.= = = = ng thng x 0= l tim cn ng.
ng thng1
y x4
= l tim cn xin.
0,25
c) Bng bin thin:
x 2 0 2 + y + 0 0 +
y
1 + +
1
0,25
d) th
0,25
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2
I.2 1,0
2
1y ' m , y ' 0
x= = c nghim khi v chkhi m 0> .
Nu m 0> th 1 21 1
y ' 0 x , xm m
= = = .
0,25
Xt du y ' x
1
m 0
1
m +
y ' + 0 || 0 +
Hm slun c cc trvi mi m 0.>
0,25
im cc tiu ca ( )mC l1
M ;2 m .m
Tim cn xin (d) : y mx mx y 0.= =
( )2 2
m 2 m md M,d .
m 1 m 1
= =
+ +
0,25
( ) 22
1 m 1d M;d m 2m 1 0 m 1.
2 2m 1= = + = =
+
Kt lun: m 1= .0,25
II. 2,0II.1 1,0
Bt phng trnh: 5x 1 x 1 2x 4 > . K:
5x 1 0
x 1 0 x 2.2x 4 0
0,25
Khi bt phng trnh cho tng ng vi
5x 1 2x 4 x 1 5x 1 2x 4 x 1 2 (2x 4)(x 1) > + > + + 0,25
2 2x 2 (2x 4)(x 1) x 4x 4 2x 6x 4 + > + + > + 2x 10x 0 0 x 10. < < <
0,25
Kt hp vi iu kin ta c : 2 x 10 < l nghim ca bt phng trnh cho. 0,25II.2 1,0
Phng trnh cho tng ng vi( ) ( )1 cos6x cos 2x 1 cos 2x 0+ + = cos6x cos 2x 1 0 =
0,25
cos8x cos 4x 2 0 + = 22cos 4x cos 4x 3 0 + = 0,25
( )
= =
cos 4x 1
3cos 4x loi .
2
Vy ( )
= = cos 4x 1 x k k .2
0,5
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3
III. 3,0III.1 1,0
V ( )1A d A t; t .
V A v C i xng nhau qua BD v B,D Ox nn ( )C t; t . 0,25
V 2C d nn 2t t 1 0 t 1. = = Vy ( ) ( )A 1;1 , C 1; 1 . 0,25
Trung im ca AC l ( )I 1;0 . V I l tm ca hnh vung nn
IB IA 1
ID IA 1
= =
= =
0,25
b 1 1B Ox B(b;0) b 0,b 2
D Ox D(d;0) d 0,d 2d 1 1
= = =
= = =
Suy ra, ( )B 0;0 v ( )D 2;0 hoc ( )B 2;0 v ( )D 0;0 .
Vy bn nh ca hnh vung l( ) ( ) ( ) ( )A 1;1 , B 0;0 , C 1; 1 , D 2;0 ,
hoc
( ) ( ) ( ) ( )A 1;1 , B 2;0 , C 1; 1 , D 0;0 .
0,25
III.2a 1,0
Phng trnh ca tham sca
x 1 t
d : y 3 2t
z 3 t.
=
= +
= +
0,25
( )I d I 1 t; 3 2t;3 t + + , ( )( )2t 2
d I, P .3
+= 0,25
( )( )t 4
d I, P 2 1 t 3t 2.
== =
= 0,25
Vy c hai im ( ) ( )1 2I 3;5;7 , I 3; 7;1 . 0,25
III.2b 1,0
V A d nn( )
A 1 t; 3 2t;3 t + + .
Ta c ( )A P ( ) ( ) ( )2 1 t 3 2t 2 3 t 9 0 t 1 + + + + = = .
Vy ( )A 0; 1;4 .
0,25
Mt phng ( )P c vectphp tuyn ( )n 2;1; 2 .=
ng thng d c vectchphng ( )u 1;2;1=
.
V ( )P v d nn c vectchphng ( )u n,u 5;0;5
= =
.
0,5
Phng trnh tham sca :
x t
y 1
z 4 t.
=
= = +
0,25
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4
IV 2,0IV.1 1,0
2
0
(2cosx 1) sinxI dx
1 3cos x
+=
+ .
0,25
t
2t 1cosx 3
t 1 3cos x3sinx
dt dx.2 1 3cos x
=
= + = +
x 0 t 2, x t 1.2
= = = =
0,25
( )1 22
2
2 1
t 1 2 2I 2 1 dt 2t 1 dt.
3 3 9
= + = +
0,25
23
1
2 2t 2 16 2 34t 2 1 .
