http://www.VNMATH.com 122 N THI TT NGHIP MN TON THPT NM HC 2010
- 2011 1I.PHN CHUNG CHO TT C TH SINH.(7 im)Cu I.(3 im) Cho hm s y =
2 11+xx c th (C).1/ Kho st s bin thin v v th (C) ca hm s.2/ Vit
phng trnh tip tuyn ca (C) ti giao im ca (C) vi trc tung.Cu II. (3
im)1/ Gii phng trnh : log3(x + 1) + log3(x + 3) = 1.2/ Tnh I =
230cos .x dx.3/ Xt s ng bin v nghch bin ca hm s y = -x3 + 3x -1Cu
III. (1 im). Cho hnh chp S.ABC c ABC l tam gic vung cn ti B, a AC ,
SA( ) ABC, gc gia cnh bn SB v y bng 600. Tnh th tch ca khi chp.II.
PHN RING (3 im).1.Theo chng trnh chun.Cu IVa. (2 im). Trong khng
gian vi h ta Oxyz , cho im M(1; 1 ; 0) v mt phng (P): x + y 2z + 3
= 0.1/ Vit phng trnh mt cu tm M v tip xc vi mp(P).2/ Vit phng trnh
ng thng (d) i qua M v vung gc vi (P). Tm ta giao im.Cu Va. (1 im).
Tnh din tch hnh phng gii hn bi cc ng y = 3 vy = x2 2x2. Theo chng
trnh nng cao. Cu IVb(2 im) Trong khng gian vi h ta Oxyz, cho im
M(-1 ; 2 ; 1) v ng thng(d): 1 22 1 1 + x y z.1/ Vit phng trnh mt cu
tm M v tip xc vi (d).2/ Vit phng trnh mt phng i qua M v vung gc vi
(d). Tm ta giao im.CuVb.(1im).Tnh dintch hnh phng gii hnbi cc
ngy=214x vy = 2132 + x x 2I.PHN CHUNG CHO TT C TH SINH.(7 im)Cu
I.(3 im). Cho hm s y = x3 3x2 + 2 c th (C).1/ Kho st s bin thin v v
th (C) ca hm s.2/ Bin lun theo m s nghim ca phng trnh: x3 3x2 m =
0.Cu II. (3 im).1/ Gii phng trnh: 3x + 3x+1 + 3 x+2 = 351.2/ Tnh I
= 10( 1) . +xx e dx http://www.VNMATH.com1http://www.VNMATH.com 122
N THI TT NGHIP MN TON THPT NM HC 2010 - 20113/ Tm gi tr ln nht v gi
tr nh nht ca hm s y = x4 2x2 + 1 trn an [-1 ; 2].Cu III. (1 im).
Tnh th tch khi t din u S.ABC c tt c cc cnh u bng a.II. PHN RING.(3
im)1.Theo chng trnh chun.Cu IV a. (2 im). Trong khng gian vi h ta
Oxyz, cho cc im A(-1 ; 2 ; 0), B(-3 ; 0 ; 2), C(1 ; 2 ; 3), D(0 ; 3
; - 2).1/ Vit phng trnh mt phng (ABC) v phng trnh ng thng AD.2/ Tnh
din tch tam gic ABC v th tch t din ABCD.Cu V a. (1 im). Tnh th tch
khi trn xoay do hnh phng gii hn bi cc ng y = tanx , y = 0, x = 0, x
= 4 quay quanh trc Ox.2. Theo chng trnh nng cao.Cu IV b.(2 im)Trong
khng gian vi h ta Oxyz, cho cc im A(-2 ; 0 ; 1), B(0 ; 10 ; 2), C(2
; 0 ; -1), D(5 ; 3 ; -1).1/ Vit phng trnh mt phng (P) i qua ba im
A, B, C v vit phng trnh ng thng i qua D song song vi AB.2/ Tnh th
tch ca khi t din ABCD, suy ra di ng cao ca t din v t nh D.Cu Vb. (1
im). Tnh th tch khi trn xoay do hnh phng gii hn bi cc ng y = 12.xx
e, y = 0, x = 0, x = 1 quay quanh trc Ox. 3 I.PHN CHUNG CHO T C TH
SINH. (7 im)Cu I. (3 im) Cho hm s y = - x3 + 3x -1 c th (C).1/ Kho
st s bin thin v v th (C) ca hm s.2/ Vit phng trnh tip tuyn ca (C)
ti im cc tiu ca (C).Cu II.(3 im)1/ Gii phng trnh:26log 1 log 2
+xx2/ Tnh I = 220cos 4 .x dx3/ Tm gi tr ln nht v gi tr nh nht ca hm
s y = ln xx trnon [1 ; e2 ]Cu III.(1 im). Cho hnh chp tam gic u
S.ABC c cnh y bng a, cc cnh bn u to vi y mt gc 600. Tnh th tch ca
khi chp.II. PHN RING. (3 im)1.Theo chng trnh chun.Cu IV a.(2 im).
Trong khng gian vi h ta Oxyz, cho mt phng (P): 2x + y z 6 = 0 v im
M(1, -2 ; 3).1/Vit phngtrnhmtphng(Q)iquaMvsongsongvi mp(P).Tnh
khang cch t M n mp(P). http://www.VNMATH.com2http://www.VNMATH.com
122 N THI TT NGHIP MN TON THPT NM HC 2010 - 20112/ Tm ta hinh chiu
ca im M ln mp(P).Cu Va. (1 im). Gii phng trnh: x2 2x + 5 = 0 trong
tp s phc C.2. Theo chng trnh nng cao.Cu IV b.(2 im). Trong khng
gian vi h ta Oxyz , cho hai mt phng (P): 3x 2y + 2z 5 = 0, (Q): 4x
+ 5y z + 1 = 0.1/ Tnh gc gia hai mt phng v vit phng tnh tham s ca
giao tuyn ca hai mt phng (P) v (Q).2/ Vit phng trnh mt phng (R) i
qua gc ta O vung gc vi (P) v (Q).Cu Vb.(1 im). Cho s phc z = x + yi
(x, y ) R. Tm phn thc v phn o ca s phc z2 2z + 4i . 4I.PHN CHUNG
CHO TT C TH SINH. (7 im)Cu I. (3 im). Cho hm s y = 21 +xx c th
(C).1/ Kho st s bin thin v v th (C) ca hm s.2/ Vit phng trnh tip
tuyn ca(C) ti im c hanh x = -2.Cu II. (3 im)1/ Gii phng trnh : 1 13
3 10+ + x x.2/ Tnh I = tan 420cosxedxx3/ Tm gi tr ln nht v gi tr nh
nht ca hm s y = 21 x .Cu III.(1 im).Cho hnh chp t gic u S.ABCD c
cnh y bng a, cnh bn hp vi y mt gc 600 .1/ Tnh th tch khi chp
S.ABCD2/ Tm tm v tnh bn knh mt cu ngai tip hnh chp.II. PHN RING. (3
im)1. Theo chng trnh chun.Cu IV a. (2 im). Trong khng gian vi h ta
Oxyz, cho im D(-3 ; 1 ; 2) v mt phng (P) i qua ba im A(1 ; 0 ; 11),
B(0 ; 1 ; 10), C(1 ; 1 ; 8).1/ Vit phng trnh ng thng AB v phng trnh
mt phng (P).2/Vit phng trnh mt cu tm D, bn knh R = 5. Chng minh rng
mt cu ny ct mt phng (P).Cu Va. (1 im). Tnh din tch hnh phng gii hn
bi cc ngy = lnx ,y = 0, x = 1e, x = e .2.Theo chng trnh nng cao.Cu
IV b.(2 im). Trong khng gian vi h ta Oxyz, cho mt phng(P): 2x + 2y
+ z + 5 = 0 v mt cu (S):x2 + y2 + z2 2x 4y + 4z = 0.1/ Tm tm v bn
knh ca mt cu (S).2/ Vit phng trnh mt phng (Q) song song vi (P) v
tip xc vi (S). Tm ta ca tip im.
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TON THPT NM HC 2010 - 2011Cu Vb.(1 im). Tm m ng thng d: y = mx + 1
ct th (C): y = 231+xx ti hai im phn bit. 5I.PHN CHUNG CHO TT C TH
SINH.(7 im)Cu I. (3 im). Cho hm s y = - x4 + 2x2 +3 c th (C).1/ Kho
st s bin thin v v th (C) ca hm s.2/ Da vo th (C), tm cc gi tr ca m
phng trnh x4 2x2 + m = 0 c bn nghim thc phn bit.Cu II. (3 im)1/ Gii
bt phng trnh:2 4log log ( 3) 2 x x2/ Tnh I = 40sin 21 cos 2+xdxx.3/
Cho hm s y = 25log ( 1) + x. Tnh y(1).Cu III.(1 im).Cho hnh chp
S.ABC c y ABC l tam gic vung ti B, cnh bn SA(ABC), bitAB = a, BC =3
a , SA = 3a.1/ Tnh th tch khi chp S.ABC theo a.2/ Gi I l trung im
ca cnh SC, tnh di ca cnh BI theo a.II. PHN RING. (3 im)1.Theo chng
trnh chun.Cu IV a. (2 im) Trong khng gian vi h ta Oxyz , cho ba im
A(1 ; 4 ; 0), B(0 ; 2 ; 1), C(1 ; 0 ; -4).1/ Tm ta im D ABCD l hnh
bnh hnh v tm ta tm ca hnh bnh hnh .2/ Vit phng trnh ng thng (d) i
qua trng tm ca tam gic ABC v vung gc vi mp(ABC).Cu V a.(1 im). Tnh
th tch ca khi trn xoay to thnh khi quay quanh trc tung hnh phng gii
hn bi cc ng y = lnx, trc tung v hai ng thng y = 0, y = 1.2. Theo
chng trnh nng cao.Cu IV b. (2 im) Trong khng gian vi h ta Oxyz, cho
hai ng thng d: 1 2 32 1 1 x y z, d: 1 51 3 ' x ty tz t1/ Chng minh
d v d cho nhau.2/ Vit phng trnh mt phng (P) cha d v song song vi
d.Tnh khang cch gia d v d.Cu V b. (1 im). Tnh th tch khi trn xoay
to thnh khi quay quanh trc hanh hnh phng gii hn bi cc ng y = lnx, y
= 0, x = 2. 6 http://www.VNMATH.com4http://www.VNMATH.com 122 N THI
TT NGHIP MN TON THPT NM HC 2010 - 2011I.PHN CHUNG CHO TT C TH SINH.
(7im)Cu I.(3 im) Cho hm s y = x(x 3)2 c th (C).1/ Kho st s bin thin
v v th (C) ca hm s.2/ Vit phng trnh ng thng i qua hai im cc tr ca
th hm s.Cu II. (3 im)1/ Gii bt phng trnh: 2 22 2log 5 3log + x x.2/
Tnh I = 220sin 2 .x dx.3/ Tm gi tr ln nht v gi tr nh nht ca hm s y
= x2e2xtrn nakhong (-; 0 ]Cu III.(1 im). Cho hnh chp S.ABC c y ABC
l tam gic vung ti A. Bit AB = a, BC = 2a, SC = 3a v cnh bn SA vung
gc vi y. Tnh th tch khi chp S.ABC theo a.II. PHN RING. (3 im)1.Theo
chng trnh chun.Cu IV a. (2 im). Trong khng gian Oxyz, cho bn im A(1
; -2 ; 2), B(1 ; 0 ; 0),C(0 ; 2 ; 0), D(0 ; 0 ; 3).1/ Vit phng trnh
mt phng (BCD). Suy ra ABCD l mt t din.2/ Tm im A sao cho mp(BCD) l
mt phng trung trc ca an AA.Cu V a. (1 im). Tnh th tch khi trn xoay
to thnh khi quay quanh trc hanh hnh phng gii hn bi cc ng y =
sinx.cosx, y = 0, x = 0, x = 2.2. Theo chng trnh nng cao.CuIVb.
(2im). Trongkhnggianvi htaOxyz, chongthngd:1 12 1 2 + x y z v hai
mt phng (P1): x + y 2z + 5 = 0, (P2): 2x y + z + 2 = 0.1/ Tnh gc
gia mp(P1) v mp(P2), gc gia ng thng d v mp(P1).2/ Vit phng trnh mt
cu tm I thuc d v tip xc vi mp(P1) v mp(P2).Cu Vb.(1 im). Tnh th tch
khi trn xoay to thnh khi quay quanh trc tung hnh phng gii hn bi cc
ng y = x2 v y = 6- | x | . 7I.PHN CHUNG CHO TT C TH SINH.(7 im).Cu
I. (3 im). Cho hm s y = 1 xx c th l (C).1/ Kho st s bin thin v v th
(C) ca hm s.2/ Tm m ng thng d: y = -x+ m ct th (C) ti hai im phn
bit.Cu II.(3 im)1/ Gii phng trnh: 4x + 10x = 2.25x.2/ Tnh I = 924(
1) dxx x3/ Tm gi tr ln nht v gi tr nh nht ca hm s y =.ln x xtrn an
[ 1; e ].Cu III.(1 im). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh
a, cnh bn SA = a 3v vung gc vi y.
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TON THPT NM HC 2010 - 20111/ Tnh th tch khi chp S.ABCD.2/ Chng minh
trung im I ca cnh SC l tm ca mt cu ngai tip hnh chp S.ABCD.II. PHN
RING. (3 im)1. Theo chng trnh chun.Cu IV a.(2 im). Trong khng gian
vi h ta Oxyz,cho hai im A(2 ; 1 ; 1), B(2 ; -1 ; 5).1/ Vit phng
trnh mt cu (S) ng knh AB.2/ Tm im M trn ng thng AB sao cho tam gic
MOA vung ti O.Cu V a. (1 im). Gii phng trnh sau trn tp s phc : z4 1
= 0.2. Theo chng trnh nng cao.Cu IV b.(2 im). Trong khng gian vi h
ta Oxyz, cho mt cu (S): x2 + y2 + z2 2x 4y 6z = 0 v hai im M(1 ; 1
; 1), N(2 ; -1 ; 5).1/ Tm tm I v bn knh R ca mt cu (S).Vit phng
trnh mt phng (P) qua cc hnh chiu ca tm I trn cc trc ta .2/ Chng t
ng thng MN ct mt cu (S) ti hai im. Tm ta cc giao im .Cu V b.(1 im).
Biu din s phc z = 1 i. 3di dng lng gic. 8http://www.VNMATH.comI.PHN
CHUNG CHO TT C TH SINH. (7 im)Cu I. (3 im). Cho hm s y = 4 21 532 2
+ x x c th l (C).1/ Kho st s bin thin v v th (C) ca hm s.2/ Vit
phng trnh tip tuyn ca (C) ti im M(1; 0).Cu II. (3 im)1/ Gii bt phng
trnh: 22 33 44 3 _ ,x x. 2/ Tnh I = 220cos 21 sin+xdxx.3/ Tm gi tr
ln nht v gi tr nh nht ca hm s y = sin2x x trn an ;6 2 1 1 ].Cu III.
