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CNG TY TNHH CHU VN BIN T 0985829393 - 0943191900

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THI TH LN 14 THPT QUC GIA NM 2017

Bi thi: TON Thi gian lm bi: 90 pht, khng k thi gian pht

Cu 1: Cho a, b l cc s hu t tha mn 62 2 21

log 360 a.log 3 b.log 52

. Tnh a b

A. a b 5 B. a b 0 C. 1

a b2

D. a b 2

Hng dn:

Ta c 3 262 2 2 2 21 1 1 1 1

log 360 .log 360 .log 2 .3 .5 .log 3 .log 56 6 2 3 6

Mt khc 62 2 21

log 360 a.log 3 b.log 52

suy ra 1

a3

v 1

b6

1 1 1

a b3 6 2

p n : C

Cu 2: Cho hm s y f x lin tc trn v c bng bin thin nh hnh v. Tm tt c

cc gi tr thc ca m phng trnh f x 2m c ng hai nghim phn bit.

x -1 0 1 y - 0 + 0 - 0 +

y

0

-3

0

A. m 0

m 3

B. m 3 C.

m 0

3m

2

D. 3

m2

Hng dn:

Da vo bng bin thin, ta thy rng phng trnh f x 2m c ng hai nghim phn

bit khi v ch khi m 0

2m 03

2m 3 m2

p n : C

Cu 3: Tm s nghim ca phng trnh: 2

3 3log x 1 log 2x 1 2

A. 2 B. 1 C. 0 D. 3

Hng dn:

Phng trnh

2

23 3

3 3

2x 1 0; x 1log x 1 log 2x 1 2

log x 1 2log 2x 1 2

2 2 2 2

3

2x 1 0; x 1 2x 1 0; x 1x 2

log x 1 . 2x 1 2 x 1 . 2x 1 9

Vy phng trnh cho c nghim duy nht.

p n : B

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

Cu 4: Mt khi nn c th tch bng 30 . Nu gi nguyn chiu cao v tng bn knh

mt y ca khi nn ln hai ln th th tch khi nn mi bng

A. 120 B. 60 C. 40 D. 480

Hng dn:

Gi h, r ln lt l chiu cao v bn knh mt y ca khi nn.

Th tch khi nn ban u l 2 2non1

V r h 30 r h 903

Th tch khi nn sau khi tng bn knh y l 2 2

s

1 4V 2r h r h 120

3 3

p n : A

Cu 5: Cho hm s 1

y lnx 1

. Hi h thc no sau y ng?

A. yxy ' 1 e B. yxy ' 1 e C. yxy ' 1 e D. yxy ' 1 e

Hng dn:

Ta c y1 1 x 1

y ln y ' ln x 1 ' x.y ' 1 1 ex 1 x 1 x 1 x 1

p n : A

Cu 6: Nguyn hm F x x sin x dx tha mn F 0 19 l

A. 21

F x x cos x 202

B. 21

F x x cos x 202

C. 21

F x x cos x 182

D. 2F x x cos x 18

Hng dn:

Ta c 2x

F x x sin x dx x dx sin x dx cos x C2

M F 0 19 C 1 19 C 20 . Vy hm s 21

F x x cos x 202

p n : A

Cu 7: Tm tt c cc gi tr thc ca tham s m bt phng trnh

5 4 xx x x 12 m.log 3

c nghim.

A. m 2 3 B. m 2 3

C. 3m 12log 5 D. 32 m 12log 5

Hng dn:

iu kin: x 0;4 . Ta thy 5 4 x

4 x 4 5 4 x 3 log 3 0

Khi bt phng trnh cho tr thnh 3m f x x x x 12 .log 5 4 x *




Vi 3 x 1

u x x x 12 u '2 2 x 12

v

3

1v log 5 4 x v '

2 4 x 5 4 x .ln 3

Suy ra f ' x 0; x 0;4 f x l hm s ng bin trn on 0;4

bt phng trnh (*) c nghim

0;4

m min f x f 0 2 3

p n : B

Cu 8: Cho hm s 3x 1

y2x 1

c th (C). Khng nh no sau y l ng?

