[dethithu.net]20 de thi thu toan 2016 co dap an
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TUYNTP
20THI THTHPT QUCGIA NM 2016MN : TON ( Phn 1)
(Ca cc trng THPT, S GD & T trong nc)
Ngisu tmv tnghp: Hu Hng Hin HaFacebook:http://fb.com/huuhunghienhoa
ng ti ln u ti website:http://dethithu.netFanPage:http://fb.com/dethithu.net
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MC LC THISTT Tn trng/ S GD STT Tn trng/ S GD
1 THPT chuyn H Vinhln 1
11 THPT Kim Lin ln 1
2 THPT chuyn H Vinh
ln 2
12 THPT L Li
3 S GD Vnh Phc (L1) 13 THPT H Huy Tp (L1)
4 THPT chuyn Vnh Phcln 3
14 THPT L Thi T (L1)
5 S GD Qung Ninh 15 THPT Vit Tr - Ph Th
6 THPT chuyn Nguyn Huln 2
16 Chuyn Nguyn Tt Thnh
7 THPT chuyn Bin Ha
ln 1
17 THPT Hm Nghi
8 THPT Ng S Lin Ln 3 18 THPT Nghn- H Tnh
9 THPT chuyn H Long ln2
19 THPT Hng Kh - H nh
10 THPT TrnPh ln 1 20 THPT Phan Thc Trc-NA
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TRNG I HC VINHTRNG THPT CHUYN
THI TH THPT QUC GIANM 2016 LN 1Mn: TON
Thi gian lm bi: 180 pht, khng k thi gian pht
b)Cho = Tnh theo
Cu 4(1,0 ).Tnh tch phn+ +
=+
Cu 5 (1,0 ).Trong khng gian vi h ta cho mt phng + + = v
ng thng+
= =
Tm ta giao im ca vi v lp phng trnh mt
phng cha ng thi vung gc vi
Cu 6(1,0 ).a) Gii phng trnh + = +
b) Nhn dp knim ngy Nh gio Vit Nam, trng THPT tuyn chn c tit mc vnngh tiu biu, trong s lp c tit mc cng din trong ton trng. Ban t chc cho
bc thm ngu nhin chia thnh hai bui cngdin, mi bui tit mc. Tnh xc sut tit mc ca lp c biu din trong cng mt bui.
Cu 7(1,0 ). Cho hnh chp c y l hnh ch nht tm vung
gc vi mt phng = = gc gia hai mt phng v
bng Tnh theo th tch khi chp v khong cch gia hai ng thng
Cu 8(1,0 ).Trong mt phng vi h ta cho tam gic c phng trnh cc ngthng cha trung tuyn v ng cao k t ln lt l + = v + = ng thng
cha trung tuyn k t i qua Tnh bit rng im c tung m v thuc
ng thng + + =
Cu 9 (1,0 ).Gii bt phng trnh + + + + +
Cu 10(1,0 ). Gi s l cc s thc khng m tha mn + + = Tm gi tr
ln nht ca biu thc = + ++ + +
------------------ Ht ------------------
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Cu 1 (1,0 ). Kho st s bin thin v v th ca hm s = +
Cu 2 (1,0 ). Vit phng trnh tip tuyn ca th hm s = + bit rng tip tuyn
song song vi ng thng + =
Cu 3 (1,0 ).
a) Gii bt phng trnh + + ++ < <
Suy ra hm sngbintrn mikhong ( ; 1)- v (3; );+ hm snghch bintrn
khong(1; 3).
* Cc tr: Hm s t cc i ti 1,x = y (1) 3y= = ;
hm s t cc tiu ti 3, (3) 1.CT
x y y= = = -
* Gii hn ti v cc:
32 3
6 9 1lim lim 1 ;x x
y xx x x- -
= - + - = -
32 3
6 9 1lim lim 1 .x x
y xx x x+ +
= - + - = +
0
Cu 1.(1,0
im)
* Bng bin thin:
3o. th:
0
H s gc ca dl3
.4
k= - Suy ra h s gc ca tip tuyn cng l3
.4
-
Ta c
( )2
3 , 1.1
y x
x
= - -
Honh tip im ca tip tuyn vi th l nghim ca phng trnh
2
2
13 3 3 ( 1) 4
34 4( 1)
xy x
xx
= -= - - = - - =
=-
0
Cu 2.(1,0
im)
* Vi 1x = - ta c1
.2
y= Suy ra tip tuyn l3 1
( 1) ,4 2
y x= - + + hay3 1
.4 4
y x= - -
* Vi 3x =
ta c
7
.2y=
Suy ra tip tuyn l
3 7
( 3) ,4 2y x= - - +
hay
3 23
.4 4y x= - +
Vy c hai tip tuyn cn tm l3 1
y x= - - v3 23
y x= - +
0
x
y
y
1- + 3
3
-
+
1-
+ 0 0 +
x
O
3
y
1-
1 3
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a) iu kin: 3.x -
t 32 0,x t+ = > bt phng trnh cho tr thnh
222 5 2 5 2 0,t t tt
+ < - + < (v 0t> )1
22
t < <
1 32 2 2 1 3 1 3 2.x x x- + < < - < + < - < - Vy bt phng trnh cho c nghim 3 2.x- < -
0
Cu 3.(1,0
im)
b) Ta c3
45453
log 75
log 75 2 log 75 2 log 45= =
23 3
233
log (3.5 ) 1 2 log 5 2 4
2 2 .2 log 5 2log (3 .5)
a
a
+ +
= = =+ + 0
t2
dln(2 1), d .
( 1)
xu x x v
x= + + =
+Suy ra
2 1d 1 d , .
2 1 1u x v
x x
= + = -
+ +
Theo cng thc tch phn tng phn ta c1 1
0 0
ln(2 1) 1 2d
1 1 (2 1)( 1)
x xI x
x x x x
+ += - + +
+ + + +
0
Cu 4.(1,0
im)
1 1
0 0
1 1 4 2 1 4 1(1 ln 3) d (1 ln 3) d
2 1 2 1 1 2 2 1 1x x
x x x x x
= - + + + - = - + + -
+ + + + +
( )1
0
1(1 ln 3) 2 ln(2 1) ln( 1)
2x x= - + + + - +
1(1 ln 3) 2 ln 3 ln 2
2= - + + -
( )3 1 1
ln 3 ln 2 3 ln 3 2 ln 2 1 .2 2 2
= - - = - -
0
Gi ( ).M d P= V M d nn ( 2 3; 4 8; ).M t t t - + - -
Suy ra ( ) ( 2 3) (4 8) ( ) 7 0 12,M P t t t t - + + - + - - = = hay ( 21; 40; 12).M - - 0Cu 5.
(1,0im)
Mt phng ( )Qcha dv vung gc vi ( )Pnn ( )Qc cp vtcp( 2; 4; 1)
(1; 1; 1)
d
P
u
n
= - -
=
Suy ra , (5; 1; 6).Q d Pn u n = = -
Ly (3; 8; 0)N d- nn ( ).N Q
Suy ra phng trnh ( ) : 5 6 7 0.Q x y z + - - =
0
a) iu kin: sin 0.x Khi phng trnh cho tng ng vi
( ) ( )
( ) ( )
cos sin sin 2 1 cot 0 cos sin 2 cos sin cos 0
cos sin4cos sin 1 2 cos 0 ( ).1
cos 22 3
x x x x x x x x x
x x x kx x x k
x x k
p
p
p
p
- + - = - + - =
= = + - - = = = +
0
Cu 6.(1,0
im)
b) Gi hai bui cng din l , .I IIS cch chia 24 tit mc thnh hai bui cng din
chnh l s cch chn 12 tit mc cho bui ,I l 1224
.C
Gi l bin c 2 tit mc ca lp 11 c biu din trong cng mt bui.Nu 2 tit mc ca lp 11 cng biu din trong bui Ith s cch chn 10 tit mc cn
li chobui Il 1022
.C Hai tit mc ca lp 11 cng c th cng biu din trong bui .II
V vy, s cch chia bin cAxy ra l 1022
2. .C
Do 1022
1224
2. 11( ) 0, 4783.
23
CP
C= =
2.C2 11 ( ) 12
0
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V( )SD BCD
DC BC
^
^ nn .SC BC^
Suy ra ( ) 0( ), ( ) 45SCD SBC BCD= = (do SCDD vung ti Dnn 090 ).SCD< V BCD l hnh ch nht nn ,O OD=
kt hp vi0 0
180 60 .OD OB = - = Suyra O DD u.
Do 0, 60 .O OD a DO= = =
Suy ra 0. tan 60 3.B D a= =
Suy ra 2. 3BCDS B D a = = v 0. tan 45 3.SD CD a = =
Suy ra 3.
1. .
3S BCD BCD V SD S a = =
0
Cu 7.(1,0
im)
K Bx// C mp( , )S Bx//C
( ) ( )
1
( , ) , ( , ) , ( , ) .2d C SB d O S Bx d D S Bx = =
(1)H , .DK Bx DH SK^ ^ V ( )Bx SDK^ nn ( , ).Bx DH DH S Bx^ ^ (2)
V 2 2BD DO a = = v 060DBK DO= = (ng v) nn 0sin 60 3.DK BD a = =
Suy ra SDKD vung cn ti2 6
.2 2 2
SK SD a D DH = = = (3)
Kt hp (1),(2) v (3) ta suy ra1 6
( , ) .2 4
ad C SB DH= =
0
T h2 0
( 4; 2).3 2 8 0
y
Cx y
+ = - - - + =
Gi ,M Nl trung im , .B BC
Ta c: 2 2 0 ( 2 2; ) ( 0)d x y a a a + + = - - <
: 2 0 ( ; 2).M CM y M m + = -
M Ml trung im Bnn6
(2 2 2; 4) 1; .2
aB a m a N a m
- -+ + - - + -
V CH B^ nn . 0 2(2 2) 3( 2) 0 2 2.CHu M a m a a m = + + + - - = = - + (1)
0
Cu 8.(1,0im)
Ta c ( 2 16; 3)K a a= - + - v12
17; .2
aKN a m
- -= + +
V , ,N Kthng hng nn K cng phng KN. Do
(2)( 2 16)( 12) 2( 3)( 17).- + - - = - + +a a a a m
5Thay (1) vo (2) ta c 2 21 65 0m m2 + - =
m a= = -
23 ( )tm
m a= - =13 28 ( )ktm
Suy ra B(4; 3), (1; 1).- -
Ta c B BC B BC = - = - - = = -(3; 2), ( 5; 1) cos , .( ) 3( 5) ( 2)( 1) 1- + - -9 4 25 1 2+ +
0
S
KB
O
D
H
a
x
045 C
C
H B M
N
K
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iu kin: 2.x -
t 2 3 , 2 ,x u x v + = + = bt phng trnh cho tr thnh
( )2 2 2 23 4 2 3 1 0u v v u u v u v u v - + + - + + - - + ( 1)( 3) 0u v u v - + + -
2 23 2 1 3 2 3 0.x x x x + - + + + + + -
(1)
Ta c2
2
2
13 2 1 1 0.