(1 im).Cho hnh chp S.ABCD c y ABCD l hnh vung, cnh bn 2 a SA v vung
gc vi y, gc gia SC v y l 450.Tnh th tch ca khi chp.II. PHN RING. (3
im)1.Theo chng trnh chun.Cu IV a. (2 im).Trong khng gian vi h ta
Oxyz, cho hai im A(3 ; 0 ; -2), B(1 ; -2 ; 4).1/ Vit phng trnh ng
thng AB v phng trnh mt phng trung trc ca an AB.2/ Vit phng trnh mt
cu tm A v i qua im B. Tm im i xng ca B qua A.Cu V a.(1 im). Tnh th
tch ca khi trn xoay c to thnh khi quay quanh trc tung hnh phng gii
hn bi cc ng y = 2 x2 v y = | x | .
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TON THPT NM HC 2010 - 20112. Theo chng trnh nng cao.Cu IV b.(2 im)
Trong khng gian vi h ta Oxyz, cho hai ng thng d: 1 1 22 3 4 + x y z
vd: 2 21 34 4 + +' +x ty tz t.1/ Chng minh d song song vi d. Tnh
khang cch gia d v d.2/ Vit phng trnh mt phng (P) cha d v d.Cu V
b.(1 im).Cho hm s y = 23 62+ ++x xx (1). Vit phng trnh ng thng d i
qua im A(2 ; 0) v c h s gc l k. Vi gi tr no ca k th ng thng d tip
xc vi th ca hm s (1). 9http://www.VNMATH.comI.PHN CUNG CHO TT C TH
SINH. (7 im).Cu I.(3 im). Cho hm s y = -x3 + 3x2 2 c th (C).1/ Kho
st s bin thin v v th (C) ca hm s.2/ Vit phng trnh tip tuyn vi (C)
bit tip tuyn c h s gc k = -9.Cu II.(3 im).1/ Gii phng trnh: 12 2log
(2 1).log (2 2) 6++ + x x 2/ Tnh I = 20sin 2.1 cos+xdxx3/ Tm gi tr
nh nht ca hm s y = x lnx + 3.Cu III.(1 im). Cho hnh chp S.ABC c SA,
AB, BC vung gc vi nhau tng i mt. Bit SA = a, AB = BC = a 3 .Tnh th
tch ca khi chp v tm tm ca mt cu ngai tip hnh chp.II. PHN RING. (3
im).1. Theo chng trnh chun.Cu IV a. (2 im). Trong khng gian vi h ta
Oxyz, cho im A(2 ; -1 ; 3), mt phng (P): 2x -y - 2z+ 1 = 0 v ng
thng d: 1 22 1 3 x y z.1/ Tm ta im A i xng ca A qua mp(P).2/ Tm ta
ca im M trn ng thng d sao cho khang cch t M n mp(P) bng 3.Cu V a.(1
im). Gii phng trnh sau trn tp s phc: z4 z2 6 = 02. Theo chng trnh
nng cao.Cu IV b. (2 im).Trong khng gian vi h ta Oxyz, cho im A(1 ;
1 ; 1), mp(P): x + y z 2 = 0 v ng thng d: 2 11 1 1 x y z.1/ Tm im A
i xng ca A qua d.2/ Vit phng trnh ng thng i qua A, song song vi
mp(P) v ct d.Cu Vb. (1 im). Gii h phng trnh:22 422 45log log 85log
log 19 ' x yx y 10 http://www.VNMATH.com7http://www.VNMATH.com 122
N THI TT NGHIP MN TON THPT NM HC 2010 -
2011http://www.VNMATH.comI.PHN CHUNG CHO TT C TH SINH. (7 im).Cu
I.(3 im). Cho hm s y = (x 1)2(x +1)2 c th (C).1/ Kho st s bin thin
v v th (C) ca hm s.2/ Tm m ng thng d: y = m ct th (C) ti ba im phn
bit.Cu II.(3 im)1/ Gii phng trnh: log(x 1) log(x2 4x + 3) = 1.2/
Tnh I = 31(1 ln ).+exdxx.3/ Cho hm s y = x3 (m + 2)x + m( m l tham
s). Tm m hm s c cc tr ti x = 1.Cu III.(1 im). Cho hnh lng tr ABC.
ABC c y l tam gic u cnh a, cnh bn bng a 3v hnh chiu ca A ln mp(ABC)
trng vi trung im ca BC.Tnh th tch ca khi lng tr .II. PHN CHUNG. (3
im)1. Theo chng trnh chun.Cu IV a.(2 im). Trong khng gian vi h ta
Oxyz, cho hai im A, B c ta xc nh bi cc h thc2 , 4 4 uuur uuurOA i k
OB j kv mt phng (P): 3x 2y + 6z + 2 = 0.1/ Tm giao im M ca ng thng
AB vi mp(P).2/ Vit phng trnh hnh chiu vung gc ca AB trn mp (P).Cu V
a.(1 im). Tnh th tch khi trn xoay tao thnh khi quay quanh trc Ox
hnh phng gii hn bi cc ng y = 12+xx, y = 0, x = -1 v x = 2.2/ Theo
chng trnh nng cao.CuIVb. (2im). Trongkhnggianvi htaOxyz,
chongthngd: 1 22 + 'x ty tz t v mt phng(P): x + 2y 2z + 3 = 0.1/
Vit phng trnh ng thng i qua gc ta O vung gc vi d v song song vi
(P).2/ Vit phng trng mt cu c tm thuc d, tip xc (P) v c bn knh bng
4.Cu Vb.(1 im).Tnh( )83 +i 11I/_ Phn dnh cho tt c th sinhCu I ( 3
im) Cho hm s ( )111+xyx c th l (C)1) Kho st hm s (1)2) Vit phng
trnh tip tuyn ca (C) bit tip tuyn i qua im P(3;1).Cu II ( 3 im)1)
Gii bt phng trnh:2.9 4.3 2 1 + + >x x2) Tnh tch phn: 15 301 I x
x dx3) Tm gi tr ln nht v gi tr nh nht ca hm s 21 + +x xyxvi 0 >
x http://www.VNMATH.com8http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 2011Cu III (1 im). Xc nh tm v bn knh mt cu
ngoi tip mt hnh lng tr tam gic u c 9 cnh u bng a.II/_Phn ring (3
im)1) Theo chng trnh chunCu IV. a (2 im) Trong khng gian cho h ta
Oxyz, im A (1; -1; 1) v hai ng thng (d1) v (d2) theo th t c phng
trnh: ( ) ( )1 23 3 0: 1 2 ; :2 1 03 + ' ' + x tx y zd y t dx yz
tChng minh rng (d1), (d2)v A cng thuc mt mt phng.Cu V. a (1 im) Tm
mun ca s phc( )22 2 + z i i2)Theo chng nng cao.Cu IV. b (2 im)
Trong khng gian cho h ta Oxyz, cho mt phng ( ) ( ) v ln lt c phng
trnh l: ( ) ( ) : 2 3 1 0; : 5 0 + + + + x y z x y z v im M (1; 0;
5).1. Tnh khong cch t M n ( ) 2. Vit phng trnh mt phng i qua giao
tuyn (d) ca ( ) ( ) vng thi vung gc vi mt phng (P): 3 1 0 + x yCu
V. b (1 im) Vit dng lng gic ca s phc1 3 + z i 12I. Phn chung cho tt
c th sinh (7,0 im)Cu I.( 3,0 im) Cho hm s 3 21 23 3 + + y x mx x
m
( )mC1. Kho st s bin thin v v th ( C) ca hm s khi m =0.2.Tm im c
nh ca th hm s( )mC.Cu II.(3,0 im)1.Tm gi tr ln nht v nh nht ca hm s
4 28 16 + y x xtrn on [ -1;3].2.Tnh tch phn7 33 20 1+xI dxx3. Gii
bt phng trnh 0,52 125log++xxCu III.(1,0 im)Cho t din S.ABC c SA
vung gc vi mt phng (ABC), SA = a; AB = AC= b, 60 BAC . Xc nh tm v
bn hnh cu ngoi tip t din S.ABC.II.Phn ring(3,0 im)Th sinh hc chng
trnh no th ch c lm phn dnh ring cho chng trnh .1. Theo chng trnh
Chun:Cu IV.a(2,0 im) Trong khng gian vi h to Oxyz:
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TON THPT NM HC 2010 - 2011a)Lp phngtrnh mt cu c tmI(-2;1;1) v tip
xc vi mt phng 2 2 5 0 + + x y z b) Tnh khong cch gia hai mt phng:0
1 2 4 8 0 12 2 4 + z y x v z y x
Cu V.a(1,0 im)Gii phng trnh : 4 23 4 7 0 + z ztrn tp s
phc.2.Theo chng trnh nng cao.Cu IV.b(2,0 im) Trongkhnggianvi
htoOxyz,chongthngdcphngtrnh:1 12 1 2 + x y zv hai mt phng0 5 2 : )
( + + z y x v0 2 2 : ) ( + + z y x . Lp phng trnh mt cu tm I thuc
ng thng d v tip xc vi c hai mt phng ( ) ( ) , .Cu V.b(1 im)Tnh din
tch hnh phng gii hn bi h cc hm s , 2 , 0 y x y x y 13I. Phn chung
cho tt c th sinh (7,0 im)Cu I.( 3,0 im)1. Kho st s bin thin v v th
ca hm s 23+xyx2.Tm trn th im M sao cho khong cch t M n ng tim cn ng
bngkhong cch t M n tim cn ngang.Cu II.(3,0 im)1. Giiphng trnh 2 13
.5 7 245 x x x.2.Tnh tch phn a) 11 ln +exI dxx Cu III.(1,0 im)Mt
hnh tr c thit din qua trc l hnh vung, din tch xung quanh l 4.1.Tnh
din tch ton phn ca hnh tr.2. Tnh th tch ca khi tr.II.Phn ring(3,0
im)Th sinh hc chng trnh no th ch c lm phn dnh ring cho chng trnh
.1. Theo chng trnh Chun:Cu IV.a(2,0 im) Trong khng gian vi h to
Oxyz:cho A(1;0;0), B(1;1;1),1 1 1; ;3 3 3 _ ,Ca)Vit phng trnh tng
qut ca mt phng ( ) i qua Ov vung gc vi OC. b) Vit phng trnh mt
phng( ) cha AB v vung gc vi ( ) Cu V.a(1,0 im)Tm nghim phc ca phng
trnh 2 2 4 + z z i 14I.PHN CHUNG CHO TT C CC TH SINHCu 1 (4,0
im):1. Kho st v v th (C) ca hm s 3 23 y x x2. Da vo th (C) bin lun
theo m s nghim ca phng trnh
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TON THPT NM HC 2010 - 20113 23 0 + x x m3. Tnh din tch hnh phng gii
hn bi th (C) v trc honh.Cu 2 ( 2,0 im) 1. Gii phng trnh: 23 5.3 6 0
+ x x2. Gii phng trnh: 24 7 0 + x xCu 3 (2,0 im)Cho hnh chp S.ABCD
c y ABCD l hnh vung cnh a, cnh bn SB vung gc vi y, cnh bn SC bng3 a
.1. Tnh th tch ca khi chp S.ABCD.2. Chng minh trung im ca cnh SD l
tm mt cu ngoi tip hnh chp S.ABCD.II. PHN DNH CHO TNG TH SINHA. Dnh
cho th sinh Ban c bn: Cu 4 (2,0 im)1.Tnh tch phn: 10( 1). +xI x e
dx2. Trongkhnggianvi htaOxyz chobaimA(5;0;4), B(5;1;3), C(1;6;2),
D(4;0;6)a. Vit phng trnh tham s ca ng thng ABb. Vit phng trnh mt
phng ( ) i qua im D v song song vi mt phng (ABC).B. Dnh cho th sinh
Ban nng caoCu 5 (2,0 im)1. Tnh tch phn:23 2 311 +I x x dx2. Trong
khng gian vi h ta Oxyz, cho im M(1;2;3) vmt phng (P) c phng trnh:x
- 2y + z + 3 = 0 a. Vit phng trnh mt phng (Q) i qua im M v song
song vi mt phng (P).b. Vit phng trnh tham s ca ng thng (d) i qua im
M v vung gc vi mt phng (P). Tm ta giao im H ca ng thng (d) vi mt
phng (P) 15I . PHN CHUNG CHO TT C TH SINH ( 7 im ) Cu 1( 3 im )Cho
hm s y = 42x 5 - 3x+ 2 2(1)1.Kho st v v th hm s (1).2.Vit phng trnh
tip tuyn ti im c honh x = 1 Cu 2 ( 3 im )1.Tnh tch phn ( )1 +1320I
= 2x xdx2. Tm gi tr ln nht v gi tr nh nht ca hm sy = 3 22 4 2 2 + +
x x x trn [ 1; 3] . http://www.VNMATH.com11http://www.VNMATH.com
122 N THI TT NGHIP MN TON THPT NM HC 2010 - 20113. Gii phng trnh:
16 17.4 16 0 + x xCu 3 ( 1 im )Cho khi chp S.ABC c ng cao SA= a, (a
> 0 ) v y l tam gic u. Gc gia mt bn (SBC) v mt dy bng 600 . Tnh
th tch ca ca khi chp S.ABC theo a.II. PHN RING(3 im)1. Theo chng
trnh Chun:Cu 4. a ( 2im)Trong khng gian vi h to Oxyz cho A(2 ; 0;
0) , B( 0; 4; 0 ) v C(0; 0; 4).1.Vit phng trnh mt cu qua4 im O, A,
B, C. Xc nh to tm I v tnh bn knh R ca mt cu. 2.Vit phng trnh mt
phng ( ABC) v ng thng d qua I vung gc vi (ABC).Cu 4. b (1 im )Tm s
phc z tho mn 5 z v phn thc bng 2 ln phn o ca n.Theo chng trnh nng
cao:Cu 4.a ( 2 im) Trong khng gian vi h to Oxyz,cho 2 ng thng c
phng trnh 11: 12 + 'x ty tz
23 1:1 2 1 x y z1.Vit phng trnh mt phng qua ng thng 1 v song
song vi ng thng 2 2.Xc nh im A trn 1 v im B trn 2 sao cho AB ngn
nht . Cu 4. b (1 im ) Gii phng trnh trn tp s phc: 2z2 + z +3 = 0
16http ://VNMATH.com I . PHN CHUNG CHO TT C TH SINH ( 7 im ) Cu 1 (
3 im )Cho hm s y = 4 2x + 2(m+1)x+ 1 (1)1.Kho st v v th hm s (1)
khi m = 1.2.Tm m hm s c 3 cc tr.Cu 2 ( 3 im )1.Tnh tch phn ( )1
+1320I = 4x .xdx3. Tm gi tr ln nht v gi tr nh nht ca hm sy = 3 22 4
2 1 + + x x x trn [ 2;3] . 3. Gii phng trnh: 2 33.2 2 2 60+ ++ + x
x xCu 3 ( 1 im )Cho khi chp S.ABC cy l tam gic u cnh a, (a >0).