A. ng thng 1

y2

l tim cn ngang ca th (C).

B. ng thng y 3 l tim cn ngang ca th (C).

C. ng thng 1

y2

l tim cn ng ca th (C).

D. ng thng 3

y2

l tim cn ng ca th (C).

Hng dn:

Ta xt xx

3x 1 3lim y lim

2x 1 2

v

11 xx 22

3x 1lim y lim

2x 1

suy ra

1 3x ; y

2 2 ln lt l ng tim

cn ng, tim cn ngang ca th (C)

p n : C

Cu 9: Tnh gi tr ca biu thc 2016 164T log 2 .2 . 2

A. 3999

T4

B. T 2016 C.

3999T

2

D. T khng xc nh

Hng dn:

Ta c: 21 3999

2016 16 2016 16 2 24 22

1 1 3999 3999T log 2 .2 . 2 log 2 .2 .2 .log 2 .

2 2 2 4

p n : A

Cu 10: Trong khng gian vi h ta Oxyz cho hai im A 1; 1;2 v B 3;1;4 . Mt

cu (S) ng knh AB c phng trnh l:

A. 2 22x 2 y z 3 3 B.

2 22x 2 y z 3 3

C. 2 22x 2 y z 3 3 D.

2 22x 2 y z 3 3

Hng dn:

Gi I l trung im ca AB suy ra I 2;0;3 v AB 2 3 R 3

Phng trnh mt cu (S) cn tm l 2 22x 2 y z 3 3

p n : B



4

Cu 11: Trong khng gian vi h ta Oxyz cho mt phng (P) i qua im M 9;1;1 ct

cc tia Ox, Oy, Oz ti A, B, C (A, B, C khng trng vi gc ta ). Th tch t din

OABC t gi tr nh nht l

A. 81

6 B.

243

2 C.243 D.

81

2

Hng dn:

Phng trnh mt phng (P) i qua A a;0;0 ,B 0;a;0 ,C 0;0;c l x y z

1a b c

Mt khc (P) i qua im 39 1 1 9 1 1

M 9;1;1 1 3. . . abc 243a b c a b c

Th tch khi t din OABC l OABC1 abc 81

V .OA.OB.OC6 6 2

. Du bng xy ra khi

a 9b 9c

p n : D

Cu 12: Trong khng gian vi h ta Oxyz cho vct

a 1;m;2 ;b m 1;2;1 ;c 0;m 2;2 . Gi tr ca m a, b, c ng phng l:

A. 2

5 B.

2

5

C.

1

5 D. 1

Hng dn:

Ta c:

2

a 1;m;2a;b m 4;2m 1;2 m m a;b .c 2 5

b m 1;2;1

ba vecto a, b,c ng phng khi v ch khi 2

a;b .c 0 2 5m 0 m5

p n : A

Cu 13: Tm tt c cc gi tr thc ca tham s m th hm s 4 2mC : y x mx m 1

ct trc honh ti bn im phn bit.

A. m 1 B. m 1

m 2

C. khng c m D. m 2

Hng dn:

Phng trnh honh giao im ca mC v d l 4 2 4 2x mx m 1 0 x 1 m x 1

2

2 2 2

22

x 1x 1 0x 1 x 1 m x 1

x m 1 *x m 1

mC ct d ti bn im phn bit (*) c hai nghim phn bit khc m 1

1m 2

p n : B

Cu 14: Nguyn hm ca hm s f x cos3x.cos x l:




A. sin 4x sin 2x

C2 2

B. sin 4x sin 2x

C8 4

C. sin 4x sin 2x

C8 8

D. sin3x.sin x C

Hng dn:

1 1 1 sin 4x sin 2x

f x cos3x.cos x cos 4x cos 2x f x dx . cos 4x cos 2x dx C2 2 2 8 4

p n : B

Cu 15: Hm s no sau y nghch bin trn ?

