Tnrmg THPT Chuyen Le Quy Don DE THI THU D~I HOC LAN 3 - MON TOAN KHOI A Tinh Ba Ria Viing Tau Thai gian him bai: 180 phut... Khao sat S\l bien thien va ve do thiJCcua ham so... i Trong m
Trang 1Tnr(mg THPT Chuyen Le Quy Don DE THI THU D~I HOC LAN 3 - MON TOAN KHOI A
Tinh Ba Ria Viing Tau Thai gian him bai: 180 phut
I pHAN CHUNG CHO TAT CATHi SINH: (7 diim)
3 Cau I (2 di8m): Cho ham s6 y x - 3x2 + 3
1 Khao sat S\1' biSn thien va ve d6 thi (C) cua ham s6
2 ViSt phuong trinh tiSp tuy~n cua db thi (C), biSt ti~p tuySn di qua di~m A(-l; -1)
Cau II(2 di8m):
1 Giai phuong trinh x3 -6x2 + 12x -7 = {j_x3 +9x2 -19x + 11
2sinx + 1 cos2x + 2cosx 7sinx +5
2 Gi1ii phuong trinh: - - - - ; = - - -
2 cos X J3 - cos 2x + 2 cos X + 1-J3(cos x + Cau III (l di~m):
' h ' h hA I liZx3~x3+8+(6x3+4x2)lnxdx
TIII bc p an sau:
Cau IV (l di~m):
Cho hlnh chop SABCD co ABCD 1a hinh binh hanh tam 0, AB = 2a, AD 2aJ3 , c~c Cl;l.llh ben b~ng nhau va bkg 3a, gOi M 1a trong di~m cua OC Tinh th~ dch kh6i chOp SABMD ya dien dch
-Ciu V (1 di~m)
Cho x, y thOa x2+ r -xy = 1 Tim GTNN va GTLN ella P = X4 + l x2y2
n pHAN RIENG (3 diim)
Thi sinh chi dU'Q'c lam m9t trong hai phAn (phAn 1 ho,"c phAn 2)
1 Theo ChU'01l~ trinh chuAn:
Cau VI.a (2 diem):
1 Trong m~t phing Oxy cho MBC nQi ti~p duang tron (T): x2 + l 4x - 2y - 8 = O Binh A thuQc tia Oy, duemg cao ve tir C n~m trcn duong th~ng (d): x + Sy = O Tim toa dQ cac dlnh A, B, C bi8t
~g C co hoanh dQ 1a m¢t s6 nguyen
2 Trong khong gian Oxyz cho hai duang th~ng Cd1): ~2 -1 72=Z~:, (d 2): {;=~::
Z 4+t
va m~t ph~g (a.): x - y + z 6 == O L~p phuong triM duang th~g (d) bi~t d II (a.) va (d) cit (d1),
(d2) l~n luQ't ~i M va N sao cho MN = 3.J6
Cau VII.a (1 di~m):
Tim t~p hQ'P cac diam biau di€n cho s6 phuc Z th6a man MtMc: Iz + 3 - 2il 12z + 1- 2il
2 Theo chU'O'D~ tdnh nang cao:
Ciu VI.b (2 diem)
1 Trong mftt ph~ng Oxy, cho tam giac ABC co dlnh A(O; 4), trong tam G [~; ~) va tf\1'C tam trimg vai g6c toa dQ Tim toa dQ cac dlnh B, C va di~n dch tam giac ABC bi~t XB < Xc
x-I Y+2 Z 2 {X 2 - t ~
2 Trong khong gian Oxyz cho hai duang thang (d!): - - ; ; ; - -=- - , (d2): y = 3 +t va m~t
z=4+t ph~ng (a.): x - y + Z 6 = O Tim tren (d2) nhfrng diem M sao cho duang th~ng qua M song song voi (d1), c~t (eL) t~j N sao cho MN = 3
Ciu VII.b (1 di6m):
' , ,{eX eY (In y In x)(l +xy)
G1a1 he phuong trmh
21nH2lny _ 3.4ln :< ;;;
Thi sinh kMng (/l.f9c sa dl,mg tai lif?u Can b9 coi thi kMng gial thlch gi them
HQ va ten thf sinh : -; So bao danh: -
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Trang 2DAp AN DE THI THU D~I
Caul
1
I
2
!
