kÌ thi thỬ thptqg lẦn 11- nĂm hỌc 2016 - 2017...
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DAO NG C HC
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K THI TH THPTQG LN 11- NM HC 2016 - 2017
Mn: TON
Thi gian lm bi: 90 pht, khng k thi gian pht
Cu 1: Cho hm s 2 2y bx cx dax c th nh hnh v bn.
Mnh no di y ng ?
A.a 0,b 0,c 0,d 0
B. a 0,b 0,c 0,d 0
C. a 0,b 0,c 0,d 0
D. a 0,b 0,c 0,d 0
Hng dn:
Da vo th hm S, ta c cc nhn xt sau
Ta thy rng x xlim y ; lim y
h s a > 0
th hm s ct trc Ox ti im AA(x ;0) vi Ax 0 chnh l im un ca th
hm s. Do 2A Ay' 3ax 2bx c y'' 6ax 2b y''(x ) 0 b 3a.x 0
th hm s ct trc Oy ti im BB(0; y ) vi B By 0 y d 0
Hm s cho ng bin trn 2y ' 0; x b 4ac 0 m a 0 c 0
p n : D
Cu 2: Tm tt c cc gi tr ca tham s m ng thng y = m ct th hm s 4 2y | x 2x 2 | ti 6 im phn bit.
A. 2 m 3 B. 2 m 4 C. m = 3 D. 0 m 3
Hng dn:
V th (C) ca hm s 4 2y | x 2x 2 |
Phn 1. Gi nguyn th hm s 4 2y x 2x 2 pha trn
trc honh
Phn 2. Ly i xng phn th hm s 4 2y x 2x 2
pha di trc honh qua trc honh
Da vo th hm s (hnh v bn) ng thng y = m
ct th (C) ti 6 im phn bit khi v ch khi 2 < m < 3.
p n : A
Cu 3: Tm gi tr ln nht ca hm s 3 2y x 2x 7x 1 trn 3;2
A. 3 B. 1 C. 4 D. 13
Hng dn:
Xt hm s 3 2y x 2x 7x 1 trn on 3;2][
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2
ta c 2x 1
y ' 7 4x 3x ; y ' 0 7x
3
Tnh cc gi tr 7 419
y( 3) 13, y(1) 3, y , y(2) 33 27
Vy gi tr ln nht ca hm s bng 3
p n : A
Cu 4: Tm tt c cc khong ng bin ca hm s 3 21
y x 2x 3x 53
A. ( ;1) (3; ) B. ( 3; )
C. ( ;1);(3; ) D. ( ;4)
Hng dn:
Xt hm s 3 21
y x 2x 3x 53
vi x , ta c 2x 3
y ' x 4x 3 0x 1
Suy ra hm s cho ng bin trn khong (3; ) v ( ;1)
p n : C
Cu 5: Tm gi tr cc tiu ca hm s 4 2y x 2x 3
A. 1 B. 2 C. 3 D. 0
Hng dn:
Xt hm s 4 2y x 2x 3 ta c 3 2y ' 4x 4x y '' 12x 4, x
Phng trnh 3x 0
y ' 0 4x 4x 0 y ''( 1) 0 x 1, x 1x 1
l im cc tiu ca
hm s
Vy gi tr cc tiu ca hm s bng y( 1) 2
p n : B
Cu 6: Cho hm s x 1
yx 1
v ng thng y 2x m. Tm gi tr ca tham s m
th hm s cho ct nhau ti 2 im phn bit A, B v trung im ca AB c honh
bng 5
2
A. 8 B. 11 C. 9 D. 10
Hng dn:
Phng trnh honh giao im ca (C) v (d) l
2
x 1x 1m 2x
x 1 2x (m 1)x m 1 0(*)
(C) ct (d) ti hai im phn bit (*) c hai nghim khc 1
2m 7
(m 1) 8(m 1) 0m 1
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Khi gi A Bx , x l honh ca hai giao im A, B suy ra A Bm 1
x x 5 m 92
p n : C
Cu 7: Cho hm s 2y cos x 1 cos x c gi tr ln nht l M v gi tr nh nht l m.