A. 3 2y x 3x 3x 2 B. 3 2y x 3x 3x 2

C. 3 2y x 3x 3x 2 D. 3 2y x 3x 3x 2

Hng dn:

Xt hm s 3 2y ax bx cx d vi x , ta c 2y ' 3a.x 2b.x c

Hm s cho nghch bin trn 2

y'

a 0y ' 0; x

' b 3ac 0

nn hm s

3 2y x 3x 3x 2 l hm s ng bin trn

p n : B

Cu 16: ng cong hnh di y l th ca mt hm s trong bn hm s c lit

k bn phng n A, B, C, D di y. Hi hm s l hm s no?

A. xy 2 B. xy 2 C. 2y log x D. 2y log x

Hng dn:

Da vo th hm s, ta c nhn xt sau:

Hm s ng bin trn tp xc nh vi h s a > 0

th hm s i qua im M 1;2

th hm s nm phi trn trc honh v nhn trc honh lm tim cn ngang.

Vy hm s cn tm l xy 2

p n : A

Cu 17: Tm s nghim ca phng trnh 3 93log x.log x.log x 8

A. 2 B. 0 C. 1 D. 3

Hng dn:



6

iu kin: x > 0. Ta c 33

log x 2.log x v 9 31

log x .log x2

Khi phng trnh 3

3 9 3 33log x.log x.log x 8 log x 8 log x 2 x 9

p n : C

Cu 18: Vi gi tr thc no ca m th phng trnh x x 24 2 m 0 c hai nghim thc

phn bit?

A. m 0 B. 0 m 4 C. m 4 D. m 0

Hng dn:

t xt 2 0 , khi 2

x x 2 x x 24 2 m 0 2 4.2 m 0 t 4t m 0 *

phng trnh c hai nghim thc phn bit * c hai nghim dng phn bit

0 m 4

p n : B

Cu 19: Tm tt c cc gi tr thc ca tham s 2y x mx m hm s ng bin trn

khong 1;

A. m 2 B. m 1 C. m 1 D. m 2

Hng dn:

Ta c 22

2x my x mx y '

2 x mx

vi mi x thuc tp xc nh

hm s ng bin trn khong

2x m 0 m 2x1; ; x 1 ; x 1 m 1

x x m 0 m x

p n : B

Cu 20: Gi s c sau mt nm din tch rng ca nc ta gim x phn trm din tch

hin c. Hi sau 4 nm dim tch rng ca nc ta s l bao nhiu phn din tch hin

nay?

A. 4

x1

100

B. 100% C. 4x

1100

D. 4

x1

100

Hng dn:

Gi S l din tch rng nc ta hin nay.

Sau nm th nht, din tch rng cn li l x

S S.x% S 1100

Sau nm th hai, din tch rng cn li l 2

x x x xS 1 S 1 . S 1

100 100 100 100

Sau nm th n, din tch rng cn li l n

xS 1

100

nn sau 4 nm din tch rng s l

4x

1100

phn din tch nc ta hin nay.

p n : D




Cu 21: Hnh chp S.ABCD c y hnh ch nht AB a,SA ABCD , SC to vi mt

y gc 045 . Mt cu ngoi tip hnh chp S.ABCD c bn knh y bng a 2 . Th tch

khi chp S.ABCD bng

A. 32a B. 32a 3 C. 3a 3

3 D.

32a 3

3

Hng dn:

Gi O l tm ca hnh ch nht ABCD v I l trung im ca SC.

Khi OI ABCD IA IB IC ID m SAC vung ti A IA IS IC

Do I l tm mt cu ngoi tip khi chp S.ABCD suy ra IA a 2 SC 2a 2

Mt khc AC l hnh chiu ca SC trn mt phng

0ABCD SC; ABCD SC;AC SAC 45

Suy ra SAC vung cn 3

S.ABCD ABCD

1 1 2a 3SA AC 2a V .SA.S .2a.a.a 3

3 3 3

p n : D

Cu 22: Trong khng gian vi h ta Oxyz, C cho mt phng

P : x y 2z 1 0, Q : x y z 2 0, R : x y 5 0 . Trong cc mnh sau mnh no

sai?