3 2
x xx x
x x
- ++ - + + = + >+ + +
Do (1) tng ng vi 2 3 2 3 0x x+ + + -
0
Cu 9.(1,0
im)
2
2
3 2 03 3 2
3 9 6 2 2
xx x
x x x
- + + - +
+ - + + +
( ) ( )
2
22 2
2 7, 8 07
6 2 8 36 2 8
x x xx
x x x x x x
- + -
+ + - + + -
( )( )2 2
1 332 2 2 2 3
21
1 4 8 0
x x
xx x x
+ - - -
= - + - -
Vy bt phng trnh cho c nghim 1x = - v 2 2 2 3.x- -
0
t 2 tan , 2 tan , 2 tan ,2 2 2
B Cx y z= = = vi 0 , , .B C p < (1)
T gi thit ta c tan tan tan tan tan tan 1.2 2 2 2 2 2
B B C C+ + =
Khi 1 tan tan
2 2tan cot tan .2 2 2 2
tan tan2 2
B C B C B C
B C
p- + +
= = = - +
Suy ra , .2 2 2
B Ck k
p
p
+= - + Hay 2 .B C kp p+ + = +
T (1) suy ra 0.k= Do .B C p+ + = Khi
0
Cu 10.(1,0
im)
21 1sin sin sin
22 2
CP B= + + 2
1.2 sin cos 1 cos
2 2 22
B B C+ -= + -
22 cos cos 12 2
C C - +
23 1 3
cos .2 2 22
C = - -
Du ng thc xy ra khi
12 2cos 22 2
2.4
C C x y
z B B
p
p
= = = -=
= = = =
Vy gi tr ln nht ca Pbng3
.2
0
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Truy cp http://dethithu.netthng xuyn cp nht nhiu Thi ThTHPT Quc Gia, ti liuhi THPT Quc Gia cc mn Ton, L, Ha, Anh, Vn, Sinh , S, a c DeThiThu.Net cp n
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TRNG I HC VINHTRNG THPT CHUYN
THI TH THPT QUC GIA NM 2016 LN 2Mn: TON
Thi gian lm bi: 180 pht, khng k thi gian pht
Cu 1(1,0 im).Kho st s bin thin v v th ( )H ca hm s1
.2
xy
x
Cu 2 (1,0 im).Tm cc im cc i, cc tiu ca hm s 4 3 2( ) 3 4 12 .f x x x x
Cu 3 (1,0 im).
a) Cho hm s 2( ) .x xf x e e Tm x '( ) 2 ( ) 3.f x f x
b)Cho s phc ztha mn 2(1 ) 2 4 .i z i Tm phn thc v phn o ca .z
Cu 4(1,0 im).Tnh tch phn1
0
3 1sin d .
5
xI x x
x
Cu 5 (1,0 im).Trong khng gian vi h ta ,Oxyz cho mt phng ( ) : 3 0P x y z v
im (1; 2; 3).I Vit phng trnh mt cu ( )S tm ,I tip xc vi mt phng ( ).P Tm ta tip
im ca ( )S v ( ).P
Cu 6(1,0 im).
a) Cho1
cos .3
a Tnh gi tr biu thcsin 3 sin
.sin 2
a aP
a
b) Nam v Hng chi bng qua li, ai thnh cng nhiu hn l ngi thng cuc. Nu
bng v tr A th xc sut thnh cng ca Nam l 0, 9 cn ca Hng l 0,7; nu bng v
tr Bth xc sut thnh cng ca Nam l 0,7 cn ca Hng l 0, 8. Nam v Hng mi ngi
u 1 qu v tr A v 1 qu v tr .B Tnh xc sut Nam thng cuc.
Cu 7(1,0im
). Cho hnh lng tr . ' ' 'ABC A B C c y ABC l tam gic u cnh ,a gc giacnh bn v mt phng y bng 045 , hnh chiu ca A ln mt phng ( ' ' ')A B C l trung im
ca ' ' .A B Gi M l trung im ca ' ' .B C Tnh th tch khi lng tr . ' ' 'ABC A B C theo a vcsin ca gc gia hai ng thng ' , ' .A M AB
Cu 8(1,0 im).Trong mt phng vi h ta ,Oxy cho hnh thang ABCDvung ti A v ,D
1.
3AB AD CD Giao im ca AC v BDl (3; 3),E im (5; 9)F thuc cnh AB sao
cho 5 .AF FB Tm ta nh ,D bit rng nh A c tung m.
Cu 9 (1,0 im).Gii phng trnh 2 1 2
2 22 log 1 4 log (3 ).x xx x x
Cu 10(1,0 im).Tm s thc mln nht sao cho tn ti cc s thc khng m , ,x y ztha mn
4x y z v 3 3 3 2 2 28 .x y z xy yz zx m
------------------ Ht ------------------
Ghi ch: 1. BTC s tr bi vo cc ngy 16, 17/4/2016. nhn c bi thi, th sinh phi np liphiu d thi cho BTC.
2. Thi th THPT Quc gia ln 3 s c t chc vo chiu ngy 07 v ngy 08/5/2016. ng kd thi ti Vn phng Trng THPT Chuyn t ngy 16/4/2016.
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TRNG I HC VINHTRNG THPT CHUYN
P N THI TH THPT QUC GIA NM 2016 LN 2Mn: TON; Thi gian lm bi: 180 pht
Cu p n im1o. Tp xc nh: \ {2}. 2o. S bin thin:* Gii hn, tim cn: Ta c
2
limx
y
v2
lim .x
y
Do ng thng 2x l
tim cn ng ca th ( ).H
V lim lim 1x x
y y
nn ng thng 1y l tim cn ngang ca th ( ).H
* Chiu bin thin: Ta c 21
' 0,( 2)
yx
vi mi 2.x
Suy ra hm s ng bin trn mi khong ( ; 2), (2; ).
0,5
* Bng bin thin:
Cu 1(1,0
im)
3o. th: th ( )H ct Ox ti (1; 0), ct Oy
ti
10; ;
2 nhn giao im (2; 1)I
ca hai ng tim cn lm tm i
xng.
0,5
Hm s xc nh vi mi .x
Ta c3 2
1 2 3'( ) 12 12 24 ; '( ) 0 1, 0, 2.f x x x x f x x x x
0,5Cu 2(1,0
im)
2''( ) 12 3 2 2f x x x .Ta li c ''( 1) 0, ''(0) 0, ''(2) 0.f f f
Suy ra 1, 2x x l cc im cc tiu; 0x l im cc i ca hm s.
Ch . Hc sinh c th lp Bng bin thin a ra kt lun.
0,5
a) Hm s xc nh vi mi x v 2'( ) 2 , .x xf x e e x Khi
2 2'( ) 2 ( ) 3 2 2 2 3 1 0.x x x x x f x f x e e e e e x 0,5Cu 3
(1,0im)
b) T gi thit ta c
2
2 4 2 4 12 2 .
2(1 )
i iz i
i ii
Vy, phn thc ca zbng 2, phn o ca zbng 1.
0,5
x
'y
2
1 1
yO
1
y
1
2
I
x
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Ta c1 1
0 0
3 1sin d d
5
xI x x x
x
+)
11
00
1 2sin cos .xdx x
0,5
Cu 4(1,0
im)
+) Tnh
1
03 1 d .5
xx
x
t 3 1 .x t
Khi 0 1; 1 2x t x t v2 1 2
.3 3
t tx dx dt
Suy ra1 2 22
20 1 1
3 1 2 2d 2 2 1
5 4 416
x tx dt dt
x t tt
2
1
2 4 ln 4 4 ln 4 2 8 ln 3 4 ln5.t t t
T ta c2
2 8 ln 3 4 ln 5.I
0,5
Ta c , ( ) 3.R d I P Suy ra 2 2 2( ) : ( 1) ( 2) ( 3) 3.S x y y 0,5Cu 5(1,0
im) Gi Hl tip im ca ( )S v ( ).P Khi Hl hnh chiu ca Iln ( ).P
Ta c (1; 1; 1).IH P
u n
Suy ra1 2 3
: .1 1 1
x y zIH
Do ( 1; 2; 3).H t t t V ( )H P nn
( 1) ( 2) ( 3) 3 0 1.t t t t
Suy ra (0; 1; 2).H
0,5
a) Ta c
2sin 3 sin 2 cos 2 sin cos 2 2 cos 1 7.
sin 2 2 sin cos cos cos 3
a a a a a a P
a a a a a
0,5Cu 6(1,0
im)
b) Gi Xl bin c Nam thng cuc; ( 0, 1, 2)i
N i l bin c Nam thnh cng i
qu; ( 0, 1, 2)i
H i l bin c Hng thnh cng iqu.
Khi
1 0 2 0 2 1X N H N H N H .Theo gi thit ta c
1 0 1 0. 0, 9.0, 3 0, 1.0, 7 0, 3.0, 2 0, 0204.P N H P N P H
2 0 2 0. 0, 9.0, 7 0, 3.0, 2 0, 0378.P N H P N P H
2 1 2 1. 0, 9.0, 7 0, 7.0, 2 0, 3.0, 8 0, 2394.P N H P N P H
Suy ra (X) 0, 0204 0, 0378 0, 2394 0, 2976.P
0,5
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a
450
K N
MH
C
B
A
C'
B'
A'
Gi Hl trung im ca ' ' .A B Khi
( ' ' ').AH A B C Suy ra
0' ( ', ( ' ' ')) 45 .AA H AA A B C
Do ' .2
aAH A H Suy ra
30
. ' ' '
1 3. . . .sin60 .
2 2 8ABC A B Ca a
V a a
0,5
Cu 7(1,0
im)
Gi Nl trung im ca .BC Khi ( ' , ') ( , ').A M AB AN AB
Trong tam gic vung 'HAB ta c2 2
2 2 2' ' .2 2 2
a a aAB AH HB
Tam gic ABCu cnh ann
3
.2
a
AN
Gi Kl trung im ca .AB Khi ' / /B K AH nn ' .B K KN Suy ra
2 2
2 2 2' ' .2 2 2
a a aB N B K KN
p dng h qu ca nh l hm s csin trong tam gic 'AB Nta c
2 2 22 3 2
4 4 4 6cos( ' , ') cos ' .
42 32. .
2 2
a a a
A M AB NAB a a
0,5
11
I
F
E
D C
BA Gi .I EF CD Ta s chng minh tam
gic EAIvung cn ti .E
t , .AB a AD b
Khi a b
v
. 0.a b
Ta c 3 .AC AD DC b a
1 5 1 5 13 3 .4 6 4 6 12
FE AE AF AC AB b a a b a
Suy ra
2 21
. 3 3 0.12AC EF b a
Do .AC EF (1)
T (1) suy ra t gic ADIEni tip. Suy ra 01 1
45 .I D (2)
T (1) v (2) suy ra tam gic EAIvung cn ti .E
0,5
Cu 8(1,0
im)
Ta c (2; 6)AC
n EF
nn : 3 12 0 (3 12; ).AC x y A a a
Theo nh l Talet ta c
3EI EC CD
EF EA AB 3 ( 3; 15).EI FE I
0,5
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Khi
2 2 3
(3 9) ( 3) 360 .9
aEA EI a a
a
V A c tung m nn ( 15; 9).A
Ta c (20; 0)AD
n AF
nn : 15 : 15.AD x CD y Do ( 15; 15).D
iu kin: 0.x Phng trnh cho tng ng vi2 1 2 3
2 22 log 1 2 log (3 ).x x xx x x
(1)
Xt hai trng hp sau:
TH1.1
0 .3
x Khi 2 1 2 3
2 22 log 1 2 0 2 log (3 ).x x xx x x
Suy ra (1) khng tha mn.
0,5
Cu 9(1,0
im)
TH2.1
.3
x Ta c 2 1x x v 3xu thuc khong [1; + ).
Xt hm s2
( ) 2 logtf t t trn khong [1; + ).
Ta c2
1'( ) 2 ln 2. log 2 . 0
ln 2t tf t t
t vi mi tthuc khong [1; + ).
Suy ra ( )f t ng bin trn khong [1; + ).
Do (1) tng ng vi 2 1 3 .x x x T y gii ra c1
.3
x
Vy, phng trnh cho c nghim duy nht1
.3
x
0,5
Gi s tn ti cc s thc , ,x y ztha mn yu cu bi ton t ra.
Khng mt tnh tng qut ta gi s ynm gia xv .zKt hp vi gi thit ta c0 2y v ( )( ) 0.x y x y z
T y ta c 22 2 2 .xy yz zx y x z
Mt khc, do ,x zkhng m nn 33 3 .x z x z
Do
3 2 3 23 38 4 8 4m x z y y x z y y y y
3 28 52 80 64y y y . (1)
0,5
Cu
10(1,0im)
Xt hm s 3 2( ) 8 52 80 64, 0 2.f y y y y y Ta c
2 2( ) 24 104 80 8 3 13 10 .f y y y y y ( ) 0, 0 2 1.f y y y
Ta c (0) 64, (1) 100, (2) 80.f f f
Suy ra ( ) (1) 100, [0; 2].f y f y (2)
T (1) v (2) ta c 100.m Khi 0, 1, 3x y z ta c du ng thc.
Vy s mln nht cn tm l 100.