Tam gic SAC cn ti S gc SAC bng 600 ,(SAC) (ABC) . Tnh th tch ca ca
khi chp S.ABC theo a.II. PHN RING (3 im) 2. Theo chng trnh Chun:Cu
4. a ( 2 im) http://www.VNMATH.com12http://www.VNMATH.com 122 N THI
TT NGHIP MN TON THPT NM HC 2010 - 2011Trong khng gian vi h to Oxyz
cho A(2 ; 4; -1) , B( 1; 4; -1 ) , C(2; 4; 3) v D(2; 2; -1).1.CMR
AB AC, AC AD, AD AB . Tnh th tch ca t din ABCD.2.Vit phng trnh mt
cu qua4 im A, B, C, D. Xc nh to tm I v tnh bn knh R ca mt cu. Cu 4.
b (1 im )Tnh T = 5 63 4+ii trn tp s phc.Theo chng trnh nng cao:Cu
4. a ( 2 im)Trong khng gian vi h to Oxyz cho A(4 ; 3; 2) , B( 3; 0;
0 ) , C(0; 3; 0) v D(0; 0; 3).1. Vit phng trnh ng thng i qua A v G
l trng tm ca tam gic BCD.2.Vit phng trnh mt cu tm Av tip xc (BCD).
Cu 4. b (1 im )Cho s phc 1 32 2 + z i , tnh z2 + z +3
17http://VNMATH.comI . PHN CHUNG CHO TT C TH SINH (7 im) Cu I.(3
im) Cho hm s 33 2 + y x x1. Kho st s bin thin v v th hm s cho.2.
Bin lun theo m s nghim ca phng trnh 33 2 + x x mCu II.(3 im)1. Gii
phng trnh: 123 63 3 80 0 x x2. Tnh nguyn hm: ln(3 1) x dx3. Tm gi
tr ln nht v nh nht hm s 3 2( ) 3 9 3 + + f x x x x trn on [ ] 2; 2
Cu 3.(1 im)Cho t din S.ABC c ba cnh SA, SB, SC i mt vung gc v SA=a,
SB=b, SC=c. Hai im M, N ln lt thuc 2 cnh AB, BC sao cho 1 1,3 3 AM
AB BN BC. Mt phng (SMN) chia khi t din S.ABC thnh 2 khi a din (H) v
(H) trong (H) l khi a din cha nh C. Hy tnh th tch ca (H) v (H)II .
PHN RING(3 im) : 1.Theo chng trnh chun :Cu IV.a(2 im) Trong khng
gian vi h ta Oxyz, cho im A(1 ; 4 ; 2) v mt phng (P) c phng trnh :x
+ 2y + z 1 = 0.1. Hy tm ta ca hnh chiu vung gc ca A trn mt phng
(P). http://www.VNMATH.com13http://www.VNMATH.com 122 N THI TT
NGHIP MN TON THPT NM HC 2010 - 20112. Vit phng trnh ca mt cu tm A,
tip xc vi (P).Cu V.a(1 im) Tnh th tch khi trn xoay c to bi php quay
quanh trc Ox hnh phng gii hn bi cc ng 22 1, 0, 2, 0 + y x x y x
x.2.Theo chng trnh nng cao :Cu IV.b(2 im) Cho mt phng (P):
2x+y-z-3=0 v ng thng (d): 2 31 2 2+ + x y z1. Tm ta giao im M ca ng
thng (d) v mt phng (P).2. Vit phng trnh hnh chiu ca ng thng (d) trn
mt phng (P).Cu Vb. (1 im) Xcnhtagiaoimcatimcnxincath hms23 12 +x
xyxvi parabol (P): 23 2 + y x x 18http://www.VNMATH.comCu I:(3
im):1/Kho st s bin thin v v th (C ) ca hm s y=11+xx 2/Vit phng trnh
tip tuyn vi(C) ti giao im ca ( C) vi trc tungCu II:(3im) 1/Tnh I=(
)cos0sin+xe x xdx2/Gii bt phng trnh log3( ) 2 + xlog9( ) 2 + x
3/Tnh cc cnh ca hnh ch nht c chu vi nh nht trong tt c cc hinh ch
nht c din tch 48m2Cu III: (2im) Trong khng gian Oxyz cho 3 im
A(2;2;3) ;B(1;2;-4) ;v C(1;-3;-1)1/Vit phng trnh mt phng ABC 2/Vit
phng trnh mt cu ngoi tip t din OABC.Tm ca mt cu c trng vi trng tm
ca t din khng? Cu IV:(1 im) Cho hnh chp t gic u S.ABCD c cnh y bng
a;gc SAB bng 300.Tnh din tch xung quanh ca hnh nn nh S, y l hnh trn
ngoi tip t gic ABCDCu V: (1 im)Tnh 2 153 2+ii
19http://www.VNMATH.comI . PHN CHUNG CHO TT C TH SINH ( 7 im ) Cu I
( 3,0 im ) Cho hm s3 23 1 + x y x c th (C)1. Kho st s bin thinv v
th (C). http://www.VNMATH.com14http://www.VNMATH.com 122 N THI TT
NGHIP MN TON THPT NM HC 2010 - 20112. Dng th (C), xc nh k phng trnh
3 23 0 + x x kc ng 3 nghim phn bit.Cu II ( 3,0 im )1. Gii phng
trnh:4.9 12 3.16 0. ( ) + x x xx2. Tnh tch phn: 2 230 1+xI dxx.3.
Tm gi tr ln nht, gi tr nh nht ca hm s: 24 4 . + y xCu III ( 1,0 im
)Cho hnh chp S.ABC c y ABC l tam gic vung ti A, , 3, AB a AC amt bn
SBC l tam gic u v vung gc vi mt phng y. Tnh theo a th tch ca khi
chp S.ABC.II . PHN RING( 3 im ) 1. Theo chng trnh chun :Cu IV.a(
2,0 im ): Trong khng gian vi h ta Oxyz , cho ng thng (d): 2 31 2 2+
+ x y z v mt phng(P):2 2 6 0 + + x y z.1. Vit phng trnh mt cu tm
(1; 2; 3) I v tip xc vi mt phng (P).2. Vit phng trnh mt phng ( )
cha ng thng (d) v vung gc vi mt phng (P).Cu V.a( 1,0 im ): Tnh mun
ca s phc 3(1 2 )3+izi.2. Theo chng trnh nng cao :Cu IV.b( 2,0 im ):
Trong khng gian vi h ta Oxyz , cho ng thng (d): 2 31 2 2+ + x y z v
mt phng (P):2 2 6 0 + + x y z.1. Vit phng trnh mt cu tm (1; 2; 3) I
v tip xc vi mt phng (P).2. Vit phng trnh hnh chiu vung gc ca ng
thng (d) trn mt phng (P).Cu V.b( 1,0 im ):Tmcn bc hai ca s phc 4 z
i 20http://www.VNMATH.comCu 1 : Cho hm s 33 2 + y x x(C) a.Kho st v
v th hm s (C)b.Da vo (C) bin lun theo m s nghim phng trnh : 33 1 0
+ x x m c.Tnh din tch hnh phng gii hn bi (C ) v trc Ox .Cu 2 :
a)Tnh o hm ca hm s sau :4 2os(1-3x)+xy e c;y = 5cosx+sinx b) Tm
GTLN, GTNN ca hm s 4 21( ) 24 + f x x x trn on [-2 ;0]c) Tnh gi tr
biu thc A = 9 21 log 4 2 log 3(3 ) : (4 )+ d) Gii cc phng trnh, bt
phng trnh sau :2 4 16log log log 7 + + x x x
e) tnh cc tch phn sau :I = 2211 +x x dx ; J = 2332cos 33 _ ,x dx
http://www.VNMATH.com15http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 2011Cu 3 : Tnh din tch xung quanh v th tch
khi chp t gic u c di cnh bn gp i cnh y v bng a ?Cu 4/ Cho 2 imA (0;
1; 2)vB (-3; 3; 1) a/ Vit phng trnh mt cu tm A v i qua Bb/ Vit phng
trnh tham s ca ng thng (d ) qua B v song song vi OAc/Vit phng trnh
mt phng ( OAB)Cu 5/ a/ Gii phng trnh sau trong tp tp s phc : x2 x +
1 = 0b/ Tm moun ca s phc Z = 3 2i 21Cu 1 : a)Kho st v v th hm s: y
= 22 1+xx th (C) b)Vit phng trnh tip tuyn ca (C) ti im c honh bng
-1c.) Tnh din tch hnh phng gii hn bi (C) ; tim cnh ngang ; x=0 ;
x=1Cu2: a) Tm GTLN GTNN ca hm s y = (x 6)24 + xtrn on [0 ; 3].
b)Tmm hm s: y = 33x - (m + 1)x2 + 4x + 5 ng bin trn R c)Tnh o hm cc
hm s sau: a/ ( )21 xy x e b/ y = (3x 2) ln2xc/ ( )2ln 1+xyxd) tnh
cc tch phn :I =( )221ln +ex x xdx ; J =1202 + dxx xe) Gii phng trnh
: a) 2 2log (- 3) +log (- 1) = 3 x xb)3.4 21.2 24 0 x xCu 3 : Thit
din ca hnh nn ct bi mt phng i qua trc ca n l mt tam gic u cnh aTnh
din tch xung quanh; ton phn v th tch khi nn theo a ?Cu 4 : Trong
khng gian Oxyza) Cho 4 3 +r r ra i j,rb = (-1; 1; 1). Tnh 12 r r rc
a bb) Cho 3 imA(1; 2; 2), B(0; 1; 0), C(0; 0; 1) + Tnh uuurAB .
uuurAC+ Chng minh A, B, C khng thng hng. Vit phng trnh mt phng (
ABC ).+ Vit phng trnh mt cu tm I ( -2;3;-1) v tip xc (ABC)Cu 5 : a/
Gii phng trnh : (3-2i)x + (4+5i) = 7+3i b/ Tm x;y bit : (3x-2) +
(2y+1)i = (x+1) (y-5)i . 22Cu1:Cho hm sy = x 3 - 3x2 + 2 (C)a).Kh o
st sbi n thin v vthhm s . b).Tm gi trc a m ph ng trnh :-x 3 + 3x2 +
m = 0 c 3 nghi mphn bi t. c) .Tnh di n tch hnh ph ng gi i h n b i
(C); Ox ; Oy ; x=2. Cu 2:a)Tm gi trl n nh t, gi trnhnh t c a hm s
:y = x+ 21 x http://www.VNMATH.com16http://www.VNMATH.com 122 N THI
TT NGHIP MN TON THPT NM HC 2010 - 2011b) nh m hm s: y = x3 + 3mx2 +
mx c hai cc tr .c) Cho hm s f(x) =ln 1+xe . Tnh f(ln2)d) Gii phng
trnh , Bt phng trnh:9x - 4.3x +3 < 0e) 220( sin ) cos +E x x
xdxCu 3 : Cho hnh chp tgic S.ABCD c y l hnh vung c nh a , c nh bn
SA vung gc v i y, c nh bn SC t o v i y m t gc 30o .a) Tnh di n tch
xung quanh v thtch kh i chp. b) Tm tm v bn knh m t c u ngo i ti p
hnh chp. Cu4:Trongkhnggianchohai ngthng(d1)v(d2)cphngtrnh: (d1) 2
12( )3 1 + + ' x ty t t Rz t 2)21 2 ( )1 + + ' +x my m m Rz ma.
Chng t d1 v d2 ct nhaub. Vit phng trnh mt phng (p) cha (d1)v (d2)c.
Vi t ph ng trnh m t c u ng knh OH v i H l giao i m c a hai ng th ng
trn Cu 5 :a. Tm ngh ch o c a z = 1+2i b. Gi i ph ng trnh : (3+2i)z
= z -1 23A. P hn chung cho th sinh c hai ban Cu 1: Cho hm s: 3 23 4
+ y x x. Vi m l tham s.1. Kho st v v th ( C ) ca hm s.2. Bin lun
theo m s nghim ca phng trnh: 3 23 2 1 0 + + + x x mCu 2: Gii h phng
trnh sau: 12 3 05 5 10 + '+ x yx yCu 3: Tm phn thc v phn o ca s
phcsau: 2 2(1 ) (2 1)1+ ++i izi iCu 4: Tnh th tch ca khi lng tr ng
c y l tam gic u cnh a, gc gia ng cho mt bn v y l 30 .B. P hn ring
cho th sinh tng ban Th sinh ban khoa hc t nhin lm cu 5a hoc 5bCu
5a:1. Tnh tch phn: 203cos 1sin +I x xdx2. Tm m hm s: 22 42+ +x mx
myx c 2 cc tr nm cng mt pha so vi trc honh.
http://www.VNMATH.com17http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 2011Cu 5b:Trong h to Oxyz cho cc im A(0,1,2),
B(2,3,1), C(2,2,-1). Lp phng trnh mt phng i qua A,B,C.Chng minh rng
im O cng nm trn mt phng v OABC l hnh ch nht. Tnh th tch khi chp
SOABC bit rng S(0,0,5)Th sinh ban khoa hcx hi lm cu 6a hoc 6bCu
6a:1. Tnh tch phn: 21( 1) ln +eI x xdx 2. Tm m hm s: 4 218 5 2008 y
x mx c 3 cc tr .Cu 6b:Trong h to Oxyz cho cc im:
A(0,1,1),B(1,2,4),C(-1,0,2).Hy lp phng trnh mt phng (Q) i qua
A,B,C.Lp phng trnh tham s ca ng thng i qua B v M vi M l giao im ca
mt phng (Q)( vi trc Oz. 24I. Ph n chung: Cu I: (3) Cho hm s y = x3
3x1) Kho st s bin thin v v th (C) ca hm s 2) Da vo th (C), bin lun
theo m s nghim ca phng trnh : x3 3x + m = 0Cu II : (3) 1) Gii phng
trnh : lg2x lg3x + 2 = 02) Tnh tch phn : I = / 20osxdxxe c3) Cho hm
s f(x) =x3 + 3x2 + 1 c th (C). Vit phng trnh tip tuyn ca (C) i qua
gc ta .Cu III : (1) Cho hnh chp t gic u, tt c cc cnh u bng a. Tnh
th tch hnh chp S.ABCDII. Phn ring: (3) Chng trnh chun:Cu IVa: Trong
khng gian Oxyz cho 4 im A(3 ;-2 ; -2), B(3 ;2 ;0),C(0 ;2 ;1),
D(-1;1;2)1) Vit phng trnh mt phng (BCD). Suy ra ABCD l 1 t din2)
Vit phng trnh mt cu tm A tip xc vi mt phng (BCD)Cu Va : Gii phng
trnh : x2 + x + 1 = 0 trn tp s phcChng trnh nng cao:Cu VIb: Cho 2
ng thng d1 : 434 + 'x ty tz,d2 : 21 2 '' +' xy tz t1) Tnh on vung
gc chung ca 2 ng thng d1 v d22) Vit phng trnh mt cu c ng knh l on
vung gc chung ca d1 v d2Cu Vb: Gii phng trnh:x2 + (1 + i)x ( 1 i) =
0 trn tp s phc 25 http://www.VNMATH.com18http://www.VNMATH.com 122
N THI TT NGHIP MN TON THPT NM HC 2010 - 2011I/ PHN CHUNG : (7im)C u
I : (3 im)Cho hm s Cho hm s y = (x 1)2 (4 x)1/ Kho st v v th (C) ca
hm s. Vit phng trnh tip tuyn ca th (C) ti A(2;2).2/ Tm m phng trnh:
x3 6x2 + 9x 4 m = 0, c ba nghim phn bit.C u II : ( 3 im)1/ Tnh tch
phn: I = 30(cos 4 .sin 6 )x x x dx2/ Gii phng trnh: 4x 6.2x+1 + 32
= 03/ Tm tp xc nh ca hm s: y = 31 log ( 2) xC u III : (1 im) Cho
hnh chp S.ABCD c y ABCD l hnh vung cnh a, mt bn SAB l tam gic u v
vung gc vi y. Gi H l trung im AB. Chng minh rng: SH vung gc mt phng
(ABCD). Tnh th tch khi chp S.ABCD theo a.II/ PHN RING: (3im)1. Theo
chng trnh chun:C u IV.a : (2 im)Trong khng gian Oxyz cho mt cu (S):
x2 + y2 + z2 2x 4y 6z = 0.1/ Xc nh tm v bn knh ca mt cu (S).2/ Gi A
; B ; C ln lt l giao im (khc gc to O) ca mt cu (S) vi cc trc Ox ;
Oy ; Oz. Tm to A ; B ; C. Vit phng trnh mt phng (ABC).C u V.a :
(1im)Gii phng trnh sau trn tp s phc: z2 + 4z + 10 = 02. Theo chng
trnh nng cao:C u IV.b : (2 im)Trong khng gian Oxyz cho ng thng (D):
2 1 12 3 5 + x y z v mt phng (P): 2x + y + z 8 = 0.1/
Chngtngthng(D)khngvunggcmp(P). Tmgiaoimca ng thng (D) v mt phng
(P).2/ Vit phngtrnhngthng(D) lhnhchiuvunggccang thng (D) ln mt phng
(P).C u V.b : (1im)Gii phng trnh sau trn tp s phc: (z + 2i)2 + 2(z
+ 2i) 3 = 0. 26http://www.VNMATH.comPHN CHUNG CHO TT C CC TH SINH
(7): Cu I (3):1. Kho st v v th (C) ca hm s 31++xyx2. CMR vi mi gi
tr ca m, ng thng (d) y = 2x + m lun ct (C) ti 2 im phn bit.