A. Q R B. P Q C. P / / R D. P R

Hng dn:

Ta xt P QP : x y 2z 1 0 n 1;1;2 , Q : x y z 2 0 n 1;1; 1

V RR : x y 5 0 n 1; 1;0 suy ra

P Q

P R

Q R

n .n 0 P Q

n .n 0 P R

Q Rn .n 0

p n : C

Cu 23: Mt hnh tr c bn knh 5cm v chiu cao 7cm. Ct hnh truh bng mt phng

(P) song song vi trc v cch trc 3cm. Din tch thit din to bi hnh tr v mt phng

(P) bng

A. 2112cm B. 228cm C. 254cm D. 256cm

Hng dn:

Thit din ct bi mt phng (P) v hn tr ABCD l hnh ch nht, c di AD h 7 cm

Gi O l tm ng trn y cha cnh AB d O; P d O; AB 3cm

Gi I l trung im ca AB 2 2 2 2AI OA OI 5 3 4 AC 8

Din tch ca hnh ch nht ABCD l 2ABCDS AB.AD 8.7 56cm

p n : D



8

Cu 24: Cho hm s y x 2 . Chn khng nh ng?

A. Hm s t cc tiu ti x = 0 B. Hm s t cc i ti x = -2

C. Hm s t cc tiu ti x = -2 D. Hm s khng c cc tr.

Hng dn:

Xt hai trng hp x 2 0

x 2 0

ph du tr tuyt i nn d thy hm s t cc tiu ti x

= -2

p n : C

Cu 25: Trong khng gian vi h ta Oxyz, cho mt phng (P) ct cc trc ta ti

M 8;0;0 , N 0;2;0 ,P 0;0;4 . Phng trnh mt phng (P) l

A. x 4y 2z 8 0 B. x 4y 2z 8 0

C. x y z

14 1 2 D.

x y z0

8 2 4

Hng dn:

Phng trnh mt phng (P) l x y z

1 x 4y 2z 8 08 2 4

p n : A

Cu 26: Trong cc hm s sau, hm no nghc bin trn khong 0;

A. 2y x log x B. 21

y x logx

C. 2 2y x log x D. 2y log x

Hng dn:

Da vo p n, ta c cc nhn xt sau:

21

y x log x y ' 1 0; x 0x.ln 2

hm s ng bin trn khong 0;

21 1

y log y ' ; x 0x x.ln 2

hm s nghch bin trn khong 0;

2 21

y x log x y ' 2x 0; x 0x.ln 2

hm s ng bin trn khong 0;

p n : B

Cu 27: gii bt phng trnh 12

log 2x 1 1

A. 3

;2

B. 3

1;2

C. 1 3

;2 2

D. 3

;2

Hng dn:

Bt phng trnh

11 1

2

2x 1 0 2x 1 0 1 3log 2x 1 1 x ;

2x 1 2 2 22x 1 2

p n : C




Cu 28: Trong khng gian vi h ta Oxyz cho mt phng (P) i qua gc ta O v

vung gc vi hai mt phng Q : 2x y 3z 0, R : x 2y z 0 . Phng trnh mt phng

(P) l

A. 7x y 5z 0 B. 7x y 5z 0

C. 7x y 5z 0 D. 7x y 5z 0

Hng dn:

Ta c:

Q

P Q R

R

Q : 2x y 3z 0 n 2; 1;3n n ;n 7;1;5

R : x 2y z 0 n 1;2;1

V mt phng (P) i qua O 0;0;0 nn phng trnh mt phng (P) l 7x y 5z 0

p n : B

Cu 29: Cho ming tn trn tm O bn knh R. Ct ming tn hnh qut OAB v g phn

cn li thnh mt hnh nn nh O khng y (AO trng vi OB). Gi S, S ln lt l

din tch ca ming tn hnh trn ban u v din tch ca ming tn cn li. Tm t s S

S '

th tch khi nn ln nht.