Tnh M m
A. 1 2 B. 2 C. 2 1 D. 2
12
Hng dn:
t t cos x 1;1][ , khi 22
t 1f (t) t 1 t f '(t) 1 ;f '(t) 0 t
21 t
Tnh cc gi tr 1
f ( 1) 1, f (1) 1, f 22
suy ra
M 2M m 2 1
m 0
p n : C
Cu 8: Tm tt c cc gi tr ca tham s m hm s 4 2 2y mx (m 1)x m 1 c ba cc
tr.
A. 1 m 0
m 1
B.
1 m 0
m 1
C.
m 1
0 m 1
D.
0 m 1
m 1
Hng dn:
Vi 2m 0 y 1 x hm s c mt im cc tr
Vi m 0, ta c 4 2 2 3 2y mx (m 1)x m 1 y ' 4mx 2(m 1)x; x
Phng trnh 2 32 2
x 0y ' 0 (m 1)x 2mx 0
2mx m 1(*)
hm s cho c ba im cc tr (*) c hai nghim phn bit khc 0 m 1
1 m 0
p n : B
Cu 9: Cho hm s y = f(x) xc nh trn khong 2; v tha mn xlim f (x) 1
. Vi gi
thit hy chn mnh ng trong cc mnh sau
A. ng thng y 1 l tim cn ngang ca th hm s y = f(x).
B. ng thng y 1 l tim cn ng ca th hm s y = f(x).
C. ng thng x = 2 l tim cn ngang ca th hm s y = f(x).
D. ng thng x 1 l tim cn ng ca th hm s y = f(x).
Hng dn:
Ta c xlim f (x) 1 y 1
l tim cn ngang ca th hm s.
p n : A
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4
Cu 10: th hm s 2x x 1
yx
c bao nhiu tim cn ?
A. 3 B. 1 C. 0 D. 2
Hng dn:
2 2x
x x x
x
1 1| x | 1 lim y 1
x x 1 x xlim y lim limlim y 1x x
th hm s c hai tim cn
ngang.
V 2
x 0 x 0
x x 1lim y lim x 0
x
l tim cn ng ca th hm s
p n : A
Cu 11: Tm s giao im ca th hm s 2x 2x 3
yx 1
vi ng thng y = 3x 6.
A. 3 B. 0 C. 1 D. 2
Hng dn:
Phng trnh honh giao im ca th (C) v ng thng (d) l 2x 2x 3
3x 6x 1
2 2 2 2
x 1 0 x 1 x 1(*)
x 2x 3 (x 1)(3x 6) x 2x 3 3x 9x 6 2x 7x 3 0
H phng trnh (*) c hai nghim phn bit nn (C) ct (d) ti hai im
p n : D
Cu 12: Tm tt c cc gi tr ca tham s m gi tr nh nht ca hm s 3m x 2
yx m
trn 1;1 bng 2.
A. m 0
m 2
B. m = 0
C. m 2 D. Khng tn ti m
Hng dn:
Ta c 3 4
2
m x 2 m 2y y ' 0; x m
x m (x m)
suy ra hm s cho nghch bin trn 1;1][
Mt khc hm s lin tc trn on 1;1][ nn 3
3
1;1]
m 2min y y(1) 2 m 2m 0 m 0
1 m[
p n : B
Cu 13: Tp xc nh ca hm s 3(x 1)
ylog(9 x)
l
A. (1;9) B. (1;9) \ 8 C. 1;9 \ 8 D. 1;9 \ 8
Hng dn:
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Hm s xc nh khi v ch khi
x 1 0;9 x 0 9 x 1 9 x 1
x 1;9) \ 8log(9 x) 0 9 x 1 x 8
[
p n : D
Cu 14: Cho hm s 2y 3ln(x x 1) c th C . Vit phng trnh tip tuyn ca
th C ti giao im ca C vi trc honh.
A. y 3x
y 3x
B.
y 3x
y 0
C.
y 3x 3
y 0
D.
y 3x 3
y 3x
Hng dn:
Phng trnh honh giao im ca (C) vi Ox l
2 2x 0
2ln(x x 1) 0 x x 0x 1
Ta c 2
y '(0) 33(2x 1)y '
y '( 1) 3x x 1
nn phng trnh tip tuyn cn tm l
y 3x
y 3x
p n : A
Cu 15: Cho ba hm s x x xy a , y b , y c c th nh hnh di y. Khng nh no
sau y ng?
