b Vi6t phuong uinh tifo tuy6n cua dd *iI tCl t4r giao diAm cria tl6 ttri C voi tryc tung... HTTOI.IG NAN CHT]NG NiSu ttri sinh lim bdi kh6ng theo c6ch nhu itrip 6n nhrmg tlung thi vdn ch
Trang 1Ciu 1 (2,0 dtdm)Cho hhm s5, y = T x-z
a) Khio s6t sg bi6n thiOn vi vE AO tfri (C) cua him s6 dd cho.
b) Vi6t phuong uinh tifo tuy6n cua dd *iI tCl t4r giao diAm cria tl6 ttri (C) voi tryc tung.
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CAU 2 (1,0 di6m)
a)Cho g6c a thodmin:
1."<n vi'sinA=1.rrnr, A=sin 2(a+
b) Cho s6 phftc z thoi mxn hQ thric: (1- 2i)z!3(1+t)t =2+7i Tim
,,i
clur so pnuc e
Cflu 3 (0,5 didm)Gidi phuong trinh: 3.4x+I -fl.2x -29 =0 .
L7't-SO GDÐA TINH
@A thi cd 0I trang)
CAu 10 (l ,0 diem) Cho Q,b,c le
Im gratri nho nhAt cua biOu thirc:
EE THr CUOI LoP 12 THPT NAM HgC 2014 - 201s
MOn thi:TOAN
Thoi gian ldm bdi:180 phfit
7t)
phan thgc,phan do
-JY=*
CAu 4 U,o Aidd Giei he phuong uinh: I ' -
':t
[t+'*llt.r/+-$ +{ffi-l) = e.
L+
-4 Cf,u5 ( 1,0 didr r)Tinh tichphAn: I = I.(1 +sin2x)dx
0
Ceg 6 1t,O drdmlCho hinh ch6p S.AB CD c6ddy li hinh thoi,c4nh a, gOc frD=600 Fllnh chi6u vu$ng g6c cira dinh S l0n (ABCD) tA di6m I/ thuQc canh AB thoa man HB=2AH Bi6t
SH = oJd ,tinh th€ tictr kh6i ch6p S.ABD vh khoing cdch tu diAm Cd6n mlt phing (SBD) CAu ? (l,O dfdm) Trong mpt phing tqa d0 Oxy, cho hinh thang ABCD voi hai ddy lh AB vit
CO ei6t hinh tirang rO aien tictr Uing 14, dinh A(/; /) vi trung di6m cira cqnh BC'lh
"(-;t) vitit phuong trinh dusng thing er Ui6t dinh D c6 hohnh dQ duong vi D nim trOn
duong thdng d c6 phuong trinh 5x - y+ I = 0
CAu 8 (1,0 apd Trong kh6ng gian voi hq tqa dQ Oxyz, cho di€m 4(1;3;0) vd mflt phing (P)' c6 phuong trinh 2x+ 2y - z+ I = 0 Tinh khoring cdch tu di6m e d6n mflt phing (P) vd tim tga
Ag ei6m A'd5i xring voi di6m e qua m{t phing (Pl'
CAu 9 (0,5 di6d Gqi S n gfln hgp c6c sb qu nhi€n c6 6 cht sd phdn ,UiQldugc lAp rft cdc cht sd 0,
1,2,3,4,5,6 Chgn ngiu nhiOn mQt s0 thuQc S Tim x6c sudt d0 s6 du-o.c chgn lcm hon 300475.
c6c sO thuc kh6ng dm, phAn biet thoa mdn az + b2 + cz - 3
F !* l-+ L.
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Thi sinh khing dryic s* tu4ng rdi ti&{ Gicim thi kh1ng giai thich gi th€m.