!
CiuII
1
i
i
NQidung Cho ham so y =xj
- 3x" + 3
Khao sat S\l bien thien va ve do thiJC)cua ham so
T~p xd va Gi61 h~
y' =3x" 6x y' ::; 0 ~x =0 hay x =2
Bang bien thien:
y" va diem uon Gia tri d~c bi~t
Do thi va nh~xet:
Viet pt tiep tuyen eua (C), biet tiep tuyen di qua diem AC-l; -1)
DuOng tMng (d) qua A va co h~ so goc k
=> Cd): y + 1 "" k(x + 1) => (d):y = kx + k - 1
(d)" '(e) {XJ-3X2+3~kx+k 1
tIepxue ~
3x2 -6x k
=> x 3 3x2 + 3 =3x3 - 6x2 + 3x2 - 6x 1
~2x3 - 6x - 4 = 0
~ X = 2 hay x =-1
, x = 2 => k = 0 => (d): y =-1
, x =-1 => k =9 => (d): y = 9x + 8
Giai phuong trinh x3 - 6x 2 + 12x 7 = ~_X3 + 9x 2 -19x + 11
Kh' d' Iota c 6 { Y ~ x 3 - 6x' + 12x-7
y3=_x3+9x2 19x+ll
=> y3 + 2y = x3 - 3x2 + 5x - 3
~ Tu (1) => y = x-I
Qx3-6x2+11x-6=O
-Di~m 2:=2d
"L=.1.25d 0.25 0.25 0.25 0.25
0.25
'L= 0.75d 0,25
0.25
0.25 2:=24
:L=ld
I
!
0.25
i
0.5
0.25
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Trang 3l x ~j
¢::> x=2
x=3
I
I
I 12 Giai phucmg trinh: 2sinx+ 1 cos 2x + 2cosx -7sinx + 5
2cosx-Ji == cos2x+2cosx+1-Ji(cosx+l) 2.: = Id
I
!
Dieu ki~n:
cos2x+2cosx+1-.y3(cosx+l):;eO<=> (cosx+l)(2cosx-.y3):;tO<=> cosx:;e _~
-0.25
(1) <=> (2sinx + 1)(cosx + 1) == cos2x + 2cosx -7sinx + 5
<=> 2sinxcosx + 2sinx + cosx + 1 = 1 - 2sin2x + 2cosx - 7sinx + 5 0.25
<=> 2sinxcosx - cosx + ~ = 0
<=> cosx(2sinx - 1) + (2sinx - 1)( sinx + 5) = 0
<=> (2sinx - 1)( cosx + sinx + 5) = 0
[ 1 [ x =-+ k21t ~
sm x + cos x = 5 x == (5 + k21t
So s8n.h diSu ki~n ta duQ'c nghi~m cua phucmg trinh 1ft x == ~: + k21t (k E Z) 0.25
Call III
rnh 'h han I fX3~X3+8+(6X3+4X2)lnxd
I
) Tinh II: Dat t = ~X3 + 8 => t2 = x3 + 8 => 2tdt = 3x2dx => x2dx = ~tdt,3
D6i c~n: x 1 =>t=3
x = 2 => t =4,
2 2 t3 2
h
do II rt.3"tdt =3"' 3" 3 == g(64-27) == - ,
$
) Tinh h D~t u = lnx => u' = !
x v' = 6x2 + 4x chon v , , == 2x3 + 2X2
h = [(2x3 +2X2)1nXr-f(2x 2 +2x)dx =
0.25
241n2- -3-+ x2 1 = 241n2
-3
J
!