A. a b c 1 B. 1 c b a C. c 1 b a D. c 1 a b
Hng dn:
Da vo th hm s, ta c cc nhn xt sau:
Hm s x xy a , y b l cc hm s ng bin trn R, hm s xy c l hm s nghch
bin trn R
Khi x x
x
a .ln a;b .ln b 0 ln a; ln b 0 z,b 1y '
0 c 1ln c 0c .ln c 0
Ta c x
x
f (x) a
g(x) b
m 0 0f (x ) g(x ) (khi
0 0x x
0x ) a b a b
Hoc c th chn x = 10 th 10 101 a b a b
Vy ta c b a 1 c 0
p n : D
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6
Cu 16: o hm y ' ca hm s 2y ln(x x 1) bng
A. 1
x 1 B.
2
2
(x 1) x 1 C.
2
1
x 1 D.
2
2x
x 1
Hng dn:
Ta c 2 2
2
2 2 2
x1ln x x 1 ' 1x 1y ln x x 1 y '
x x 1 x x 1 x 1
p n : C
Cu 17: Tm tp nghim ca bt phng trnh 2 22log (x 1) log (5 x) 1
A. [3;5] B. (1;5) C. (1;3] D. [-3;3]
Hng dn:
Bt phng trnh
2 2 2 2
2 2
5 x 1 5 x 12log (x 1) log (5 x)
log (x 1) log 2(5 x) (x 1) 2(5 x)
2 2
5 x 1 5 x 1 5 x 13 x 1 S (1;3]
3 x 3x 2x 1 10 2x x 9
p n : C
Cu 18: Cho a, b 0;a, b 1 v x y, l hai s dng. Tm mnh SAI trong cc mnh
sau
A. 2 21 aa
log 4.log x B. a a alog (xy) log x log y
C. 2016a alog x 2016.log x D.
ba
b
log xlog x
log a
Hng dn:
Ta c 12 2 2 2 2
1 aa
a
log x log x 4.(log x) p n A ng v cc cng thc p n B, C, D
c gii thiu SGK GII TCH 12
p n : A
Cu 19: Bit rng phng trnh x 1 3 x5 5 26 c hai nghim l 1 2x , x . Tnh tng 1 2x x
A. 2 B. 4 C. -2 D. 5
Hng dn:
Ta c x
x 1 3 x x 2 x x x
x
5 1255 5 26 26 (5 ) 130.5 625 0 (5 125)(5 5) 0
5 5
x x 3
1
1 2x x 12
x 35 125 5 5x x 4
x 15 5 5 5
p n : B
Cu 20: Tm tp nghim ca bt phng trnh x x x3.4 5.6 2.9 0
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A. ;0 B. 2
;15
C. 2
0;3
D. (0;1)
Hng dn:
Bt phng trnh 2
x x
x x x x 2 x x x 2 2 22.4 5.6 2.9 0 3.(2 ) 5.2 .3 2.(3 ) 0 3. 5. 2 03 3
2x 2 x x
2 2 2 2 23. 5. 2 0 1 . 0 1 x 0 S (0;1)
3 3 3 3 3
Tham kho. S dng bng TABLE (Mode 7) kho st hm s x x xf (X) 3.4 5.6 2.9
V nhp cc gi tr Start ? = a, End ? = b, Step ? = 0.05 vi (a,b) l cc khong p
n
Nu tt c gi tr f(X) nhn c trn khong (a,b) mang gi tr th ta s chn khong
V d: Start 0
20; 2
3 End3
Nh thy trn khong 2
0;3
th f(X) < 0, tuy nhin ta cn p n D cha khong
nn cu xt thm trn (0;1) la chn c p n ng. Kinh nghim c a ra l ta
s kho st trn khong ln nht loi tr p n.