Cảm ơn thầy Đào Trọng Xuân (trongxuanht@gmail.com ) đã chia sẻ đến www.laisac.page.tl
Trang 2SO GD&DTIIA TTNH Ki'THI cuOI Lop 12 THpr NAMHoc z0L4 -201s
n{!n tni:rOm nudnc nAN cnAvr rnr
(Bdn hutng ddn ndy gim 06 trang)
I HTTOI.IG NAN CHT]NG
NiSu ttri sinh lim bdi kh6ng theo c6ch nhu itrip 6n nhrmg tlung thi vdn cho thi s6 Ai6m tmg
phennhuhudng d6n.
Di6m toan bii kh6ng quy tdn.
n DAP Ax vA THANG orE*r
r
Ciu 1
?
(2.0 dihm)
3
L) (1.0 iliam)
o Tapx6cdintr: P=R\{2\
o Gi6i han ve tiem cfn:
lirn y @; lim y-*o ; lim l=2;
x+2-n ' x+zin r+<-suy ra dO thi c6 mQt tiem cAn dung h
ngang ld duong thang ! =2.
lg} !=2.
duong thang x=2 ve mOt tigm c?n
0.25
o SU bii5n thi6n:
- Chi0u biOn thiOn: Y'=- ( 0, Vx € D
(x- 2)' Him s6 nghich bii5n trOn mdi khoang (-*; 2) va Q; + @)
0.25
- Ben
x
bi€n thi€n
-@2*m
0.25
-v
2
-@
*m
2
\ D6 thi
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Trang 3b (1.0 di6m
Gqi M(0;%) li
_1
MQ;;)
giao di6m cria (C) vd tryc turg, ta c6 2.0 +l -1
!o= 0-Z= Z suY ta
0.25
H9 sd g6c cta ti€p Qyen tai M le /'(0) :+ 0.25
Phuong trintr tiOp tuyi5n cfia dO thi tai M Ld y=+(x-o)-;
0.25
- L
V_
Cflu 2
QQdiam)
Cfiu 3
(0.5 dihm)
Cfiu 4
(1.0 diem)
a (0.5 diem)
k= sw2(or+ r\-= sin(2a + 2r) = sffia-=2-sirurcosru@ -.
0.25 Tac6cos'e-l-sin2 e-1 -16 25 =L
25
Do { a < rr ndn cos q <0, k6t hqp vdi (z)ta c6
2
Thay (3) viro (1) tac6 ,{=-1.+ =!! .
55 2s
cos d 1
5
(2)
b (0,5 rli6m)
Ddt z=a+bi(a,b eR),tac6 z=a-bi
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e (4a + 5b - 2) + (a -2b - 7)i - 0
0.25
[a-2b=7 LD= -2 VOy phAn thuc ctia z li 3, phAn io ctia z ld-2
0.25
Tac6 3.4x+r-l7.zx
D?t t:2' (t > 0) -29-o<+ lz.4x -l7.zx -29-0
Phuong trintr dd cho tr& thanh I2t2 -l7t -29 =0 e
t - -1(z)
t_2
T2
0.25
Vdi f =2,tac6 2'=Lex=lc 29
VAy nghiem cria phuong trinh li: x: 1og, 29
l2
0.25
DiAu kien {i = t
Lo< y<16(*)
1-13 + vJr+1> o
Vdi di6u kiQn (*) ta c6:
0.25
2
Trang 4L- x3
do do (1) e -,tffi=n+Jy* '-*_x,
*r[ffi
.[-;*,[m) =o<+ Jy=-x
t:-,IYJY+I>o)
-x+,1 y
o(x+6)('
(do x'-*Ji +
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Vi
" =7 khdng phni h nghigm cria (3) ndn (4) e J r+q_+{lEr+e -;fr;-l = 0 0.25
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-!=!!.r,?-L-+
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Vx)-4rx7L-t-l
2J,+4 W $x$)2' \' YJv' -r)*7' 4
trOn (a;+€)rt?l
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[ = I r(t + s in}x)dx - I **+ I x sin 2xdx
4.25
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L
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-t'
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,' -8I \ .' -F K
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d (c,(,sBD)) = d (t,(sBD)) =
KC HK L BO,HM J- S/( ( K thuOc BO, M thuQc SK)
Ta c6 BO L(Srrq + BO L HM do tl6 HM L (SBD) + d(H,(SBD)) = HM (2)
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fiSt trq,p (1), (2), (3) ta c6 d(C,(SBDD:og ,,A
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K6o dei AII cAt CD tai E Do ABCD
hinh thang (ABI/CD) va H trung di6m
BC n6n OE ttr6y LruB - MnEC
* S-o, = S eaco =14
Ta c6 AE =ilH = Jii
vi phuong trintr dudrng thAng AE:
2x-3y * 1 :0.