VayI= 241n2+ -=241n +
I J
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Trang 4- - ~""" r -~ -
ABCD 1ft hinh binh hanh tam
0, AB = 2a, AD = 2afj, cac
cl;U1h ben bkg nhau va bkg
3a, gQi M 1ft trung di~m cua
2:=ld
OC Tinh th~ tich kh6i chop
SABMD va di~n tich cua hinh
~ Ta co SA = SB =SC =SO nen SO 1 (ABC D)
, ~ /), SOA = .= /), SOD nen OA ::;; OB OC = 00 => ABCD 1ft hinh chu nhat 0.25
• => SABCD = AB.AD = 4a2fj
~ Ta co BD ~AB2 + A02 ~4a2 + 12a2 =4a
=> SO = ~SB2 -OB2 ::;; ~9a2 -4a2 == a.J5
0.25
V~y VSABCD - "3SABCD'SO == 3 Do do VSABMD - "4 VSABCD = a .;15
~ GQi G la trQng tam /), OCD, vi /),
D\ffig dUOng thkg d qua G va song song SO thi d 0.25
/)"OCD
i Trong mp(SOG) d\ffig dUOng trung trgc Clla SO, c~t d t~i K dt SO ~I
· Ta co: 01 la trung tT\Ic cua SO => KO::;; KS rna KO = KC = KD nen K 18 tam
mi},t cAu n o~i tiS ill di~n SOCD
Taco: GO::;; CD =
0.25
R=KO= ~OI2+0G2
i Do do S A = 47tR2 =47t. = -
CauV! Cho x, y thca x + y - xy = 1 Tim GTNN va GTLN Clla P = x
2:=ld
•
0.25
0.25
Ta co: f(t) = -4t +
f(t) = 0 ¢::> t =
f(-,,!,) = ! fO) = 1 va f( !)'::;;~.
Yay MaxP = maxf(t) = l va Min P =:: minf(t) =
0.25 I
-+ -::-::: -:-'- '
~-~~' -~ -+ ~=~2~d -Vl.a
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Trang 5I 1 : Dinh A thuoc tia Oy, duemg cao ve tir C nfun tren duemg thfutg (d): llUUg Ulp vxy cno L.\ fin\ " nQl nep Quang l:ron \1): x -r Y - Ll-X - Ly X + 5y lS = u O I=ld
i
Tim tQa do cac dinh A, B, C bi~t rfutg C co hO<lnh do h\ mot s6 nguyen
» A thuoc tia Oy nen A(O; a) (a> 0)
, Vi A E (T) nen a' 2a - 8 ~0 "" [a 4 => a ~ 4 => A(O; 4) 0.25
I
I » C thuqc (d): x + 5y = 0 nen C(-5y; y)
: C E (T) :::::> 25y2 + y2 + 20y 2y - 8 0
<=> 26y2 + 18y - 8 0
0.25
<=> y = ~:::::> x = _ 20 :::::> C(5; -1) (Do Xc E Z)
» (AB) 1-(d) nen (AB): 5x - y + m = 0 rna (AB) qua A nen 5.0 4+m 0 :::::>m==4
V~y (AB): 5x -y + 4 O
B E (AB) :::::> B(b; 5b + 4)
i
B E (T) <=> b2+ (5b + 4)2 - 4b - lOb 8 - 8 == 0 <=> 26b2 + 26b == 0 <=> -
b=-l
I Khi b = 0 :::::> B(O; 4 ) (lo~i vi trimg vOi A)
i
2
r 2 t
Trang kg Oxyz cho (d}): x 1 = +2 1 = z 2 -2 ,(d2): y=:03+t vam~tphang, _ ,:
2
(a):x-y+z-6 0, L~p phuong trinh duemg thkg (d) bi~t d II (a) va (d) d.t
i (d!), (d2) iAn luQ"t t~i M va N sao chc MN = 3.J6
ME (d!):::::> M(1 + 2m; -2 + m; 2 2m)
I
NE(d2):::::>N(2 n; 3 + n; 4 + n) :::::> NM ==(2m+n-l;m-n
-5'-2m-n -2)' , , n =(1'-1'1)a ' ,
0.25
-MN II (a) :::::> na .NM =0 <=> 2m + n - 1 -em n 5) 2m-n-2=0
<=> -m + n + 2 = 0 ¢:;> n = m _ 2
=> NM =(3m 3' -3' -3m), ,
:::::> NM = )(3m-3)2 +(_3)2 +9m 2 =3J2m2 -2m+2
NM 3.J6 <=> 2m2 - 2m + 2 = 6 <=> m 2 m-2 o<=> m -1 hay m =2 0.25 i
» m = -1: M(-I;;Zj 4) va NM .~ -3(Ui::-1) ~Ed):· 3 -"" , 1 -~-.-~:;::"~ -1 0.25
_ ' _ lli ~ 1 ~;~ x 5 _ Y z+2
2 M(5, 0, -2)va NM - ~.J\J, r,-r.rt (d) - - 0.25
I VIl.a Tim t~p hqp cac dibm bi~u dien cho s6 phuc z th6a: Iz + 3 2il = 12z + 1- 2il I=ld
!