http://www.VNMATH.com19http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 20113. Gi A l giao im ca (C) vi trc Ox. Vit
phng trnh tip tuyn ca (C) ti A.Cu II (3): 1. Gii phng trnh: 32 log3
81xx1) Tm gi tr ln nht v gi r nh nht ca hm s: y = 2sin2x + 2sinx 1
Cu III (1):Cho t din SABC c cnh SA vung gcvi mt phng (ABC) v c SA =
a, AB = b, AC = c v 090 BAC . Tnh din tch mt cu v th tch khi cu
ngoi tip t din SABC.PHN RING (3):1.Theo chng trnh chun: Cu IV.a
(2):TrongkhnggianOxyz. ChoimM(-3;1;2)vmt phng(P)cphng trnh: 2x + 3y
+ z 13 = 0 1) Hy vit phng trnh ng thng (d) i qua M v vung gc vi mt
phmg (P). Tm ta giao im H ca ng thng (d) v mt phng (P).2) Hy vit
phng trnh mt cu tm M c bn knh R = 4. Chng t mt cu ny ct mt phng (P)
theo giao tuyn l 1 ng trn.Cu V.a (1):Tnh din tch hnh phnggii ha n
bi cc ng (P): y = 4 x2, (d): y = -x + 22.Theo chng trnh Nng cao: Cu
IV.b (2):Trong khng gian Oxyz cho 4 im A(-2;1;2), B(0;4;1),
C(5;1;-5), D(-2;8;-5) v ng thngd: 5 11 93 5 4+ + x y z. 1) Vit phng
trnh mt cu (S) ngoi tip t din ABCD.2) Tm ta giao im M, N ca (d) vi
mt cu (S).3) Vit phng trnh cc mt phng tip xc vi mt cu (S) ti M,NCu
V.b (1): Tnh din tch hnh phnggii han bi cc ng (P): y = x2 + 1, tip
tuyn ca (P) ti M(2;5) v trc Oy 27http://www.VNMATH.comCuI: ( 3
im)1/Kho st s bin thin v v th(C ) ca hm s y= -x3+3x2-3x+2.2/Tnh din
tch hnh phng gii hn bi (C ) v 2 trc ta .Cu II: (3 im)1/Cho hm s y=
xsinx .Chng minh rng :xy-2( ) ' sin y x+xy=0 2/Gii phng trnh: log3(
)3 1 x.log3( )13 3+x= 6. 3/Tnh I=33 201 +x xdxCu III( 2 im)
http://www.VNMATH.com20http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 2011Trong khng gian Oxyzcho 2 mt phng() v ('
) c phng trnh: () :2x-y+2z-1=0 v ():x+6y+2z+5=0 1/Chng t 2 mt phng
cho vung gc vi nhau. 2/Vit phng trnh mt phng() i qua gc ta v giao
tuyn ca 2 mt phng() , (' )Cu IV: (1 im):Chokhi hpABCD.ABCD
cthtch2009cm3.Tnhthtchkhi tdin CABCCu V:(1 im)Tnh mun ca s phc z
bitZ =( )2 3 i132 _+ ,i 28http://www.VNMATH.comI. PHN CHUNG (7,0 im
)Cu 1 ( 3,0 im ) Cho hm s 3 22 3 2 + y x x c th (C)1.Kho st s bin
thin v v th (C).2.Vit phng trnh tip tuyn ca (C) ti im c honh 2
ox.Cu 2 ( 3,0 im )1. Gii phng trnh 13 18.3 29+ + x x.2. Tnh tch phn
20cosI x xdx3. Tm GTLN, GTNN ca hm s 29 7 y xtrn on [-1;1].Cu 3 (
1,0 im ) Cho t din u ABCD c cnh bng 2a1. Tnh chiu cao ca t din
ABCD.2. Tnh th tch ca t din ABCD.II. PHN DNH CHO TH SINH TNG BAN (
3,0im ) Cu 4a( 2,0 im ) Cho bn im A(1;0;0), B(0;1;0), C(0;0;1),
D(-2;1;-1)1. Chng minh A, B, C, D l bn nh ca mt t din.2. Tnh th tch
ca t din .3. Lp phng trnh mt cu ngoi tip t din ABCD.Cu 5a ( 1,0 im
) Gii phng trnh27 0 + + x xtrn tp s phc. 29http://VNMATH.comI. PHN
CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 23 4 + y x x c th
(C)1.Kho st s bin thin v v th (C).2.Vit phng trnh tip tuyn ca (C)
ti tm i xng.Cu 2 ( 3,0 im )1.Gii phng trnh 6 33. 2 0 + x xe e.
http://www.VNMATH.com21http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 20112.Tnh tch phn 220sin 2 .sinI x xdx3.Tm
GTLN, GTNN ca hm s 3 22 3 12 10 + y x x x trn on [-3;3].Cu 3 ( 1,0
im )Cho hnh chp tam gic u S.ABC c cnh y bng 2a, cnh bn bng a1.Tnh
chiu cao ca hnh chp S. ABC.2.Tnh th tch ca hnh chp S.ABC.II. PHN
DNH CHO TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho mt cu (S) c
ng knh AB, bit A(6;2;-5), B(-4;0;7).1.Lp phng trnh mt cu (S).2.Lp
phng trnh mt phng (P) tip xc mt cu (S) ti im A.Cu 5a ( 1,0 im ) Gii
phng trnh22 7 0 + + x xtrn tp s phc. 30http://www.VNMATH.comI. PHN
CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 23 4 + y x x c th
(C)1.Kho st s bin thin v v th (C).2.Dng th (C), bin lun theo m s
nghim ca phng trnh 3 23 4 + + x x m.Cu 2 ( 3,0 im )1.Gii phng
trnh94log log 3 3 + xx.2.Tnh tch phn 10ln(1 ) +I x dx3.Tm GTLN,
GTNN ca hm s 5 4 y x trn on [-1;1].Cu 3 ( 1,0 im )Cho hnh chp
S.ABCD c y ABCD l hnh ch nht, cnh bn SA vung gc vi mt phng y. SA =
3a, SB = 5a, AD = a1.Tnh di AB.2.Tnh th tch ca hnh chp S.ABCD.II.
PHN DNH CHO TH SINH TNG BAN ( 3,0im )Cu 4a( 2,0 im ) Cho bn im
A(-2;6;3), B(1;0;6), C(0;2;-1), D(1;4;0)1. Vit phng trnh mt phng
(BCD). Suy ra ABCD l mt t din.2. Tnh chiu cao AH ca t din ABCD.3.
Vit phng trnh mt phng (Q) cha AB v song song vi CD.Cu 5a ( 1,0 im )
Gii phng trnh25 0 + + x xtrn tp s phc. 31I.PHN CHUNG (7,0 im )Cu 1
( 3,0 im ) Cho hm s 3 23 1 + + y x xc th (C)1.Kho st s bin thin v v
th (C).2.Vit phng trnh tip tuyn ca (C) ti im c honh 2 ox.
http://www.VNMATH.com22http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 2011Cu 2 ( 3,0 im )1.Gii bt phng trnh 24 61
13 27 + _ ,x x.2.Tnh tch phn 21ln eI x xdx3.Tm GTLN, GTNN ca hm s
1xyx trn on [-2;-1].Cu 3 ( 1,0 im )Cho hnh chp S.ABCD c y ABCD l
hnh bnh hnh. ( ) SA ABCD .SA =2a, AB = 2a, AD = 5a, gc BAD c s o
30oTnh th tch ca hnh chp S.ABCD.II. PHN DNH CHO TH SINH TNG BAN (
3,0im ) Cu 4a( 2,0 im ) Cho mt phng ( ) : 3 5 2 0 + x y z v ng thng
12 4( ) : 9 31 + +' +x td y tz t.1. Tm giao im M ca ng thng (d) v
mt phng ( ) .2. Vit phng trnh mt phng ( ) cha im M v vung gc vi ng
thng (d).Cu 5a ( 1,0 im ) Gii phng trnh22 7 0 + + x xtrn tp s phc.
32 I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 23 1 + + y x x c
th (C)1.Kho st s bin thin v v th (C).2.Vit phng trnh tip tuyn ca
(C) ti im c honh 1 ox.Cu 2 ( 3,0 im )1.Gii phng trnh log( 1) log(2
11) log 2 x x.2.Tnh tch phn ln 330 ( 1)+xxeI dxe3.Tm GTLN, GTNN ca
hm s 3 212 3 43 + + y x x x trn on [-4;0].Cu 3 ( 1,0 im )Cho hnh
chp t gic u S.ABCD c cnh y bng 2a, cnh bn bng 3a1.Tnh chiu cao ca
hnh chp S.ABCD.2.Tnh th tch ca hnh chp S.ABCD.II. PHN DNH CHO TH
SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho hai ng thng11( ) : 2 23
+'x td y tz t v //21( ) : 3 21 + 'x td y tz.Chng minh rng (d1) v
(d2) cho nhau.Cu 5a ( 1,0 im ) Gii phng trnh22 3 7 0 + + x xtrn tp
s phc. http://www.VNMATH.com23http://www.VNMATH.com 122 N THI TT
NGHIP MN TON THPT NM HC 2010 - 2011 33I. PHN CHUNG (7,0 im )Cu 1 (
3,0 im ) Cho hm s 3 23 4 + y x x c th (C)1.Kho st s bin thin v v th
(C).2.Vit phng trnh tip tuyn ca (C) ti im c ta ( 1; 2) .Cu 2 ( 3,0
im )1.Gii phng trnh 16 17.4 16 0 + x x.2.Tnh tch phn 2322( 1) x xI
x e dx3.Tm GTLN, GTNN ca hm s 1 + y xx trn khong ( 0 ; + ).Cu 3 (
1,0 im )Cho hnh chp S.ABCD c y ABCD l hnh ch nht. Cnh bn SA vung gc
vi mt phng y. SB = 5a, AB = 3a , AC= 4a. 1.Tnh chiu cao ca
S.ABCD.2.Tnh th tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN (
3,0im ) Cu 4a( 2,0 im ) Cho mt cu 2 2 2( ) : 10 2 26 170 0 + + + +
+ S x y z x y z.1. Tm to tm I v di bn knh r ca mt cu (S).2.
Lpphngtrnhngthng(d) qua imI vunggc vi mt phng( ) : 2 5 14 0 + x y
z.Cu 5a ( 1,0 im ) Gii phng trnh22 4 7 0 + x xtrn tp s phc. 34I.PHN
CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 26 9 + y x x x c th
(C)1.Kho st s bin thin v v th (C).2.Vit phng trnh tip tuyn ca (C)
ti im cc i ca n.Cu 2 ( 3,0 im )1.Gii phng trnh 1 39 4.3 3 0+ + x
x.2.Tnh tch phn ln 5 2ln 2 1xxeI dxe3.Tm GTLN, GTNN ca hm s 3 28 16
9 + y x x x trn on [1;3].Cu 3 ( 1,0 im )Cho t din u ABCD c cnh bng
32a1.Tnh chiu cao ca t din ABCD.2.Tnh th tch ca t din ABCD.II. PHN
DNH CHO TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho ba im
A(1;0;-1), B(1;2;1), C(0;2;0). Gi G l trng tm tam gic ABC.1. Vit
phng trnh ng thng OG.2. Vit phng trnh mt cu (S) i qua bn im O, A,
B, C. http://www.VNMATH.com24http://www.VNMATH.com 122 N THI TT
NGHIP MN TON THPT NM HC 2010 - 20113. Vit phng trnh cc mt phng vung
gc vi ng thng OG v tip xc vi mt cu (S).Cu 5a ( 1,0 im ) Gii phng
trnh23 9 0 + x xtrn tp s phc. 35I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im
) Cho hm s 33 y x x c th (C)1.Kho st s bin thin v v th (C).2.Dng
(C), tmcc gi tr ca m phng trnh sau c ba nghimthc 33 2 0 + x x m.Cu
2 ( 3,0 im )1.Gii phng trnh 2 2 3+ x x.2.Tnh tch phn 120ln(1 ) +I x
x dx3.Tm GTLN, GTNN ca hm s 4232 2 +xy x trn on [-1/2;2/3].Cu 3 (
1,0 im )Cho t din u ABCD c cnh bng 23b1.Tnh chiu cao ca t din
ABCD.2.Tnh th tch ca t din ABCD.II. PHN DNH CHO TH SINH TNG BAN (
3,0im )Cu 4a (2,0 im) Cho ng thng2 1 1( ) :1 2 3 + x y zdv mt phng
( ) : 3 2 0 + + x y z.1. Tm to giao im M ca ng thng (d) v mt phng (
) .2. Vit phng trnh mt phng cha (d) v vung gc vi mt phng ( ) .Cu 5a
( 1,0 im ) Gii phng trnh25 0 + + x xtrn tp s phc. 36http://ww
w.VNMATH.com I. PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 23 4
2 + + y x x x c th (C)1.Kho st s bin thin v v th (C).2.Vit phng
trnh tip tuyn ca (C) ti im c honh 1 ox.Cu 2 ( 3,0 im )1.Gii phng
trnh 1 15 5 24+ x x.2.Tnh tch phn 251(1 ) I x x dx3.Tm GTLN, GTNN
ca hm s 23 61 +x xyx trn khong (1 ; + ).Cu 3 ( 1,0 im )Cho hnh chp
t gic u S.ABCD c cnh y bng 2b, cnh bn bng 2b1.Tnh chiu cao ca
S.ABCD.2.Tnh th tch ca S.ABCD.
http://www.VNMATH.com25http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 2011II. PHN DNH CHO TH SINH TNG BAN ( 3,0im )
Cu 4a( 2,0 im ) Cho mt phng ( ) : 2 4 0 + x y z v im M(-1;-1;0).1.
