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8/8/2019 [vnmath.com]- De Thi Hoc Ki 1 Toan 12 (co loi giai) nam hoc 2010-2011
1/24
s 1
THI HC K 1 Nm hc 2009 2010Mn TON Lp 12 Nng cao
Thi gian lm bi 90 pht
Bi 1 (3 im)
a) Kho st v v th hm s: y f x x x x C 3 21
( ) 2 3 1 ( )
3
= = + ( 2 im)
b) Tm m ng thng d y mx( ) : 2 1= ct C( ) ti 3 im phn bit? ( 1 im)
Bi 2 (3 im)
a)Tm gi tr ln nht, gi tr nh nht ca hm s:
f x x x1 2
( ) cos 2 2 sin2 3
= + , vi x 0;2
( 1 im)
b) Gii phng trnh: x x21 93
log 6 log 1 0 = ( 1 im)
c) Gii h phng trnh: x y xx y
2
3 2 0
27 3 .9 0
+ =
=( 1 im)
Bi 3 (1 im) Cho hm s mx m x m
y Cx
2( 1) 1
( )1
+ + + +=
+, m l tham s.
Chng minh rng vi m , th ( )mC lun c cc i, cc tiu. Tm m khong cch t
im cc i ca th ( )mC n ng thng x y( ) : 3 4 2 0 + = bng 4? ( 1 im)
Bi 4 (3 im) Cho hnh chp S ABC. c SA ABC ( ) , y l ABC vung cn ti A .
Bit SA a AB a AC a2 , 3, 3= = = .
a) Tnh th tch ca khi chp S ABC. . (1,5 im)b) Xc nh tm I v tnh bn knh R ca mt cu ngoi tip hnh chp S ABC. . Suy ra dintch mt cu ngoi tip hnh chp S ABC. v th tch khi cu ngoi tip hnh chp S ABC. .
(1 im)
c) Gi M N P, , ln lt l trung im ca SB SC AC , , . Mt phng MNP( ) ct AB ti Q .
Tnh din tch ton phn ca khi a din MNPQBC. ( 0,5 im)
===========================
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s 1
P N THI HC K 1Mn TON Lp 12 Nng cao
Thi gian lm bi 90 pht
Bi 1 (3 im)
a) Kho st v v th hm s: y f x x x x C 3 21
( ) 2 3 1 ( )3
= = +
Tp xc nh D R= ( 0,25 im) Gii hn
x xy ylim ; lim
+
= + = ( 0,25 im)
x yy x x y x xx
y
2 21
1' 4 3; ' 0 4 3 0 33
1
= == + = + = = =
( 0,25 im)
Bng bin thin ( 0,5 im)
Hm s nghch bin trn (1;3) , ng bin trn ( ;1) v (3; )+
im cc tiu I1(3; 1) , im cc i I21
1;3
Ta c y x y x x'' 2 4; '' 0 2 4 0 2= = = = . im un I1
2;3
(0,25 im)
th: ( 0,5 im)
im c bit: ( )A 0; 1 , B1
4;3
.
th hm s nhn im un I1
2;3
lm tm i xng.
1
3
1
+ 0
1 3
0
+
+
+-
x
( )'f x
( )f x
0-2
A
2-1x
y
I1
-2
34
1
3 1
3 .
.
.
..
...
B2
1
.-1
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b) Tm m ng thng d y mx( ) : 2 1= ct ( )C ti 3 im phn bit?
Phng trnh honh giao im ca C( ) v d( ) l:
xx x x = 2mx x x x m
x x m
3 2 22
01 1
2 3 1 1 2 3 2 0 12 3 2 03 3
3
= + + = + =
t ( )g x x x m31 2 3 23
= + ( 0,5 im)
PT cho c 3 nghim phn bit th PT g x( ) 0= c 2 nghim phn bit khc 0
mm
mgm
101 (3 2 ) 0
0 3 33(0) 0
22
> > >
( 0,5 im)
Bi 2 ( 3 im)
a) Tm gi tr ln nht, gi tr nh nht ca hm s:
f x x x1 2( ) cos 2 2 sin2 3
= + , vi x 0;2
Ta c ( )f x x x x x x2 21 2 1
( ) 1 2 sin 2 sin sin 2 sin , 0;2 3 6 2
= + = +
(0,25 im)
t t x t g t t t t 21
sin , 0 1 ( ) 2 , 0;16
= = + . (0,25 im)
g t t g t t t ( ) 2 2, ( ) 0 1, 0;1 = + = = . (0,25 im)
Ta c: g g1 5
(0) ; (1)
6 6
= =
Gi tr ln nht l:[ ]
g t g khi t f x khi x0;1
0;2
5 5max ( ) (1) 1 max ( )
6 6 2
= = = = =
Gi tr nh nht l: ( )g t g khi t f x khi x0;1
0;2
1 1min ( ) (0) 0 min 0