A. 1

4 B.

6

3 C.

2

3 D.

1

3

Hng dn:

Gi gc AOB rad suy ra di dy cung AB l ABL .R

Nn di dy cung cn li l cL 2R R R 2 l chu vi ca ng trn y ca

hnh nn.

Bn knh ng trn y hnh nn l

22 20 0

R 2 1 1 R R 1 V .R .h .R . 1 .h

2 2 3 3 2

Mt khc

2 2

2 2 2

0

R 2 2 h OA R R R 1

2 2

Khi 2 2

2 3

0

1 1 2 2 V .R .h .R . 1

3 3 2 2

. Vi 0

R2 t

2 R

, ta xt

2 2f t t . 1 t

Ta c 3

2

2t 3t 6 6f ' t ; f ' t 0 t f

3 31 t

t gi tr nh nht



10

Din tch xung quanh ca hnh nn l 2 xq 0 0S S r l rR R

Din tch ming tn ban u l 21S R suy ra 01

2

RS 6

S R 3

p n : B

Cu 30: Cho hm s y f x c o hm trn on a;b . Ta xt cc khng nh sau:

1)Nu hm s f x t cc i ti im 0x a;b th 0f x l gi tr ln nht ca f x

trn on a;b .

2)Nu hm s f x t cc i ti im 0x a;b th 0f x l gi tr nh nht ca f x

trn on a;b .

3)Nu hm s f x t cc i ti im 0x v t cc tiu ti im 1 0 1x x , x a;b th

ta lun c 0 1f x f x

Gi n l khng nh ng. Tm n ?

A. n 1 B. n 3 C. n 2 D. n 0

Hng dn:

Khng nh trn u 1, 2 u sai. V ta c th xt hm s 4 2y x 2x 1 trn on 2;2

3 sai v n ch ng trong 1 s trng hp nh hm trng phng hm bc 3.

p n : D

Cu 31: Trong khng gian vi h ta Oxyz cho mt cu (S) c tm I 2; 1;3 v ct mt

phng P : 2x y 2z 10 0 theo mt ng trn c chu vi bng 8 . Phng trnh mt cu

(S) l:

A. 2 2 2

x 2 y 1 z 3 5 B. 2 2 2

x 2 y 1 z 3 5

C. 2 2 2

x 2 y 1 z 3 25 D. 2 2 2

x 2 y 1 z 3 25

Hng dn:

Bn knh ca ng trn l C 2r 8 r 4




Khong cch t tm I 2; 1;3 n mt phng (P) l

2 22

2.2 1 2.3 10d I; P 3

2 1 2

Suy ra bn knh mt cu (S) l 2 2 3 2R r d I; P 3 4 5

Phng trnh mt cu cm tm l 2 2 2

x 2 y 1 z 3 25

p n : C

Cu 32: Cho hm s 3y log 2x 1 . Chn khng nh ng ?

A. Hm s ng bin trn khong 0;

B. Trc Oy l tim cn ngang ca th hm s.

C. Hm s ng bin trn khong 1

;2

D. Trc Ox l tim cn ng ca th hm s.

Hng dn:

Hm s 3y log 2x 1 c

2 1y ' 0; x

2x 1 .ln 3 2

hm s ng bin trn

1;

2

p n : C

Cu 33: Cho hnh lp phng cnh a v mt hnh tr c hai y l hai hnh trn ni tip

hai mt i din ca hnh lp phng. Gi 1S l din tch ca su mt ca hnh lp phng,

2S l din tch xung quanh ca hnh tr. T s 1

2

S

S bng

A.