Cch 2: Nhp x x xf (X) 3.4 5.6 2.9 CALC cc gi tr 2 2
x 10, x ; x 0; x ; x 15 3
t
suy ra p n cn chn
p n : D
Cu 21: Bit rng phng trnh: x
xx 1x 3 c hai nghim phn bit 1 2x , x . Tnh gi tr biu
thc 1 2x xP 3
A. 9 B. 5 C. 1 D. 6
Hng dn:
Ta c 1 2x
x xxx 1
x322
x 1 0x 0x 1
2 3 3 6xx log 6x x(x 1) log 3log 3
x 1
p n : D
Cu 22: Cho x 0 tha mn 2 8 8 2log (log x) log (log x) . Tnh2
2(log x)
A. 3 B. 3 3 C. 27 D. 9
Hng dn:
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8
t 2t log x, ta c 38 221 t
log x log x .log x3 3
suy ra 2 8 2 2t t 1
log log t log log t3 3 3
2 23 3
2 2 2
t tlog log t t t 3 3 (log x) t 27
3 3
p n : C
Cu 23: Cho a,b 0 v a,b 1 . t alog b , tnh theo biu thc
2
3
baP log b log a
A. 22 5
P
B. 2 12
P2
C.
24 3P
2
D.
2 3P
Hng dn:
Ta c
22
3 3
a b a b abaa
1 1 1 6 12P log b log a .log b 2log a .log b 6.log a .log b
2 2 2 log b 2
p n : B
Cu 24: Mt hnh nn c thit din qua trc l tam gic vung cn c din tch S . Hy
tnh th tch ca khi nn cho
A. 36
( S)3 B. 3
2( S)
3 C. 3
2( S)
3 D. 3
1( S)
3
Hng dn:
Thit din qua trc l tam gic ABC vung cn ti A c
21S .AB AB 2S BC 2 S2
Bn knh ng trn y ca khi nn l BC
r S2
v chiu cao ca khi nn l
BCh S
2
Vy th tch ca khi nn cn tnh l 2 2 31 1 1
V . r h ( S) . S ( S)3 3 3
p n : D
Cu 25: Cho hnh chp S ABCD . c y ABCD l hnh ch nht vi AB a,AD 2a SA
vung gc vi mt y, SA 3a. Tnh th tch ca khi chp S ABCD
A. 36a B. 33a C. 3a D. 32a
Hng dn:
Th tch ca khi chp S.ABCD l 3ABCD1 1
V .SA.S .3a.a.2a 2a3 3
p n : D
Cu 26: Cho tam gic ABC u cnh a , ng cao AH . Tnh th tch ca khi nn sinh
ra khi cho tam gic ABC quay xung quanh trc AH.
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A. 3a 6
12
B.
3a 3
12
C.
3a 2
24
D.
3a 3
24
Hng dn:
p n : D
Khi quay tam gic ABC quanh trc AH ta c khi nn c bn knh BC a
r2 2
V chiu cao ca khi nn l a 3
h AH2
. Vy th tch khi nn cn tnh l
21V . r h3
3a 3
24
Cu 27: Cho hnh chp S ABCD . c y l hnh ch nht, AB a,AD 2a ; cnh bn SA =
a v vung gc vi y. Tnh khong cch A ti mt phng SBD .
A. a B. 2a
3 C.
a
3 D.
a
2
Hng dn:
Gi K l hnh chiu ca A ln BD nn AK BD
Ta c SA (ABCD) SA BD BD (SAK)
T A k AH BD(H BD) m BD (SAK) BD AH
AH (SBD) d(A;(SBD)) AH
K SAK vung ti A, ng cao AH khi
2 2 2
1 1 1
AH SA AK
Mt khc
2 2 2 2 2 2 2 2
1 1 1 1 1 1 1 9
AK AB AD AH SA AB AD 4a
Suy ra 2a
AH3
, vy khong cch cn tnh l
2as(A;(SBD))
3
p n : B
Cu 28: Cho lng tr tam gic u ABC.A'B'C' c AB = a; gc gia hai mt phng ABC
v ABC l 60o . Tnh th tch khi chp ABCCB'