6 6
= = = = =
Vy f x khi x0;
2
5max ( )
6 2
= = , ( )f x khi x0;
2
1min 0
6
= = ( 0,25 im)
b) Phng trnh x x21 9
3
log 6 log 1 0 = x x23 34 log 3 log 1 0 = (0,25 im)
t t x3log= , ta c phng trnh: (0,25 im)
xxtt t
xt x
32
43
3log 1114 3 1 0 1 1
log4 34
= = = =
== =
(0,5 im)
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C
S
A
B
KE
M
N
Q
P I
H
d
c) Gii h phng trnhx x
x y
y2
3 2 0 (1)
27 3 .9 0 (2)
+ =
=
x y x y x x y2 2 2
(2) 27 3 .9 3 3 = = = , thay vo phng trnh (1) ta c:
y
y y xy yy xyy
2
1
1 1 13 2 02 42
2
= =
= = + = = == =
( 0,5 im)
Vy h phng trnh c nghim (1;1); (1; 1); (4; 2); (4; 2) ( 0,5 im)
Bi 3 (1 im)
Tp xc nh { }D R\ 2= ( 0,25 im)
x m x x m x m x x
yx x
2 2
2 2
(2 1)( 1) ( 1) 1 2'
( 1) ( 1)
+ + + + + + + + = =
+ +
x y my x x x y m2 0 1' 0 2 0 2 3
= = += + = = = ( 0,25 im)
Da vo BBT im cc i l: I m1( 2; 3) (0,25 im)
Khong cch tim cc i I m1( 2; 3) n ng thng x y( ) : 3 4 2 0 + = l:m m
d I mm1
8 4 3( ,( )) 4 2 5
75
= = = = =
(0,25 im)
Bi 4 (3 im)
V hnh ng (0,5 im)
Do SA ABC ( ) nn SA l ng cao
ca hnh chp S ABC. .
ABCV SA S 1
.3
= (0,25 im)
M ABC vung cn ti C
ABC
aS AC AB a a
21 1 3
. 3. 32 2 2
= = =
( 0,25 im)
Suy ra V a a a2 31 3
2 .3 2
= = . ( 0,5 im)
2
3m
1m +
+ 0
0
0
+
+
+-
x
( )'f x
( )f x
-
1
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b) Gi H l trung im BC. Ta c: HA HB HC = = (do ABC vung ti A )
TH dng ng thng d ABC( ) . Suy ra d l trc mt cu ngoi tip hnh chp S ABC. .
Dng mt phng trung trc ca cnh SA i qua trung im E ca SA , ct d ti im I.
Ta c IA IS (1)=
Tng t, dng mt phng trung trc cc cnh SB SC , . Ta c: IC IB IS (2)= =
T (1),(2) suy ra I l tm mt cu ngoi tip S ABC. . Bn knh R IA= .
Ta ca
IA IH AH 2 2 10
2= + = (0,5 im)
Din tch mt cu l: S R a2 24 10 = = .
Th tch khi cu l: V R a3 34 5 10
3 3 = = (0,5 im)
c) Mt phng MNP( ) ct ABC( ) theo giao tuyn PQ song song vi BC, vi Q l trung imca AB . (0,25 im)
Din tch ton phn ca khi a din MNPQBC bng:
( ) ( ) ( ) ( ) ( )dt MNPQ dt BMQ dt PNC dt BCPQ dt MNBC
a a a a aa
2 2 2 2 226 3 3 9 3 33 6 3 9 3 33
2 4 4 8 8 2 2 8 8
+ + + + =
= + + + + = + + +
(0,25 im)
=============================
s 2
THI HC K 1 Nm hc 2009 2010Mn TON Lp 12 Nng cao
Thi gian lm bi 90 pht
Bi 1 (3 im)
a) Kho st v v th hm s: y f x x x x C 3 21
( ) 2 3 1 ( )3
= = + + (2 im)
b) Tm m ng thng d y mx( ) : 1= + ct C( ) ti 3 im phn bit? (1 im)
Bi 2 (3 im)
a) Tm gi tr ln nht, gi tr nh nht ca hm s:
f x x x1 4
( ) cos 2 2 sin3 3
= + , vi x 0;2
(1 im)
b) Gii phng trnh: xx
22 2 22
4log 5 log 13 log 4 0 + = (1 im)
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c) Gii h phng trnhyx
xy1
1
2
16 4 3 0
= =
(1 im)
Bi 3 (1 im)
Cho hm s ( )mx m x m m
y Cx
2 22( 1)
2
+ + + +=
+, m l tham s.
Tm m hm s ( )mC c cc i, cc tiu v khong cch gia hai im cc i, cc tiu
bng 5 ? (1 im)
Bi 4 (3 im)
Cho hnh chp S ABC. c SA ABC ( ) , y l ABC vung ti C.