Vit phng trnh mt phng ( ) qua M v song song vi ( ) .2. Vit phng
trnh ng thng (d) qua M v vung gc vi ( ) .3. Tm to giao im H ca (d)
v ( ) .Cu 5a ( 1,0 im ) Gii phng trnh22 0 + + x xtrn tp s phc.
37http://www.VNMATH.comI.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm
s 3 22 3 1 + y x x c th (C)1.Kho st s bin thin v v th (C).2.Vit
phng trnh tip tuyn ca (C) ti im cc i ca n.Cu 2 ( 3,0 im )1.Gii phng
trnh 21 22log log 2 + x x.2.Tnh tch phn 312 ln I x xdx3.Tm GTLN,
GTNN ca hm s 33 1 + y x x trn on [0;2].Cu 3 ( 1,0 im )Cho hnh chpu
S. ABC c cnh SA = AB = 321.Tnh chiu cao ca S.ABC.2.Tnh th tch ca
S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho
bn im A(1;-1;2), B(1;3;2), C(4;3;2), D(4;0;0)1. Lp phng trnh mt
phng (BCD). T suy ra ABCD l mt t din.2.Tnh th tch t din.3. Lp phng
trnh mt phng ( ) qua gc to v song song mt phng (BCD).Cu 5a ( 1,0 im
) Gii phng trnh22 2 0 + + x xtrn tp s phc.
38http://www.VNMATH.comI.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm
s 3 23 4 + y x x c th (C)1.Kho st s bin thin v v th (C).2.Tnh din
tch hnh phng gii hn bi th (C) , trc honh v hai ng thng x = 0 v x
=1.Cu 2 ( 3,0 im )1.Gii bt phng trnh 23142 _ ,x x.2.Tnh tch phn
120xI x e dx3.Tm GTLN, GTNN ca hm s 3 23 9 35 + y x x x trn on
[-4;4].Cu 3 ( 1,0 im ) http://www.VNMATH.com26http://www.VNMATH.com
122 N THI TT NGHIP MN TON THPT NM HC 2010 - 2011Cho hnh chp S. ABC
c y ABC l tam gic vung ti A. Cnh bn SA vung gc vi mt phng y. SA =
AB = 2a, BC = 3aTnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN
( 3,0im ) Cu 4a( 2,0 im ) Cho bn im A(0;-1;1), B(1;-3;2),
C(-1;3;2), D(0;1;0)1. Lp phng trnh mt phng (ABC). T suy ra ABCD l
mt t din2. Lp phng trnh ng thng (d) qua trng tm G ca tam gic ABC v
i qua gc ta . Cu 5a ( 1,0 im ) Gii phng trnh29 0 + + x x trn tp s
phc. 39http://www.VNMATH.comI.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im )
Cho hm s 3 23 2 + y x x c th (C)1.Kho st s bin thin v v th
(C).2.Tnh din tch hnh phng gii hn bi th (C) , trc honh v hai ng
thng x = -2 v x =-1.Cu 2 ( 3,0 im )1.Gii bt phng trnh 232 913 25 _
,x x2.Tnh tch phn 2sin0.cosxI e xdx3.Tm GTLN, GTNN ca hm s 3 22 3 1
+ y x x trn on 12;2 1 1 ]Cu 3 ( 1,0 im )Cho hnh chp S. ABC c y ABC
l tam gic vung ti B. Cnh bn SA vung gc vi mt phng y. SA = AB = 2a,
BC = 3aTnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0im
) Cu 4a( 2,0 im ) Cho im A(0;-1;1) v mt phng ( ) : 2 3 7 0 + x y
z1. Lp phng trnh ng thng (d) cha A v vung gc vi mt phng ( ) .2. Tnh
khong cch t A n mt phng( ) .Cu 5a ( 1,0 im ) Gii phng trnh28 0 + +
x xtrn tp s phc. 40http://VNMATH.comI.PHN CHUNG (7,0 im )Cu 1 ( 3,0
im ) Cho hm s 33 4 + y x x c th (C)1.Kho st s bin thin v v th
(C).2.Vit phng trnh tip tuyn ca (C) tai dim c honh xo l nghimca
phng trnh //( ) 6 oy xCu 2 ( 3,0 im )1.Giiphng trnh 25 6.5 5 0 + x
x.2.Tnh tch phn 1ln eI x xdx
http://www.VNMATH.com27http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 20113.Gii bt phng trnh 20, 2 0, 2log 5log 6 x
xCu 3 ( 1,0 im )Cho hnh chp S. ABC c y ABC l tam gic vung ti C. Cnh
bn SA vung gc vi mt phng y. SA = AB = 5a, BC = 3aTnh th tch ca
S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho
ba im A(1;0;4), B(-1;1;2), C(0;1;1)1. Chng minh tam gic ABC vung.2.
Lp phng trnh ng thng (d) qua trng tm G ca tam gic ABC v i qua gc ta
. Cu 5a ( 1,0 im ) Tnh gi tr biu thc: 22( 3 )( 3 )+iPi 41I.PHN
CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 4 22 2 + y x x c th
(C)1.Kho st s bin thin v v th (C).2.Dng th (C), bin lun theo m s
nghim ca phng trnh 4 22 2 + x x mCu 2 ( 3,0 im )1.Giiphng trnh22 26
43log 2 log+ x x.2.Tnh tch phn 32041+xI dxx3.Tnh gi tr biu thc 2009
2009log(2 3) log(2 3) + + A Cu 3 ( 1,0 im )Cho hnh chp S. ABC c y
ABC l tam gic vung ti A. Cnh bn SB vung gc vi mt phng y. SA = 5a,
AB = 2a, BC = 3aTnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN
( 3,0im ) Cu 4a( 2,0 im ) Chohai im A(1;2;-1), B(7;-2;3) v ng thng
1 3( ) : 2 22 2 + ' +x td y tz t1. Lp phng trnh ng thng AB.2.
ChngminhngthngABvngthng(d)cngnmtrongmt mt phng. Cu 5a ( 1,0 im )
Gii phng trnh22 9 0 + + x x trn tp s phc. 42I.PHN CHUNG (7,0 im )Cu
1 ( 3,0 im ) Cho hm s 3 2123 + y x x c th (C)1.Kho st s bin thin v
v th (C).2.Vit phng trnh tip tuyn ca (C) ti tm i xng ca n.Cu 2 (
3,0 im )1.Giiphng trnh2 4log log ( 3) 2 x x.
http://www.VNMATH.com28http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 20112.Tnh tch phn 2213 +I x x dx3.Tm GTLN,
GTNN ca hm s 3 23 7 1 + y x x x trn on [0;3].Cu 3 ( 1,0 im )Cho hnh
chp S. ABC c y ABC l tam gic vung ti C. Cnh bn SA vung gc vi mt
phng y. SA = BC, bit CA = 3a, BA = 5aTnh th tch ca S.ABC.II. PHN
DNH CHO TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho ba im
A(0;2;1), B(3;0;1), C(1;0;0)1. Lp phng trnh mt phng (ABC). 2. Lp
phngtrnhng thng(d) qua M(1;-2;1/2)v vunggc mtphng (ABC).3. Tnh
khong cch t im M n mt phng (ABC). Cu 5a ( 1,0 im ) Tnh gi tr ca biu
thc25 3 31 2 3 _+
,iPi 43I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 4 214 + y x
x c th (C)1.Kho st s bin thin v v th (C).2.Dng th (C), tm cc gi tr
ca m phng trnh sau c bn nghim thc 422 04 + xx m.Cu 2 ( 3,0 im
)1.Giiphng trnh 1 22log (2 3) log (3 1) 1 + + + x x.2.Tnh tch phn
21lnexI dxx3.Gii bt phng trnh 2 13 3 28+ + x x .Cu 3 ( 1,0 im )Cho
hnh chpS. ABC c y ABC l tam gic vung cn ti A. Cnh bn SA vung gc vi
mt phng y. SA = AB = 2a.Tnh th tch ca S.ABC.II. PHN DNH CHO TH SINH
TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho hai im A(1;0;-2), B(0;1;1)1.
Lp phng trnh ng thng i hai A v B.2. Lp phng trnh mt cu (S) c ng knh
l AB. Cu 5a ( 1,0 im ) Tnh gi tr ca biu thc20101 _ + ,ii 44I.PHN
CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 4 22 3 + + y x x c th
(C)1.Kho st s bin thin v v th (C).2.Dng th (C) , bin lun theo m s
nghim ca phng trnh4 22 0 x x mCu 2 ( 3,0 im )
http://www.VNMATH.com29http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 20111.Gii phng trnh 1 14 6.2 8 0+ + + x
x.2.Tnh tch phn 22 302. +I x x dx3.Tm GTLN, GTNN ca hm s 3 23 9 + y
x x x trn on [-2;2].Cu 3 ( 1,0 im )Cho hnh chp S. ABC c y ABC l tam
gic vung ti B. Cnh bn SC vung gc vi mt phng y. SC = AB = a/2, BC =
3aTnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0im ) Cu
4a( 2,0 im ) Cho hai imM(3;-4;5), N(1;0;-2)1. Lp phng trnh cu i qua
M v c tm l N.2. Lp phng trnh mt phng qua M tip xc vi mt cu. Cu 5a (
1,0 im ) Gii phng trnh22 3 11 0 + + x xtrn tp s phc. 45I.PHN CHUNG
(7,0 im )Cu 1 ( 3,0 im ) Cho hm s 4 2112 + y x x c th (C)1.Kho st s
bin thin v v th (C).2.Lp phng trnh tip tuyn ca (C) ti im c honh
bng2 . Cu 2 ( 3,0 im )1.Gii bt phng trnh 262 55 2 _ _ , ,x x.2.Tnh
tch phn 201 3cos .sin +I x xdx3.Gii phng trnh 3 3log log ( 2) 1 + +
x x Cu 3 ( 1,0 im )Cho hnh chp S. ABCD c y ABCD l hnh vung cnh a.
Cnh bn SA vung gc vi mt phng y, SA = 2a.Tnh th tch ca S.ABCD.II.
PHN DNH CHO TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho im
H(1;0;-2) v mt phng ( ) : 3 2 7 0 + + x y z1. Tnh khong cch t H n
mt phng ()2. Lp phng trnh mt cu c tm H v tip xc vi mt phng ()Cu 5a
( 1,0 im ) Tnh gi tr ca 2010(1 ) +i 46http://www.VNMATH.comI.PHN
CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 4 21 34 2 + y x x c th
(C)1.Kho st s bin thin v v th (C).2.Dng th (C), bin lun theo m s
nghim phng trnh 4 22 3 + x x mCu 2 ( 3,0 im )1.Giiphng trnh 24 2.5
10 x x x.2.Tm nguyn hm ca hm s 3cos .sin y x x
http://www.VNMATH.com30http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 20113.Tm GTLN, GTNN ca hm s 22 5 42+ ++x xyx
trn on [0;1].Cu 3 ( 1,0 im )Cho hnh chpS. ABCD c y ABCD l hnh ch
nht. Cnh bn SA vung gc vi mt phng y. SA = AC , AB = a, BC = 2AB.Tnh
th tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0im ) Cu 4a(
2,0 im ) Cho im M(1;4;2) v mt phng ( ) : 1 0 + + x y z1. Lp phng
trnh ng thng (d) qua M v vung gc vi mt phng ( ) 2. Tm to giao im H
ca (d) v mt phng ( ) Cu 5a ( 1,0 im ) Tnh gi tr ca biu thc( ) ( )2
23 3 + + P i i 47http://www.VNMATH.comI.PHN CHUNG (7,0 im )Cu 1 (
3,0 im ) Cho hm s 11+xyx c th (C)1.Kho st s bin thin v v th
(C).2.Lp phng trnh tip tuyn ca (C) ti im c honh 2 ox.Cu 2 ( 3,0 im
)1. Gii phng trnh 2.4 17.2 16 0 + x x.2.Tnh tch phn 11 ln +exI
dxx3.Tm GTLN, GTNN ca hm s 115 + +y xx (x > 5 )Cu 3 ( 1,0 im
)Tnh th tch ca khi t din u c cnh bng aII. PHN DNH CHO TH SINH TNG
BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho mt phng ( ) : 3 5 2 0 + x y z v
ng thng12 9 1( ) :4 3 1 x y zd1. Tm to giao im H ca (d) v mt phng (
) .2. Lp phng trnh mt cu (S) qua H v c tm l gc ta .Cu 5a ( 1,0 im )
Gii phng trnh22 11 0 + x x trn tp s phc. 48http://VNMATH.comI.PHN
CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 22 1 ++xyx c th (C)1. Kho
st s bin thin v v th (C).2. Tnh din tch hnh phng gii hn bi th (C),
trc honh v cc ng thng x = 0 v x = 2. Cu 2 ( 3,0 im )1. Giiphng trnh
2 1 22log (1 3 ) log ( 3) log 3 + x x.
http://www.VNMATH.com31http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 20112.Tnh tch phn 522 ln( 1) I x x dx3.Tnh th
tch vt th trn xoay, sinh bi mi hnh phng gii hn bi cc ng sau y khi n
quay quanh trc Ox: 20; 2 y y x x .Cu 3 ( 1,0 im )Cho hnh chp t gic
u S. ABCD c cnh y bng 3cm, cnh bn bng 5cm. Tnh th tch ca S.ABCD.II.
PHN DNH CHO TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho ba im
A(2;-1;-1), B(-1;3;-1), M(-2;0;1). 1. Lp phng trnh ng thng (d) i
qua A v B.2. Lp phng trnh mt phng( ) cha M v vung gc vi ng thng
AB.3. Tm to giao imca (d) v mt phng ( ) Cu 5a ( 1,0 im ) Gii phng
trnh 213 02+ + x x trn tp s phc. 49http://www.VNMATH.comI.PHN CHUNG
(7,0 im )Cu 1 ( 3,0 im ) Cho hm s 3 22++xyx c th (C)1. Kho st s bin
thin v v th (C).2.Tm trn th (C)nhng im c to l cc s nguyn.Cu 2 ( 3,0
im )1. Giiphng trnh 2 24. 3 x xe e.2.Tnh tch phn221ln I x xdx3.Tm
GTLN, GTNN ca hm s 23 1xyx trn on [-1;-1/2].Cu 3 ( 1,0 im
)Chohnhhpchnht ABCD.A/B/C/D/cchiudi 6cm, chiurng5cm, chiu cao 3cm.
1. Tnh th tch ca khi hp ch nht.2. Tnh th tch ca khi chp A/.ABD.II.
PHN DNH CHO TH SINH TNG BAN ( 3,0im )Cu4a(2,0im) Chomt cu 2 2 2( )
: 4 8 2 4 0 + + + + S x y z x y zvmt phng ( ) : 3 5 1 0 + + x y z1.