A. 3a 3
8 B.
33a
4 C.
3a 3
4 D.
33a 3
8
Hng dn:
Gi M l trung im ca BC, ABC u nn AM BC
Tam gic ABC u nn AM BC BC (A 'AM)
Ta c (A 'AM) (A 'BC) A 'M
(A 'BC);(ABC) (A 'M,AM) A 'MA(A 'AM) (ABC) AM
Xt 'MAA vung ti A, c tan 0AA ' a 3 3a
A 'MA AA ' tan 60 .AM 2 2
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T gic BCC'B' l hnh ch nht c din tch 2
BCC'C
3aS BB'.BC
2
M AM BC a 3
AM (BCC'B') d(A;(BCC'B')) AMAM BB' 2
Th tch khi chp ABCC'B' l 3
ABCC'B' BCC'B'
1 a 3V d(A;(BCC'B')).S
3 4
p n : C
Cu 29: Cho t din OABC c ba cnh OA, OB, OC i mt
vung gc vi nhau, OA = a, OB = 2a, OC = 3a. Tnh din tch ca
mt cu ngoi tip hnh chp O.ABC
A. 2S 11 a B. 2S 14 a C. 2S 12 a D. 2S 10 a
Hng dn:
Bi ton tng qut: Cho t din OABC c ba cnh OA, OB, OC
i mt vung gc vi nhau, OA= a, OB=b, OC=c th bn knh
mt cu ngoi tip hnh chp l 2 2 2a b c
R2
Vi a 14
OA a,OB 2a,OC 3a R2
din tch mt cu cn tnh l 2 2S 4 R 14 a
p n : B
Cu 30: Hnh lng tr tam gic u c bao nhiu mt phng i xng?
A. 3 B. 4 C. 5 D. V s
Hng dn:
Hnh lng tr tam gic u c bn mt phng i xng
p n : B
Cu 31: Cho hm s 2
1f (x)
sin x Nu Fx l mt nguyn hm ca hm s v F 0
6
th F x l
A. 3 cot x B. 3
cot x3
C. 3 cot x D. 3
cot x3
Hng dn:
Ta c 2
dxF(x) cot x C
sin x m
F 0 C 3 0 C 3 f (x) 3 cot x6
p n : A
Cu 32: Tm h cc nguyn hm ca hm s 2xf (x) 2
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A. x
1C
4 .ln 4 B. x4 C C. x4 .ln 4 C D.
x4C
ln 4
Hng dn:
Ta c x
2x x 4f (x)dx 2 dx 4 dx Cln 4
p n : D
Cu 33: Tm h cc nguyn hm ca hm s 1
f (x)2x 1
A. 1
2x 1 C2
B. 2 2x 1 C C. 1
C2x 1
D. 2x 1 C
Hng dn:
Ta c 1 1
2 2dx
f (x)dx (2x 1) dx (2x 1) C 2x 1 C2x 1
p n : D
Cu 34: Cho 11
7
f (x)dx 10 . Tnh 5
3
I 2. f (2x 1)dx
A. 10 B. 20 C. 5 D. 30
Hng dn:
t t 2x 1 dt 2dx v i cn x 1 t 7
x 5 t 11
. Khi
11
7
I f (t)dt 10
p n : A
Cu 35: Bit 3
6
dx 1I (ln a ln b)
sin x 2
. Tnh S a b
A. S 10 4 3 B. 22
S 4 33
C. S 10 4 3 D. 22
S 4 33
Hng dn:
t t = cosx dt sin dxx v 2 2sin x 1 t , i cn 3 1
x t ;x t6 2 3 2
Khi
3
3 2
3 3 2
2 2
11
6 6 22
dx sin x 1 1 t 1 1 1I dx dt .ln ln(7 4 3) ln 2
s 1 c x 1 t 2 t 1 2 2inx os
Suy ra 1 1 a 7 4 3
I ln(7 4 3) ln 3 (ln a ln b) a b 10 4 32 2 b 3
p n : A
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12
Cu 36: Bit 1
ax
0
dxI log b
2 1
. Tnh S a 3b
A. S 4 B. 8
S3
C. 20
S3
D. S 6
Hng dn:
