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Chohinhch6pd6u S.ABCc6tASB =30o.MOtmatphingthayddiquaA citcdccanh SB vi SC lAn lrrot tai M virN.[r]
(1)sd co vA or euANc rnr
TRUoNG THPT cnuy0m r,t euf oON
Cffu 1 Ydi alh s6 thgc drrong t(uy j,,lo1,tja bHng v6i I
A.)logra B.ll1logra C.logya
22 Ciu 2 N6u
Ir(x)dx = 3 thi
[,ra)-x)dx=
00
A -1 8.2.
Cffu 3 NghiQm cria phrrong trinh 2*-s = 4li.
A.x=1 B.x=9. C.x = -1.
.1 u ,
oE rnr rHU TOr Ncrrr0p rHpr Niu zozo MON ToAN
Thdi gian ldm bdt 90 phfit, kh1ng kdthdi gian giao di (Dd thi gim c6 trang)
Ma dA thi 23s
D.2logra.
c.1 D.5.
D.x = 5
Cffu 4 COng thrlc nio du6i tldy dtng ttd tfnh thd tich V cria kh6i ch6p v6i di6n tfch ddy B vit chiOu cao h?
d,.v =f,an n.v = !an c.v - Bh o.v -Latr.
3
Cffu 5 Tiong k*r6ng gian Oxyz, hinh chi6u vudng g6c cria tlidm A(-1;2;l) tr6n trgc Oy c6 tga d0le
A (-1;2;0) B (-1;0; 1) C (0;2;0) D (0;0; 1)
Cdu 6 Tflp xric tlinh eta him s6 y = log(x - 1) le
A (-1; +oo) B (0; +oo) C (1; +m) D [1; +oo)
Ciu 7.D6 thi him s6 nio drr6i itdy c6 dang nhu dudng cong trong y hinh b6n?
A.y = -xa +2x2. c.y = xa -2*.
B.Y = x4 +2x2.
D.y = -f +2* - l. o x
Ciu 8 Trong khdng gian Oxyz, cho dudn g thilng d :
phrrong cta dlir
A.iq = (1;3; -1) B.it = (1;3; 1) C.iz = (-1;3; -1) D.it = 0;2;5). Cf,u 9 MOt kh6i l6ng trg c6 di6n tich dity bXng vh c5 thd tich bing 6 thi c5 chi6u cao blng
4.4 B 6 C.2 D.3.
x=l-t
! = 2 + 3t (r e R) M6t vec-td chi
o-\-t L_ J
(2)Ciu 10 Cho hdm s6 y = f(x) c6 bing bi6n thi6n nhu sau
H6i ham s6 de cho d6ng bi6n tr6n khoing nio du6i ddy?
A.(-3;2) B (-oo;-3) C.(2;3) D (2; +m).
CAu 11 TiQm cdn ngang cria dd thi hhm s6 y = '-=1i drrdng thing
3-X
A.y=-1 S.y=2 C.y=3 D.y=-z.
J
Cdu12 Gii srl F'(x) ln m6t nguy6n him cira him s6 /(x) vi G(x) li mQt nguy0n him cria hhm s6 S(.r) H6i kh&ng dinh nio dttdi diy sai?
A.kF(x) lir m6t nguy6n him ctia kf (x) (v6i k li mOt hing s6 thuc)
B F{x)G(x) th mQt nguyOn him cria f(x)S@). C F(x) - G(x) li mQt nguyOn him ctra f (x) - s@).
D F(x) + G(x) li m6t nguy0n hlm cira f (x) + g(x) Ciu13 N6uZ -2-3ithiz=
A -2 - 3i B -2 + 3i C.2 + 3i D.2 - 3i.
CAu 14 M6t cAp s6 c6ng c6 hai s6 hang dAu ti6n lAn lrrot li I vi 4, h6i s6 hang thrl 5 bing bao nhi6u?