G9i M(x; y) la di~m bi~u di~n cho s6 phuc z =x + yi (x; y E R)
Taco: Iz+ 3 - 2il= 12z + 1- 2il
<=> Ix + yi + 3 - 2i\ 12(x+yi)+1- 2il <=>1(x+3)+(y 2)il == IC2x + 1) + (2y 2)il 0.5
<=> (x+3i+(y 2)2=(2x+1i+(2y-2t<=> 3x2 -:3y-2x-4y 8=0 0.25 i
I
Trong m,t phing Oxy cho tam giac ABC cO dinb A( 0.4 ttrong tim G ( ~; ~) I=ld
,
1
I
I
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Trang 6I
bi~t r~ng XB < Xc
3 ( 4 )
,
Xl-O=- 0
Tac6 AI -AG :=;> 2 ( ) ¢> _ => 1(2; -1)
Y -4 ~ ~ 4 YI 1
I 2 3
BC qua I va c6 VTPT la OA = (0;4) = 4(0;1) :=;> BC: y =-1
GQi B(b; -1), vi I la trung di~m BC nen C(4 b; -1)
-
OB.AC:::: 0 ~ 4b b2 + 5 = 0 ~ b2- 4b 5 ::::: 0 ~b =-1 hay b 5
I
I : * b=-l :=;>B(-1;-l)vaC(5;-I)(nh~)
I * b = 5:=;> B(5; -1) va C(-I; -1) (lo~i) 0.25
» BC:::: ( 6;0) :=;> BC 6; d(A; BC) =5 :=;> SABC = 15 I 0.25
2
-t
Trong kgOxyz cho (d1): -2-::::T == =2 ' (d2): y 3 + t va m~t phang
I=ld
z=4+t
I (a): x y + Z - 6 O Tim tren (d2) nhiing di~m M sao cho dlI6ng thAng qua M
song song v6i (dl ), c~t (a) 4ti N sao cho MN 3
M E (d2):=;> M(2 m; 3 + m; 4 + m)
r=2-m+2t (d) qua M va II (dJ) nen (d): y::::: 3+m+t
0.25
z 4+m-2t
N (d) n (a) nen tQa dQ N th6a M:
x 2 m+2t
y ==3+m+t
:=;> 2 - m + 2t - 3 - m t+4+m 2t 6=0
~t -3 -m:=;> N(-3m 4; 0; 3m + 10) i :=;> NM = (6 + 2m; 3 + m; -2m 6)
0.25
I I :=;> NM2 (2m + 6)2 + (m + 3)2 + (-2m-6i
I Do d6 MN == 3 ~ 9(m + 3i = 9 ~ m + 3 = ± 1 ~ m =-2 hay m = -4
!
VII.h
G···h" h .nh{e' e'=(lny Inx)(1+xy)
inx+2lny _3.4lnx ::::: 421ny Dieu ki~n: x, y > o
I
Ta c6: 1 + xy > O
* x> y: VT (1) > 0 va VP(I) < O:=;> VT(I) > VP(l) (voU)
Do d6 tir (1) :=;> x = y
Thay vao (2) ta duQ'c:
231nx _3.4 lnx =4.inx ¢::::>inx[{inxi_3.21nx 4] 0 0.25
21n x ::::: _ 1 <=> lnx =2 ¢::::> x :::: e2 I
, _ _1 0.25 I
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