6 B.

2 C.

3 D.

Hng dn:

Din tch ton phn ca hnh lp phng l 21S 6a

Gi O, O ln lt l tm ca hnh vung ABCD, ABCD, khi trh OO' a

Gi r l bn knh ng trn y ca hn tr suy ra a

r2

. Suy ra 22 xqS S 2rh a

Vy t s 2 21 1

2 2

S S6 66a : a

S S

p n : A

Cu 34: Ngi ta mun xy mt b cha nc dng hnh hp ch nht khng np c th

tch bng 2500

m3

y b l hnh ch nht c chiu di gp i chiu rng. gi thu nhn

cng xy b l 500.000 ng/ 2m . Chi ph thu nhn cng thp nht l:

A. 150 triu ng B. 75 triu ng

C. 60 triu ng D. 100 triu ng

Hng dn:



12

Gi chiu rng ca hnh ch nht y b l x m suy ra chiu di ca hnh ch nht l

2x m

Gi h l chiu cao ca b nn ta c 2 22

500 250 250V S.h 2x .h x .h h

3 3 3x

Din tch ca b l 2 2 2 22

250 500S 2.h.x 2.2h.x 2x 2x 6.hx 2x 6. .x 2x

3x x

p dng bt ng thc AM GM, ta c 2 2 23500 250 250 250 250

2x 2x 3 2x . . 150x x x x x

Du = xy ra khi 2 3250

2x x 125x

chi ph thp nht thu nhn cng l 1

150. 752

triu ng

p n : B

Cu 35: Tm cc gi tr thc ca tham s m th hm s 3 2y x 3x mx 2 c hai

im cc tr A v B sao cho ng thng AB song song vi ng thng d : y 4x 1

A. m 1 B. m 3 C. m 0 D. khng c m tha mn

Hng dn:

Xt hm s 3 2y x 3x mx 2 , ta c 2y ' 3x 6x m y'' 6x 6

hm s c hai im cc tr khi v ch khi y = 0 c hai nghim phn bit

9 3m 0 m 3

Ta c: 2

3 23x 6x m 6x 6y '.y '' 2m 6 6 m

y x 3x mx 2 x18 18 3 3

2m 6 6 m

AB : y x3 3

l ng thng i qua hai im cc tr ca th hm s.

Mt khc (AB) song song vi (d) suy ra

2m 64

2m 6 123m

6 m 6 m 31

3

p n : D

Cu 36: Mt hnh nn nh O c din tch xung quanh bng 260 cm , di ng cao bng 8cm. Khi cu (S) c tm l nh hnh nn, bn knh bng di ng sinh ca

hnh nn. Th tch khi cu (S) bng

A. 32000cm B. 34000cm C. 3288cm D. 34000

cm3

Hng dn:

Din tch xung quanh ca hnh nn l 2 2 2xqS rl r h r 60 r r 64 60 r 6

di ng sinh 2 2l r h 10cm . Th tch ca khi cu (S) l

3 3 34 4 4000V l cm3 3 3

p n : D




Cu 37: Hm s ln 2xF x e x 0 l nguyn hm ca hm s no sau y ?

A. ln 2x

ef x

x B. ln 2xf x e C.

ln 2xe

f x2x

D. ln 2xf x 2e

Hng dn:

Ta c ln 2x

ln 2x ln 2x ef x F ' x e ' ln 2x '.e

x

p n : A

Cu 38: Mt cng ty d kin lm mt ng ng thot nc thi hnh tr di 1km, ng

knh trong ca ng (khng k lp b tng) bng 1m; dy ca lp b tng bng 10cm.

Bit rng c mt khi b tng phi dng 10 bao xi mng. S bao xi mng cng ty phi

dng xy dng ng ng thot nc gn ng vi s no nht?

