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hay, dùng cho các bạn học sinh năm 1997 và giáo viên toán trường chuyên
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1
S GIO DC V O TO BNH NH KSCL KHI 12, THNG 01, NM 2015 TRNG THPT CHUYN L QU N Mn: TON. -------------------------- Ngy kho st:24/01/2015
i gian lm bi:180 pht, khng k thi gian pht ----------------------------------------
Cu 1 (2,0 im). Cho hm s . 4 22 1y x x a) Kho st v v th (C) ca hm s.
2 2
x b) Vit phng trnh tip tuyn d vi th (C) ti im M c honh . Tm ta cc giao im ca tip tuyn d vi th (C). Cu 2 (1,0 im).
a) Gii bt phng trnh 2 3 2lo log 32 1g log (2 1)2x x .
b) Mt ban vn ngh chun b c 3 tit mc ma, 5 tit mc n ca v 4 tit mc hp ca. Nhng thi gian bui biu din vn ngh c gii hn, ban t chc ch cho php biu din 2 tit mc ma, 2 tit mc n ca v 3 tit mc hp ca. Hi c bao nhiu cch chn cc tit mc tham gia biu din?
1 tancot 2 1 tan
xxx
Cu 3 (1,0 im). Gii phng trnh .
Cu 4 (1,0 im). Tnh tch phn 5
1
1 3 1
I dxx
x .
(2;1; 1), (1;0;3)A AB Cu 5 (1,0 im). Trong khng gian vi h ta Oxyz, cho im . Chng minh ba im A, B, O khng thng hng. Xc nh ta im M thuc ng thng OA sao cho tam gic MAB vung ti M. Cu 6 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh ch nht, hnh chiu vung gc ca nh S ln mp(ABCD) trng vi giao im O ca hai ng cho AC v BD. Bit 52, 2 ,
2SA a AC a SM a , vi M l trung im cnh AB. Tnh theo a th tch khi chp
S.ABCD v khong cch gia hai ng thng SM v AC. Cu 7 (1,0 im). Trong mt phng vi h ta Oxy, cho hnh thang cn ABCD (AD // BC) c phng trnh ng thng : 2 3AB x y 0 v ng thng . Gi I l giao im ca hai ng cho AC v BD. Tm ta cc nh ca hnh thang cn ABCD, bit
: 2AC y 02IB IA ,
honh im I: v nm trn ng thng BD. 3Ix 1;3M 2 3
32 3
(1 )( 3 3) ( 1) . ( , )
2 4 2( 2)
y x y x y x x y
x y x y
Cu 8 (1,0 im). Gii h phng trnh .
Cu 9 (1,0 im).
------ Ht ------
Cho x, y l hai s thc dng tha mn 2 3x y 7 . Tm gi tr nh nh t cbiu thc 2 2 2 2324 8(x y 2 5( ) ) ( 3)P xy y x y x y .
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2
S GIO DC V O TO BNH NH KSCL KHI 12, THNG 01, NM 2015TRNG THPT CHUYN L QU N P N TON. Ngy thi:24/01/2015 Cu Ni dung im
Kho st v v th (C) ca hm s 4 22 1y x x . 1,00 TX: Gii hn: lim , lim
x xy y
0,25
/ 0 10 1 2
x yy
x y
/ 34 4 ,y x x x S bin thin: Hm s nghch bin trn mi khong ( 1;0) v (1; ) , hm s ng bin trn mi khong v (0 ( ; 1 ) ;1)
0,25
Bng bin thin x -1 0 1 y + 0 - 0 + 0 - y 2 2 1
0,25 1.a
th c im cc i A(-1;2), B(1;2) v im cc tiu N(0;1). V th (C). 0,25
Vit phng trnh tip tuyn d vi th (C) ti im M c honh 22
x . Tm ta cc giao im ca tip tuyn d vi th (C).
1,00
Ta c 2 7; (2 4
)M C
. V / 2( ) 22
y 0,25
Pttt (d) c dng / 2 2 74
2 2
y y x
32 4
y x 0,25
Pt h giao im ca d v (C): 4 2 4 232 1 2 4 8 4 2 14
x x x x x x 0 0,25
1.b
2 22 4 4 2 2 02x x x 2 2 2 2, ,2 2 2x x x 2 .