Do dinh D c6 hoanh d0 duong va D lr^r.?
n[m tr0n cludrng thdng (d) c6 phuong
trinh 5x- y+1 =0
n6n D(d; 5d+l) vdi d > 0
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0.25
Tt ct6 D(2; 11)
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I(hoang circh fh A(I;3;0) d6n
fl= lz,t+z.z - o +tl
3
mflt phang (P) li:
0.25
4
Trang 6arct
Duong thaog AA' qua A nhan vecto ph6p tuytin cria mp(P) n iQ;Z;-t) lem
vecto chiPhuong'
rx=r+Zt
I
Ta c6 phuong tinh tham sti cria rtulng th6ne AA':{ y =3+2t
I
lz= -t
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Do I thuQc du0mg thang AA'n6n l(l+2t;3 + Zt;-t) 4.25
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2(I + 2t) + 2(3 + 2t) - (-t) + I - 0 e f 1 =) I (-1;1;1)
Vi I ld trung ttiOm cria AA'nOn ta c6 A'(-3;-!;2) 0.25 Ciu 9
(0.5 diim)
56 phen fl}cria kfiOng gran m6u li s6 phdn tu cria tap hqp S
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c6ch chgn
iai '6iiur" "t u s n a.at = 41 2)
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4.25
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ar c6 5 c6ch chen; ao c6 4 cilch cho.n; a, c6 3 cdch chQn; a, c6 2 cfuchchgn.
suy ra c63.6! sd arararanosaa)300475 md ar24 Tt12: at =3 Ta th6y sO IOO+ZS c6 2 cht s6 0 nen Rhi chgn mot sii W
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Do tt6 a, c6 6 c6ch chgn a, c6 5 c6ch cho.n; ao c6 4 cdchchgn; a, c63 c6ch cho.n; au c62 cdch chgn.
suy ra c6 6! s6
"rqr"r"r"">300475 md a, = l.
Viy c6 4.6! sO
"rtr"r"""r"" thuQc S mi tr"r"rq"rrr>300475 .
X6c suat can tim li t3 p=-2
At>3
Cfiu 10
Q,0 dihm)
0.25
0,25
Trang 7= r * 2* = *( @- v)'* r)'.
(x-y)" ( xy )
DAt @- Y)' +z= r (r > 2); ra c6
xy
suyra G-t'+ 2 -r
t-2
f'(t)=Zt- (t ' 2*
2)'
xy_
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2(t - lxrz -3r + 1)
(t -2)'
3 +.6
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2
=J
1- f(t)
z(tt -4tz +4t-t) (t -2)'
(do t>2)
c-0;arb>0
2
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0.25
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v4y F >t 1+ 5tF , c6 K-,,khi
6
3+V5
*'+y'_l+.6
2
xy
a'+b2 +c2 -3
c-0;a,b>0 e = A; a,b>0
a2 +b2 3+.6
= + e lab =+ Ta th6y hg nny tu6n c6 nghiQmphdn biQt.
g -3.,6 ab
2
=l
Vay siatri nho nhdt cria F h ll.!}E
6
a
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Cảm ơn thầy Đào Trọng Xuân (trongxuanht@gmail.com ) đã chia sẻ đến www.laisac.page.tl