A 13 B 16 c.7. D 10
CAu 15 Cho cdc s6 thgc a vi, b th6a m6n tog, (S' 6') = 10916 5 Khing dlnh nho drl6i ildy
iting?
A.2a+ b = 4
x -oo _J 2 *oo
v' 0+0
v
+oo J
2 -oo
B.2a+b=1. C.2a+4b=4 D.a+4b=4
CAu 16 Trong kh6ng gian Oxyz,khoang cr{ch tt di6m M(l;2;D d6nmdt phing (P) : x - 3y +
z- 1 = 0bXng
A 1 8.4{3 c s Vll \n2
11 3 t' tr ' D' 3 '
Cffu 17 C6 bao nhiOu c6ch chia g6i quh gi6ng cho 3 diatr6, sao cho cflng c6 quh?
4.20 8.6 C 15 D 10
Cdu18 GoiDldhinhphinggi6ihanbdicr{cdrrdng ! = ,,,1i,} = 0, x= | vix = 3 Khiquay D quanh truc hohnh, ta thu duoc kh6i trdn xoay vdi thd tichV dugc tfnh bdi c6ng thrlc
3333
A.v= l*a* B.v= nl,l*o* c.v= lr/*o* D.v= nl*a*.
J J' J' J
(3)Ciu 19 Cho hai s6 phfc Zr = 2- i vi Zz = l+ 21 Khi d5, phAn io cira s6 phrlc azzbdng
A.3' B -2. c -2i
Ciu 20 Trongkh6ng gianOxyz,mitciu (S) : x2 +y2 +* -Sx+ 10y *62+25 = 0 c6b6nkinh
bXrg
A \ffi 8.25 c.5 D.ls,
Cia23 Cho hl,rn s6 U4c b6n trtng phtrong y = f (.r) c6 d6 thi nhtt hinh vE b6n 56 nghiQm cria phrtong trinh
?
f (x) - ;tit
4.4 8.3.
c.2 D.1.
Ciu24 Tdp nghiOm cria b6t phrrong trinh logl x - 2log2.r - 3 > 0li
v
1
B. ,2
C tli at
-1 0
D
^llat
0
x
1
A.
C
oo U (8; +oo) D. -oo;-l) U (3; +oo)
CAu 25 NOu mdt hinh n5n c6 chidu cao vi bdn kinh ddy ctng bing2thi c6 diQn tich xung quanh
bing
A.4tl-zr B.8tl-2n C.8zr D.hr.
cda26.Xdt tich ,n* ,B dx Bing crich tldi bi6n s6 r = hx, tich phdn dang x6t trd
thanh
leel
1
)
;3
0;
(-1
U (8;+oo)
) (
A.
[, a, B. [, u,
0
Ciu 27 TAp nghiOm cria bAt phrrong trinh ln x < 1 li
A (-oo; 1) B (1;e) C (0; e) D (-oo;e).
CAu 28 Trong kh6ng gian oxyz, cho hai didm A(1; 2;-3) vi B(-3; 2;9).Mifiphing trung tnJc
cta doan th&ng AB c6 phrtong trinh li
A.x-3y+ 10 = 0 B.x-3e+ 10 = 0
C.-4x+ l2z- 10 = 0 D r+ 3z + 10 = 0
D.3.
Cffu 21 Trong kh6ng gian Oxyz, cho hai tlidmA(O;2; 1) vi 8(2;-2;-3) Phuong trinh m{t cAu
drrdng kfnh AB h
A.(x- l)2 +y2 +(z+l)2 =9 B.* +O-D2 +k-l)2 =3.
C.(x- 2)2 +0+2)2 +(z+312 =36 D.(x+ I)z +y2 +(z-l)2 =6.
Cia22 GeiMlngir{rrildnnh6tcriahims6 f @)=+.*-2x-l tr6ndo4n[0; 2].Tinhgi6 tri cria bidu thrlc P - 6M + 2020.
(4)Ciu29 Di6n tfch cfia mit cAu ngoai ti6p hinh h6p chrl nhAt c6 c6c kfch thrrdc ld 1, 2 vh 3
bHng
4.6r 8.28r C.l4r D.49r.
