[r]
(1)Ma da thi 468
Cflu MOt kh6i Hng trg c6 diQn rtch d6ry bHng vir c6 thd tictr bXng thi c5 chi6u cao bing
A 8.4 c.3 D.2.
Cfiu 2 Gii sfi F(x) ld mQr nguyOn hhrn cfra hhm s6 /(x) vi G(x) li mQt nguy6n hdm cria hi,rn s6
S(x) H6i khing dinh nio dtr6i ddy sai?
A.kF(x) th mQt nguy6n hl,rn cria kf (x) (v6i k ld mQt hing s6 thUc) B F(x)G(x) lh mQt nguyOn him cria f(x)s@).
C f(x) + G(.r) li mQt nguyOn hi,rn cfia /(x) + S(x). D F(x) - G(x) li mQt nguyOn hi,rn cta f(x) - S(x).
sd cu vA or eUANG rRI
TRUoNG THPT CHUYoN r,f, QU'f OoN
i c- i
uE rm rn0 r0r NGHT0P rHPT NAM 2020 IUON TOAN
Thdi gian ldm bdi 90 phfi, kh6ng kd thdi gian giao di (EA thi g6m cd tang)
D -1.
D.y=3'
2
Irn'
0
Cdu 3 N€u )dx=3thi
[,ro)-x)dx=
0
A 1 B.5 c.2.
CAu 4 TiQm cfln ngang cira d6 thi hhm s6 y =
Tli thrbng thing
2
A.y=t B.y=-2 C.Y=-1
CAu 5 Cho hhm s6 y - f(x) c6 bing biOn thi6n nhrt sau
x -oo _Ja
2 too
v 0+0
v
+oo
J
2 -oo
H6i hem s6 Ca cho d6ng biOn tr6n khoang nAo du6i day?
A.(2;3) B (2; +oo) C (-3;2) D (-oo; -3).
Cffu 6 Trong khOng gian Oxyz, hinh chi6u vu6ng g6c cfia tlidm A(-1;2;l) trOn tryc Oy c6 tga
d6la
A (0;0; 1) B (0;2;0) C (-1;2;0) D (-1;0; 1) Cfru 7 Thp xdc dinh cria hhm s6 ) = log(x - 1) 1I
(2)phudng clfr'a dlit
A.iq = (1;3;-1) B.iz = (-1;3;-l). Cffu 9 Nghi0m cfra phrlong trinh 2*-s = 4Id,
A.x=5 B.x=9.
C.it = Q;2;5) D.ir = (1;3; 1)
C.x='7 D.x=-1.
v Cfiu 8 Trong kh6ng gian Oxyz, cho dudn g th&ng d :
x I-t
y = 2 + 3t (r e R) M6t vec-to chi
1-J-t
x
Cffu 10 Ydi a li s6 thrrc dttong t.iu,y i,,log.lrc bhng v6i
A.log2a. n'f,,,s" C.',lllogra D.2logra
CAu11 NOuZ -2-3i thiz=
A -2 - 3i. 8.2 - 3i C.2 + 3i D *2 + 3i.
Cial} C6ng thtlc nio drr6i ddy dtng dd tinh thd tich V cria kh6i ch6p v6i diQn tich dity B vir chidu caoh?
1
A.V = -Bh
-J
1
6 o.v - lnt2
B.V= Bh C.V = Bh
CAu 13 D6 thi hdm s5 nho drr6i ddy c6 dang nhu drrdng cong hinh bdn?
A.y = -i +2x2 - l.
C.Y = x4 +2x2.
B.y = -x4 +2x2.
D.y = xa *2x2. o
CAu 14 Tiong kh6ng gian Oxyz, khoing c6ch til di€m M0;2;D dlnmdt phing (P) : x - 3y +
z-l=0blng
^+ ,# ".+ ".+
Ciu 15 Tkong kh6ng gian Oxyz,mlt ciu (S) : ./ +yz + z2 -8x+ 10y - 6z+25 = O c6 br4n kinh
bbng
A.5 B "t,ffi C.25 D.75.