0,5
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S GD&T VNH PHC KSCL N THI THPT QUC GIA LN 1NM HC 2015-2016
MN THI: TONThi gian lm bi: 180 pht, khng k thi gian pht
Cu 1(1,0 im). Kho st s bin thin v v th ca hm s2 1
2y
Cu 2 (1,0 im). Tm cc im cc tr ca th hm s 3 23 6y x x
Cu 3 (1,0 im).
a) Gii bt phng trnh 22 2log log 44x
x
b) Gii phng trnh 5.9 2.6 3.4x x x
Cu 4(1,0 im). Tnh nguyn hm 2 sin 3I x xdx Cu 5 (1,0 im). Cho hnh chp .S ABC c 0, 90 , , 3, 2SA ABC ABC AB a BC a SA a .
Chng minh trung im I ca cnh SCl tm ca mt cu ngoi tip hnh chp .S ABCv tnh din
tch mt cu theo a.
Cu 6(1,0 im).
a) Gii phng trnh: 22cos sin 1 0x x .
b) i vn ngh ca nh trng gm 4 hc sinh lp 12A, 3 hc sinh lp 12Bv 2 hc sinh lp
12C. Chn ngu nhin 5 hc sinh t i vn ngh biu din trong l b ging nm hc. Tnh xc sut
sao cho lp no cng c hc sinh c chn v c t nht 2 hc sinh lp 12A.
Cu 7 (1,0 im). Cho hnh chp .S ABCD c y l hnh vung cnh a,3
2
aSD . Hnh chiu vung
gcHca nh Sln mt phng (ABCD) l trung im ca onAB . Gi Kl trung im ca on
AD . Tnh theo ath tch khi chp .S ABCD v khong cch gia hai ng thng HKv SD .
Cu 8 (1,0 im). Trong mt phng vi h to Oxy cho hnh thang ABCD vung ti A v D c
AB AD CD , im (1;2)B , ng thng BD c phng trnh l 2 0y . ng thng qua B
vung gc vi BCct cnh AD ti M. ng phn gic trong gcMBCct cnh DCti N. Bit
rng ng thng MNc phng trnh 7 25 0x y . Tm ta nh D .
Cu 9 (1,0 im). Gii h phng trnh:
2
2
2 1 11 ,
3 8 3 4 1 1
xx y x y
x x y
x x x y
Cu 10 (1,0 im). Cho ,x y tha mn2
2
2
2 3
y x
x x
. Tm gi tr nh nht ca biu thc:
4 4
2
2P x y
y
-------------HT------------
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S GD&T VNH PHC HNG DN CHM KSCL N THI THPT QUC GIA LN 1NM HC 2015-2016
MN THI: TON
I. LU CHUNG:- Hng dn chm ch trnh by mt cch gii vi nhng c bn phi c. Khi chm bi hc sinh lmtheo cch khc nu ng v th vn cho im ti a.- im ton bi tnh n 0,25 v khng lm trn.- Vi bi hnh hc khng gian nu th sinh khng v hnh hoc v hnh sai th khng cho im tngng vi phn .II. P N:
Cu Ni dung trnh by im1
Kho st s bin thin v v th ca hm s2 1
2y
1,0
2 1
2
xy
x
1. Tp xc nh: \ {2}D
2. S bin thin.
2
3' 0,
( 2) x D
x
Suy ra hm s nghch bin trong cc khong ( ;2) v (2; ) Hm s khng c cc tr
0,5
Cc gii hn2 2
lim 2; lim 2; lim ; limx x x x
y y y y
Suy ra 2x l tim cn ng, 2y l tim cn ngang ca th.0,25
Bng bin thin
0,25
3. th: Giao vi trc Oxti1
;02
, giao vi trc Oyti1
0;2
, th c tm i
xng l im (2;2)I
0,25
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2 Tm cc im cc tr ca th hm s 3 23 6y x x 1,0
* Tp xc nh: 0,25
2 0' 3 6 , ' 02
xy x x y
x
0,25
Bng xt du o hm
0 2 + 0 - 0 +
0,25
T bng xt u o hm ta cHm s t cc i ti 0x v gi tr cc i 6y ; t cc tiu ti 2x v gi trcc tiu 2y .
Vy im cc i ca th hm s l M 0;6 , im cc tiu ca th hm s l
N 2;2
0,25
3 aGii bt phng trnh 22 2log log 44
xx (1) 0,5
+) iu kin ca bt phng trnh (1) l: 0x (*)+) Vi iu kin (*),
2 2
2 2 2 2 2(1) log log log 4 4 log log 2 0x x x x
2 2(log 2)(log 1) 0x x
0,25
2
2
4log 2
1log 1 0
2
xx
x x
+) Kt hp vi iu kin (*), ta c tp nghim ca bt phng trnh (1) l
1
0; 4;2
S
0,25
b Gii phng trnh 5.9 2.6 3.4x x x (1) 0,5
Phng trnh cho xc nh vi mi x
Chia c hai v ca phng trnh (1) cho 4 0x ta c :2
3 35.9 2.6 3.4 5. 2. 3
2 2
x xx x x
23 3
5. 2. 3 02 2
x x
3 31 5. 3 0
2 2
x x
(2)
0,25
V3
5. 3 02
x
x
nn phng trnh (2) tng ng vi
31 0
2
x
x
.
Vy nghim ca phng trnh l: 0x
0,25
4 Tnh nguyn hm 2 sin 3I x xdx 1,0
t2
sin3
u x
dv xdx
0,25
ta c cos33
du dx
v
0,25
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Do : 2 cos 3 1
cos33 3
x xI xdx
0,25
2 cos 3 1sin3
3 9
x xC
0,25
5 Cho hnh chp .S ABC c 0, 90 , , 3, 2SA ABC ABC AB a BC a SA a .Chng minh trung im I ca cnh SC l tm ca mt cu ngoi tip hnh chp
.S ABCv tnh din tch mt cu theo a.
I
A C
B
S
1,0
V SA ABC SA BC
Mt khc theo gi thit AB BC , nn BC SAB v do BC SB 0,25Ta c tam gic SBC vung nhB; tam gic SABvung nhAnn
2
SCIA IB IS IC (*)
Vy imIcch u bn nh ca hnh chp, do Il tm mt cu ngoi tip cahnh chp .S ABC
0,25
T (*) ta c bn knh ca mt cu l2
SCR
Ta c 2 2 2AC AB BC a 2 2 2 2 2SC SA AC a R a
0,25
Din tch mt cu l 2 24 8R a 0,256 a Gii phng trnh 22cos sin 1 0x x . 0,5
Ta c: 2 22cos sin 1 0 2sin sin 3 0 (sin 1)(2sin +3)=0x x x x x x 0,25sin 1x (do 2sin 3 0x x )
sin 1 22
x x k k
Vy nghim ca phng trnh cho l: 22
x k k
0,25
b i vn ngh ca nh trng gm 4 hc sinh lp 12A, 3 hc sinh lp 12Bv 2 hcsinh lp 12C. Chn ngu nhin 5 hc sinh t i vn ngh biu din trong l b
ging nm hc. Tnh xc sut sao cho lp no cng c hc sinh c chn v c tnht 2 hc sinh lp 12A.
0,5
Gi khng gian mu ca php chn ngu nhin l S phn t ca khng gian mu l: 59 126C
Gi A l bin c Chn 5 hc sinh t i vn ngh sao cho c hc sinh c ba lp vc t nht 2 hc sinh lp 12A.Ch c 3 kh nng xy ra thun li cho bin c A l :+ 2 hc sinh lp 12A, 2 hc sinh lp 12B, 1 hc sinh lp 12C+ 2 hc sinh lp 12A, 1 hc sinh lp 12B, 2 hc sinh lp 12C+ 3 hc sinh lp 12A, 1 hc sinh lp 12B, 1 hc sinh lp 12C
0,25
S kt qu thun li cho bin c A l: 2 1 2 2 2 1 3 1 14 3 2 4 3 2 4 3 2. . . . . . 78C C C C C C C C C . 0,25
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Xc sut cn tm l78 13
126 21P .
7Cho hnh chp .S ABCD c y l hnh vung cnh a,
3
2
aSD . Hnh chiu vung
gcHca nh Sln mt phng (ABCD) l trung im ca onAB . Gi Kl trungim ca on AD . Tnh theo ath tch khi chp .S ABCD v khong cch giahai ng thng HKv SD .
1,0
T gi thit ta c SHl ng cao ca hnh chp S.ABCDv
2 2 2 2 2 2 2 23( ) ( ) ( )2 2
a aSH SD HD SD AH AD a a
0,25
Din tch ca hnh vungABCDl 2a ,3
2.
1 1. .
3 3 3S ABCD ABCDa
V SH S a a 0,25
T gi thit ta c / / / /( )HK BD HK SBD
Do vy: ( , ) ( ,( ))d HK SD d H SBD (1)
GiEl hnh chiu vung gc caHlnBD, F l hnh chiu vung gc caHln SETa c , ( )BD SH BD HE BD SHE BD HF m HF SE nn suy ra
( ) ( , ( ))HF SBD HF d H SBD (2)
0,25
+) 02
.sin .sin 452 4
a aHE HB HBE
+) Xt tam gic vung SHE c:
2 2
2.. 4. .
32( )
4
aaSH HE a
HF SE SH HE HFSE a
a
(3)
+) T (1), (2), (3) ta c ( , )3ad HK SD .
0,25
8 Trong mt phng vi h to Oxy cho hnh thang ABCD vung ti A v D cAB AD CD , im (1;2)B , ng thng ng thng BD c phng trnh l
2 0y .. ng thng qua B vung gc vi BC ct cnh AD ti M . ngphn gic trong gc MBC ct cnh DC ti N. Bit rng ng thng MN cphng trnh 7 25 0x y . Tm ta nh D .
1,0
0,25
E
O
K
H
B
A D
C
S
F
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T gic BMDCni tip 045BMC BDC DBA BMC vung cn tiB,BNl
phn gic trongMBC,M C i xng quaBN
4( , ) ( , )
2AD d B CN d B MN 0,25
Do 2 4AB AD BD AD 0,25
: 2 0 ( ; 2)BD y D a ,
5 5; 24
3 3;2 ( )
a DBD
a D loai cung phia B so voi MN
Vy c mt im tha mn l: (5;2)D
0,25
9 Gii h phng trnh:
2
2
2 1 11 ,
3 8 3 4 1 1
xx y x y
x x y
x x x y
1,0
iu kin:1
1
x
y
33 2 11 2 1 1 2 1
1 1 1
x x xx x xx y y y
x x x
3
31 11 1
x x yx x
.
0,25
Xt hm s 3f t t t trn c 23 1 0f t t t suy ra f(t)ng bin
trn . Nn 1 11 1
x xf f y y
x x
. T y suy ra 0x Thay
vo (2) ta c 23 8 3 4 1x x x .
0,25
22
2 1 2 1x x x
2
2
1
6 3 0 3 2 32 1 1
1 5 2 132 1 1 3
3 99 10 3 0
x
x x xx x
xx x x
x x
0,25
Ta c2
11
xy
Vi4 3 3
3 2 32
x y
. Vi5 2 13
( 0)9
x loai do x
.
0,25
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KL: H phng trnh c mt nghim 4 3 3
; 3 2 3;2
x y
.
10Cho ,x y tha
2
2
2
2 3
y x
x x
. Tm gi tr nh nht ca biu thc
4 4
2
2P x y
x y
1,0
T gi thit ta c 0y v
2
2 62 3 02 5x x x x v
22 2 2 2 2 22 3 2 2 6 5x y x x x x x x
Xt hm s 2 26
( ) 2 2 6 5 ; 0;5
f x x x x x ta c
60;
5
Max
f(x) = 2
2 2 2x y
0,25
22 22 22 2 2 2 2 2
2 2 2
2 22
2
x yP x y x y x y
yx y
t 2 2t x y 2
2 , 0 22tP tt
0,25
Xt hm s:
2 2
( ) , 0;22
tg t t
t
33
2 2
2 2'( ) ; '( ) 0 2
tg t t g t t
t t
0,25
Lp bng bin thin ta cMin3 63 4 16
2 2P khi x y
0,25
------------Ht------------
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TR NG THPT CHUY N V NH PH C
CHNH THC
THI THPT QUC GIA NM HC 2015-2016-LN 3Mn: TON
Thi gian lm bi: 180 pht, khng k thi gian pht .