9 3 9 3 3 27
= + = + + =
0,25
IV.2 1,0
Ta c ( )2n 1 0 1 2 2 3 3 2n 1 2n 1
2n 1 2n 1 2n 1 2n 1 2n 11 x C C x C x C x ... C x+ + +
+ + + + ++ = + + + + + x . 0,25
o hm hai vta c
( )( ) ( )2n 1 2 3 2 2n 1 2n
2n 1 2n 1 2n 1 2n 12n 1 1 x C 2C x 3C x ... 2n 1 C x+
+ + + ++ + = + + + + + x . 0,25
Thay x 2= ta c:( )1 2 2 3 3 4 2n 2n 12n 1 2n 1 2n 1 2n 1 2n 1C 2.2C 3.2 C 4.2 C ... 2n 1 .2 C 2n 1.
+
+ + + + + + + + + = + 0,25
Theo githit ta c 2n 1 2005 n 1002+ = = . 0,25
V 1,0
Vi a,b 0> ta c : 21 a b 1 1 1 1
4ab (a b) .a b 4ab a b 4 a b
+ + +
+ +
Du " "= xy ra khi v chkhi a b= .
0,25
p dng kt qutrn ta c:
1 1 1 1 1 1 1 1 1 1 1 1 1(1).
2x y z 4 2x y z 4 2x 4 y z 8 x 2y 2z
+ + + = + +
+ + +
Tng t
1 1 1 1 1 1 1 1 1 1 1 1 1
(2).x 2y z 4 2y x z 4 2y 4 x z 8 y 2z 2x
+ + + = + + + + +
1 1 1 1 1 1 1 1 1 1 1 1 1(3).
x y 2z 4 2z x y 4 2z 4 x y 8 z 2x 2y
+ + + = + +
+ + +
0,5
Vy1 1 1 1 1 1 1
1.2x y z x 2y z x y 2z 4 x y z
+ + + + =
+ + + + + +
Ta thy trong cc bt ng thc (1), (2), (3) th du " "= xy ra khi v chkhi
x y z.= = Vy ng thc xy ra khi v chkhi3
x y z .4= = =
0,25
-------------------------------Ht-------------------------------
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BGIO DC V O TO-------------------------
CHNH THC
THI TUYN SINH I HC, CAO NG NM 2005Mn: TON, khi B
Thi gian lm bi: 180 pht, khng kthi gian pht --------------------------------------------------
Cu I (2 im)
Gim
(C ) l thca hm s( )2x m 1 x m 1
y
x 1
+ + + +=
+
(*) ( m l tham s).
1) Kho st sbin thin v vthca hm s(*) khi m 1.= 2) Chng minh rng vi mbt k, th
m(C ) lun lun c im cc i, im cc tiu
v khong cch gia hai im bng 20.
Cu II (2 im)
1)
Gii hphng trnh( )2 39 3
x 1 2 y 1
3log 9x log y 3.
+ =
=
2) Gii phng trnh 1 sin x cos x sin 2x cos2x 0.+ + + + =
Cu III(3 im)1) Trong mt phng vi hta Oxycho hai im A(2;0) v B(6;4) . Vit phng trnh
ng trn (C) tip xc vi trc honh ti im A v khong cch ttm ca (C) nim B bng 5.
2) Trong khng gian vi h ta Oxyz cho hnh lng tr ng 1 1 1ABC.A B C vi
1A(0; 3;0), B(4;0;0), C(0;3;0), B (4;0;4).
a)Tm ta cc nh 1 1A , C . Vit phng trnh mt cu c tm l A v tip xc vi
mt phng 1 1(BCC B ).
b)
Gi M l trung im ca 1 1A B . Vit phng trnh mt phng (P) i qua hai im
A, M v song song vi 1BC . Mt phng (P) ct ng thng 1 1A C ti im N.
Tnh di on MN.
Cu IV(2 im)
1) Tnh tch phn2
0
sin2x cosxI dx
1 cosx
=+
.
2)
Mt i thanh nin tnh nguyn c 15 ngi, gm 12 nam v 3 n. Hi c bao nhiu
cch phn cng i thanh nin tnh nguyn vgip 3 tnh min ni, sao cho mitnh c 4 nam v 1 n?
Cu V(1 im)
Chng minh rng vi mi x , ta c:x x x
x x x12 15 20 3 4 55 4 3
+ + + +
.
Khi no ng thc xy ra?
--------------------------------Ht--------------------------------
Cn bcoi thi khng gii thch g thm.
Hv tn th sinh .................................................. Sbo danh ...............................