Bit SA a AB a AC a3, 2 ,= = = .
a) Tnh th tch ca khi chp S ABC. . (1,5 im)
b) Gi H K, ln lt l hnh chiu vung gc ca A xung SC SB, . Xc nh tm I v tnhbn knh ca mt cu ngoi tip hnh chp H ABC. . Suy ra din tch mt cu ngoi tip hnh
chp H ABC. v th tch khi cu ngoi tip hnh chp H ABC. . (1 im)c) Tnh t s th tch ca hai khi chp A BHK. v A BCH. ? (0,5 im)
===============================
s 2
P N THI HC K 1 Nm hc 2009 2010Mn TON Lp 12 Nng cao
Thi gian lm bi 90 pht
Bi 1 (3 im)
a) Kho st v v th hm s: y f x x x x (C)3 21( ) 2 3 13= = + +
Tp xc nh D R= (0,25 im)
Gii hnx x
y ylim ; lim+
= = + (0,25 im)
x yy x x y x xx
y
2 21
1' 4 3; ' 0 4 3 0 33
1
= = = + = + = = =
(0,25 im)
Bng bin thin (0,5 im)
1
3
1
- 0
1 3
0 +
+
-+x( )'f x
( )f x
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Hm sng bin trn (1;3) , nghch bin trn ( ;1) v (3; )+
im cc i I1(3;1) , im cc tiu I21
1;3
Ta c y x y x x'' 2 4; '' 0 2 4 0 2= + = + = = .
im un I1
2;3
( 0,25 im)
im c bit: ( )A 0;1 , B1
4;3
.
th hm s nhn im un I1
2;3
lm tm i xng. (0,5 im)
b) Tm m ng thng d y mx( ) : 1= + ct C( ) ti 3 im phn bit?
Phng trnh honh giao im ca (C) v (d) l:
xx x x = mx x x x m
x x m
3 2 22
01 1
2 3 1 1 2 3 0 12 3 03 3
3
= + + + + + = + + =
(0,5 im)
t g x x x m21
( ) 2 33
= + + .
PT cho c 3 nghim phn bit th PT g x( ) 0= c 2 nghim phn bit khc 0
( )( ) mm
g mm
1' 0 01 3 030 0 3
3
>
(0,5 im)
Bi 2 ( 3 im)
a) Tm gi tr ln nht, gi tr nh nht ca hm s:
f x x x1 4
( ) cos 2 2 sin3 3
= + , vi x 0;2
Ta c ( )f x x x x x x2 21 4 2
( ) 1 2 sin 2 sin sin 2 sin 1, 0;3 3 3 2
= + = +
(0,25 im)
t t x t g t t t t 22
sin , 0 1 ( ) 2 1, 0;13
= = + . (0,25 im)
0-2
A
2-1
x
y
I1
-2
3 4
1
3
1
3
..
.
.
.. ...
..
.
.
B2
1
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( )
m mm m
g m m
2
2
' 0 3 00 3
2 0 3 0
> > < >
+ (0,25 im)
Vy ( ) ( )m ; 0 3; + th hm s cho c cc tr.
Vi ( ) ( )m ; 0 3; + , gi hai im cc tr l ( ) ( )I x x m I x x m1 1 1 2 2 2; 2 2 2 ; ; 2 2 2+ + + +
( ) ( ) ( )( ) ( )
I I I I x x x x x x
x x x x
2 2 221 2 1 2 2 1 2 1 2 1
2
2 1 1 2
5 5 2 2 5 5 5
4 1 *
= = + = =
+ =
p dng h thc Viet, ta cx x
x x m m
1 22
1 2
4
3 4
+ =
= + +. (0,25 im)
Thay vo (*) ta c phng trnhm
m m
m
2
3 10
24 12 1 03 10
2
=
= +
=
(0,25 im)
Bi 4 (3 im)V hnh ng ( 0,5 im)
a) Do SA ABC ( ) nn SA l ng cao ca hnh chp S ABC. .
Th tch ca khi chp l:ABC
V SA S 1
.3
= (0,25 im)
M ABC vung ti C nn:
ABC
aS AC BC a a
21 1 3
. . 32 2 2
= = = (0,25 im)
Suy ra aV a a3
21 33.3 2 2
= = . (0,5 im)
b) Ta c: BC SAC ( ) ( do BC AC BC SA; )
Suy ra BC AH .
Mt khc, SC AH .
T, AH SBC AH HB( ) .
AHB vung ti H.
Gi I l trung im ca AB , ta c IA IB IH (1)= =
ACB vung ti C, ta c IA IB IC (2)= =
T (1), (2) suy ra I l tm mt cu ngoi tip hnhchp H ABC. .
Bn knhAB
R IA a2
= = = . (0,5 im)
Din tch mt cu l: S R a2 24 4 = = . Th tch khi cu l: V R a3 34 4
3 3 = = (0,5 im)
c) T s th tch 2 khi chp A BHK. v A BCH.
B
S
A
C
I
K
H
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Ta cA BCH B AHC ACH
aV V BC S BC AH HC a a
32
. .
1 1 1 1 3. . . 3. .