t x x xdt
t 2 dt 2 .ln 2dx 2 dxln 2
, i cn x 0 t 1
x 1 t 2
Khi 21 1 2 2x
x x x
0 0 1 1 1
dx 2 dx 1 dt 1 1 1 1 tI dt .ln
2 1 2 .(2 1) ln 2 t(t 1) ln 2 t t 1 ln 2 t 1
2
4ln
1 2 1 43I . ln ln logln 2 3 2 ln 2 3
m
a
a 2
I log b S a 3b 64b
3
p n : D
Cu 37: Tnh din tch hnh phng gii hn bi th hai hm s 3y x x v 2y x x
A. 37
12 B.
9
4 C.
155
12 D.
17
12
Hng dn:
Phng trnh honh giao im ca 1 2(C ), (C ) l
3 2 3 2x x x x x x 2x 0 x 2;0;1
Khi din tch hnh phng gii hn cn tnh l 1 1
3 2 3 2
2 2
S x x (x x ) dx x x 2x dx
Cch 2: Bm my tnh CASIO
Xt biu thc 3 2x x 2x trn on 2;1 ta thy
3 2
3 2
x x 2x 0; x 2;0
x x 2x 0; x 0;1
Khi 0 1
3 2 3 2
2 0
S x x 2x dx x x 2x dx 2F(0) F( 2) F(1)
Vi 4 3
3 2 2x x 8 5 37F(x) (x x 2x)dx x S 2.F(0) F( 2) F(1)4 3 3 12 12
p n : A
Cu 38: Tnh th tch V ca vt th trn xoay sinh ra khi cho hnh phng gii hn bi
th hm s y x ln x , trc honh v ng thng x = e quay quanh Ox
A. 32e 1
V9
B.
32e 1V
3
C.
32e 1V
9
D.
32e 1V
3
Hng dn:
Phng trnh honh giao im ca (C) vi trc Ox l x ln x 0 x 1
Th tch khi trn xoay cn tnh l 4
2
1
V x ln xdx. t 3
2
u ln x dx xdu ;v
x 3dv x dx
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44 43 2 3 3 3 3 3
11 1
x .ln x x x .ln x x e e 1 2e 1V dx
3 3 3 9 3 9 9 9
p n : A
Cu 39: Tnh din tch hnh phng gii hn bi ng cong 2(C) : y 1 x 0 v hai ng
thng x 0 , x 3
A. 14
3 B.
28
3 C.
7
3 D.
32
3
Hng dn:
Ta c 2 2y 1 x 0 y x 1 y x 1 nn din tch hnh phng gii hn bi ng
cong (C) v hai ng thng x = 0, x = 3 l 33
3
0 0
2 16 2 14S x 1dx (x 1)
3 3 3 3
p n : A
Cu 40: Tnh m un ca s phc z tha mn z.z 3(z z) 4 3i
A. | z | 2 B. | z | 3 C. | z | 4 D. | z | 1
Hng dn:
Ta c z a bi(a,b ) z a bi v 2 2 2z.z | z | a b
Khi 2 2
2 2a 0a b 4
z.z 3.(z z) 4 3i a b 6bi 4 3i | z | 2b 26b 3
p n : A
Cu 41: Tm s phc lin hp ca s phc 2z (2 i)( 1 i)(2i 1)
A. z 15 5i B. z 1 3i C. z 5 5i D. z 5 15i
Hng dn:
Ta c 2z (2 i)( 1 i)(2i 1) (i 3)(4i 3) 5 15i z 5 15i
Cch 2: Chuyn sang ch Mode 2 (CMPLX) v bm my
p n : A
Cu 42: Cho s phc z tha mn z 1
1z 1
. Tp hp cc im biu din s phc z trong
mt phng l
A. ng trn B. trc thc C. trc o D. mt im
Hng dn:
t z x yi(x, y ), ta c z 1 x (y 1)i v z i x (y 1)i
Ch 1 1
2 2
z | z |
z | z | suy ra 2 2 2 2
z i1 | z 1| | z 1| x (y 1) x (y 1) y 0
z 1
Vy tp hp im biu din s phc z l ng thng y = 0 hay trc thc.
p n : B
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14
Cu 43: Cho s phc z a bi(a,b ) tha mn (1 i)(2z 1) (z 1)(1 i) 2 2i .Tnh P = a
+ b
A. P = 0 B. P = 1 C. P 1 D. 1
P3
Hng dn:
t z a bi(a,b ) z a bi. Ta c (1 i)(2z 1) (z 1)(1 i) 2(1 i)z (1 i)z 2i
Suy ra 2(1 i)z (1 i)z 2 2(1 i)(a bi) (1 i)(a bi) 2
3a 3b 2 02a 2b a b (a b)i 2 3a 3b 2 (a b)i 0 P 0
a b 0
Tham kho. S dng my tnh CASIO tm z nh v d di y (cu 43 cc bn test bng
cch ny nh).