Ciu 30 Cho hai s6 phrlc phin bi6t 4 vd Zz H6i trong mit phing phrfc, t{p hgp c6c didm bidu
di6n cfia s6 phrlc z li mQt dudng thing n6u didu ki6n nho dudi ddy drtgc th6a mdn?
A.l.- zl+lz*zzl=lzr- zzl B.l.- zzl=7.
C,lr-zi= l D.l.- zil=lz- zzl
Cffu 31 Tdng m6-dun c6c nghiOm phrlc cria phrtong trinh z2 - 6z + 25 = 0 bXng
A 14 B 10 C.8 D.6.
CAu 32 Cho hdm s6 /(x) c6 bing x6t dAu cia f'(x) nhrr sau
x -@ -2 *oo
f'(x) +0 0+
56 didm ctlc tri cira hhm s6 da cho la
A.3 8.1. c.2
CAu 34 Trong kh6ng gian Oxyz, cho dtrdng thlng d :
2x + y + z - | = 0 Gei A ld dudng thfrng nlim (P),
d0y lh m6t vec-td chi phrrong cria A?
1l = !=
= ! rimlt phing (p) : 21-3
a
c[t vh vu6ng g6c v6i d.Yec-td nho du6r
D.0
CAu 33 Cho him s6 bac 4 trtng phrrong y = f (x) c6 d6 thi nhu trong
hinh vE b6n H6i d6 thi him s6 y = lf (x)l c6 tAt chbao nhi6u didm crrc tri?
4.2 B 3
c.4 D.5.
v
o x
A.it = (r,-rt-f,) B.it = (r,i,-l) c.iz= (1;-2;0) D.ir= (-1 ;-2;o)
Ciu 35 f,k* hixh *h*p $fu #*d&fi} *S *ph ke$ bnrry *ryk S*y" I{*i g&c Si{*il *#i mryt ph$*e {;He
vd (SAD) gAn nh6t v6i k6t qui nho du6i d6y?
A.89'31', B.l0g"2g' C.61"29' D.7032'. Ciu 36 Tinh tdng titcitcdc nghiOm ctia phuong trinh
x + 1 = 2log2(2' + 3)- logz (zozo - 2'-').
A.log, 2020 B 13 C.log, 13 D.2020.
Cffu 37 56 giao didm cria dd thi him s6 y - x2lx2 - 3lvh tltrdng thing ! = 2li,
(5)Ciu 38 MQt hinh n6n vi mQt hinh tru c5 ctng chidu cao h vh b6n kfnh ddy r, hon nfra diOn tfch xung quanh cria chring cung bing nhau Khi d6, ti s6
^ 16 B 18 c.i.
3'
Ciu 39 Cho kh6i l6p phttong I vi gqi B la kh6i b6t
Ti s6 thd tfch cira B vd Z li
1
A. B.
2
Cffu 40 Tr0n m{t phing toa d6, cho drldng cong (C) | y = x4 - 4x2 + Zvd,hai tlidm A(- ",11;O),
n (fi;O) CO tdt chbao nhiOu didm trOn (C) mI tdng khoang crich tr) didm d6 d6n c6c didm A vh
B bhng 216?
A.3 8.7 C.6 D 1
Cffu 41 Cho hhm s6 /(x) th6a mdn,f(0) = O, f(2) = 2vd,lf'@)l32,Yx e R Bi6t ring tQp t6t
2
ch cdc gi6 tri cfra tich ph0n
Ir(x) dx llr khoing (a;b), tinh b - a
a.4 8.3 c.1
Ciu 42.Sdn vudn nhh 6ng An c6 d4ng hinh chfr
nhat, v6i chidu di,i vi chi6u r6ng Hn lugt li 8 m6t vi 6 m6t Trdn d6, 6ng dho m6t c6i ao nu6i c6 hinh br{n nguyQt c6 b6n kinh bing Z met (ttfc li lbng ao c6 dpg m6t nrla cria kh6i tnl cit bdi mit phing
qua tryc, tham kh6o th6m hinh v6 b6n)
PhAn d6t dho l6n, 6ng san bing trOn phAn vrrdn cdn lai, vi lhm cho mdt ndn cria vudn drtoc ndng 16n 0,1 m6t H6i sau khi hodn thinh, ao cd c6 d6 siu bhng bao nhi6u? (K6t qui tinh theo don vi m6t, liryn trdn d6n hhng phAn r[m.)