Cu 1 (1,0 im). Kho st s bin thin v v th ca hm s : 3 23 2y x x
Cu 2 (1,0 im).Tm gi tr ln nht v gi tr nh nht ca hm s : 2 1
1f x
x
trn on 3; 5
Cu 3 (1,0 im).a) Cho ;
2
v
1sin
3 . Tnh gi tr biu thc sin 2 cos 2P
b) Gii phng trnh : 2sin 2 2sin sin cosx x x x
Cu 4 (1,0 im).Tnh tch phn sau : 4
2 2
0
2 2 ln 9I x x x dx
Cu 5 (1,0 im).a) Gii bt phng trnh : 2 2log 3 2 log 6 5 0x x .
b) Cho tp hp 1;2;3;4;5;6E vMl tp hp tt c cc s gm hai ch s phn bit lp t
E. Ly ngu nhin mt s thucM. Tnh xc sut tng hai ch s ca s ln hn 7 .
Cu 6 (1,0 im). Trong khng gian vi h ta Oxyz , cho cc im 1; 2;0 , 3;4; 2M N v
mt phng : 2 2 7 0P x y z . Vit phng trnh ng thng MNv tnh khong cch t
trung im ca on thng MNn mt phng P .
Cu 7 (1,0 im). Cho hnh chp .S ABCc y ABCl tam gic u cnh a GiIl trung imcnhAB .Hnh chiu vung gc ca nh S trn mt phng y l trung im H ca ,CI gc
gia ng thng SA v mt y bng 060 . Tnh theo a th tch khi chp .S ABC v khongcch t im Hn mt phng SBC .
Cu 8 (1,0 im)..
Trong mt phng vi h to Oxy , cho hai ng thng 1 2:3 4 8 0 , :4 3 19 0d x y d x y .Vit phng trnh ng trn C tip xc vi hai ng thng 1d v 2d , ng thi ct ng
thng :2 2 0x y ti hai im ,A B sao cho 2 5AB .Cu 9 (1,0 im).
Gii bt phng trnh : 2
2 2 1
26 2 4 2 2
x
x x x
Cu 10 (1,0 im).
Cho cc s thc dng ,x y tha mn iu kin 2016x y .Tm gi tr nh nht ca biu thc
2 2 2 2 2 2 2 25 3 3 5 2 2P x xy y x xy y x xy y x xy y
--------Ht-------
Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm.
H v tn th sinh:.......; S bo danh:
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TR NG THPT CHUY N V NH PH C HNG DN CHM THI THPT QUC GIA LN 3NM HC 2015-2016
Mn: TON ( Gm 5 trang)
Cu p n im
Cu 1. Kho st s bin thin v v th ca hm s : 3 23 2y x x 1,0
Tp xc nh: D .
Ta c 23 6y' x x. ;0
02
xy'
x
0,25
1 (1,0 ) - Hm s ng bin trn cc khong ( ;0) v (2; ) ; nghch bin trn khong (0;2) .
- Cc tr: Hm s t cc i ti 0, 2CDx y ; t cc tiu ti 2, 2CTx y
- Gii hn: lim , limx xy y
0,25
Bng bin thin:x 0 2 y' + 0 - 0 +y 2
-2
0.25
th:
f(x)=(x^3)-3*(x)^2+2
-8 -6 -4 -2 2 4 6 8
-5
5
x
y
0,25
Cu2.Tm gi tr ln nht v gi tr nh nht ca hm s : 2 1
1f x
trn on 3; 5 1,0
Hm s xc nh v lin tc trn 3; 5D 0,25
2 (1,0 ) Ta c
23
0, 3;51
f x xx
0,25
Do hm s ny nghch bin trn on 3; 5 0,25
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Suy ra
3;53;5
7 11max 3 ; min 5
2 4xxf x f f x f
0,25
Cu 3a. Cho ;2
v
1sin
3 . Tnh gi tr biu thc sin 2 cos 2P 0,5
V ;2
nn cos 0 , suy ra 2
2 2cos 1 sin
3 0,25
3.(1,0) Do 2sin 2 cos 2 2sin cos 1 2sinP
21 2 2 1 7 4 2
2 1 23 3 3 9P
0,25
Cu 3b) Gii phng trnh : 2sin 2 2sin sin cosx x x 0,5
Phng trnh cho 2sin sin cos sin cosx x x x x
sin cos 0 1
2sin 1 2
x x
x
0,25
1 tan 1 ,x x k k
1 5
2 sin 2 2 ,
2 6 6
x x k x k k
0,25
Vy phng trnh c ba h nghim5
, 2 , 24 6 6
x k x k x k
vi k
Cu 4. Tnh tch phn sau : 4
2 2
0
2 2 ln 9I x x x dx 1,0
4 4
3 21 2
0 0
4 2 ln 9I x dx x x dx I I 0,25
443 4
1 00
4 256I x dx x 0,25
4 .(1,0 )
42
2
0
2 ln 9I x x dx . t 2
2
2
2ln 9 92 9
xdu dxu x xdv xdx v x
4
4 4 42 2 2 2 22 00 0
0
9 ln 9 2 9 ln 9I x x xdx x x x
2 25ln 25 9ln 9 16 50ln 5 18ln 3 16I
0,25
Vy 1 2 240 50ln 5 18ln 3I I I 0,25
Cu 5 a) Gii bt phng trnh : 2 2log 3 2 log 6 5 0x x . 0,5
Bt phng trnh cho 2 2
3 2 0
log 3 2 log 6 5 6 5 0
3 2 6 5
x
x x x
x
0,25
2
36 6
15 51
x
x x
x
.Vy nghim ca bt phng trnh l :6
15
x 0,25
5 (1,0 ) Cu 5 b) Cho tp hp 1;2;3;4;5;6E v M l tp hp tt c cc s gm hai ch s
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phn bit thuc E. Ly ngu nhin mt s thuc M . Tnh xc sut tng hai ch sca s ln hn 7.
S phn t ca tp Ml 26 30A 0,25
Cc s c tng hai ch s ln hn 7 gm 26,62,35,53,36,63,45,54,46,64,56,65
C 12 s nh vy . Suy ra xc sut cn tm l12 2
30 5P
0,25
Cu 6. Trong khng gian vi h ta Oxyz , cho cc im 1; 2;0 , 3;4; 2M N
v mt phng : 2 2 7 0P x y z . Vit phng trnh ng thng MN v tnhkhong cch t trung im ca on thng MN n mt phng P .
1,0
ng thng MNc vect ch phng 4;6;2MN
hay 2;3;1u
0,25
6 .(1,0 ) Phng trnh ng thng MN:1 2
2 3 1
y z
(c th vit di dng pt tham s) 0,25
Trung im ca on thng MNl 1;1;1I 0,25
Khong cch t In mt phng P l :
2 2 1 7
, 24 4 1d I P
0,25
Cu 7. Cho hnh chp .S ABC c y ABC l tam gic u cnh a GiI l trungimcnhAB .Hnh chiu vung gc ca nh Strn mt phng y l trung im Hca
,CI gc gia ng thng SA v mt y bng 060 . Tnh theo a th tch khi chp
.S ABC v khong cch t im Hn mt phng SBC .
1,0
7. (1,0 )
K
H
KH
S
A
B
C
A
B C
I
I
A'I'
H'
E
H'
0,25
Ta c 2 23
2
aCI AC AI
Do 2 27
4
aAH AI IH , suy ra 0
21.tan60
4
aSH AH .
Vy3
.
1 7.
3 16S ABC ABC a
V SH S
0,25
Gi ', ', 'A H I ln lt l hnh chiu ca , ,A H ItrnBC;El hnh chiu caHtrn SH'
th ( ) ;( )HE SBC d H SBC HE . Ta c1 1 3
' ' '2 4 8
aHH II AA
0,25
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T2 2 2
1 1 1
'HE HS HH , suy ra
21
4 29
aHE . Vy
21;( )
4 29
ad H SBC . 0,25
Cu8.Trong mt phng vi h to Oxy, cho hai ng thng 1 :3 4 8 0d x y ,
2 :4 3 19 0d x y .Vit phng trnh ng trn C tip xc vi hai ng thng 1d
v 2d, ng thi ct ng thng :2 2 0x y ti hai im ,A B sao cho 2 5AB
1,0
Gi ;I a b l ta tm v R l bn knh ng trn C .
Do ng thng ct ng trn C ti hai im ,A B sao cho 2 5AB nn ta c
2 22 2, 5 2 5 *5
a bd I R R
0,25
8 .(1,0 )ng trn C tip xc vi 1d , 2d khi :
1
2
,
,
d I d R
d I d R
3 4 83 4 8
55
4 3 194 3 19 3 4 8
5
a ba bR
R
a ba b a bR
7 27
5 20
7 11
5 5
b a
R a
a b
R b
0,25
-Vi 7 275 20
b aR a
thay vo * ta c 2 95 5 5 20 5 3
2a a a a
Vy phng trnh ng trn l
2 2
: 3 6 25C x y hoc 2 2
9 9 25:
2 2 4C x y
0,25
-Vi7 11
5 5
a b
R b
thay vo * ta c
2 35 3 4 5 5 5 2
2b b b b
Vy phng trnh ng trn l
2 2
: 3 2 25C x y
hoc
2 21 3 25
: 2 2 4C x y
0,25
Cu 9.Gii bt phng trnh : 2
2 2 1
26 2 4 2 2
x
x x x
1,0
iu kin : 2x
Ta c
2
2
2
2 2 46 2 4 2 2 0, 2
6 2 4 2 2
x xx x x x
x x x
0,25
9 .(1,0 )
Do bt phng trnh 22 2 2 6 2 4 2 2x x x x
2
2 2 2 12 2 6 1x x x x
Nhn xt 2x khng l nghim ca bt phng trnhKhi 2x chia hai v bt phng trinh 1 cho 2 0x ta c
0,25
2
2 2 12 6 22 2
x x
x x
. t
2t
x
th bt phng trnh 2 c
2
22 2
12 2 02 2 12 6 2
4 8 4 12 6 2 2 0
ttt t t
t t t t
0,25
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Lu khi chm bi:
- p n ch trnh by mt cch gii bao gm cc bt buc phi c trong bi lm ca hc sinh. Khi chm
nu hc sinh b qua bc no th khng cho im bc .
- Nu hc sinh gii cch khc, gim kho cn c cc trong p n cho im.
- Trong bi lm, nu mt bc no b sai th cc phn sau c s dng kt qu sai khng c im.- Hc sinh c s dng kt qu phn trc lm phn sau.
- Trong li gii cu 7 nu hc sinh khng v hnh th khng cho im.
- im ton bi tnh n 0,25 v khng lm trn.
2
02 2 2 2 3
4 8 02
xxt x
x xx
. Bt phng trnh c nghim duy
nht 2 2 3x . (Ch bi ny c nhiu cch gii khc nh dng vc t, dng btng thc ,dng php bin i tng ng)
0,25
Cu 10.Cho , 0x y tha mn iu kin 2016x y .Tm gi tr nh nht ca biu thc
2 2 2 2 2 2 2 25 3 3 5 2 2P x xy y x xy y x xy y x xy y
1,0
P A B . Trong 2 2 2 25 3 3 5A x xy y x xy y v2 2 2 22 2B x xy y x xy y
0,25
10.(1,0) 2 2 2 26 180 36 108 108 36 180A x xy y x xy y
2 2 2 2
11 7 59 11 7 59y x y y x y x
11 7 11 7 18y y x x y
3 3 2016 6048 *A x y du ng thc xy ra khi v ch khi 1008x y
0,25
2 2 2 24 16 16 32 32 16 16B x xy y x xy y
2 2 2 2
3 5 7 3 5 7x y x y y x y x
3 5 3 5 8y y x x y
2 2 2016 4032 **B x y du ng thc xy ra khi v ch khi 1008x y
0,25
T * v ** ta c 6048 4032 10080P A B ,du ng thc xy ra khi v
ch khi 1008x y . Vy min 10080 1008P x y 0,25
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SGIO DC V O TOTNH QUNG NINH
CHNH THC
THI KHO ST LP 12 NM HC 2015-2016Mn: TON
Thi gian lm bi: 180 pht khng kthi gian pht
(thi gm 01 trang)
Cu 1 (1,0 im)Kho st sbin thin v vthhm s 3 26 9 1= + y x x x
Cu 2 (1,0 im)Tm gi trln nht, gi trnhnht ca hm s
9
1= +
y x
xtrn on [2;5]
Cu 3 (1,0 im)a) Gi
1 2,x x l hai nghim trn tp sphc ca phng trnh 2 2 5 0x x+ + = . Tnh 1 2x x+ .
b) Gii phng trnh 22 12
log ( 2 8) 1 log ( 2)x x x = + .