3 3 2 3 8 8= = = = = (0,25 im)
( )H ABK B AHK a
V V BK dt AHK BK AH HK 3
. .
1 1 1 3. . .
3 3 2 14= = = =
Suy ra A BHKA BCH
a
VV
a
3
.
3.
3
12141 7
8
= = (0,25 im)
=================================
S GD & T s 3
THI HC K 1 Nm hc 2009 2010Mn TON Lp 12
Thi gian lm bi 90 pht
I. PHN CHUNG CHO TT C TH SINH (7 im)
Cu I (3 im)
1. Kho st s bin thin v v th hm s y x x4 25 4= + .
2. Tm m phng trnh x x m4 25 4 + = c 4 nghim phn bit.
Cu II (1 im)
Gii phng trnh: x x2 4 21
2(log 1) log log 04
+ + = .
Cu III (3 im)Cho tam gic ABCu cnh a. Trn ng thng di qua A v vung gc vi mt phng
(ABC) ly imD sao choAD = 2a.
1. Tnh th tch khi chpD.ABC.2. Tnh din tch ca mt cu ngoi tip hnh chpD.ABC.
3. Mt phng i quaB, trung im caAD v tm ca mt cu ngoi tip hnh chp chia khi
chp thnh hai phn. Tnh t s th tch ca hai phn .
II. PHN TCHN (3 im)Th sinh chc chn mt trong hai phn: Theo chng trnh Chun hoc Nng cao1. Theo chng trnh Chun
Cu IVa (3 im)
1. Tm gi tr ln nht, gi tr nh nht ca hm sy x x1 9= + + .2. Gii bt phng trnh: x x2 31 2 2
4
log log (2 ) log 0
.
3. Tm m hm sy =x3 6x2 + 3(m + 2)x m 6 c hai cc tr v hai gi tr cc tr cng
du.
2. Theo chng trnh Nng caoCu IVb (3 im)
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1. Tm gi tr ln nht, gi tr nh nht ca hm sy x x24= + .
2. Gii h phng trnh:
x y
y x
x y x y3 13
4 32
log ( ) 1 log ( )
+ =
= + +
3. Tm m phng trnh x xm m m2 2
2( 2)2 2( 1)2 2 6 0 + + = c nghim thuc on
0; 2 .
--------------------Ht-------------------
H v tn th sinh: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . SBD :. . . . . . . . . .
s 3 P N THI HC K 1 Nm hc 2009 2010Mn TON Lp 12
Thi gian lm bi 90 pht
Cu Ni dung imI.1 Kho st hm sy x x4 25 4= + 2,00
1) Tp xc nh : R2) Sbin thin:
a) Gii hn :x x
y ylim , lim +
= + = + 0,50
b) Bng bin thin: y x x34 10 = ;x
yx
0
0 10
2
= = =
x 10 / 2 0 10 / 2 +
y' 0 + 0 0 +
y
+ 4 +
9/4 9/4
0,50
Hm sng bin trn cc khong10 10
; 0 , ;2 2
+
Hm s nghch bin trn cc khong10 10
; , 0;2 2
Hm st cc i ti x = 0, yC = y(0) = 4
Hm st cc tiu ti x 102
= , yCT = y10 9
2 4 =
0,50
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3) th: th (C) ca hm s c hai im un U5 19
;6 36
nhn Oy lm
trc i xng, giao vi Ox ti 4 im ( 1; 0); ( 2; 0) (Hnh 1)
(Hnh 1) (Hnh 2)
0,50
I.2 Tm m phng trnh x x m4 25 4 + = (1) c 4 nghim phn bit 1,00
Gi (C1) l th hm sy x x4 2
5 4= + . (C1) gm hai phn:+) Phn th (C) nm trn trc Ox+) i xng ca phn th (C) nm di Ox qua Ox
0,25
V th(Hnh 2) 0,25S nghim ca (1) bng s giao im ca (C1) vi ng thng y = m. Theo
th ta c (1) c 4 nghim phn bit khi v ch khi m = 0 v m9
44
< < 0,50
II Gii phng trnh x x2 4 2 12(log 1) log log 04
+ + = (1) 1,00
iu kin: x > 0
(1) x x x x22 2 2 2(log 1) log 2 0 log log 2 0 + = + = 0,5
xx
x x2
2
2log 1
1log 2
4
= = = =
0,5
III.1 Tnh thtch khi chp D.ABC. 1,00
O1 2
x-1-2
4
y
y = m
(C1)
9/4
O1 2
x-1-2
4
y
y = m
(C1)
9/4
O1 2
x-1-2
4
-9/4
y
(C)
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Th
tch kh
i chp
D ABC ABC
a aV AD S a
2 3
.
1 1 3 3. 2 .