Cffu 16 Cho hai s6 phrlc Zr = 2- i vh Zz = | + 21 Khi d6, phAn io ctia s6 phtlc azzbdng
A -2i B 3i C 3 D -2.
Cffu 17 C6 bao nhi0u c6ch chia g6i qui gi6ng cho 3 drtatr6, cho cfrng c6 quh?
A.6 B 15 C 10 D.20.
Ciu18 GoiDlihinhph&nggi6ihanb6ic6ctludng y= li,) = 0, x=L vix= 3.Khiquay D quanh truc hoinh, ta thu drtoc kh6i trbn xoay v6i thd tichV drrgc tinh b6i c6ng thrlc
3 J
fi
fi ^[
I
(3)Ciu 19 M6t c6p s6 c6ng c5 hai s6 hang dAu ti0n lAn lrrot li I vI4, h6i s6 hang thrl5 bXng bao nhi6u?
A 13 B 16 c.7. D.10.
Ciu 20 Cho c6c s6 thrrc a vd, b th6a m5n tog, (S, fSr) = log f 5 Khing dinh nho drl6i ddy dfng?
A.2a*b=4 C.2a+4b=4 D.a+4b=4.
cda2l Tdng m6-dun cdc nghiOm phrlc cria phuong tr\nh z2 - 6z + 25 = 0 bXng
A.8 8.6 c 10 D.14.
Ciu22 Cho hhm s6 bac b6n trtng phrrong y = f (x) c6 dd fti nhrr hinh vE b6n 56 nghi6m cfla phrlong trinh
f (x) = " lit
+
4.4 B 3
c.2 D.1.
Cffu 23 Cho hai s6 phrlc ph6n bi6t 71vd Zz H6i trong mit phing phrlc, t4p hqp cric didm bidu di6n cria s6 phrlc z li mQt drrdng thing n6u di6u ki6n ndo drr6i ddy drroc th6a m6n?
B.2a+b-l
B. li at. C
v
1
-1 0
B l - zi + lz - zzl = ln - zzl
D.l.- zzl=1.
ln x, tich phAn dang x6t trd
x
ciu24.X6t tich ,n*
';ffi dx B[ng crich ddi bi6n s6 r =
A.lr- zi = l.
C.lr-arl=lz-zzl.
thinh
ft at
e
A. ,li at. D.
0
Ciu 25 Tdp nghi0m cria b6t phrrong trinh ln.r < 1 li
A (-oo; e) ' B (-oo; 1) C (0; e) D (1;e).
Chu 26 Trong ktr6ng giar_ Oxyz, cho hai didm A(0; 2; l) vit B(2; -2;-3) phuong trinh mat cAu dudng kinh AB h
A.(x- t)2 +y2 +(z+!)2 =9 B.(x+ 1)2 +y2 +(z-l)2 =6.
C.(x- 2)2 + (y+2)z +(z+312 =36 D.x2 +(J-Z)2 + k-1)2 =3.
Cda27 Cho hi,rn s6 /(x) c6 bing x6t d6u ctn f,(x) nhu sau
[, o,
x -oo -2 0 2 *oo
f'(x) +0 0+
S6 aidm cgc tri cria him s6 oa cho li
A.3 B 1 c.2. D.0 .
I
I
(4)CAu 28 Tiong kh6ng gran Oxyz, cho hai didm A(1 ;2; -3) vi B(-3; 2;9) Min phing ffung trtlc cira doan thing AB c6 phrrong trinh lh
A -4x+l2z- 10 = 0 B.x- 3y+ 10 = 0
C.x+ 3z+ 10 = 0 D.x- 3z+ 10 = 0
Cdu29 Ggi Mli giri tri 16n nhAt ctra hhm s6 f @) = + 32 + -2x -1 tr6n doan [0; 2] .Tinh gi6 tri cia bidu thfc P = 6M + 2020
A.P=2019 B.P=2007 C.P=2018 D.P=2014.