Xc nh ta tm I v di bn knh r ca mt cu (S).2. Lp phng trnh ng thng
(d) qua im I v vung gc vi mt phng ( ) .Cu 5a ( 1,0 im ) Tnh gi tr
ca biu thc ( )( )2233+iPi 50http://www.VNMATH.comI.PHN CHUNG (7,0
im )Cu 1 ( 3,0 im ) Cho hm s 12+xyx c th (C)1. Kho st s bin thin v
v th (C). http://www.VNMATH.com32http://www.VNMATH.com 122 N THI TT
NGHIP MN TON THPT NM HC 2010 - 20112. Lp phng trnh tip tuyn ca (C)
ti giao im vi trc honh.Cu 2 ( 3,0 im )1. Giiphng trnh 1 15 5 26+ +
x x.2. Tnh tch phn221ln(1 ) +I x x dx3. Tm GTLN, GTNN ca hm s 2 11
3+xyx trn on [-1;0].Cu 3 ( 1,0 im )Cho hnh lng tr ngABC.A/ C/ B/ c
y ABC l tam gic vung ti A. AB = 4cm,BC = 5cm,AA/ = 6cm.1. Tnh th
tch ca khi lng tr .2. Tnh th tch ca khi chp A/ .ABC.II. PHN DNH CHO
TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho ba im A(3;0;4),
B(1;2;3), C(9;6;4) 1. Tm to im D sao cho ABCD l hnh bnh hnh.2. Lp
phng trnh mt phng (BCD).Cu 5a ( 1,0 im ) Tnh gi tr ca biu thc ( )(
)231 3+iPi 51http://www.VNMATH.comI.PHN CHUNG (7,0 im )Cu 1 ( 3,0
im ) Cho hm s 312 ++yx c th (C)1. Kho st s bin thin v v th (C).2.
Tm trn th (C)nhng im c to l cc s nguyn.Cu 2 ( 3,0 im )1. Giiphng
trnh47log 2 log 06+ + xx2. Tnh tch phn220( sin ) cos +I x x xdx3.Tm
GTLN, GTNN ca hm s 3 23 4 y x x trn on [-1;1/2].Cu 3 ( 1,0 im )Cho
hnh chpS. ABCD c y ABCD l hnh ch nht. Cnh bn SA vung gc vi mt phng
y. SA = 2a , AB = 3a, BD = 5a.Tnh th tch ca S.ABCD.II. PHN DNH CHO
TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho im I(-2;1;1) v mt
phng ( ) : 2 2 5 0 + + x y z1. Tnh khong cch t imI n mt phng ( ) 2.
Lp phng trnh mt cu (S) c tm l I v tip xc vimt phng ( ) Cu 5a ( 1,0
im ) Tnh gi tr ca biu thc 341 3 _
+ ,iPi 52 http://www.VNMATH.com33http://www.VNMATH.com 122 N THI
TT NGHIP MN TON THPT NM HC 2010 - 2011http://VNMATH.comI.PHN CHUNG
(7,0 im )Cu 1 ( 3,0 im ) Cho hm s 32+yx c th (C)1. Kho st s bin
thin v v th (C).2. Lp phng trnh tip tuyn ca (C) ti giao im vi trc
tung.Cu 2 ( 3,0 im )1. Giiphng trnh 2 2log log 3 x x.2. Tnh tch
phn420sin ( )4 I x dx3.Tm GTLN, GTNN ca hm s 24 y xCu 3 ( 1,0 im )
Cho hnh chpS. ABC c y ABC l tam gic vung ti B. Cnh bn SC vung gc vi
mt phng y. SC = AB = a/3, BC = 3aTnh th tch ca S.ABC.II. PHN DNH
CHO TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0 im ) Cho im M(-2;3;1) v ng
thng 2 1 2( ) :2 2 3 + + x y zd1. Lp phng trnh tham s ca ng thng
(d/) qua M v song songvi ng thng (d).2. Tm to im M/ l hnh chiu vung
gc ca M trn (d). Cu 5a ( 1,0 im ) Tnh gi tr ca biu thc 20041 _
+ ,iPi 53I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 21 xyx c
th (C)1.Kho st s bin thin v v th (C).2.Tnh din tch hnh phng gii hn
bi th (C), trc honh v cc ng thng x = -3 v x = -2. Cu 2 ( 3,0 im )1.
Giiphng trnh 0,5 0,5 2 14 3 3 2 + x x x x.2. Tnh tch phn210.xI e
xdx3.Tm GTLN, GTNN ca hm s 11 +y xx trn khong (1; ) +.Cu 3 ( 1,0 im
) Cho hnh chpS. ABCD c y ABCD l hnh ch nht. Cnh bn SA vung gc vi mt
phng y. SA = 2a , AB = a, AC = 3a.1). Tnh th tch ca S.ABCD.2). Chng
minh ( ) BC SABII. PHN DNH CHO TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0
im ) Cho mt phng ( ) : 1 0 + + x y zv ng thng 2( ) : 13 ' +x td y
tz t http://www.VNMATH.com34http://www.VNMATH.com 122 N THI TT
NGHIP MN TON THPT NM HC 2010 - 20111. Tm to giao im H ca (d) v mt
phng ( ) .2. Lp phng trnh mt phng trung trc ca on OH.Cu 5a ( 1,0 im
) Gii phng trnh 38 0 + x trn tp s phc. 54I.PHN CHUNG (7,0 im )Cu 1
( 3,0 im ) Cho hm s 1221+xyx c th (C)1. Kho st s bin thin v v th
(C).2. Lp phng trnh tip tuyn ca (C) ti giao im vi trc honh.Cu 2 (
3,0 im )1. Giibt phng trnh 20,5 0,5log log 2 0 + x x.2. Tnh tch
phn21lnexI dxx3.Tm GTLN, GTNN ca hm s 33 3 + y x x trn on
[-3;3/2].Cu 3 ( 1,0 im )Cho hnh chpS. ABCD c y ABCD l hnh ch nht.
Cnh bn SA vung gc vi mt phng y. SA = AC , AB = 5cm, BC = 2AB.Tnh th
tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN ( 3,0im ) Cu 4a( 2,0
im ) Cho bn im A(6;-2;3), B(0;1;6), C(2;0;-1), D(4;1;0) 1. Lp phng
trnh mt phng (BCD). T suy ra ABCD l mt t din.2. Tnh th tch ca t
din.Cu5a(1,0im)Tnhdintchhnhphnggii hnbith cahms 2 24; 2 y x y x x
55I.PHN CHUNG (7,0 im )Cu 1 ( 3,0 im ) Cho hm s 4 12 3++xyx c th
(C)1.Kho st s bin thin v v th (C).2.Tm GTLN, GTNN ca hm s trn on 5;
22 1 1 ]Cu 2 ( 3,0 im )1.Giibt phng trnh 20,5log ( 5 6) 1 + x x.2.
Tnh tch phn22sin 2 .sin 7 I x xdx3. Tnh din tch hnh phng gii hn bi
th ca hm s21; 3 + + y x x yCu 3 ( 1,0 im )Cho hnh chpS. ABC c y ABC
l tam gic vung cn ti A. Cnh bn SA vung gc vi mt phng y. SA = AB =
5a/2.Tnh th tch ca S.ABC.II. PHN DNH CHO TH SINH TNG BAN ( 3,0im )
Cu 4a( 2,0 im ) Cho im H(2;3;-4) v im K(4;-1;0)
http://www.VNMATH.com35http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 20111.Lp phng trnh mt phngtrung trc ca on
HK.2.Lp phng trnh mt cu (S) c ng knh l HK. Cu 5a ( 1,0 im ) Tnh gi
tr ca biu thc( ) ( )2 23 3 + P i i 56I.PHN CHUNG (7,0 im )Cu 1 (
3,0 im ) Cho hm s 1 22 4xyx c th (C)1. Kho st s bin thin v v th
(C).2. Tm trn th (C)nhng im c to l cc s nguyn.Cu 2 ( 3,0 im )1.
Giiphng trnh 1 2 32 2 2 448 + + x x x.2.Tm nguyn hm ca hm s21cos (3
2)+yx3.Tm cc tr ca hm s 21 + y x x Cu 3 ( 1,0 im ) Cho hnh chp t
gic u S.ABCD c cnh y bng 3a, cnh bn bng 3a1.Tnh chiu cao ca
S.ABCD.2.Tnh th tch ca S.ABCD.II. PHN DNH CHO TH SINH TNG BAN (
3,0im ) Cu 4a( 2,0 im ) Cho im I(-2;1;0) v mt phng ( ) : 2 2 1 0 +
+ x y z1. Lp phng trnh ng thng (d) qua I v vung gc vi mt phng ( )
2. Tm to hnh chiu vung gc ca I trn mt phng ( ) Cu 5a ( 1,0 im ) Tnh
din tch hnh phng gii hn bi th ca cc hm s; 2; 1 xy e y x
57http://www.VNMATH.comI. PHN CHUNG CHO TT C TH SINH (7 im)Cu I (3
im)Cho hm s 22 ++xyx.1. Kho st s bin thin v v th ca hm s cho.2. Vit
phngtrnhtiptuynca(C), bit nvunggcvi ngthng 1422 y xCu II (3 im).1.
Gii phng trnh :6.4 13.6 6.9 0 + x x x2. Tnh tch phn : 23 3 213 4.
+I x x dx3. Tm gi tr ln nht v gi tr nh nht ca hm s : 2( ) cos cos 3
+ + f x x x.Cu III(1 im)Cho khi chp tam gic u S.ABCc y ABC l tam
gic u cnh a v cc cnh bn to vi y mt gc 600. Hy tnh th tch ca khi chp
.II. PHN RING (3 im) http://www.VNMATH.com36http://www.VNMATH.com
122 N THI TT NGHIP MN TON THPT NM HC 2010 - 2011Th sinh hc chng
trnh no th ch c lm phn dnh ring cho chng trnh (phn 1 hoc 2)1. Theo
chng trnh Chun :Cu IVa (2 im)Trong khng gian vi h trc ta Oxyz, cho
cc im A(1 ; 0 ; 2), B(-1 ; 1 ; 5), C(0 ; -1 ; 2) v D(2 ; 1 ; 1)1.Lp
phng trnh mt phng (P) cha AB v song song vi CD.2. Vit phng trnh mt
cu (S) i qua 4 im A, B, C, D.Cu Va (1 im)Tm mun ca s phc 8 31 izi2.
Theo chng trnh Nng cao :CuIVb(2 im)Trong khng gian vihtrc ta
Oxyz,cho ng thng (d)v mt phng()ln lt c phng trnh :5 3 1( ) :1 2 3 +
x y zd, ( ) : 2 2 0 + x y z1. Vit phng trnh mt phng ()i qua giao im
Ica (d)v ()v vung gc (d).2. Cho A(0 ; 1 ; 1). Hy tm to im B sao cho
() l mt trung trc ca on AB.Cu Vb (1 im)Tm s phc z sao cho 31++z iz
iv z + 1 c acgumen bng 6. 58http://www.VNMATH.comI.PHN CHUNG (7
im)Cu I (3 ) Cho hm s y = x3 +(m -1) x2 (m +2)x -1 (1)a) Kho st v
th (C) ca hm s khi m = 1b) Vit phng trnh ng thng (d) vung gc vi ng
thng y = 3xv tip xc vi th (C) ca hm s a) Tm ta hnh chiu vung gc v t
im A(2; 0; -1) ln ng thng (d).b) Tm ta giao im B i xng ca A qua ng
thng (d).Cu II (3 ) 1) Gii phng trnh 16x -17.4x +16 = 0;2) Tnh tch
phn( )202 1 sin+x xdx3) Tm gi tr ln nht ca biu thc( )2sin0, 5xCu
III (1) Cho hnh chp tam gic S.ABC c SA, SB, SC i mt vung gc nhau v
SA = a, SB = b,SC = c. Tnh di ng cao v t S ca hnh chp S.ABC.II.PHN
RING (3 im)1 THEO CHNG TRNH CHUN Cu IV.a (2) Trong khng gian Oxyz
cho ng thng (d): 1 213 + +' x ty tz t
http://www.VNMATH.com37http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 2011a) Vit phng trnh mt phng (P) i qua A(2;
0; 0) v vung gc vi ng thng (d)b) Tm ta giao im ca (d) vi mt phng
(P).Cu IV.b (1) Gii phng trnh sau trn tp s phc( )2 3 2 3 2 2 + + i
x i i2. THEO CHNG TRNH NNG CAOCu IV.a (2) Trong khng gian Oxyz cho
ng thng (d): 1 213 + +' x ty tz tCu IV.b (1) Tm gi tr ln nht ca biu
thc 3 +x xy 59I- PHN CHUNG CHO TT C TH SINH (7.0 im)Cu I (3.0 im):
Cho hm s 4 22( 1) 2 1 + + y x m x m , c th (Cm) 1) Kho st v v th(C)
khi 0 m 2) Vit pttt vi (C) ti im c honh 2 xCu II (3.0 im): 1) Gii
bt phng trnh:232 3log 01 + x 2) Tnh tch phn sau:21n 1. nx x+l
lexdx3) Tm GTLN, GTNN ca hm s cos 2 1 y xtrn on [0;].Cu 3 (1 im)Cho
hinh cho p S.ABCD coa y lahinh chnh t, canh BC = 2a, SA = a, SA
vung go c v i mp(ABCD),SB ta o vi m t ay 1 go c 450. Ti nh thtich
cua kh i c u ngoa i ti p hinh cho pS.ABCD ?II. PHN RING ( 3 im)
Thisinh cho n m t trong hai ph n sau y :
http://www.VNMATH.com50http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 2011A. Theo chng trnh chun:Cu 4a (2 im) Trong
khng gian vi h ta Oxyz, cho hai ng thng:1 21 1 2 21 21 2 2 3: 3: 11
2 2 + + ' ' + x t x td y t d y tz t z t1) Ch ng minh r ng hai ng
thng trn che o nhau2) Vit phng trnh mt ph ng (P) ch a d1vasong song
v i d2Cu 5a (1 im) Ti m sphc z tho a : z4 + z2 12 = 0B. Theo chng
trnh nng cao:Cu 4b (2 im) Trong khng gian vi h ta Oxyz, cho 1 1:2 1
2 + x y zd1) Vi t phng tri nh ng thng nm trong mt ph ng Oxy, vung
go c v i dvact d2) Vit phng trnh m t ph ng () ch a d vah p vi Oxy m
t go c benh t.Cu 5b (1 im)Gia i phng tri nh sau trn t p hp ca c
sphc : z2 (1+5i)z 6 + 2i = 0. 75http://www.VNMATH.comI. PHN CHUNG
CHO TT C TH SINH (7 im)Cu 1 (3 im) Cho hm s3 52 2++xyx.1)Kho st s
bin thin v v th (C) ca hm s cho.2)Vi t phng tri nh ti p tuy n v i
(C) tai im cohoa nh bng 1.Cu 2 (3 im) 1) Gii b t phng tri nh:3log 2
21 14 28log 5log 3 0 + + x x 2) Tnh tch phn sau:20cos . 3sin 1.+x x
dx3) Tm GTLN, GTNN ca hm s 24 1 + y xtrn on [0;1].u 3 (1 im) Cho
hinh cho p S.ABC coay ABC tam gia c u ca nh a, canh SA vung go c v
i mp(ABC), go c ASC b ng 600 . Ti nh thti ch cua kh i cho p S.ABC
theo a ?II. PHN RING ( 3 im) Thisinh cho n m t trong hai ph n sau y
:A. Theo chng trnh chun:Cu 4a (2 im) Trong khng gian vi h ta Oxyz,
cho im A (1;2;3) vang th ng d cophng tri nh tham s: 1 21 2 ' +x td
y tz t1) Vi t phng tri nh m t cu (S) tm A vai qua O2) Vit phng trnh
mt phng qua A vavung go c v i ng th ng d. Xa c i nh khoa ng ca ch
tA n ng thng d ?Cu 5a (1 im) Ti m moun cua sph c z v i z = 36 22
3++iiB. Theo chng trnh nng cao:
http://www.VNMATH.com51http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 2011Cu 4b (2 im) Trong khng gian vi h ta
Oxyz, cho i m A (1;2;3) vang th ng 1 1:1 2 2 + x y zd
1) Vi t phng tri nh mt cu (S) tm A vati p xuc v i mp() : 2x y 2z
+1 = 02) Xa c i nh khoa ng ca ch t A n ng th ng d ?Cu 5b (1 im) Go
i z1 vaz2 la nghi m cu aphng tri nh z2 + z + 1=0. Ha y xac i nhA =
1 21 1+z z 76I.PHN CHUNG CHO TT C TH SINH (7,0 im)Cu I : ( 3 im
)Cho hm s y = f(x) = - x4 2(m 1)x2 + 2m 11) nh m th hm s ct trc
honh ti 3 im phn bit.2) Kho st s bin thin v v th (C) ca hm s khim =
0.3) Xc nh a phng trnh sau c 4 nghim thc phn bit : x4 2x2 + a = 0Cu
II: ( 3 im )1. Gii cc phng trnh v bt phng trnh sau:2. a) 2 22 9.2 2
0+ + x xb) 2 2log ( 3) log ( 2) 1 + x x2. Tnh tch phn a) I = 120(2
1) xx e dxb) J = 2201 x dx3. Tm GTLN, GTNN ca hm s y = 24 +xx .Cu
III : ( 1 im )Cho hnh lng tr ABC.ABC cy ABC l tam gic u cnh bng a,
gc gia cnh bn v mt y bng 600, Hnh chiu ca nh A ln mt phng (ABC)
trng vi tm ca tam gic ABC. Tnh th tch khi lng tr trn.II. PHN RING1.