3 3 4 6= = =
1,00
III.2 Tnh din tch ca mt cu ngoi tip hnh chp D.ABC. 1,00Gi O l trng tm ca tam gic ABC, gi l ng thng i qua O v vung
gc vi (ABC), suy ra // DA v l trc ca ng trn ngoi tip tamgic ABC. Trong mt phng (d, ) kng thng trung trc ca AD ct ti I, khi I cch u A, B, C, D nn I l tm ca mt cu ngoi tipD.ABC
0,25
Gi M, N l trung im ca BC v AD. T gic AOIN l hnh ch nht nn
IA = ON = AN AO2 2+ . AN =a a
DA a AO AM 1 2 2 3 3
,2 3 3 2 3
= = = = 0,25
a aIA a
2
2 3 2 3
3 3
= + =
.
Mt cu c bn knha
R IA2 3
3= = nn S
a aR
22
2 2 3 164 4
3 3
= = =
0,50
III.3 Tnh tsthtch... 1.00Gi E = DM IN, F = BE DC khi tam gic BNF l thit din ca hnh
chp ct bi mt phng (BNI).0,25
Do N l trung im ca DA, NE // AM nn E l trung im ca DMGi K l trung im ca FC MK l ng trung bnh ca tam gic BFC MK // BF EF l ng trung bnh ca tam gic DMK F l trung im
ca DK DC = 3 DF SDBC = 3SDBF.
0,25
D
A C
B
O
N
M
I
d
F
E
K
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Gi h l khong cch t A n mt phng (DBC), do N l trung im ca DAnn khong cch t N n (DBC) bng h/2.
Gi th tch khi chp D.ABC l V, th tch khi chp D.NBF l V1, th tchphn cn li l V2.
Ta cDBF DBC
hV S h S V V V V V V V 1 2 1
1 1 1 1 5. .
3 2 6 6 6 6= = = = = =
0,25
Do ta c t s th tch: 12
VV
15
= hoc 21
VV
5=
Ch th sinh cng c thlm theo cch sau:V DN DF DB
V DA DC DB1 1. .
6= =
0,25
IVa.1 Tm gi trln nht, gi trnh nht ca hm sy x x1 9= + + . 1,00
Tp xc nh D = [1; 9]
yx x
1 1'
2 1 2 9=
, y x x x' 0 1 9 5 = = 0,50
y(1)= y(9) = 2 2 , y(5) = 4
y y y y ymax (5) 4, min (1) (9) 2 2 = = = = = 0,50
IVa.2 Gii bt phng trnh... 1,00
x x x x2 3 2 31 2 2 2 2
4
log log (2 ) log 0 log (2 ) log 1
(iu kin: x > 0) 0,25
xx x x x
x2 2 2
2 2 2 22
log 1(1 log ) 3 log 1 0 log log 0
log 0
+
0,50
x
x
2
1
. Vy bt phng trnh c tp nghim S (0;1] [2; )= + 0,25
IVa.3 Tm m hm sy = x3 6x2 + 3(m + 2)x m 6 c hai cc trcng du. 1,00y = 3x2 12x + 3(m +2). iu kin hm s c cc tr l y c hai nghim
phn bit m m' 36 9( 2) 0 2 = + > < Gi x1, x2 l hai im cc tr ca hm s, khi theo nh l Viet ta c
x x
x x m1 2
1 2
4
2
+ =
= +
0,25
Do y x y m x2
' ( 2)(2 1)3 3
1= + +
v y(x1) = y(x2) = 0
nn y x m x1 1( ) ( 2)(2 1)= + , y x m x2 2( ) ( 2)(2 1)= +
0,25
C CTy y y x y x m x x m x x x x2 2
1 2 1 2 1 2 1 2( ) ( ) ( 2) (2 1)(2 1) ( 2) [4 2( ) 1]= = + + = + + +
m m m m2 2( 2) [4( 2) 2.4 1] ( 2) (4 17)= + + + = + 0,25
Do hai gi tr cc tr cng du khi
C CT
my y m m
m
22
. 0 ( 2) (4 17) 0 17
4
> + > >
Kt hp vi iu kin ta c m17
24
< <
0,25
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IVb.1 Tm gi trln nht, gi trnh nht ca hm sy x x24= + 1,00
Tp xc nh: D = [ 2; 2]
x x xy
x x
2
2 2
4' 1
4 4
= + =
0,25
xy x x xx x
2 2 20' 0 4 0 24
= = = =
0,25
y(2) = 2, y (2) = 2, y( 2) 2 2= 0,25
y y y ymax ( 2) 2 2, min ( 2) 2= = = = 0,25
IVb.2 Gii hphng trnh
x y
y x
x y x y3 13
4 32 (1)
log ( ) 1 log ( ) (2)
+ =
= + +
1,00
iu kin: x y > 0, x + y > 0, x 0, y 0
x y x y x y x y2 2 2 23 3 3(2) log ( ) log ( ) 1 log ( ) 1 3 (3) + + = = = 0,25
(1)
x y
y x x y
y x
25
2 2 2 5
+
= + =
.