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3
Vy c 3 im: / / /2 7 2 2 1 2 2 1; , , 2 , , 2M 2 4 2 4 2 4
M M
0,25
Gii bt phng trnh 2 32 1log log (2 1) log 32x x 2 . 0,50
KX 12 1 0 2
x x (*) Vi k (*), pt 2 3log (2 1) log (2 1) 1 log 3x x 2
2 3 3 2log 3.log (2 1) log (2 1) 1 log 3x x
0,25 2.a
2 3log 3 1 log (2 1) 1 log 3x 2 3log (2 1) 1x 2 1 3 1x x i chiu (*), tp nghim: 1 ;1
2S
0,25
Mt ban vn ngh chun b c 3 tit mc ma, 5 tit mc n ca v 4 titmc hp ca. Nhng thi gian bui biu din vn ngh c gii hn, ban t chcch cho php biu din 2 tit mc ma, 2 tit mc n ca v 3 tit mc hp ca.Hi c bao nhiu cch chn cc tit mc tham gia biu din?
0,50
Mi cch chn 2 tit mc ma trong 3 tit mc ma l mt t hp chp 2 ca3, suy ra s cch chn 2 tit mc ma: 2C 3 3. Mi cch chn 2 tit mc n ca trong 5 tit mc n ca l mt t hp chp 2ca 5, suy ra s cch chn 2 tit mc n ca: 2C 5 10.Mi cch chn 3 tit mc hp ca trong 4 tit mc hp ca l mt t hp chp 3ca 4, suy ra s cch chn 3 tit mc hp ca: 3C 4 4.
0,25
2.b
Theo quy tc nhn, s cch chn cc tit mc tham gia biu din: 3.10.4 = 120 0,25 Gii phng trnh 1 tancot 2
1 tanxxx
. 1,00
K: sin 2 0
2cos 0 tan 1
4
x x kx
x kx
0,25
Vi K pt tan 2 tan2 4
x x 0,25
22 4
x x k 0,25
3
Kt hp K, ta c nghim: ,4
x k k 0,25
Tnh tch phn 5
1
1 3 1
I dxx x .
1,00 4
t 2 13 1, 0 3
tt x t x 23
dx tdti cn: 1 2; 5x t x t 4. 0,25
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4
4
22
1I 2 1
dtt
4
2
1 1( )1 1
I dtt t
0,25
42ln 1 ln 1I t t 0,25 2 ln 3 ln 5I 0,25
Cho im . Chng minh ba im A, B, O khng thnghng. Xc nh ta im M thuc ng thng OA sao cho tam gic MAB vung ti M.
(2;1; 1), (1;0;3)A AB 1,00
Ta c (3;1;2) (3;1;2)OB OA AB B 0.25
* khng cng phng: O, A, B khng thng hng. (2;1; 1), (1;0;3)OA AB 0.25 Ta c v (2 ; ; ) (2 ; ; )OM t OA t t t M t t t
2; 1; 1), (2 3; 1; 2t t t BM t t t
(2 )AM
Tam gic MAB vung ti M th . 0 (2 2)(2 3) ( 1)( 1) ( 1))( 2) 0AM BM t t t t t t
2 56 11 5 0 1, 6
t t t t .
0.25
5
A (loi) v 1 (2;1; 1)t M 5 5 5 5; )6 6
( ;6 3
t M tha bi ton. 0,25 Cho hnh chp S.ABCD c y ABCD l hnh ch nht, hnh chiu vung gcca nh S ln mp(ABCD) trng vi giao im O ca hai ng cho AC v BD. Bit 52, 2 ,
2SA a AC a SM a , vi M l trung im cnh AB. Tnh
theo a th tch khi chp S.ABCD v khong cch gia hai ng thng SM vAC.
N
MO
A
B C
D
S
H
K
1,00
T gi thit , ( ) ,SO ABCD SO AC OA a 2 2SO SA OA a 0,25
6
2 2 1: 2
OSM O OM SM SO a Ta c 2 2: 2 , 3A BC B BC MO a AB AC BC a
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5
3.