Theo chng trnh Chun :Cu IV.a (2 im)Trong khng gian Oxyz choim
A(1;1;3) v ng thng (d) : 11 1 2 y x z1) Tm ta im H l hnh chiu vung
gc ca A ln ng thng (d) .2) Lp phng trnhmt cu tm A v tip xc ng thng
(d) .3) Tm im M thuc ng thng (d) sao cho tam gic OAM cn ti nh O.Cu
Va : ( 1 im ) 1.Xc nh tp hp cc im trong mt phng phc biu din cc s
phc z tha mn iu kin : 2 z i2.Gii phng trnhtrn tp s phc: z2- 2z + 5
= 02.Theo chng trnh nng caoCu IV.b (2 im)Trong khng gian Oxyz
chohai ng thng 12 2: 11 + +'x ty tzv 21: 1 '3 ' +' xy tz t
http://www.VNMATH.com52http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 20111.CMR:1 cho2. Tnh khong cch gia hai ng
thng1, 2.2. Vit phng trnh ng thng d qua im A(2,-1,0) vung gc1 v
ct2.Cu V.b (1 im) Gii phng trnhtrn tp s phc:z2 (3+4i) z + (-1+5i)
=0 77I.PHN CHUNG CHO TT C TH SINH (7,0 im)Bi 1 : ( 3 im )Cho hm s
:3 22 (3 ) 2 + + y x m x mx;m l tham s.1./ nh m :a. Hm s ng bin tng
khong trn tp xc nh.b. Hm s c cc tr.2./ Kho st v v th (C) ca hm s
khim= 0.3./ nh a phng trnh :3 222 3 log 0 x x ac 3 nghim phn bit.Bi
2 : ( 3 im )1./ V th ca hm s :2log ( 2) y x.2./ Tnh cc tch phn :2
520 3ln( 2).4 + dxA B x dxx3./ Tm gi tr ln nht v gi tr nh nht ca :
2( ) sin cos 2 + + f x x x.Bi 3 : (1 im )Hnh chp S.ABCD c y l hnh
vung cnh a. Cnh SA vung gc vi y. Cnh SC hp vi y gc 450.1./ Tnh th
tch khi chp theo a.2./ Tnh din tch mt cu ngoi tip hnh chp theo a
.II. PHN RING1. Theo chng trnh Chun :Bi 4 : (2 im )Trong khng gian
Oxyz cho A(-4;-2;4) v ng thng d:3 211 4 + ' +x ty tz t1./ Tm to im
H l hnh chiu vung gc ca A ln ng thng d.2./ Vit phng trnh ng thng d1
qua A , vung gc vi d v ct d.Bi 5 : (1 im)Tnh din tch hnh phng gii
hn bi :215 ' +y xy x 78I.PHN CHUNG CHO TT C TH SINH (7,0 im)Cu 1 :
(3 im )Cho hm s3 23 4 + x x y c th (C)1. Kho st s bin thinv v th
(C).2. Cho h ng thng ( ) : 2 16 +md y mx m vi m l tham s. Chng minh
rng ( )md lun ct th (C) ti mt im c nh I .Cu 2 : (3 im) 1. Gii phng
trnh4 2log log (4 ) 5 + x x. 2. Gii bt phng trnh : 32.4x 18.2x + 1
< 0.3. Tnh tch phn :I = 10( ) +xx x e dx
http://www.VNMATH.com53http://www.VNMATH.com 122 N THI TT NGHIP MN
TON THPT NM HC 2010 - 20114. Tm GTLN, GTNN ca hm s y = 222+ ++x xx
trn on [-1 ; 3].Cu 3 : (1 im) Cho hnh chp S.ABC c y ABC l tam gic u
cnh a, SA = a 3, SA vung gc vi mt phng (ABC). Gi J l trng tm tam
gic SBC. Tnh th tch khi chp J.ABC?II. PHN RING1. Theo chng trnh
Chun :Cu 4: ( 2 im) Trong khng gian vi h trc ta Oxyz cho ba im
A(1;0;0), B(0;-2;0), C(0;0;3).a) Vit phng trnh mt phng (ABC)b) Xc
nh ta im D sao cho t gic ABCD l hnh bnh hnh.c) Cho S(-3;4;4) . Vit
phng trnh ng cao SH ca khi chp S.ABCD, suy ra ta chn ng cao H.Cu 5:
( 1 im)Cho hm s 21xyxc th (C).Tnh din tch hnh phng gii hn bi (C),
trc Ox v x = -3. 79I.PHN CHUNG CHO TT C TH SINH (7,0 im)Bi 1: Cho
hm s 44yx(C)a. Kho st s bin thin v v th (C)b. Vit phng trnh tip
tuyn (d) ca (C) ti im thuc (C) c honh l 3c. Tm din tch hnh phng gii
hn bi (C), tip tuyn (d) v trc Oy.d. Bin lun theo k s giao im ca (C)
v ng thng () i qua A(-4, 0), c h s gc k.Bi 2: a. Gii phng trnh: 4
10 2.25 + x x xb. Gii bt phng trnh: 5 15log ( 1) log ( 2) 0 + x xc.
Tm GTLN, GTNN ca hm s: 24 + y x xBi 3: Mt bn ca mt hnh nn c cun t
mt na hnh trn c bn knh r. Tm th tch ca hnh nn theo r.II. PHN RING1.
Theo chng trnh Chun :Bi 4: TrongkhnggianOxyzchoD(-3, 1, 2) v() i
qua3imA(1,0,11), B(0,1,10), C(1,1,8).a. Vit phng trnh ng thng ACb.
Vit phng trnh mt phng ()c. Vit phng trnh mt cu (S) tm D, bn knh R =
5. CMR () ct (S).Bi 5: Tm 2 s phc bit tng ca chng l 2 v tch ca chng
l 3 80http://www.VNMATH.comI.PHN CHUNG CHO TT C TH SINH (7,0 im)Bi
1: ( 3 im )Cho hm s y = ( 2 x2 )2 C th (C) .1/. kho st v th ( C )
ca hm s . http://www.VNMATH.com54http://www.VNMATH.com 122 N THI TT
NGHIP MN TON THPT NM HC 2010 - 20112/. Da vo th ( C ) , bin lun
theo m s nghim ca : x4 -4x2 m = 03/. Gi A l giao im ca ( C ) v Ox ,
xA > 0 . Vit phng trnh tip tuyn vi ( C ) ti im A .Bi 2: ( 3 im
)1/. Gii phng trnh - bt phng trnh :a/. 4x 2.2x+1 + 3 = 0 b/. 3
5131log+xx2/. Tnh cc tch phn :a/. I = 016 2.24 4 1 + xdxx xb/. I =
2( 1).sin .0+x x dx3/. Tm GTLN , GTNN ca cc hm s :a/. y = x4 2x2 +1
trn [ ] 0; 2b/. y = cos2x + sinx +2Bi 3: ( 1 im )Trong khng gian
cho hnh vung ABCD cnh 2a . Gi M,N ln lt l trung im ca AB v CD . Khi
quay hnh vung ABCD xung quanh trc MN ta c hnh tr trn xoay . Tnh th
tch khi tr trn xoay c gii hn bi hnh tr ni trn. II. PHN RING1. Theo
chng trnh Chun :Bi 4: ( 2 im )Trong khng gian Oxyz cho 2 im
A(5;-6;1) ; B(1;0;-5) .1/. Vit phng trnh chnh tc ca ng thng () qua
B v c VTCP(3;1; 2) u . Tnh cosin ca gc to bi () v ng thng AB.2/.
Vit phng trnh mt phng (P) quaA v cha().Bi 5: ( 1 im )1/. Gii phng
trnh trong tp phc : x2 6x + 10 = 0 2/. Tnh gi tr biu thc : P = ( )
( )2 21 3 1 3 + + i i . 81I.PHN CHUNG CHO TT C TH SINH (7,0 im)Bi
1: ( 3 im )Cho (Cm) : y = 12+xx m1/. Tm m (Cm) ct Ox ti im c honh
xo =12.2/. Kho st v v th ( C ) khi m = - 1.3/. Tnh din tch hnh phng
gii hn bi ( C ) ; Ox ; Oy.Bi 2: ( 3 im )1/. Gii phng trnh - bt phng
trnh :a/. 16.16 33.4 2 0 + x xb/. ( ) ( )3 9log 2 log 2 + > + x
x 2/. Tnh cc tch phn :a/. I = 13 20. . x x x dxb/. I = 1ln(2 1).0+x
dx3/. a/. Tm GTLN , GTNN ca cc hm s : y =13sin3x + cos2x-3b/. Tnh
gi trbiu thc P =521log 22+.
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TON THPT NM HC 2010 - 2011Bi 3: ( 1 im )Hnh lng tr ng ABC.ABC c y
ABC l tam gic vung ti A , AC = a, 60oC. ng cho BC ca mt bn (BBCC)
to vi mt phng (AACC) mt gc 30o . Tnh th tch khi chp C.ABCII. PHN
RING1. Theo chng trnh Chun :Bi 4: ( 2 im ) Trong khng gian Cho
A(1;0;-2), B(-1;-1;3) v mt phng (P) : 2x y + 2z + 1 = 0 .a/. Tm to
hnh chiu vung gc ca A ln (P).b/. Vit phng trnh mt phng (Q) cha A,B
v vung gc (P). Bi 5: ( 1 im )1/. Tm s phc z bit : 2. 1 6. + z z
i2/. Gii phng trnh trn tp s phc : z4 - z2 - 6 = 0 82I.PHN CHUNG CHO
TT C TH SINH (7,0 im)Cu1( 3): Cho hm s : y=3 21+xx1. Kho st s bin
thin v v th (C) ca hm s cho . 2. Chng minh rng ng thng y = -2x-m
lun ct (C) ti hai im phn bitCu2( 3):1. Gii bt phng trnh :
log120,52log (4 11) log ( 6 8)+ < + + x x x2. Tnh tch phn :
120100( 1) x xdx.3. Tm GTLN , GTNN ca hm s y= 6 3 xtrn on [ ] 1;1
.Cu 3 ( 1):Cho mt hnh tr c bn knh y R=5 v khong cch hai y l 7.1.
Tnh din tch xung quanh v th tch khi tr.2. Ct khi tr bi mt mt phng
song song trc v cch trc mt khong l 3.Tnh din tch thit din.II. PHN
RING1. Theo chng trnh Chun :Cu 4 ( 2): Cho 2 ng thng d1:1 21 35 +
+' +x ty tz t v ng thng d2:2 2 12 1 3 + x y z1. Chng minh rng d1 ct
d2 . T m to giao im .2. Vi t phng trnh mt ph ng (p) song song vi 2
ng th ng d1 , d2 v ti p x c vi m t cu tm O bn k nh bng 2 .Cu 5 (
1):Cho hnh phng (H) gii hn bi cc ng , 2, 0 xy xe x y. Tnh th tch vt
th trn xoay khi (H) quay quang Ox. 83I. PHN CHUNG CHO TT C TH SINH
(7,0 im)Cu 1(3 im). Cho hm s 33 + y x xc th (C)1. Kho st v v th
(C)2. Dng (C) bin lun theo m s nghim phng trnh 33 0 + x x m
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TON THPT NM HC 2010 - 20113. Vit phng trnh tip tuyn ca (C) vung gc
vi ng thng (d): x - 9y + 3 = 0Cu 2(3 im).1. Tnh tch phn :a) 1 230
2+xI dxxb) J = 20(2 1) ln x xdx.2. Gii phng trnh : a)2.16 17.4 8 0
+ x xb) log4(x + 3) log4(x1) = 123. Tm gi tr ln nht v gi tr nh nht
ca hm s
( )3 212 3 73 + f x x x xtrn [ 1; 2]
II. PHN RING1. Theo chng trnh Chun :Cu 3(1im). Cho hnh chp u
S.ABCD c cnh y AB = a v cnh bn SA = a. AC ct BD ti 0.a/ Chng minh
rng 0 l tm ca mt cu (S) i qua 5 im S, A, B, C, D v tnh bn knh R ca
n.b/ Tnh th tch ca khi chp S.ABCD.Cu 4.(2 im).Trong khng gian Oxyz,
cho ng thng (d) c phng trnh 1 1 22 3 1+ x y z v mt phng (P) c phng
trnh 2 3 0 + x y z1) Tm to giao im A ca ng thng (d) v mt phng
(P).2) Vit phng trnh mt cu (S) c tm I thuc (d), bn knh 66 R v tip
xc vi mt phng (P).Cu 5 (1im).a)Tnh : ( ) ( )2 23 3 + i ib) Gii phng
trnh 24 7 0 + x x trn tp s phc 84http://www.VNMATH.comI.PHN CHUNG
CHO TT C TH SINH (7,0 im) Cu 1 ( 3 im) Cho hm s3 23 1 + + y x x c
th (C)a. Kho st v v th (C).b. Vit phng trnh tip tuyn ca th (C) ti
A(3;1).c. Dng th (C) nh k phng trnh sau c ng 3 nghim phn bit 3 23 0
+ x x k .Cu 2 : ( 3 im) 1. a/.Gii phng trnh sau : 2 22 2 2log ( 1)
3log ( 1) log 32 0 + + + x x.b/.Gii bt phng trnh14 3.2 8 0+ + x
x
2. Tnh tch phn sau : a/.230(1 2sin ) cos+ x xdx I. b/. I = 10( )
+xx x e dx 3. Tm MAX , MIN ca hm s ( )3 212 3 73 + f x x x x trn on
[0;2]Cu 3( 1 im):Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a . SA
(ABCD) v SA = 2a .1. Chng minh BD SC.2. Tnh th tch khi chp S.BCD
theo a . http://www.VNMATH.com57http://www.VNMATH.com 122 N THI TT
NGHIP MN TON THPT NM HC 2010 - 2011II. PHN DNH CHO HC SINH TNG BAN
Cu IV.a (2 im)Trong Kg Oxyz cho im A(2;0;1), mt phng (P): 2 1 0 + +
x y z v ng thng (d): 122 + ' +x ty tz t.1. Lp phng trnh mt cu tm A
tip xc vi mt phng (P).2.Vit phng trnh ng thng qua im A, vung gc v
ct ng thng (d).Cu V.a ( 1 im)Cho s phc1 3 + z i .Tnh 2 2( ) + z z.