Cu 4 (1,0 im) Tnh tch phn
= +2
2
0
( sin ) cos .I x x xdx
Cu 5 (1,0 im)Trong khng gian vi h ta Oxyz, cho hai im A(2; 3; 5), B(6; 1; 3) v mt phng (P) c
phng trnh 2 2 13 0x y z+ + = . Vit phng trnh ng thng AB v phng trnh mt cu c tm ltrung im ca on thng AB ng thi tip xc vi mt phng (P).
Cu 6 (1,0 im)
Cho hnh hp ngABCD.ABCDc y l hnh thoi cnh a, gc 060ACB= , mt phng (ABD)
to vi y mt gc 060 . Tnh theo athtch khi hp v khong cch gia hai ng thng CD,BD.
Cu 7 (1,0 im)
a) Cho 2sin3
= , vi2
< < . Tnh 2cos3
A = +
b) Gii bng chuyn VTV Cup gm 12 i bng tham d, trong c 9 i nc ngoi v 3 i caVit Nam. Ban tchc cho bc thm ngu nhin chia thnh 3 bng A, B, C mi bng 4 i. Tnh xcsut 3 i bng ca Vit Nam ba bng khc nhau.
Cu 8 (1,0 im)Trong mt phng vi h ta Oxy, cho tam gic nhn ABC ni tip ng trn (C): 2 2 25x y+ = ,
ng thng AC i qua im K(2; 1). Gi M, N ln l!t l chn ng cao k"t#$nh B v C. Tm ta cc $nh ca ABC bit phng trnh ng thng MN l 4 3 10 0x y + = v im A c honh m.
Cu 9 (1,0im) Gii phng trnh
2 2
1 2 9 18 14 33+ + = + +
x x x x x trn tp sthc.Cu 10 (1,0 im)
Chox, y, z l cc sthc tha mn 2 2 2 25 2 2 8 4 5 4 2x xy y x xz z x y z+ + + + + = + + v [0;5]x
Tm gi trln nht, gi trnhnht ca biu thc 2 21 10P z xy x z xy= + + + .
===============Ht===============
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uy cp http://dethithu.netthng xuyn cp nht nhiu Thi ThTHPT Quc Gia, ti liu HPT Quc Gia cc mn Ton, L, Ha, Anh, Vnnh , S, a c DeThiThu.Net cp nht hng ngy phc vst!
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SGIO DC V O TOTNH QUNG NINH
HNG DN CHM THI KHO ST LP 12NM HC 2015-2016
Mn: TON
Cu P N V BIU IM imCu 1
1,0
Kho st sbin thin v vthhm s 3 26 9 1= + y x x x
* Tp xc nh: D = R* Sbin thin
Chiu bin thin: )34(39123' 22 +=+= xxxxy
Ta c
>
1
30'
x
xy , 310'
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Phna)
0,5
Tnh !c hai nghim phc 1 1 2x i= ; 2 1 2x i= + 0,25
1 2 1 25 2 5x x x x= = + = 0,25
Phnb)
0,5
b) &K:x> 4
PT cho tng ng vi 22 2 2log ( 2 8) log 2 log ( 2)x x x = + +
2
2 2log ( 2 8) log 2( 2)x x x = +
0,25
2
2 0
2 8 2( 2)
x
x x x
+ >
= +
2
2 06
4 12 0
xx
x x
+ > =
=
0,25
Cu 4
1,0 Tnh tch phn
= +2
2
0
( sin ) cos .I x x xdx
= + = +
2 2 22 2
0 0 0
( sin ) cos cos sin cos .
M N
I x x xdx x xdx x xdx 0,25
TnhM
&tcos sin
u x du dxdv xdx v x
= =
= =
2
0
sin sin cos 1.2 22 2
0 0
M x x xdx x
= = + =
0,25
TnhN
&t sin cost x dt xdx= =
&i cn1
2
0 0
x t
x t
= =
= =
1 32
0
1 1.
03 3
tN t dt= = =
0,25
Vy2
.2 3
I M N
= + = 0,25
Cu 5
1,0
Trong khng gian vi hta Oxyz, cho hai im A(2;3;5), B(6;1;3) v mt phng (P) cphng trnh 2 2 13 0x y z+ + = . Vit phng trnh ng thng AB v phng trnh mt cu ctm l trung im ca on thng AB ng thi tip xc vi mt phng (P).
Ta c (8; 2;8) 2BA u= =
vi (4;1;4)u =
Suy ra u
l VTCP ca ng thng AB 0,25
Phng trnh ng thng AB l:
2 4
3
5 4
x t
y t
z t
= +
= + = +
0,25
Gi I l trung im ca on thng AB. Ta c I(-2;2;1)V mt cu cn tm tip xc vi (P) nn bn knh R = d(I,(P))= 3
0,25
Phng trnh mt cu (S) cn tm l: 2 2 2( 2) ( 2) ( 1) 9x y z+ + + = 0,25
Cu 6 Cho hnh hp ng ABCD.ABCD c y l hnh thoi cnh a, gc 060ACB= , mt phng(ABD) to vi y mt gc 060 . Tnh theo athtch khi hp v khong cch gia hai ngthng CD, BD.
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1,0
Tnh thtch:
T# 060ACB = suy ra ABC u suy ra AC = a2
0 3. .sin 602
ABCD
aS AC CB = =
Gi O = ACBD. T#githit suy ra gc gia (ABD) vi mt
y l 0' 60A OA = O
A' B'
D'C'
A
B
D C
H
0,25
0 3' . tan 60 2
aA A OA = = Suy ra
33
. ' 4ABCD
aV S A A= = 0,25
Khong cch gia hai ng thng CD, BD
Trong AAO hAH AO. Do ( ' )'
BD ACBD A AO BD AH
BD A A
T# suy ra ( ' )AH A BD . Ta c CD//AB '/ /( ' )CD A BD
( ', ) ( ,( ' )) ( ,( ' ))d CD BD d C A BD d A A BD AH = = =
0,25
Trong AHO vung ti H c 03
.sin604
aAH OA= = . Vy
3( ', )
4
ad CD BD = 0,25
Cu 7Phna)
0.5
Cho
2
sin 3 = vi 2
< < . Tnh
2
cos 3A
= +
Ta c2 2 5 5cos 1 sin cos
9 3 = = = (v
2
< < nn cos 0 < ) 0,25
2 2 5 2 3cos .cos sin .sin
3 3 6A
= = 0,25
Cu 7Phnb)
0.5
Gii bng chuyn VTV Cup gm 12 i bng tham d, trong c 9 i nc ngoi v 3 i caVit Nam. Ban tchc cho bc thm ngu nhin chia thnh 3 bng A, B, C mi bng 4 i.Tnh xc sut 3 i bng ca Vit Nam ba bng khc nhau.Tnh scch chn 3 bng, mi bng 4 i:
B1) 12 i chn 4:4
12C B2) 8 i cn li chn 4: 48C
B3) 4 i cn li chn 4: 1Scch chn l: ( )4 4 4 412 8 12 8C C n C C =
0,25
Gi A l bin c Chn 3 bng, mi bng 4 i trong c ng 1 i Vit Nam.Tnh n(A):
B1) Chn 1 trong 3 i Vit Nam: c 3 cch, ri chn 3 trong 9 i nc ngoi: c3
9C3
93.C cch
B2) cn li 8 i (6 i nc ngoi v 2 i VN): Chn 1 trong 2 i VN: 2 cch, ri
chn 3 trong 6 i nc ngoi: 3 36 62.C C cch
B3) cn li 4 i (3 nc ngoi v 1 VN): c 1 cch
Scch chn l: ( )3 3 3 39 6 9 63 2 3 2C C n A C C = ( )3 3
9 6
4 4
12 8
6 16
55
C CP A
C C = =
0,25
Cu 8
1,0
Trong mt phng vi hta Oxy, cho tam gic nhn ABC ni tip ng trn (C): 2 2 25x y+ = ,ng thng AC i qua im K(2;1). Gi M, N ln l!t l chn ng cao k"t#$nh B v C. Tmta cc $nh ca ABC bit phng trnh ng thng MN l 4 3 10 0x y + = v im A c
honh m.
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T#githit suy ra tgic MNBC ni tip ng trn. Suy ra
ABC AMN= (1) (cng b vi NMC).Gi D l giao im thhai ca AO vi ng trn (C). Khi ABC ADC= (2)
T#(1) v (2) suy ra ADC AMN= .Mt khc
0 090 90ADC DAC DAC AMN OA MN+ = + = D
N
MO
A
B C
0,25
Khi phng trnh OA l 3 4 0x y+ =
Ta A l nghim ca hPT2 2
3 4 0
25
x y
x y
+ =
+ =A(-4;3) hoc A(4;-3) (loi)
0,25
Khi AC i qua A(-4;3) v K(2;1) nn c PT: 3 5 0x y+ =
Ta C l nghim ca hPT2 2
3 5 0 ( 4;3) A
C(5;0)25
x y C
x y
+ =
+ =
Ta M l nghim ca hPT3 5 0
( 1; 2)4 3 10 0
x yM
x y
+ =
+ =
0,25
Phng trnh BM: 3 5 0x y + =
Ta B l nghim ca hPT2 2
3 5 0 ( 3; 4)
(0;5)25
x y B
Bx y
+ =
+ =
Th)li ta thy A(4;3), B(0;5), C(5;0) loi v gc B tVy A(4;3), B(3;4), C(5;0)
0,25
Cu 9
1,0
Gii phng trnh 2 21 2 9 18 14 33+ + = + +x x x x x (1)
&K:3
11
x
x
PT ( ) ( )2 21 2 9 18 14 33 1x x x x x x + = + +
(2)
& r*ng hai phng trnh 2 9 18 0x x x + + = v ( )2 14 33 1 0x x x + + + = v
nghim nn nhn lin h!p hai vca (2) ta c:
( ) ( )
( )2 218 2 16 2
9 18 14 33 1
x x
x x x x x x
=
+ + + + +
0,25
( )2 2
2
9 8(3)
9 18 14 33 1
x
x x x x x x
=
= + + + + +
Pt (3)
2 2
8 9 18 9 14 33 9x x x x x + + = + (4)
0,25
Kt h!p (1) v (4) ta c h2 2
2
2 2
8 9 18 9 14 33 95 14 33 3 13
2 9 18 14 33 1
x x x x xx x x
x x x x x
+ + = + + =
+ + =
0,25
2
1317 5 5
32
17 41 0
xx
x x
+
= + =
Th)li thy tha mn
Vy PT cho c 2 nghimx= 2 v17 5 5
2x
+=
0,25
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Cu10
1,0
Chox, y, z l cc sthc tha mn 2 2 2 25 2 2 8 4 5 4 2x xy y x xz z x y z+ + + + + = + + (*) v
[0;5]x . Tm gi trln nht, gi trnhnht ca biu thc 2 21 10P z xy x z xy= + + + .