1 3. .3 3S ABCD
V AB BC SO a 0,25
Gi N trung im BC / / ( , ) ( , ( )) ( , ( ))M N AC d SM AC d AC SMN d O SMN OMN O : : , (OMN O OH MN SO MN MN SOH )
, ( ): ( ) (SOH O OK SH OK SMN OK d O SMN 0,25
OMN O : 3 3, ,2 2
aON a OM OH MN OH a 4
2 2
. 5: ( , ) 19
OS OHSOH O d SM AC OK a OS OH
7 0,25
Cho hnh thang cn ABCD (AD // BC) c phng trnh ng thng v ng thng: 2 3AB x y 0 0: 2AC y . Gi I l giao im ca hai
ng cho AC v BD. Tm ta cc nh ca hnh thang cn ABCD, bit 2IB IA , honh im I: v 3Ix 1;3M nm trn ng thng BD.
E
I
A D
B C
F M
1,00
Ta c A l giao im ca AB v AC nn 1;2A . 0,25 Ly im . Gi 0;2E AC 2 3;F a a AB sao cho EF // BD. Khi EF 2 2EF AE BI EF AE
BI AI AE AI
2 2 1
2 3 2 2 11.5
aa a
a
0,25
Vi th l vtcp ca ng thng BD. Nn chn vtpt ca BD l
1a
EF 1; 1 1; 1 BD xn . Pt : 4y 0 2;2 BD AC I
5; 1BD AB BTa c 3 32 2;
2 2IB IBIB ID ID ID DID IA
2 .
1 3 2 2;22
IA IAIA IC IC IC C IC IB
.
0,25
7
Vi 115
a th 7 1;5 5
EF
l vtcp ca ng thng BD. Nn chn vtpt ca
BD l . Do , 1; 7n : 7 22BD x y 0 8;2I (loi). 0,25
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6
Gii h phng trnh. 2 3
32 3
(1 )( 3 3) ( 1) . (1) ( , )
2 4 2( 2) (2)
y x y x y x x y
x y x y
(I) 1,00
KX: 2 20
0, 1 1, 1x y x yx y x y
Nhn xt 1, 1x y khng l nghim ca h. Xt th pt (1) ca h (I) 1y 2 2( 1) 3( 1) ( 1) ( 1)x x y y y x y 0
2
3 01 1 1
x x xy y y
0,25
,1
xt ty
0.
. Khi , pt (1) tr thnh 4 2 3 23 0 1 2 3 0 1t t t t t t t t 0,25
Vi t = 1, th 11
x y xy
1 , th vo pt(2), ta c
3 32 3 2 3
22
2 233 33
22
2 233 33
1 2 4 2 1 1 2 4 1 0
11 6 04 1 4 1
6 11 1 04 1 4 1
x x x x x x x x
x xx xx x x x
x xx xx x x x
0,25
8
2 1 51 0 12
x x x x . 1 5 3 5 .
2 2x y Vi
i chiu K, h phng c nghim 1 5 3 5; ;2 2
x y
.
0,25
Cho x, y l hai s thc dng tha mn 2 3x y 7 . Tm gi tr nh nht ca biu thc 2 2 2 2324 8(x y 2 5( ) ) ( 3)P xy y x y x y .
1,00
Ta c 22 2 3 36( 1)( 1) (2 2)(3 3) 36 5
2x yx y x y x y xy . 0,25
9
Ta c 22 2 2 25( ) 2 5( ) 2x y x y x y x y0
v
2 2 2
2 2
( 3) 9 2 6 62( 3) 8( ) ( 3)
x y x y xy x yx y xy x y x y
0,25
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7
Suy ra 32( ) 24 2( 3)P xy x y x y xy t , 0;t x y xy t 5 , 3( ) 2 24 2 6P f t t t Ta c 23/
2 23 3
(2 6) 824.2( ) 2 2 0, 0;53 (2 6) (2 6)
tf t t
t t
Vy hm s f(t) nghch bin trn na khong 0;5 . Suy ra 3min ( ) (5) 10 48 2f t f .
0,25
Vy 3 2min 10 48 2, 1
xP khi
y 0,25
Ch : Mi cch gii khc ng u cho im ti a. ------ Ht ----
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cng trao i hc tp,h tr ln nhau!