KÌ THI THỬ THPTQG LẦN 11- NĂM HỌC 2016 - 2017 TOÁNchuvanbien.vn/tai-lieu/tai-xuong/106/dap-an-mon-toan-lan-11ngay... · KÌ THI THỬ THPTQG LẦN 11- NĂM HỌC 2016 - 2017

DAO NG C HC

Website hc trc tuyn: chuvanbien.vn Group hc tp: https://www.facebook.com/groups/chuvanbien.vn/

CNG TY TNHH CHU VN BIN T 0985829393 - 0943191900

Email: [email protected] Fanpage: https://www.facebook.com/chuvanbien.vn/

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][

NI N LUYN THI THPT QG MN VT L l nhc n THY CHU VN BIN

CNG TY TNHH CHU VN BIN T 0985829393- 0943191900 Email: [email protected] Fanpage: https://www.facebook.com/chuvanbien.vn/

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




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



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:




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



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




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:



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.




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



10

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




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



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




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



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




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



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:




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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