85I/ PHN DNH CHUNG CHO TT C CC TH SINH (7,0 im)Cu I: (3,0im) Cho hm
s : 3 2.1xyx1/ Kho st s bin thin v v th hm s cho.2/ Tm tt c cc gi
tr ca tham s m ng thng y= mx+2 ct th hm s cho ti hai im phn bit .Cu
II:(3,0im)1/ Gii bt phng trnh: 2 1log 0112x x b) Tnh tch phn :
20cos1 sin+xdxxc) Tm gi tr ln nht v gi tr nh nht ca hm s 4 22 6 1 +
y x x trn [-1;2] Cu 3 (1.0 im): Cho hnh chp S.ABCD c ABCD l hnh
vung cnh a, ( ) SA ABCD, gc to bi SC v mt phng (ABCD) l 060. Tnh th
tch khi chp S.ABCDII. PHN RING (3 im)A. Th sinh theo chng trnh
chun:Cu 4a: (1,0 im) Gii phng trnh sau trn tp s phc: 2x4 + 7x2 + 5
= 0.Cu 5a. ( 2,0 im)Trong khng gian Oxyz, cho 4 im A(3; 1; 2); B(1;
1; 0); C(-1;1;2); D(1; -1; 2)1. Chng minh rng 4 im A, B, C, D to nn
1 t din. Vit phng trnh mt cu (S) ngoi tip t din .2. Vit phng trnh
mt phng (MNP) bit M,N, P ln lt l hnh chiu ca im A ln cc trc ta Ox,
Oy, Oz.B. Th sinh theo chng trnh nng cao:Cu 4b. (1,0 im)Tnh th tch
khi trn xoay khi quay quanh trc honh phn hnh phng gii hn bi cc ng y
= lnx, y=0, x = 2.Cu 5b. (2,0 im)Trong khng gian Oxyz, cho im A(3;
2; 1) v ng thng d: 32 4 1+ x y z1. Vit phng trnh ng thng (d) qua A
vung gc vi (d) v ct (d).2. Tm im B i xng ca A qua (d). 87A- PHN
CHUNG CHO TT C TH SINH (7,0 im)Cu I (3 im)Cho hm s y = x3 3x + 4 c
th (C)a- Kho st s bin thin v v th ca hm sb- Vit phng trnh tip tuyn
ca th (C) song song vi ng thng y = 15x + 2010Cu II (3 im)a- Gii
phng trnh:22x + 3 + 7.2x + 1 4 = 0
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TON THPT NM HC 2010 - 2011b- Tnhtch phn: I = 4 11xedxxc- Tm gi tr
ln nht v gi tr nh nht ca hm s y = x 2.lnx trn on [1 ; e]Cu III (1
im)Cho hnh chp S.ABC c cnh bn SA vung gc vi mt y v SA = a, SB = a.
5 . Tam gic ABC l tam gic u. Tnh th tch ca khi chp S.ABC theo aB-
PHN RING (3,0 im)1. Theo chng trnh ChunCu IVa (2 im)Trong khng gian
Oxyz cho hai im A(1 ; 3 ; 1), B(0 ; 2 ; 6) v 2. + uuur r r rOG i j
ka- Vit phng trnh mt phng (P) i qua G v vung gc vi ng thng AB.Tm to
im C sao cho G l trng tm ca tam gic ABCb- Vit phng trnh mt cu (S) c
tm l im A v i qua im BCu Va (1 im)Cho s phc z = (1 + i)3 + (1 + i)4
. Tnh gi tr ca tch. z z2. Theo chng trnh Nng caoCu IVb (2 im)Trong
khng gian Oxyz cho bn im A(1 ; 2 ; 2), B(3 ; 0 ; 2), C(2 ; 3 ; 5),
D(5 ; 1 ; 4) a). Vit phng trnh mt phng (ABC). Chng minh A, B, C, D
l bn nh ca mt t dinb). Vit phng trnh mt cu (S) tm D v tip xc vi mt
phng (ABC).Tnh th tch ca t din ABCDCu Vb (1 im)Tnh din tch hnh phng
gii hn bi th (C) ca hm s 23 2 12 1 +x xyx, tim cn xin ca th (C), ng
thng x = 1 v trc tung. 88http://www.VNMATH.comI/ PHN CHUNG CHO TT C
CC TH SINH ( 7 im )Cu I (3 im)Cho hm s y = 3x2 x3 c th l ( C).1.
Kho st s bin thin v v th ( C) ca hm s.2. Vit phng trnh tip tuyn vi
( C) ti im A thuc ( C) c honh x0 = 3.Cu II ( 3 im)1. Gii phng trnh
sau: 4x- 2. 2x + 1 + 3 = 0 2. Tnh tch phnI = 1(2 2) ln +ex xdx.3.
Tm gi tr ln nht v gi tr nh nht ca hm s 1 + y xx trn on [12; 2].Cu
III ( 1 im) Cho t din u ABCD c cnh bng a, tnh th tch khi t din ABCD
theo a.II. PHN RING ( 3 im)1. Theo chng trnh chun:
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TON THPT NM HC 2010 - 2011Cu IV.a ( 2 im)Trong khng gian vi h ta
Oxyz, cho bn im A( 1; 0; 0), B(0; 1; 0), C(0; 0; 1), D(1; 1; 1).1.
Vit phng trnh mt phng (ABC).2. Vit phng trnh ng thng d i qua D v
vung gc vi mt phng (ABC).Cu Va. ( 1 im) Gii phng trnh sau trn tp s
phc: z2 2z + 3 = 02. Theo chng trnh nng cao:Cu IV.b ( 2 im)Trong
khng gian vi h ta Oxyz cho ng thng d c phng trnhx = 1 + t d: y = 2
-t z = t v mt phng () c phng trnh x + 3y + 2z 3 = 0.1. Vit phng
trnh ng thng d l hnh chiu ca d trn mt phng ().2. Vit phng trnh mt
cu tm I(1; 2; 3) v tip xc vi mt phng ().Cu V.b ( 1 im)Gii phng trnh
sau trn tp s phc: z4 + z2 - 6 = 0 89I. PHN CHUNG CHO TT C TH SINH (
7 im ) Cu I ( 3,0 im )Cho hm s4 22 1 x x y c th (C)a) Kho st s bin
thinv v th (C).b) Dng th (C ), hy bin lun theo m s nghim thc ca
phng trnh 4 22 0 x x mCu II ( 3,0 im ) a) Tm gi tr ln nht v gi tr
nh nht ca hm s y =3 22 3 12 2 + + x x x trn [ ] 1; 2 . b) Gii phng
trnh: 20.2 0.2log log 6 0 x xc) Tnh tch phn 40tancosxI dxx Cu III (
1,0 im ) Cho hnh chp tam gic u c cnh y bng 6 v ng cao h = 1.Hy tnh
din tch ca mt cu ngoi tip hnh chp .II. PHN RING( 3 im ) 1. Theo
chng trnh chun : Cu IV.a ( 2,0 im ):Trong khng gian vi h ta Oxyz ,
cho hai ng thng:11 2( ) : 2 2 + ' x ty tz t v22 '( ) : 5 3 '4 +'x
ty tz
a) Chng minh rngng thng1( ) v ng thng 2( ) cho nhau . b) Vit
phng trnh mt phng ( P ) cha ng thng1( ) v song song vi ng thng 2( )
. Cu V.a( 1,0 im ):Tnh gi tr ca biu thc2 2(1 2 ) (1 2 ) + + P i i
2. Theo chng trnh nng cao :
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TON THPT NM HC 2010 - 2011Cu IV.b( 2,0 im ): Trong khng gian vi h
ta Oxyzcho im M(2;3;0), mt phng (P) : x + y + 2z +1 = 0 v (S) : x2
+ y 2 + z2 - 2x + 4y - 6z +8 = 0 . a)Tm im N l hnh chiu ca im M
lnmt phng (P) . b) Vit phng trnh mt phng (Q) song song vi (P) v tip
xc vi mt cu(S) .Cu V.b( 1,0 im ): Tm s phc z bit 2 z z , trong zl s
phc lin hp ca s phc z . 90I- PHN CHUNG CHO TT C TH SINH( 7 im)Cu
I.( 3 im) Cho hm s y = 11+xx1. Kho st s bin thin v v th (C) ca hm
s.2.Vit phng trnh tip tuyn vi (C) ti im thuc (C) c honh x0 = -23.Gi
(H) l hnh phng gii hn bi (C) v 2 trc ta . Tnh din tch hnh phng
(H).Cu II.( 3 im) 1. Gii phng trnh : 1124 4.2 4 0+ xx 2.Tnh tch phn
: I = 20sin 2 .cosx xdx3.Tm GTLN v GTNN ca hm s : y = 3 22 3 12 10
+ x x xtrn on [ 3, 3] Cu III.( 1 im) Cho hnh chp S.ABC . c ng cao
SI = a vi I l trung im ca BC .y ABC l tam gic vung cn ti A v BC =
2a.1.Tnh th tch khi chp S.ABC.2.Tnh din tch mt cu ngoi tip hnh chp
S.ABCII- PHN DNH CHO HC SINH TNG BAN( 3 im)1.Theo chng trnh chun.Cu
IV.a ( 2 im) Trong khng gian ta Oxyz cho bn im
A(1;0;0),B(0;1;0),C(0;0;1),D(-2;1;-1)1.Vit phng trnh mt phng
(ABC),suy ra ABCD l t din.2.Vit phng trnh mt cu tm D v tip xc mt
phng (ABC)3.Gi H l chn ng cao ca t din ABCD i qua D. Vit PTTS ng
cao DH.Cu V.a ( 1im) Gii phng trnh : 27 0 + x xtrn tp s phc.2.Theo
chng trnh nng cao.Cu IV.b ( 2 im)Trong khng gian ta Oxyz cho bn im
A(1;0;0),B(0;1;0),C(0;0;1),D(-2;1;-1)1.Vit phng trnh mt phng
(ABC),suy ra ABCD l t din.2.Gi H l chn ng cao ca t din ABCD i qua
D. Vit PTTS ng cao DH.3.Vit phng trnh mt cu tm D v tip xc mt phng
(ABC). Tm ta tip imCu V.b ( 1im)Tm s phc z sao cho . ( ) 4 2 + z z
z z i 91http://www.VNMATH.comI . PHN CHUNG CHO TT C TH SINH ( 7 im
) http://www.VNMATH.com62http://www.VNMATH.com 122 N THI TT NGHIP
MN TON THPT NM HC 2010 - 2011 Cu 1 ( 3 im )Cho hm s4 21 532 2 + y x
x (1)a.Kho st v v th hm s (1).b.Vit phng trnhtip tuyn ca th hm s
(1) ti ti im c honh x = 1 . Cu 2 ( 3 im )a. Tnh tch phn 1 231 2 +xI
dxxb.Tm gi tr ln nht v gi tr nh nht ca hm s3 212 5 23 + x y x x trn
[ 1; 3] c. Gii phng trnh:32 22216 0log loglog + x xCu 3 (1im) Cho
hnh chp u S.ABCD c cnh y bng a, cnh bn SA bng 2 aa. Chng minh rng (
) AC SBD.b. Tnh th tch hnh chp S.ABCD theo a.II .PHN RING1.Theo
chng trnh chunCu4a ( 2im)Trong khng gian vi h ta Oxyz , cho tam gic
ABC vi cc nh l A(0;-2;1) , B(-3;1;2) , C(1;-1;4). a. Vit phng trnh
chnh tc ca ng trung tuyn k t nh A ca tam gic . b. Vit phng trnh mt
cutm C ,bit rng mt cu tip xc vi mt phng (OAB).Cu 5a (1 im ) Gii
phng trnh : 2z2 + z +3 = 0 trn tp s phc2.Theo chng trnh nng cao:Cu
4b.( 2 im)Trong khng gian vi h trc Oxyz, cho 2 ng thng c phng trnh
' + 2111zt yt x
1 21132z y xa.Chng minh1 v 2 cho nhau .b.Vit phng trnh mt phng
cha1 v song song vi 2 . Cu 5 b(1im ) Gii phng trnh :2(3 4 ) 5 1 0 z
i z i + + trn tp s phc 92http://www.VNMATH.comI. PHN CHUNG CHO TH
SINH C HAI BAN (7 im)Cu 1 (3 im) Cho hm s 3 26 9 + y x x x, c th
(C)1. Kho st s bin thin v v th (C) ca hm s.2. Tnh din tch hnh phng
gii hn bi th (C) v ng thngy = x.
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TON THPT NM HC 2010 - 2011Cu 2 (3 im)1. Gii phng trnh 1 39 18.3 3 0
x x2. Tnh tch phnln 6 20 3++x xxe eI dxe3. Tm gi tr ln nht v gi tr
nh nht ca hm s 2 1+xeyx trn on [0; 2]Cu 3 (1 im)Cho hnh chp S.ABCD
c y ABCD l hnh vung, cnh bn SA vung gc vi y, cnh bn SC to vi mt bn
SAB mt gc 030 , SA = h. Tnh th tch ca khi chp S.ABCDII. PHN DNH CHO
TH SINH TNG BAN (3 im)A. Theo chng trnh Chun:Cu 4a. Trong khng gian
vi h ta Oxyz, cho hai im A(2;3;4), B(0; 1; 2)1. Vit phng trnh ng
thng AB2. Gi I l trung im ca on AB. Vit phng trnh ca mt cu (S) c tm
l I v bn knh bng 2. Xt v tr tng i ca mt cu (S) vi cc mt phng ta .Cu
5a. Gii phng trnh 2(1 ) (3 2 ) 5 0 + + ix i x trn tp s phc B. Theo
chng trnh Nng caoCu 4b. Trong khng gian vi h ta Oxyz cho d: 1 2 11
2 3 + x y z v mt phng (P):2x 3y z + 6 = 0.1. Vit phng trnh mt phng
(Q) i qua d vvung gc vi (P)2. Tnh th tch phn khng gian gii hn bi
(Q) vcc mt phng ta .Cu 5b. Tm phn thc, phn o ca s phc ( )953(1
)+izi 93A.PHN CHUNG CHO TT C CC TH SINH ( 7im)Cu I:(3,0 im) Cho hm
s 32xyx c th ( C )1) Kho st s bin thin v v th ( C ) ca hm s.2) Tm
tt c cc gi tr ca tham s m ng thng d:y=mx+1ct th (C) ti hai im phn
bit .Cu II: (3,0 im) 1) Gii bt phng trnh:0,53 5log 01
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