A. 3456 bao B. 3450 bao C. 4000 bao D. 3000 bao

Hng dn:

Bn knh ca ng trn y hnh tr khng cha b tng bn trong ng ng l

100 10.2 : 2 40cm

Th tch ca ng ng thot nc l 2

2 31V r h . .1000 250 m2

Th tch ca khi tr khng cha b tng (rng ) l

2

2 3

1

2V r l . .1000 160 m

5

Vy s bao xi mng cng ty cn phi dng xy dng ng ng l 3456 bao

p n : A

Cu 39: Cho lng tr ng ABC.ABC c y ABC l tam gic

vung ti B; AB a;BC a 2 ; mt phng A'BC hp vi y ABC

gc 030 . Th tch ca khi lng tr l

A. 3a 6 B. 3a 6

12 C.

3a 6

3 D.

3a 6

6

Hng dn:

Ta c AA' ABC AA' BC m AB BC BC AA'B'B

Mt khc

0A'BC AA'B'B A 'B

ABC AA'B'B AB A'BC ; ABC A'B;AB A'BA 30

BC A'BC ABC

Xt A 'AB vung ti A, c 0AA ' a

tan A 'BA AA ' tan 30 .ABAB 3

Th tch khi lng tr l 3

ABC.A'B'C' ABC

a 1 a 6V AA '.S . .a.a 2

2 63



14

p n : D

Cu 40: Hnh chp S.ABCD c y ABCD l vung cnh a, hnh chiu vung gc ca S

trn mt phng ABCD trng vi trung im ca AD; M trung im CD; cnh bn SB hp

vi y gc 060 . Th tch ca khi chp S.ABM l:

A. 3a 15

3 B.

3a 15

4 C.

3a 15

6 D.

3a 15

12

Hng dn:

Gi H l trung im ca AD nn S.ABM ABM1 1

SH ABCD V .SH.V .SH.AB.BC3 6

Ta c HB l hnh chiu ca SB trn mt phng

0ABCD SB; ABCD SB;HB SBH 60

Xt SHB vung ti H, c 0SH a 5 a 15

tan SBH SH tan 60 .BH 3.BH 2 2

Vy th tch ca khi chp S.ABM l 3

2

S.ABM

1 a 15 a 15V . .a

6 2 12

p n : D

Cu 41: Hm s no sau y khng c gi tr ln nht?

A. y cos 2x cos x 3 B. 4 2y x 2x

C. 3y x x D. 2y 2x x

Hng dn:

Xt hm s 3y x x , ta thy rng x x

lim y , lim y

nn hm s khng c gi tr ln

nht

p n : C

Cu 42: Cho hnh chp S.ABCD y ABCD l hnh ch nht; AB 2a,AD a . Hnh chiu

ca S ln mt phng ABCD l trung im H ca AB; SC to vi y gc 045 . Khong

cch t A n mt phng SCD l

A. a 6

4 B.

a 3

3 C.

a 6

3 D.

a 3

6

Hng dn:

Gi M l trung im ca CD suy ra HM CD CD SHM

K HK SM vi K SM CD HK HK SCD d A; SCD d H; SCD HK Ta c HC l hnh chiu ca SC trn mt phng

0ABCD CD; ABCD SC;HC SCH 45

Khi SCH vung cn ti H m HC a 2 SH a 2

Xt SHM vung ti H c ng cao HK suy ra 2 2

SH.HM a 6HK

3SH HM




p n : C

Cu 43: Trong khng gian vi h ta Oxyz cho hai im A 1;1;2 ,B 3; 1;1 v mt

phng P : x 2y z 1 0 . Mt phng (Q) cha A, B v vung gc vi mt phng (P) c

phng trnh l:

A. 4x 3y 2z 0 B. 2x 2y z 4 0

C. 4x 3y 2z 11 0 D. 4x 3y 2z 11 0

Hng dn:

Ta c A 1;1;2 ;B 3; 1;1 AB 2; 2; 1 v Pn 1; 2;1 nn Q Pn AB;n 4;3;2

Phng trnh mt phng (Q) i qua A 1;1;2 v c Qn l 4x 3y 2z 11 0

p n : D

Cu 44: Bit 1

0

f x dx 2 v f x l hm s l. Khi 0

1

I f x dx

c gi tr bng

A. I 1 B. I 0 C. I 2 D. I 2

Hng dn:

t t x dx dt v x 1 t 1

x 0 t 0

nn

0 0 1

1 1 0

I f t dt f t dt f t dt 2

p n : C

Cu 45: Tch phn 1

2

0

I x x 1dx c gi tr bng

A. 2 2 1

I3

B.