Cho s phc z (2 i)z 1 9i . Tnh phn thc v phn o ca s phc z bng
t z X Yi z X Yi. Khi X Yi (2 3i)(X Yi) 1 9i 0(*)w
Thao tc trn my tnh Mn hnh hin th
n 2w a v tnh s phc
Nhp v tri ca phng trnh (*) l X Yi (2 3i)(X Yi) 1 9i
Sau , gn gi tr X = 100, Y = 0,01.
n
1 0 0 r 0 . 0 1w
Khi 10103 29097
i 101,03 290,97i100 100
w m 101,03 100 1 0,03 X 3Y 1
290,97 300 9 0,03 3X Y 9
X 3Y 1 X 2(X 3Y 1) (3X 3Y 9)i 0 z 2 i
X Y 3 Y 1w
p n : A
Cu 44: Cho ba s phc 1 2 3z , z , z tha mn 1 2 3
1 2 3
z z z 0
| z | | z | | z | 1
.Mnh no di y ng?
A. 2 2 21 2 3 1 2 2 3 3 1| z z z | | z z z z z z | B.
2 2 2
1 2 3 1 2 2 3 3 1| z z z | | z z z z z z |
C. 2 2 21 2 3 1 2 2 3 3 1| z z z | | z z z z z z | D.
2 2 2
1 2 3 1 2 2 3 3 13 | z z z | . | z z z z z z |
Hng dn:
Ta c 2 2 2 2 2 2 2
1 2 3 1 2 3 1 2 2 3 3 1 1 2 3 1 2 2 3 3 1(z z z ) z z z 2(z z z z z z ) z z z 2(z z z z z z )
Mt khc 21 1 1 1| z | 1 | z | 1 z .z 1 , tng t 2 2z .z 1 , 3 3z .z 1 nn
1 2 3
1 2 3
1 1 1z z z
z z z
Khi
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2
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3
1 2 3
1 1 1z z z 2z z z 2z z z (z z z ) 2z z z (z z z ) 0
z z z
Vy 2 2 21 2 3 1 2 2 3 3 1| z z z | | z z z z z z |
p n : A
Cu 45: Cho 3 im A(1; 1;1),B(0;1;2),C(1;0;1) Tm ta im D sao cho t gic ABCD
l hnh bnh hnh
A. D(2;2;0) B. D(2; 2;0) C. D( 2; 2;0) D. D(2;0;0)
Hng dn:
ABCD l hnh bnh hnh nn AB DC m AB ( 1;2;1) nn
1 x 1 x 2
0 y 2 y 2 D(2; 2;0)
1 z 1 z 0
p n : B
Cu 46: Mt cu tm I 1; 2;3 , bn knh AB vi A4; 3;7 v B2;1;3 c phng trnh
l
A. 2 2 2(x 1) (y 2) (z 3) 36 B. 2 2 2(x 1) (y 2) (z 3) 4
C. 2 2 2(x 1) (y 2) (z 3) 6 D. 2 2 2(x 1) (y 2) (z 3) 36
Hng dn:
Ta c A(4; 3;7)
AB ( 2;4; 4) AB 6 R 6B(2;1;3)
l bn knh mt cu (S)
Phng trnh mt cu tm I(1;2;3), bn knh R = 6 l 2 2 2(x 1) (y 2) (z 3) 36
p n : A
Cu 47: Trong khng gian vi h ta Oxyz cho tam gic ABC bit
A(5;1;3),B(1;6;2),C(5;0;4) Ta trng tm G ca tam gic l
A. 11
G ;3;73
B. 11 7
G ; ;33 3
C.