tx
ty
= ta c tt t ttt
21 22 5 2 5 2 0
1/ 2
=+ = + = =
0,25
+) Vi t = 2 x y2 = th vo (3) ta c y y y2 24 3 1 = = Khi y x 2 (thoa man)1= = Khi y x loai1 2 ( )= =
0,25
+) Vi t =1
2 y x2 = th vo (3) ta c x x vo nghiem2 24 3 ( ) =
Vy h phng trnh c 1 nghim (x, y ) = (2; 1)0,25
IVb.3 Tm m phng trnh c nghim thuc on [0; 1] 1,00
x xm m m2 22
( 2)2 2( 1)2 2 6 0 + + = (1)
t t = x2
2 , do x x nen t 0; 2 [1; 4]
(1) trthnh m t m t m2( 2) 2( 1) 2 6 0 + + = (2)
t tt t m t t m f t
t t
22 2
2
2 2 6( 2 2) 2 2 6 ( )
2 2
+ + + = + + = =
+
0,25
Xt hm s
f(t) trn [1; 4] t loait t
f t f t t t tt t
22
2 2
2 ( )6 4 16
'( ) , '( ) 0 6 4 16 0 4( 2 2)
3
= + = = + =
= +
0,25
f(1) = 10, f(4) =23
5, f
4
3
= 11
f t f f t f [1;4] [1;4]
4 23max ( ) 11, min ( ) (4)
3 5
= = = =
0,25
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(1) c nghim thuc [0; 2 ] (2) c nghim thuc [1; 4] m23
115
Vy: m23
115
0,25
s 4
THI HC K 1 Nm hcMn TON Lp 12
Thi gian lm bi 90 pht
I. PHN CHUNG CHO TT C CC TH SINH (7 im)
Cu I. (3 im) Cho hm sy x x x3 26 9 4= + + + c th (C).a) Kho st s bin thin v v th (C) ca hm s.b) Vit phng trnh tip tuyn vi th (C) ti im M(2; 2).
c) Da vo th (C), tm m phng trnh x x x m3 2 26 9 4 log+ + + = c 3 nghim phn bit
Cu II. (1 im) Tm gi tr ln nht, gi tr nh nht ca hm sy x x2 cos 2 4 sin= +
trn on0;
2
.
Cu III. (2 im) Gii cc phng trnh sau:
a) x x2 15 5 6++ = b) x x x2 1 22
log ( 1) log ( 3) log ( 7)+ + = +
Cu IV. (1 im) Bit 2 10 < . Chng minh:2 5
1 12
log log + > .
II. PHN RING (3 im)Th sinh chc chn mt trong hai phn: Theo chng trnh Chun hoc Nng cao
1. Theo chng trnh ChunCu Va. (2 im) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, cnh bn SA vung
gc vi mt phng y, cnh bn SB = a 3 .a) Tnh th tch ca khi chp S.ABCD.b) Xc nh tm v tnh bn knh ca mt cu ngoi tip hnh chp S.ABCD.
Cu VIa. (1 im) Gii bt phng trnh:x x22 3
5 6
6 5
.
2. Theo chng trnh Nng cao
Cu Vb (2 im) Trn mt phng (P) c gc vung xOy , on SO = a vung gc vi (P). Cc
im M, N chuyn ng trn Ox, Oy sao cho ta lun c OM ON a+ = .
a) Xc nh v tr ca M, N th tch ca t din SOMN t gi tr ln nht.b) Khi t din SOMN c th tch ln nht, hy xc nh tm v tnh bn knh mt cu ngoitip t din SOMN.
Cu VIb. (1 im) Gii h phng trnh: x y
xy
2 2 25log log log 2
22
= =
--------------------Ht-------------------H v tn th sinh: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . SBD :. . . . . . . . . .