CAu 30 Di6n tich cua mlt cAu ngoai ti6p hinh h6p chrl nhAt c6 c6c kich thu6c li 1, 2 vh 3
b}ing
A.49n 8.6n C.lLn D.28n.
CAu 31 N6u m6t hinh non co chiAu cao vi brin kinh riliy ctrng blng2 thi c6 diQn tich xung quanh bXng
A.4tr B.8zr C fir D.8"12n.
CAu 32 TAp nghi0m cta b6t phrrong trinh log] x - Zlogrx - 3 > 0lh
A.(-*; -1) u (3;+m; B. -oo: '2-I U (8; +m)
".(o,i) , ,', **, D (-1;3).
CAu 33 56 giao didm cria dd thi hhm s6 y = x2lx2 - 3l vd tludng thing ! = 2lb,
4.4 8.3 c.6 D.2
CAu 34 Tinh tdng tdt cic6c nghiOm cira phrtong trinh
x * 1 = 2log2(2' + 3) - logz(zOzo - 2'-.)
A.1og2 2020 8.2020 C.log, 13 D.13.
Ciu 35 Cho kh6i 16p phuong L vh goi B h kh6i brlt diQn d6u c6 c6c dinh lir tdm cic m\t clfra L.
Ti s6 thd tich cria B vi L li
I
x
1
6 D. C
A.
4 J
Ciu 36 Cho hhm s6 bac 4 trtrng phuong y = f (x) c6 d6 thi nhtt trong hinh v6 bOn H6i d6 tfri hhm s6 y = lf (x)l c6 t6t citbao nhi6u didm crrc tri?
4.2 B 3
c.4 D s
Cfiu 37.elc hinh *h*p *fu A:xrm ** ewh ll&: bfu€ ,nsrlh dry H*i g** *i*s tlei rx6: plrtuW
{,S,{$} vh (SAD) gAn nh6t v6i k6t qui nlo dudi tl6y?
A.109"29' 8.7032' C.89"31' D.6t"28' chobirt*tr$pd&x,x,rar.r
1
v
(5)ddy lir mdt vec-td chi phrrong cira A?
A.it =(r,-,-f,) s.i,= (-1; -z;0) c.i+ =(r,-i,-l)
CAu 38 Trong kh6ng gtan Oxyz, cho drrdng thhng d :
2x + y + z - I = 0 Ggi A li drrdng thing nim trong (P),
A.3 B i c.4.
Cia 44.T1 diQn ABC D c6 cdc canh AB, AC vir AD ddi mQt vu6ng g6c vi c6 t16 dhi ldn lrrot li 2,2 vd Gqi M lduung
didm cfia DC.Tinhkhoing cdch gifra hai drrdng thlngAM
vi BC
\m p 3 1122
A' 6 ' o' tr '
n {1 ^2{1
L' 2' D' 3 '
4 = = '*=2 udmdt phing (P) :
2t-3
cit vi vu6ng g6c v6i d Vec-to nio dtt6i
D,iz = Q;-2;0).
B.me(2;+m) C.me(-*;01 D.me(0;+m).
D.6.
Cffu 39 MOt hinh n6n vi m6t hinh tru c6 ctng chi6u cao h vi b6n kinh tl6y r, hdn nfra di6n tich xung quanh cria chring cfrng bhng Khi d6, ti s6 3 UHng
r
A 16 B.z c.r , +
Cffu 40 Tim t6t ch cdcgi6 trf thrrc cria tham s6 m d€ hims6 y " - i?::-: d6ng bi6n tr0n
cosx-m
khoing (t,;) A.m e U;2).