Vi mix, y, zta c2 2 2 2 25 2 2 (2 ) ( ) (2 ) 2 2x xy y x y x y x y x y x y+ + = + + + = + +
2 2 2 2 28 4 5 4(x z) ( 2 ) 4(x z) 2 x z 2(x z)x xz z z x+ + = + + + = + +
Suy ra 4 2VT x y z + +
Gt Du = xy ra 2
0
x y
z x
x
=
=
0,25
Thay vo biu thc ta c 2 24 21 3 10 ( )P x x x x f x= + + + + = lin t(c trn [0;5]
C2 2
2 3 2'( )
4 21 2 3 10
x xf x
x x x x
=
+ + + +
0,25
1'( ) 0
3f x x= = 0,25
Ta c (0) 21 10f = ; 1 23
f = ; (5) 4f =
Vy max 4P = khi 5; 10x y z= = = ; min 2P = khi1 2
;3 3
x y z= = =
0,25
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Cu 1 (2,0 im) .
a ) Kho sthm s v v th ( C ) ca hm s3 23 1
2 2y x x
b) Tm ta ca im M trn ( C ) sao cho tip tuyn ca ( C ) ti M song song ving thng ( d ) : x6 4 0y Cu 2 (1,0 im).
a). Cho hm s 2( 1)xy e x x .Tnh1
'(ln )2
y
b) Gii bt phng trnh sau3 1
3
2 log (4x 3) log (2x 3) 2 .
Cu 3 (1,0 im).Tnh tch phn0
(2 1) sinI x xdx
.
Cu 4(1,0 im). Trong khng gian Oxyz,cho mp(P) v mt cu (S) ln lt c phngtrnh ( ) : 2 2 1 0P x y z v 2 2 2( ) : 4 6 6 17 0S x y z x y z .Chng
minh mt cu(S) ct mt phng(P) .Tm ta tm v bn knh ca ng trn giaotuyn ca mt cu v mt phng.
Cu 5(1,0 im).
a)Cho tan 3 . Tnh3 3
3sin 2 cos
5sin 4 cosA
.
b)Cho a gic u 20 nh ni tip ng trn tm O. Chn ngu nhin 4 nh ca agic .Tnh xc sut sao cho 4 nh c chn l 4 nh ca mt hnh ch nht.
Cu 6(1,0 im). Cho hnh chp S.ABCc mt phng (SAC) vung gc vi mt phng(ABC), , 2SA AB a AC a v 090 .ASC ABC Tnh th tch khichp S.ABCvcosin ca gc gia hai mt phng (SAB), (SBC).Cu 7 (1,0 im). Trong mt phng vi h trc ta Oxy, cho hnh bnh hnh ABCD
c 0135B AD , trc tm tam gic ABD l H(-1;0).ng thng i qua D v H c
phng trnh 3 1 0x y . Tm ta cc nh ca hnh bnh hnh bit im G(5
;23
)
l trng tm tam gic ADC.Cu 8 (1,0 im).Gii h phng trnh sau
3 3 2
33 3 6 4 0
( 2 3 7 13 ) 3( 1)x y y x yy x y x
Cu 9 (1,0 im).Cho x, , 0y z v 2 2 25( ) 9( 2 x)x y z xy yz z .
Tm gi tr ln nht ca biu thc2 2 3
1
( )
xP
y z x y z
.
TRNG THPT CHUYNNGUYN HU
KHO ST CHT LNG LN 2MN TON NM HC 2015-2016
Thi gian lm bi: 180 pht,khng k thi gian pht http://dethithu.net
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TRNG THPT CHUYNNGUYN HU
KHO ST CHT LNGLN TH 2
NM HC 2015 2016P N V BIU IM MN: TON
Cu 1 Ni dung 1,01a.
(1,0
im)
Hm s 3 23 1
2 2y x x
TX: D = R S bin thin:
- Chiu bin thin: 2' 3 3y x x ,0
' 01
xy
x
0,25
Hm s ng bin trn cc khong ( ;0) v (1;+ ) , nghch bin trn khong (0;1)
- Cc tr: Hm s t cc i ti 1
0; 2Cx y , t cc tiu ti x 1, 0
CTy .
- Gii hn:
lim ; limx x
y y
0,25
- Bng bi n thin:
x 0 1
y + 0 0 +
y
1
2
0
0,25
th:
y
x
1
2
O 1
0,25
1b.
(1 )+ ng thng 6xy4=0 c h s gc bng 6+Gi M0( x0; y0) l im m ti tip tuyn song song ng thng6x - y- 4=0 0'( ) 6f x
0,25
0,25http://dethithu.net
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2
0 0
0
0
3 3 6
1
2
x x
x
x
+Vi x0=2 y0= 5/2M0( 2; 5/2)
x0= -1y0= -2 M0( -1 ; -2 )+ Kim tra liM0( 2,5/2) tip tuyn ti M0c pt l y= 6(x2)+5/2( nhn)M0(-1;-2)tip tuyn ti M0c pt l 6( 1) 2y x =6x+4(nhn)
0,25
0,25
Cu
2(1,0
im)2a(0,5
im)
TX: D=R2 2 2
2
( ) ( 1) ( 1) ( 1) (2 1)
( 3 )
x x x x
x
y e x x e x x e x x e x
e x x
21'(ln ) 2( ln 2 3 ln 2)2
y
0,25
0,25
2b(0,
5
im)
iu kin3
4x
Bt phng trnh tng ng2
3
(4x 3)log 2
2x 3
216x 42x 18 0
33
8x
Kt hp iu kin ta c tp nghim ca bt phng trnh l S= 3
;34
0,25
0,25
Cu
3(1)
t2 1 2.
sin cos
u x du dx
dv xdx v x
0,25
0 0(2 1)cos ( 2 cos )I x x x dx
0,25
= 0(2 1) 1 2 sin x
0,25
=2 2 0,25
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Cu
4(1,0
)
Mt cu (S) c tmI(2;3;3), bn knh 2 2 22 ( 3) ( 3) 17 5R 0,25
Khong cch t tm I n mp(P):
2 2 2
2 2( 3) 2( 3) 1
( ,( )) 11 ( 2) 2d d I P R
V ( ,( ))d I P R nn (P) ct mt cu (S) theo giao tuyn l ng trn (C)
0,25
Gi dl ng thng qua tmIca mtcu v vung gc mp(P) th dc vtcp
(1; 2;2)u
nn c PTTS
2
: 3 2
3 2
x t
d y t
z t
(*). Thay (*) vo pt mt phng (P)
ta c1
(2 ) 2( 3 2 ) 2( 3 2 ) 1 0 9 3 0 3t t t t t
Vy, ng trn (C) c tm5 7 11
; ;3 3 3
H
0,25
Bn knh 2 2 5 1 2r R d 0,25
Cu 5
a(0,5
im) 3 3 2 33sin 2 cos 3 tan 2
5sin 4 cos cos 5 tan 4A
0,25
2
3
3 tan 2 701 tan5 tan 4 139
0,25
Cu
5b(0,
5)
-C 10 ng knh ca ng trn c ni bi 2 nh ca a gic u. - Mt hnh ch nht c 4 nh l nh ca a gic c to bi 2 ng knh ni trn.
-S cch chn 4 nh ca a gic l: 420
4845C
-S cch chn 4 nh ca a gic to thnh hnh ch nht l 210
45C
-Xc sut cn tm l : P= 45 3
4845 323
0,25
0,25http://dethithu.net
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A
S
C
B
M
H
Cu
6(1,0
)
+ K SH vung gc AC (HAC) SH (ABC)
3
3, ,2
aSC BC a SH
2 3
2ABC
aS
3
.
1.
3 4S ABC ABC
aV S SH
0,25
0,25
Gi M l trung im SB v l gc gia hai mt phng (SAB) v (SBC).
Ta c: SA = AB = a, SC BC a 3
AM SB v CM SB
cos cosAMC
0,25
+ SAC = BAC3 6
2 2
a aSH BH SB
AM l trung tuyn SAB nn:2 2 2 2
2 2 2 10
4 16
AS AB SB aAM
10
4
aAM
Tng t:42
4
aCM
2 2 2AM CM AC 105cosAMC
2.AM.CM 35
Vy:105
cos35
0,25
Cu
7(1,0
)
Ta c 0 0180 45B AD BH D BH D .
Gi ( ; )n a b
2 2( 0)a b l VTPT ca ngthng HB
Do ngthng HB to vi ng thng HD gc 045 nn
cos 0 2 2
2 2
2345 2a 3a 2 0
2a. 10
a ba bb b
ba b
0,25
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Nu a=-2b. Chn a=2,b=-1. Phng trnh ng thng HB: 2x-y+2=0B(b;2b+2), D(3d-1;d)
Do G l trng tm tam gic ADC nn BG=2GD 1
2 D1
bGB G
d
B(1;4),
D(2;1)
Phng trnh ng thng AB: 3x+y-7=0; phng trnh ng thng AD:x+2y-4=0Suy ra A(2;1)(loi)Nu b=2a. Phng trnh HB: x+2y+1=0
B(-2b-1;b), D(3d-1;d)2
2 D2
bGB G
d
B(-5;2), D(5;2)
Phng trnh AB: 3x+y+13=0; Phng trnhAD:2x-y-8=0. Suy ra A(-1;-10)
Do ABCD l hnh bnh hnh suy ra DA BC
suy ra C(1;14)
Th li: cos DAB =cos ( ; )AB AD
= 01
D 452
BA (LOI)
0,25
0,25
0,25
Cu
8(1,0
)
iu kin3
x2
T phng trnh (1) ta c 3 3x 3 ( 1) 3( 1)x y y
Xt hm s3
2
( ) 3
'( ) 3 3
f t t t
f t t
'( ) 0f t vi mi t suy ra hm s f(t) ng bin trn R.( ) ( 1) 1f x f y x y
0,25
0,25
Th x=y+1 vo phng trnh (2) ta c:3
( 1)( 2x 3 7x 6) 3( 1) (3)x x Ta c x=1 khng l nghim phng trnh.T
3 3( 1)( 2x 3 7x 6 )1
x
x
Xt hm s 3 3( 1)
( ) ( 2x 3 7x 6)1
xg x
x
TX: 3
; \ 12
D
223
1 7 6'( )
( 1)2x 3 3 (7x 6)g x
x
3'( ) 0 ; 1
2g x x x ,
3'( )
2g khng xc nh.
Hm s ng bin trn tng khong3
( ;1)2
v (1; ) .
Ta c g(-1)=0; g(3)=0. T phng trnh g(x)=0 c ng hai nghim x= -1 v x=3.Vy h phng trnh c hai nghim (-1;-2) v (3;2)
0,25
0,25
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Cu 9t y+z=t (t>0);
2 22 2 ;
2 4
t ty z yz .
2 2 2
2 2
2 2 2
5( ) 9( 2 z)
5x 5( ) 9x( ) 285x 5 9x 7
(5x )(x 2 ) 0
2
x y z xy yz x
y z y z yzt t t
t t
x t
0,25
0,25
2 3 3
2x 1 4 1
27 27P
t t t t
Xt hm s3
4 1( )
27f t
t t vi t>0
2 4
4 1'( )
9
'( ) 0 1
0 6
f tt t
f tt
t
Lp bng bin thin t suy ra GTLN ca P bng 16 t c ti1 1
x ;3 12
y z
0,25
0,25
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TRNG THPT CHUYN BIN HA THI TH THPT QUC GIA LN 1Nm hc 2015 - 2016
Mn: TONThi gian lm bi : 180 pht, khng k thi gian pht
Cu 1 ( 1,0 im). Kho st s bin thin v v th ca hm s 3 3 1y x x .
Cu 2(1,0 im). Tm cc im cc tr ca th hm s 2
1f x x x .Cu 3(1,0 im).
a) Cho s phc z tha mn
31 4 1
1
i iz
i
. Tm modun ca s z.
b) Gii bt phng trnh 2 1 2 13 2 5.6 0x x x
Cu 4(1,0 im). Tnh tch phn 21
2lne xI dx
x
.
Cu 5(1,0 im).Trong khng gian vi h to Oxyz , cho im 4;1;3A v ng thng
1 1 3: 2 1 3x y zd . Vit phng trnh mt phng ( )P i qua A v vung gc vi ng
thng d. Tm ta im B thuc dsao cho 5AB . Cu 6(1,0 im).
a)
Gii phng trnh sin 2 1 6sin cos 2x x x b) cho mng ngy 26/03, trng t chc cm tri. Lp 10 A 19 hc sinh nam, 16 hc
sinh n. Gio vin cn chn 5 hc sinh trang tr tri. Tnh xc sut trong 5 hc sinhc chn c t nht 1 hc sinh n bit rng hc sinh no trong lp cng c kh nngtrang tr tri.