11 7G ; ;3
3 3
D. 11 7
G ; ;33 2
Hng dn:
Trng tm ca tam gic ABC c ta l 5 1 5 1 6 0 3 2 4 11 7
G ; ; G ; ;33 3 3 3 3
p n : C
Cu 48: Trong khng gian vi h trc ta Oxyz , cho hai ng thng 1 2d ,d c phng
trnh ln lt l
x 1 2tx y 1 z 2
, y 1 t (t )2 1 1
z 3
. Phng trnh ng thng vung gc
vi (P) 7x y 4z 0 v ct c hai ng thng 1 2d ,d l
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16
A. x y 1 z 2
7 1 4
B.
x 2 y z 1
7 1 4
C. x 1 y 1 z 3
7 1 4
D.
1 1x z
y 12 2
7 1 4
Hng dn:
Gi s 1 1d d A A d nn A(2u;1 u;u 2)
2 2d d B B d nn B(2t 1; t 1;3)
V th AB (2t 2u 1; t u;5 u) l vecto ch phng ca d.
Do d (P) nn AB|| n (7;1; 4) y n l vecto php tuyn ca mp(P)
T c h phng trnh 2t 2u 1 7t 7u2t 2u 1 t u 5 u
4(t u) u 57 1 4
t 2AB ( 7; 1;4)
u 1
v ng thng d i qua im A(2;0; 1) nn
x 2 y z 1(d) :
7 1 4
p n : B
Cu 49: Trong khng gian vi h trc ta Oxyz , phng
trnh ca mt cu i qua ba im A(2;0;1),B(1;0;0),C(1;1;1) v
c tm thuc mt phng (P) : x y z 2 0 l
A. 2 2 2(x 1) y (z 1) 1 B. 2 2 2(x 1) y (z 1) 4
C. 2 2 2(x 3) (y 1) (z 2) 1 D. 2 2 2(x 3) (y 1) (z 2) 4
Hng dn:
Gi I(x;y;z) l tm ca mt cu (S) suy ra IA IB IC v I (P) x y z 2 0
Mt khc AI (x 2;y;z 1),BI (x 1;y;z),CI (x 1;y 1;z 1) nn ta c h phng trnh
I (P) x y z 2 0 z 1
IA IB x z 2 y 0 I(1;0;1)
IA IC y z 1 z 1
v R IA 1
Vy phng trnh mt cu (S) l 2 2 2(x 1) y (z 1) 1
Cch 2: Loi p n, thay B(1;0;0) vo 4 phng n (Loi c B, C, D)
p n : A
Cu 50: Trong khng gian vi h ta Oxyz , cho hnh lng tr ng ABC.A'B'C' c
A(a;0;0),B( a;0;0),C( a;0;b) vi a b, l cc s dng thay i tha mn a b 4. . Khong
cch ln nht gia hai ng thng B C' v AC' l
A. 1 B. 2 C. 2 D. 2
2
Hng dn:
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Ta c A(a;0;0),B( a;0;0),C(0;1;0),B'( a;0;b)
V ABC.A'B'C' l hnh lng tr ng nn BB' CC' C'(0;1;b)
ng thng AC c vecto ch phng 1u ( a;1;b) v i
qua A
ng thng BC c vecto ch phng 2u (a;1; b) v i
qua B
Khi 1 21 b b a a 1
u ;u ; ; ( 2b;0; 2a)1 b b a a 1
V 1 2AB' ( 2a;0;b) u ;u . AB' ( 2b)( 2a) 2ab 2 | ab |
Vy khong cch gia hai ng thng AC, BB l
1 2
1 2
u ;u . AB'd
u ;u
2 2 2 2
2 | ab | ab ab 1 a bd 2
2ab 2 2.2 ab 2 24a 4b a b
maxd 2 . Du = xy ra a b 2 (nh gi trn p dng bt ng thc Cosi.
p n : C
p n
1-D 2-A 3-A 4-C 5-B 6-C 7-C 8-B 9-A 10-A
11-D 12-B 13-D 14-A 15-D 16-C 17-C 18-A 19-B 20-D
21-D 22-C 23-B 24-D 25-D 26-D 27-B 28-C 29-C 30-B
31-A 32-D 33-D 34-A 35-A 36-D 37-A 38-A 39-A 40-A
41-A 42-B 43-A 44-A 45-B 46-A 47-C 48-A 49-A 50-C
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