Cau 41 TrOn m4t ph8ng toa d6, cho tludng cong (C) l ! = x4 - 4x2 + Zvi,hai didm a (- f; O), n (,,11;O) CO tdt chbao nhi6u didm tron (O mn tdng khoang cdch tr) tlidm t16 d6n c6c didm A vi B bXng zlot
A.3 8.7 c 1
Cia 42.Sdn vrrdn nhi 6ng An c6 d4ng hinh chfr nhaL vdi chidu dhi vi chidu r6ng Dn hldt li 8 m6t vh m6t TrOn d6, 6ng tlio mQt c6i ao nu6i cf hinh brin nguyQt c6 brin kinh bhng 2 mdt (trlc li lbng ao c6 d4ng mQt nfta cfla kh6i try c1t bdi mdt phfing qua truc, tham khio th6m hinh v6 b0n)
' Phin dAt dho l6n, 6ng san bing tr6n phAn vrJdn cbn lai, vi thm cho m{t ndn cria vrrdn drrgc
ndng l6n 0,1 m6t H6i sau khi hoan thhnh, ao c6,c6 dQ sdu bing bao nhi6u? (KOt qui tinh theo don vi m6t, ll,rn trdn tl6n hdng phAn tr[m.)
A.0,71 m6t B 0,76 m6t C.0,66 m6t D.0,81 mdt
Cfru 43 Cho hirn s6 /(x) th6a mdn,f(0) = O, f(2) = 2vd,lf'@)l < 2,Yx e R Bi6trhng tap t6t
ch cdc gi6tri cria tich phdn
k,x) dx li khoing (a;b), tinh b - a
D.2.
C
B D
M
(6)cau 45' c6 3 h6p dung bi, h6p thrl nhAt dung 10 bi xanh,
hQp thrl hai dung bi xanh vi 5 bi d6, h6p thrl ba drrng 10 bi d6' Ngrrdi ta chon ng6u nhion m6t h6p, sau
d6 b6c ngau nhi6n vi6n bi
tt h6p d5 thi drloc c6 bi miu xanh H6i n6u ti6p tuc b6c th6m vi6n bi nfra d hop d5 (hai bi da
o:t ti{u" d6 kh6ng dugc tri l4i vho h6p) thi
xr{c su6t b6c dudc bi xanh b}ing bao nhi6u?
A-A G n.il c.;
".r;
cfiu 46' Cho a li m6t s6 nguy6n khdc kh6n g vir b li m6r s5 thuc drrong th6a
man ab2 = log2b.
H6i s6 nio ti s6 trung vi trong ddy 0, t,a,b,|.t
A.a- B.b c.t
o.1
b cda 47 ' cho hinh ch6p ddu s
'ABC c6 tAS B = 30o M6t mdt phing thay ddi qua A c6t c6c canh 's'B vi 'sc lAn hldt tai M vd'N Tinh ti s6 ttrd tich cria cric kh6i ch6p s.AMN vit s.ABCkhi chu
vi tam gidc AMN dat gi|tri nh6 nh6t.
:(r5- r)
4 cAu 4g Cho him s5 f(xl - x - m2 + m
gid,t4Ldnnh6t.u,;;;';;,r.,i,il;::::'1,3'il:;:;,il.#;,H,'#;::^Y"
trl cria tdp hop S
E -1 B r
c o o 1.
2 u r u.u
4 CAu 49 C0 tiltci bao nhi6u cdp c6cs5 nguyOn (x;y) th6a mdn logr(* - 2x +
a:+ 11
4.2.
8.2(V3 - ,)
B.6.
o.z(z- 16)
v\+t <1?
c.4. 1og2@2D.5.+y2-l) C6u 50 Cho him s6 /(x) c6 bing bi6n thi6n nhu sau.
56 nghi.m thu6c nrla kho,ng (-m; 2,2orcria phuong trinh zf (f(2x- 1)) + = 0li
A.5 8.4 c.3
D.2.
x -oo -2 J foo
0
+0
f(x) I