Cu 7(1,0 im). Cho hnh chp .S ABCD c y l hnh thang vung ti A v B. Cc mtbn (SAB) v (SAD) cng vung gc vi mt phng y. Cho AB = 2a, AD > a. SA = BC = a,
2 5CD a . Gi H l im nm trn on AD sao cho AH = a . Tnh th tch ca khi chpSABCD v khong cch gia hai ng thng BH v SC theo a.Cu 8(1,0 im). Trong mt phng vi h to Oxy cho tam gic ABC c AC=2AB, im
91;
2M
l trung im ca BC, D l im thuc cnh BC sao cho BAD CAM . Gi E l trung
im ca AC, ng thng DE c phng trnh: 2 11 44 0x y , im B thuc ng thng dc phng trnh: x + y 6 = 0. Tm ta 3 im A, B, C bit honh ca im A l mt snguyn.
Cu 9(1,0 im). Gii h phng trnh 2 2 2 2
2 2
2 5 2 4
3 2 2 9 0
xy y y xy y y xy
y x x x x y
Cu 10(1,0 im).Cho cc s , ,a b c khng m sao cho tng hai s bt k u dng. Chng
minh rng:9
6a b c ab bc ca
b c a c a b a b c
.
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P N THI TH THPT QUC GIA NM 2015-2016 LN 1
MN : TON
Ni dung imCu 1(1,0 im)
3 3 1x x TX: D R
2' 3 3y x , ' 0 1y x
0.25
limx
y
, limx
y
* Bng bin thinX -1 1 + y - 0 + 0 -
y+ 3
-1 -
0.25
Hm s nghch bin trn cc khong ; 1 v 1; , ng bin trn khong 1;1
Hm s t cc i ti 1 , 3CDy , t cc tiu ti 1x , 1CTy
0.25
th: th hm s i qua cc im (0;1), (-2;3), (2; -1). 0.25
Cu 2 (1,0 im)
:TXD
2
2 1'
2 1
xf x
x x
0.25
2
2
4
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1
' 02
f x x
0,25
Bng bin thinX 1/2 + y 0 +
y
3
2
0.25
th hm s c im cc tiu l1 3
;2 2
. th hm s khng c im cc i.0.25
Cu 3( 1,0 im)
Cu 3a) (0,5 im)
3 2 3
2
1 4 1 1 4 1 3 3 ( 1 2 )(1 )
1 1 (1 )(1 )
1 2 2 1 3
2 2
i i i i i i i iz i i i i
i i i i
0,25
2 21 3 10
2 2 2z
0,25
Cu 3b (0,5 im)
b)2
3 3 1 33 2 5 0 2
2 2 3 2
x x x
bpt
0,25
3 1
2 3
x
(lun ng)
3
2
32 log 2
2
x
x
.
Vy tp nghim ca bt phng trnh l 32
, log 2
0,25
Cu 4( 1,0 im)
2 2 211 1 1 1
1 ln ln ln2 ln 2 1 2e e e e
e x xI dx dx x dx dxx x x x
0.25
Tnh2
1
lneJ dx
x
t2
1ln ,u x dv dx . Khi
1 1,du dx v
x
0.25
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Do 2
1 1
1 1ln
e e
J x dxx x
1
1 1 21
e
Je x e
0.25
Vy4
1I
e
0.25
Cu 5( 1,0 im)
ng thng d c VTCP l 2;1;3du
V P d nn P nhn 2;1;3du
lm VTPT
0,25
Vy PT mt phng P l : 2 4 1 1 3 3 0x y z
2 3 18 0x y z
0,25
V B d nn 1 2 ;1 ; 3 3B t t t
5AB 2 22 25 3 2 6 3 5AB t t t 27 24 20 0t t
0,25
210
7
tt
Vy 5;3;3B hoc 27 17 9; ;7 7 7
B
0,25
Cu 6 ( 1,0 im)
Cu 6a(0,5 im)sin 2 1 6sin cos 2x x x
(sin 2 6sin ) (1 cos2 ) 0x x x 22sin cos 3 2sin 0x x x
2sin cos 3 sin 0x x x
0.25
sin 0
sin cos 3( )
x
x x Vn
x k . Vy nghim ca PT l ,x k k Z
0. 25
Cu 6b(0,5 im)
Gi A l bin c :trong 5 hc sinh c chn c t nht mt hc sinh n
535n C
0.25
S cch chn 5 hc sinh trong c t nht 1 hc sinh n l n(A)=5 535 19C C
Do xc sut 5 hc sinh c chn trong c t nht 1 hc sinh n l5 535 19
535
0,96C C
PC
0,25
Cu 7(1,0 im)
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Do (SAB) v (SAD) cng vung gc vi ynn ( )SA ABCD . AHCB l hnh bnh hnh,suy ra CH=AB=2a,
2 2 4 5 .HD CD CH a AD a
0.25
2
3
15 2 6
21
. 2 .3
ABCD
SABCD ABCD
S a a a a
V SA S a
0.25
Trong mt phng (ABCD), k CE//BH (E thuc AD), ta c
, , , ,
1
2BH SC BH SCE H SCE A SCE d d d d
K
,
,
A SCE
AF CE AJ SF AJ SCE
d AJ
0.25
Gi K l giao im ca BH v A F.
2 2 2
2 2 2
, ,
1 1 1 2 4
5 5
1 1 1 421
1 2
2 21BH SC A SCE
a aAK AF
AK AH AB
aAJAJ AS AF
ad d
0,25
Cu 8(1,0 im)Gi I l giao im ca BE v AD, G l giaoim ca AM v BE. ( . . )ABI AEG g c g suy ra BI=GE m BG=2GE( do G l trng tmca tam gic ABC) suy ra BI=IG=GE.
K EH //BC ( H thuc on AD). Chng minhc CD=2HE, HE=2BD, suy ra CB=5BD
0,25
K
B C
D
S
H E
F
J
G
E
B CD
H
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92 5 , ( ;6 ), (22 11 ;2 ), 1;
2
9 11 18;55 3 108
5 5 510 3 27
3 3;3
BM BD B b b D d d M
d Dd b
d bb B
0,25
91;2
M
l trung im ca BC suy ra C(-1;6). 0,25
Gi E(22-11e;2e), E l trung im ca AC suy ra A(45-22e; 4e-6)
2
2( ) (1;2)2 75 278 256 0 128
( )75
e tm AAC AB e e
e l
Vy A(1;2), B(3;3), C(-1;6).
0,25
Cu 9(1,0 im).
: 4 2 0DK y x y
Vi y=0 th x=0.
2 2 2 22
0, (1) 2 5 2 4 0
2 5 1 2 4 0
y x xy y y xy y y xy
x x x x
y y y y
0.25
t 2;4x
t ty
22 5 1 2 4 0 2 3 2 2 1 1 4 0
3 2 32 3 0 3 3
2 1 1 4
t t t t t t t t t
t t tt t t x y
t t
0,25
Thay 3x y , thay vo (2) ta c :
2 2
2
2 2 0
1 2 1 2
x x x x x x
x x x x
Xt hm s 2
2 2
21 2 , ' 1 2 0
2
tf t t t f t t t
t
0.25
0 0
11
3
x yf x f x x x
x y
0.25
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Vy h phng trnh c nghim 1
0;0 , 1;3
Cu 10 (1,0 im) .
t9a b c ab bc ca
Pb c a c a b a b c
Gi s a b c , khi . .ab ac b b c c b ca c a b b c c b
0,25
Suy rab c b c
a c a b a
.
0,25
t t b c th9a t at
Pt a a t
.0,25
Ta c9 9
6a t at a t at
t a a t a t at
(AM-GM). Do 6P (pcm).
ng thc xy ra khi 3a t at v chng hn mt b ( , , )a b c tha mn l
7 3 5( ; ; ) ;1;0
2a b c
0,25
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ruy cp http://dethithu.netthng xuyn cp nht nhiu Thi ThTHPT Quc Gu n thi THPT Quc Gia cc mn Ton, L, Ha, Anh, Vninh , S, a c DeThiThu.Net cp nht hng ngy phc vst!
ike Fanpage Thi ThTHPT Quc Gia - Ti Liu n Thi:ttp://facebook.com/dethithu.netcp nht nhiu thi thv ti liu n thi hn
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S GD&T BC GIANG THI TH THPT QUC GIA LN 3TRNG THPT NG S LIN Mn: Ton 12
- Thi gian: 120 pht, khng k thi gian pht CHNH THC ( gm01trang)
Cu 1 (2,0 im).Cho hm s y=2x + m
x 1 (C)(vi ml tham s thc).
1)Kho st s bin thin v v th hm s (C)vi m= 1.2)Tm m ng thng d : y = x+ 2ct th hm s(C)ti hai imA, Bphn bit.
Cu 2 (1,5 im). 1)Gii phng trnh: sinx + 2sin3x= sin5x
2)Gii phng trnh: log3(x 1)2 + log3(2x 1) = 2
Cu 3 (1,0 im).Tm nguyn hm sau: I=
(x+cosx)xdx
Cu 4 (1,0 im). Trong khng gian Oxyz, cho 4 im A(1;1;3), B(2;0;4), C(2;3;5),D(0;4;5Chng minh rng 4 im cho khng ng phng v tnh th tch ca t din ABCD.
Cu 5 (1,0 im).Cho hnh chpS.ABCDc y ABCDl hnh ch nht AB =a, AD=a
2. Hnh chiu vung gc ca nh Sln (ABCD)trng vi trng tm ca tam gic ABC, ngthng SD to vi mt y ABCDmt gc 450. Tnh th tch ca hnh chp S.ABCDv tnhkhong cch gia hai ng thng SCvBD theo a.
Cu 6 (1,0 im). chun b cho L k nim 70nm thnh lp ca mt trng THPT, nhtrng cn lp mt i tnh nguyn vin gm 40 em hc sinh thng qua n ng k. Qua ngk c 150 em hc sinh mun tham gia i tnh nguyn vin, bit rng trong 150 em c 60 emc hc lc gii. m bo cng bng nh trng quyt nh chn ngu nhin 40 hc sinh t 150hc sinh ni trn. Tnh xc sut trong s 40 em hc sinh c chn c ng 80%hc sinh chc lc gii.
Cu 7 (1,0 im).Trong mt phng Oxy, cho hnh ch nht ABCDc AB = 2BC. Gi Hl hnh chiu ca Aln ng thng BD. Gi E, F ln lt l trung im ca cc on thngCDvBH. Bit im A(1;1), phng trnh ng thngEF : 3x y 10 = 0v im Ec tung m. Tm ta cc nh B,C,Dca hnh ch nht.
Cu 8 (1,0 im).Gii h phng trnh sau:(x +
x2 + 1)(y+
y2 + 1) = 1
3x2 + y+ 3 =
3x + 1 +
4 5y, (x, y R)
,Cu 9 (0,5 im).Cho cc s a,b, cl cc s thc dng v tha mn: a + b+ c= 3. Tm gi
tr ln nht ca biu thc:
P = 2
3 + ab + bc + ca+ 3
abc
(1 + a)(1 + b)(1 + c)
Ht Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm.H v tn th sinh: . . . . . . . . . . . . . . . . . . . . . . . ;S bo danh: . . . . . . . . . . . . . . . . . . . .
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S GD&T BC GIANG P N - THANG IMTRNG THPT NG S LIN THI TH THPT QUC GIA LN 3
NM HC 2015 - 2016 CHNH THC Mn: Ton 12
(p n gm 6 trang)
CU NI DUNG IM
Cho hm s y=2x + m
x 1 ... (1) 1,00
Tp xc nh: D= R\{1}
Gii hn: limx1
y= , limx1+
y= +, limx
y= 2, limx+
y= 2.
th hm s c TC:x = 1, TCN: y= 2
0,25
Cu 1
S bin thin: y = 3(x 1)2 ; y
< 0,x D.
Hm s nghch bin trn cc khong (; 1)v(1;+).