A.0,76 m6t B 0,71 m6t C.0,81 m6t D.0,66 m6t
Ciu 43 Trl diQn ABCD c6 cdc canh AB,AC virAD d6i mQt
vu6ng g6c vi c6 dQ dai lin luot li 2,2vd,3 Gqi M ldtrung didm cria DC.Tinhkhoing cr{ch gifra hai tludng thhng AM vit BC
A. ,ln B. 3122
11
I uXne
r
D.2.
diQn t10u c6 cic dinh h t6rrr.cdc mpt cfia Z
D.1.
3
C
D.2.
l
I
i
t
I
D
M
C
6
\6
2
C D. 2\B B
3
Cdu 44 C6 hQp drlng bi, h6p thrl nh6t dung 10 bi xanh, h6p thrl hai il{ng 5 bi xanh vi 5 bi d6,
h6p thrl ba tlgng 10 bi d6 Ngudi ta chon ng6u nhi6n mQt hQp, sau d6 b6c ngSu nhi6n vi6n bi
tir hQp d6 thi duqc ci 2 bi mhu xanh H6i n6u ti6p tuc b6c th6m vi6n bi nfra hQp d5 (hai bi da
b6c tru6c d6 khdng duoc tri lai vho h6p) thi x6c su6t b6c ttrroc bi xanh bhng bao nhi6u?
(6)C6u 45 Tim tAt c6, c6rc gi6, trithrlc cria tham s6 ru dd him s6 y * cos x - 2 d6ng bi6n tr6n
cosx-m
khoAng 0;
A a
b
B 1 7t
2
A.m e (2; +oo) B m e ll;2).
Ciu 46 Cho a th m6t s6 nguy6n khric kh6ng
H6i s6 nio li s6 trung vi trong ddy 0, l,o,b,l
C.m e (-*; 01 D.m e (0; +m1
vlr b 1)r m6t s6 thuc dudng th6a min ab2 = logrb.
?
c.1 D.b.
B
Cia47 Cho hdm s6 71x; - x - m2 : m Gqi S li tdp tdt cit cdc gi|tri thrrc cria tham s6 m d€
x+ L
giri tri l6n nh6t cria him s6 g(x) = l,f(x)l tr6n doan [1; 2] dat gi6 tri nh6 nhAt Tinh tdng cric phAn trl cria tdp hop S
A.1.
4 c.0. D.
I
2
Ciu 48 Co 6t ci bao nhi6u cdp citc s6 nguyOn (x;y) th6a mdn
L.2. 8.5 c.6
logr(x2*2x+y\+l
logr(*+y2-l)
D.4.
.ry :(r5-t)
D.
<1?
Cffu 49 Cho hdm s6 /(x) c5 bing bi6n thi6n nhu sau
x -oo -2 *oo
f'(x) +0
f (x)
5
J
-oo
2
56 nghiOm thu6c nta khoing (-*; 2O2Ol cria phrrong trinh 2f (f(zx - 1)) + = 0ld
4.4 8.2 C.5 D.3.
Ciu50 Chohinhch6pd6u S.ABCc6tASB =30o.MOtmatphingthayddiquaA citcdccanh SB vi SC lAn lrrot tai M virN Tfnh ti s6 thd tich ctia c6c kh6i ch6p S.AMN vi S.ABC khi chu
vi tam gi6c AMN dqt gi|tri nh6 nh6t
^.2(€ - ,) 8.2(r-fi)