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s 4
P N THI HC K 1 Nm hc 2009 2010Mn TON Lp 12
Thi gian lm bi 90 pht
Cu Ni dung imI.a Kho st hm sy x x4 25 4= + 2,00
1) Tp xc nh : R2) Sbin thin:
a) Gii hn :x x
y ylim , lim +
= + = +
0,50
b) Bng bin thin: y x x23 12 9 = + + ; xyx
10
3
= = =
x 3 1 + y + 0 0 +
y4 +
0
0,50
Hm sng bin trn cc khong ( ) ( ); 3 , 1; +
Hm s nghch bin trn khong ( 3; 1) Hm st cc i ti x = 3, yC = y(3) = 4Hm st cc tiu ti x 1= , yCT = y( 1) 0 =
0,50
) th: thi qua cc im (2; 2), (0; 4), (1; 0), (3; 4), (4; 0)
-6 -5 -4 -3 -2 -1 1 2
-4
-3
-2
-1
1
2
3
4
5
6
x
y
0,50
I.b Phwong trnh tip tuyn 0,50
Phng trnh tip tuyn vi (C) ti im M(2; 2): y f x f ( 2)( 2) (2)= + + 0,25
y x3 4= 0,25
I.c Tm m PTx x x m3 2 26 9 4 log+ + + = c 3 nghim phn bit0,50
S nghim ca PT l s giao im ca (C) v d: y m2log= 0,25
Da vo th PT c 3 nghim phn bit m m20 log 4 1 16< < < < 0,25II
Tm GTLN, GTNN ca hm sy x x2 cos 2 4 sin= + trn on 0;2
.1,00
( )y x x x x2 2 sin 2 4 cos 4 cos 1 2 sin = + = 0,25
Trn 0;2
, ta c: y x x02 4
= = =
0,25
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y y y4 2; 2 2; (0) 22 4
= = =
0,25
Vy: y y y y0; 0;
2 2
min (0) 2; max 2 24
= = = =
0,25
III.a Gii phng trnh x x2 15 5 6++ = 1,00
t t = x5 , t > 0 0,25
PT trthnh t loait tt
2 6 ( )5 6 0
1
= + = =
0,50
Vi t = 1 th x x5 1 0= = 0,25
III.b Gii phng trnh x x x2 1 22
log ( 1) log ( 3) log ( 7)+ + = + 1,00
iu kin x 1> 0,25
PT x x x x x x2 2log ( 1)( 3) log ( 7) ( 1)( 3) 7+ + = + + + = +
x x x x loai2 3 4 0 1 4 ( )+ = = =
0,50
Vy PT c nghimx = 1 0,25IV
Chng minh:2 5
1 12
log log + >
1.00
Ta c: 2
2 5
1 1log 2 log 5 log 10 log 2
log log
+ = + = > =
1,00
Va.a Thtch khi chp 1,000,25
ABCDS a2= 0,25
SA SB AB a2 2
2= = 0,25
V Bh a a a2 31 1 2
2.3 3 3
= = = 0,25
Va.b Tm v bn knh mt cu ngoi tip 1,00
Gi O l tm hnhg vung ABCD O l tm ng trn ngoi tip hnh vung 0,25
Qua O k d // SA d l trc ca ng trn (ABCD), d ct SC ti trung im Ica SC.
SAC vung ti A IA = IC = IS =SC
2
IS = IA = IB = IC = ID I l tm mt cu ngoi tip hnh chp S.ABCD.
0,50
Bn knh R = IA =SC
2= a
0,25
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VIaGii bt phng trnh
x x22 35 6
6 5
1,00
x x22 3 1
5 5
6 6
0,25
x x22 3 1 0 + 0,50
x1
12
0,25
Vb.a Vtr ca M, N 1,000,25
V =SOMNV Bh OM ON OS a OM ON
1 1 1 1. . . . .
3 3 2 6= = =
0,25
V OM ON
a a
231 1
6 2 24
+=
0,25
max
aV a khi OM ON 3
1
24 2= = =
0,25
Vb.b Xc nh tm v bn knh mt cu 1,00
Gi I l trung im ca MN I l tm ng trn ngoi tip OMN.
Mt phng trung trc ca OS ct trc It ca OMN ti J.Ta c: JS = JO = JM = JN J l tm mt cu ngoi tip t din SOMN.
0,50
Bn knh R = JO =a 3
4
0,50
VIbGii h phng trnh: x y
xy
2 2 25log log log 2 (1)
22 (2)
= =
1,00
iu kin: xy
0
0
>
>
0,25
(1) x y x y25
(log log )(log log ) log 22 + =
x x x
xyy y y
52 2 2
5 5log . log( ) log 2 log . log 2 log 2 log log 2
2 2= = = =
x
y
5
22=
0,50
Kt hp (2) ta c
xyx
x
y y
7
45
32
4
22
22
= =
=
=
0,25
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s 5
THI HC K 1 Nm hc 2009 2010Mn TON Lp 12
Thi gian lm bi 90 pht
I. PHN CHUNG CHO TT C CC TH SINH (7 im)Cu 1: (2,5)
Cho hm s: y x x3 23 1= + .
1) Kho st s bin thin v v th (C) ca hm s.2) Vit phng trnh tip tuyn vi (C) ti im c honh l nghim ca phng trnh
y" 0= .Cu 2: (1)
Tm gi tr ln nht v nh nht ca hm s: y x x x3 21
2 3 13
= + + trn on [1;2]
Cu 3: (1)
Gii phng trnh:x x
1 1
2 24 4 3+
=
Cu 4: (2,5)
Cho hnh chp t gic u S.ABCD c cnh y bng 2a, cnh bn hp vi y mt gc .a) (1,25) Tnh th tch ca khi chp S.ABCDb) (1,25) Xc nh tm v bn knh ca mt cu ngoi tip hnh chp S.ABCD
II. PHN RING (3 im)Th sinh chc chn mt trong hai phn: Theo chng trnh Chun hoc Nng cao1. Theo chng trnh Chun
Cu 5a:
1) (1) Tm cc tim cn ca th hm sx
yx x
2 1
(1 )
+=
2) (1) Gii bt phng trnh: xx x2 42log 8 log log 32+ <
3) (1) Ct mt xung quanh ca mt hnh tr theo mt ng sinh, ri tri ra trn mt mt
phng, ta c mt hnh vung c din tch 100cm2. Tnh th tch ca khi tr gii hn bihnh tr.