0,25
- Bng bin thin: 0,25
th: th ct trc Oyti im (0;1), ct trc Oxti im(12
; 0) 0,25
2)Tm m ... 1,00
Phng trnh honh giao im: 2x + m
x 1 =x + 2(iu kin: x = 1) (2)(2) x2 x 2 m= 0 (3)
0,5
1
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ng thng dct th(C)ti hai im A, Bphn bit khi v ch khi phng trnh(3) c hai nghim phn bit khc 1.
> 0
1 1 2 m = 0
m > 94
m = 2
0,5
1) Gii phng trnh: sinx+2sin3x= sin5x (1) 0,5(1) 2sin3xcos2x + 2sin3x= 0 2sin3x(cos2x + 1)= 0 0,25Gii phng trnh tch ta tm c nghimx= k
3 vx=
2+ l, (k, l Z)
0,25
Cu 2 2) Gii phng trnh: log3(x 2)2 + log3(2x 1) = 2 1,0
iu kin:
x 1 = 02x 1> 0
x = 1x >
1
2
0,25
PT log3|x 1| + log
3(2x 1) = 1
log3[|x 1| (2x 1)] = 1 |x 1| (2x 1) = 3 ()
0,25
Vi 12
< x 1. PT (*) (x 1)(2x 1) = 3 2x2 3x 2 = 0 x= 1
2, x= 2
i chiu iu kin ta c nghim x = 2
0,25
Tm nguyn hm: I=
(x+cosx)xdx 1,00
Ta c: I=
x2dx +
xcosxdx= I1+ I2 0,25
Cu 3
Tnh I1=
x2dx= x3
3 + C1 0,25
Tnh I2=
xcosxdx. Bng cch s dng TPTP ta tnh c
I2= xsinx+cosx + C2
0,25
Kt lun: I= x3
3 + xsinx+cosx + C 0,25
Trong khng gian Oxyz, cho 4 im A(1;1;3), B(2;0;4), C(2;3;5), D(0;4;5).Chng minh rng 4 im cho khng ng phng v tnh th tch ca t dinABCD.
1,00
Ta c:AB= (3;1;1),AC= (1;4;2),AD = (1;3;8) 0,25
2
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Cu 4 Tnh:
AB,AC
= (2; 7;13) =
AB,
AC
AD= 85 = 0.
Suy ra 4 im A, B,C,D khng ng phng.0,5
Suy ra: VABCD =85
6 (vtt) 0,25
Cho hnh chp S.ABCD c yABCD l hnh ch nhtAB = a,AD = a
2. Hnh
chiu vung gc ca nhSln(ABCD)trng vi trng tm ca tam gicABC, ngthngS Dto vi mt yABCDmt gc450. Tnh th tch ca hnh chp S.ABCD
v tnh khong cch gia hai ng thngSCvBD theo a.
1,00
S
A D
CB
E
F I
HO
K
Cu 5 - GiHl trng tm ca ABC, nn theo gi thit:S H (ABCD)v SDH= 450.Suy ra: SH=H D=
2
3
BD =2a
3
3-ABCDl hnh ch nht nn: SABCD = a22
0,25
- Th tch ca khi chp: V =1
3SH.SABCD=
2a3
6
9 (vtt). 0,25
- Gi El im i xng vi A qua B, ta c:
BD EC
BD (SC E)SC (SC E)
d(BD,SC) =d(BD, (SC E)) =d(H, (SC E)) (1)- Gi I , Kln lt l hnh chiu vung gc ca Hln EC,SIta c:
EC SHEC HI
HK ECKH SI
HK (SEC) d(H, (SE C)) =H K (2)
0,25
- Gi Fl hnh chiu ca B ln E C, ta c:B F =H Iv 1
BF2 =
1
BE2+
1
BC2, trong
HSC: 1HK2
= 1
HS2+
1
HI2 =
1
HS2+
1
BE2+
1
BC2
HK=2a3
(3)
T (1), (2), (3) ta suy ra: d(BD,SC) =2a
3
0,25
3
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Cu 6 chun b cho L k nim70nm thnh lp ca mt trng THPT, nh trng cnlp mt i tnh nguyn vin gm 40 em hc sinh thng qua n ng k. Qua ng k
c 150 em hc sinh mun tham gia i tnh nguyn vin, bit rng trong 150 em c
60 em c hc lc gii. m bo cng bng nh trng quyt nh chn ngu nhin
40 hc sinh t 150 hc sinh ni trn. Tnh xc sut trong s 40 em hc sinh c
chn c ng80% hc sinh c hc lc gii.
1,00
GiAl bin c ngu nhin: "Chn c80% hc sinh c hc lc gii".Chn ngu nhin 40 em hc sinh t 150 em hc sinh c n() =C40150
0,25
Chn c80%hc sinh c hc lc gii, tc l chn c 32 em. Chn 32 em trong 60em c hc lc gii c C3260 cch.Chn 8 em cn li trong 90 em c C890cch.Khi : n(A) =C890C3260
0,5
Theo cng thc tnh xc sut, ta tnh c: P(A) = C890C
3260
C40150 1, 823.109 0,25
Cu 7 Trong mt phngOxy, cho hnh ch nhtABCDc AB = 2BC. GiH l hnh chiuca A ln ng thngBD. GiE, F ln lt l trung im ca cc on thngCD
vB H. Bit imA(1;1), phng trnh ng thngE F : 3x y 10 = 0v imEc tung m. Tm ta cc nhB , C, D ca hnh ch nht.
1,00
A G B
CE
F
H
D
GiE , F , Gln lt l trung im ca cc onCD, BH, AB. Ta chng minh: AFEF.
Ta thy cc t gic ADEG, ADF Gni tip, nn t gic ADEFcng ni tip, do :AF EF. T ta suy ra: AF :x + 3y 4 = 0
0,25
- Tm ta ca F =AF EF, ta c F(175
;1
5), suy ra: AF =
32
5.
- Ta cAF E DC B EF =12
AF = 2
2
5.
- Gi E(a; 3a 10) EF EF2
= (a 17
5)2
+ (3a 51
5)2
=
8
5
5a2 34a + 57 = 0 a= 3 t= 195
. Hay E(3;1)hoc E(195
;7
5)
0,25
4
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Theo gi thit ta c E(3;1), phng trnh: AE : x+ y 2 = 0. Gi D(x; y),ADEvung, cn ti D nn:
AD= DE
AD DE
(x 1)2 + (y 1)2 = (x 3)2 + (y+ 1)2
(x 1)(x 3) = (y 1)(y+ 1)
y = x 2(x 1)(x 3) = 0
x= 1y= 1
x= 3y= 1 .
T D(1;1)hoc D(3;1)
0,25
V D, Fnm v hai pha so vi ng thng AEnn D(1;1).Khi : C(5;1), B(1;5). Vy B (1;5), C(5;1), D(1;1)
0,25
Cu 8 Gii h:
(x +
x2 + 1)(y+
y2 + 1) = 1 (1)
3x2 + y+ 3 =
3x + 1 +
4 5y (2), (x, y R)
1,00
- iu kin:
x 13
y45
Vix, y R, ta c:
(x +
x2 + 1)(x +x2 + 1) = 1(y+
y2 + 1)(y+ y2 + 1) = 1
.
Kt hp vi phng trnh (1) ta c:
y+
y2 + 1 = x +x2 + 1x +
x2 + 1 = y+ y2 + 1
. Cng hai v ca h ny ta c x = y
0,25
Thx=
y vo phng trnh (2) ta c:
3x2 x + 3 = 3x + 1 + 5x + 4 ()Ta c:() 3x2 x + 3 3x + 1 5x + 4 = 0 3x2 3x + (x + 1 3x + 1) + (x + 2 5x + 4) = 0 3(x2 x) + x
2 xx + 1 +
3x + 1
+ x2 x
x + 2 +
5x + 4= 0
(x2 x)(3 + 1x + 1 +
3x + 1
+ 1
x + 2 +
5x + 4) = 0
0,50
Vx 13
nn3 + 1
x + 1 +
3x + 1+
1
x + 2 +
5x + 4 >0.
Do vy:x2 x= 0 x= 0 x= 1.Kt lun h c hai nghim: (x; y) = (0;0), (1;1)
0,25
Cu 9 Cho cc sa,b, cl cc s thc dng v tha mn: a + b + c= 3. Tm gi tr ln nht
ca biu thc:
P = 2
3 + ab + bc + ca+ 3
abc
(1 + a)(1 + b)(1 + c)
0,5
5
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Ta c: (ab + bc + ca)2 3abc(a + b + c) = 9abc >0 = ab + bc + ca 3abc.Chng minh c: (1 + a)(1 + b)(1 + c)(1 + 3abc)3, a,b,c >0. Du "=" xy rakhia= b = c. Khi :
P 23(1 +
abc)
+3
abc
1 + 3
abc=Q.
t 6abc= t. V a,b,c >0nn 0 < abc ( a + b + c3 )3 = 1
0,25
Xt hm s: Q = 2
3(1 + t3)+
t2
1 + t2, t (0;1] Q(t) = 2t(t 1)(t
5 1)(1 + t3)2(1 + t2)2
0, t (0; 1].
Do hm s ng bin trn (0; 1]nn P Q(1) =56
. V t ta tm c MaxP =5
6,
khia= b = c = 1
0,25
Ch : Hc sinh lm theo cch khc nu ng vn cho im ti a theo tng bi.
6
ruy cp http://dethithu.netth
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cp nht nhiu
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TRNG THPTCHUYN H LONG
THI TH THPT QUC GIA LN 2 NM 2015 - 2016Mn: TON
Thi gian lm bi: 180 pht, khng k thi gian pht
Cu 1 (1,0 im).Kho st s bin thin v v th hm s3
.2 1
xy
x
Cu 2 (1,0 im). Cho hm s 3 3 2y x x c th l ( ).C Vit phng trnh tip tuyn ca
th ( )C ti cc giao im ca n vi ng thng c phng trnh 2.y x Cu 3 (1,0 im).
a) Cho s phc z tha mn (2 3 ) 1 9 .z i z i Tm mun ca s phc 2 1.w z z
b) Gii phng trnh 2 23 3 82.x x
Cu 4 (1,0 im). Tnh tch phn1
0
2( )d .
1xI x e x
x
Cu 5 (1,0 im). Trong khng gian vi h to ,Oxyz cho ba im (1;1;1), (3;5;2), (3;1; 3).A B C Lp phng trnh ng thng di qua gc ta ,O vung gc vi mt phng ( )ABC v lp
phng trnh mt cu ( )S ngoi tip t din .OABC
Cu 6 (1,0 im).
a) Tnh gi tr biu thc 2 2sin ( ) cos ( ),4 3
A
bit3
cos =5
v3
.2
b) Chng trnh To Qun nm 2016 (Gp nhau cui nm) c mt tr chi tn l Vng quay k diudnh cho cc To tng t nh tr chi truyn hnh Chic nn k diutrn knh VTV3. Chic nnc hnh trn c chia u thnh cc hnh qut, trong c 10 c tn Tham nhng, 4 c tnTrong schv 2 c tn Phn thng. C 4 To (Kinh t, X hi, Gio dc v Tinh thn) cngtham gia tr chi ny, mi To ch c quay ngu nhin mt ln. Tnh xc sut c 4 To uquay vo Trong sch.
Cu 7 (1,0 im). Cho hnh chp .S ABCc y ABCl tam gic u cnh ,a mt bn SACl tamgic cn ti Sv nm trong mt phng vung gc vi mt phng ( ),ABC ng thng SB to vi
mt phng ( )ABC mt gc 060 , Ml trung im cnh .BCTnh theo a th tch khi chp .S ABCv khong cch gia hai ng thng , .SM AC
Cu 8 (1,0 im). Trong mt phng vi h to ,Oxy cho hnh vung ABCD c (4;6).A Gi
,M Nln lt l cc im nm trn cc cnh BCv CD sao cho 045 , ( 4;0)MAN M v ngthng MNc phng trnh 11 2 44 0.x y Tm ta cc im , , .B C D
Cu 9 (1,0 im). Gii h phng trnh2 2 2 21 97 1 97 97( )
( , ).27 8 97
x y y x x yx y
x y
Cu 10 (1,0 im). Cho , ,a b c l cc s thc dng tha mn
2
4 .2016
a b cabc
Tm gi tr ln
nht