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BGIO DC V O TO---------------------CHNH THC
P N THANG IMTHI TUYN SINH I HC, CAO NG NM 2005
----------------------------------------Mn: TON, Khi B
(p n thang im gm 4 trang)
Cu Ni dung imI 2,0
I.1 1,02x 2x 2 1
m 1 y x 1 .x 1 x 1
+ += = = + +
+ +
a) TX: \{ }.1
b) Sbin thin:( ) ( )
2
2 2
1 x 2xy ' 1
x 1 x 1
+= =
+ +y ' 0 x 2, x 0., = = =
0,25
yC ( ) ( )CTy 2 2, y y 0 2.= = =1
=
ng thng l tim cn ng.x= ng thng l tim cn xin.y x 1= +
0,25
Bng bin thin:
x 2 1 0 + y + 0 0 +
y
2 + +
2
0,25
c) th
0,25
1
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I.2 1,0
Ta c:1
y x mx 1
= + ++
.
TX: \{ }.1
( ) ( )( )2 2
x x 21y ' 1 , y ' 0 x 2, x 0.x 1 x 1
+= = = = =+ +
0,25
Xt du y '
x 2 1 0 + y + 0 || 0 +
thca hm s(*) lun c im cc i l ( )M 2;m 3 v im cc tiu l
.( )N 0;m 1+
0,50
( )( ) ( ) ( )( )2 2
MN 0 2 m 1 m 3 20.= + + = 0,25
II. 2,0II.1 1,0
( )
2 39 3
x 1 2 y 1 (1)
3log 9x log y 3 (2)
+ =
=
K:x 1
0 y 2.
<
0,25
( ) ( )3 3 3 32 3 1 log x 3log y 3 log x log y x y. + = = = 0,25
Thay vo (1) ta cy x=
( )( )x 1 2 x 1 x 1 2 x 2 x 1 2 x 1 + = + + =
( )( )x 1 2 x 0 x 1, x 2. = = =
Vy hc hai nghim l ( ) ( )x; y 1;1= v ( ) ( )x; y 2;2 .=
0,50
II.2 1,0
Phng trnh cho tng ng vi2sin x cos x 2sin x cos x 2cos x 0+ + + =
( )sin x cos x 2cos x sin x cos x 0 + + + =
( )( )sin x cos x 2cos x 1 0. + + =
0,50
sin x cos x 0 tgx 1 x k 4
+ = = = + ( )k . 0,25
1 22cos x 1 0 cos x x k22 3
+ = = = + ( )k . 0,25
2
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III. 3,0III.1 1,0
Gi tm ca (C) l ( )I a;b v bn knh ca (C) l R.
(C) tip xc vi Ox ti A va 2 = b R.= 0,25
( ) ( )2 2 2IB 5 6 2 4 b 25 b 8b 7 0 b 1,b 7.= + = + = = = 0,25
Vi ta c ng trna 2,b 1= =
( ) ( ) ( )2 2
1C : x 2 y 1 1. + = 0,25
Vi ta c ng trna 2,b 7= =
( ) ( ) ( )2 2
2C : x 2 y 7 49. + = 0,25
III.2a 1,0
( ) ( )1 1A 0; 3;4 ,C 0;3;4 . 0,25
( ) ( )1BC 4;3;0 ,BB 0;0;4= =
Vectphp tuyn ca ( )1 1mp BCC B l ( )1n BC,BB 12;16;0 = = .
Phng trnh mt phng ( )1 1BCC B :
( )12 x 4 16y 0 3x 4y 12 0. + = + =
0,25
Bn knh mt cu:
( )( )1 1 2 212 12 24
R d A, BCC B53 4
= =
+.=
0,25
Phng trnh mt cu:
( )2
2 2
576x y 3 z 25+ + + = .
0,25
III.2b 1,0
Ta c ( )13 3
M 2; ;4 , AM 2; ;4 , BC 4;3;4 .2 2
= =
0,25
Vectphp tuyn ca (P) l ( )P 1n AM,BC 6; 24;12 = =
.
Phng trnh (P): ( )6x 24 y 3 12z 0 x 4y 2z 12 0. + + = + + =
Ta thy Do i qua v song song viB(4;0;0) (P). (P) A, M 1BC .
0,25
Ta c ( )1 1A C 0;6;0=
. Phng trnh tham sca ng thng l1 1A C
x 0
y 3
z 4.
=
t= + =
( )1 1N A C N 0; 3 t; 4 . +
V ( )N P nn ( )0 4 3 t 8 12 0 t 2+ + + = = .
Vy ( )N 0; 1;4 .
( ) ( )2
2 23 1MN 2 0 1 4 4 .
2 2