2. Theo chng trnh Nng cao
Cu 5b:
1) (1) Tm cc tim cn ca th hm s: y x x2 1= +
2) (1) Gii bt phng trnh:x
x x2
3 93
5log 18 log log
3 2+ >
3) (1) Ct mt xung quanh ca mt hnh nn theo mt ng sinh, ri tri ra trn mt mtphng, ta c mt na hnh trn c ng knh bng 10cm. Tnh th tch ca khi nn giihn bi hnh nn .
Ht
H v tn th sinh: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . SBD :. . . . . . . . . .
P N THI HC K 1 Nm hc 2009 2010
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s 5
Mn TON Lp 12Thi gian lm bi 90 pht
Cu Ni dung im1
(2,5)1
(1,75)TX: D = R
y x x
x yy
x y
2' 3 6
0 1' 0
2 3
=
= == = =
x xx x x x3 2 3 2lim ( 3 1) , lim ( 3 1)
+
+ = + + =
+ 00
2
-
+
-3
1
0
y
y'
+-x
Hm sng bin trn cc khong ( ; 0) v (2; )+ Hm s nghch bin trn khong (0;2)Hm st cc i ti im x = 0; yC=1, t cc tiu ti im x = 2; yCT =
3 th:
x
y
2
3
1
-3
-1 1
0,25
0,25
0,25
0,25
0,25
0,50
2(0,75)
y x x y" 6 6 0 1 1= = = = y '(1) 3= Phng trnh tip tuyn l: y x y x3( 1) 1 3 2= = +
0,250,25
0,252
(1)y = x2 4x +3
y = 0x
x
1
3 1;2
= =
y(1) =13
3 , y(2) =
5
3, y(1) =
7
3
y y1;2 1;2
7 13max min
3 3 = =
0,25
0,25
0,25
0,25
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3(1)
x xx
x
1 1
2 21
4 4 3 2.4 2. 34
+
= =
t t = x4 , t>o tt
22 3 = 2t2 3t 2 = 0
t loai
t
1( )
22
=
=
t = 2 x4 = 2 x1
2=
Vy nghim ca phng trnh l x1
2=
0,25
0,25
0,25
0,25
4(2,5)
1(1,25)
Gi O l tm ca y th SO (ABCD)SAO AC a OA a SO a, 2 2 2 2 tan = = = =
Th tch ca khi chp S.ABCD l: ABCDa
V S SO3
1 4 2 tan.
3 3
= =
0,25
0,5
0,5
2(1,25) Gi H l trung im SA, trong mt phng (SAC) dng ng trung trc caSA ct SO ti I th I l tm mt cu ngoi tip hnh chp S.ABCDHai tam gic vung SHI v SOA ng dng , nn ta c:
SI SH SA SH SI
SA SO SO
.= =
a a aSA SH SO a SI
cos cos
2 2 2, , 2 tan
2 sin 2
= = = =
Vy bn knh mt cu ngoi tip hnh chp l:a
r2
sin2=
0,25
0,25
0,5
0,25
5B
(3) 1(1) Tp xc nh D= R\{0;1}
xx
x x
x x x x
2 2
00
1 1lim , lim
(1 ) (1 ) +
+ += = +
; ng thng x = 0 l tim cn ng
xx
x x
x x x x
2 2
11
1 1lim , lim
(1 ) (1 ) +
+ ++ =
; ng thng x = 1 l tim cn ng
x
x
x x
21
lim 1(1 )
+=
; ng thng y = 1 l tim cn ngang
0,250,25
0,25
0,25
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2(1)
xx x2 42
log 8 log log 32
+ < (1)
iu kin x > 0x x x
x x x
2 2 2 4 4
2 2 2
(1) log 8 log 2 log log log 2 3
1 13 log 2 log log 3
2 2
+ + (1)
iu kin: x > 0
x x x3 3 3 3
1 5(1) log 18 log 2 log log
2 2 + + + >
x3 3log 18 2 log 2 + >
0,25
0,25
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x2
3log 18 2 >
x218 9 >
( )x x v x21 1
02 2
> > >
0,25
0,25
3
(1)BOA
Gi l, rl ng sinh v bn knh y ca hnh nn.
T gi thit, ta suy ra l =10
2= 5
Din tch xung quanh ca hnh nn l: rl r5 =
Din tch ca na hnh trn l: 21 2552 2
=
Theo gi thit ta c: r r25 5
52 2
= =
Gi h l ng cao ca hnh nn th: h l r2 225 5
25 34 2
= = =
Vy th tch ca khi tr l V = r h2
21 1 5 5 125 3. . 3
3 3 2 2 24
= =
0,25
0,25
0,25
0,25
=============================
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