2I

3 C.

2 2I

3 D.

2I

3

Hng dn:

t 2 2 2u x 1 u x 1 u du x dx v x 0 u 1

x 1 u 2

Khi 2 3 2

2

11

u 2 2 1I u du I

3 3

p n : A

Cu 46: Bit tch phn 1

x

0

I 2x 1 e dx a be a ;b . Khi tch a.b c gi tr bng:

A. 1 B. -1 C. 2 D. 3

Hng dn:

Xt hm s xF x mx n e , ta c xF' x mx m n e m F x l mt nguyn hm ca

hm s x x 10m 2 m 2

f x 2x 1 e I 2x 1 e e 1 a be a b 1m n 1 n 1



16

Cch 2: t 11 1

x x x

x x 00 0

u 2x 1 du 2dxI 2x 1 e 2e dx 3e 1 2e e 1 a b 1

dv e dx v e

p n : A

Cu 47: Cho tch phn 3

0

xI dx

1 x 1

nu t t x 1 th

2

1

I f t dt trong

A. 2f t t t B. 2f t 2t 2t

C. 2f t t t D. 2f t 2t 2t

Hng dn:

t 22

dx 2t dtt x 1 t x 1

x t 1

v i cn

x 0 t 1

x 3 t 2

Khi 2 2 22

2 2

1 1 1

t 1I 2t. dt 2t. t 1 dt 2t 2t dt f t 2t 2t

t 1

p n : D

Cu 48: Khng nh no sau y sai ?

A. 2017 2016

3 1 3 1 B. 2 1 32 2

C.

2016 2017

2 21 1

2 2

D.

2017 2016

2 1 2 1

Hng dn:

Hm s xy a l hm s ng bin trn khi a > 1 v l hm s nghch bin trn khi 0

< a < 1. Khi xt vi 1 2x x th 1 2x xa a khi a > 1 v 1 2x xa a khi 0 < a < 1

Da vo cc p n, ta thy rng 2017 2016

3 1 3 1 v 1 2

0 a 3 1 1

x 2017 x 2016

p n : A

Cu 49: Tm s tim cn ngang ca th hm s 2x 1 x

A. 2 B. 1 C. 3 D. 0

Hng dn:

Ta xt 22x x xx

2

1 1lim y lim x 1 x lim lim 0 y 0

1x 1 xx 1 1

x

l tim cn

ngang

p n : B

Cu 50: Trong khng gian vi h ta Oxyz cho mt cu (S) i qua hai im

A 1;1;2 ,B 3;0;1 v c tm thuc trc Ox. Phng trnh mt cu (S) l




A. 2 2 2x 1 y z 5 B.

2 2 2x 1 y z 5

C. 2 2 2x 1 y z 5 D.

2 2 2x 1 y z 5

Hng dn:

Gi I m;0;0 l tm mt cu (S) m 2 22 2 2A,B S IA IB x 1 1 2 x 3 1

2 2 2x 1 I 1;0;0 R IA 5 S : x 1 y z 5

p n : A

p n

1-C 2-C 3-B 4-A 5-A 6-A 7-B 8-C 9-A 10-B

11-D 12-A 13-B 14-B 15-B 16-A 17-C 18-B 19-B 20-D

21-D 22-C 23-D 24-C 25-A 26-B 27-C 28-B 29-B 30-D

31-C 32-C 33-A 34-B 35-D 36-D 37-A 38-A 39 40-D

41-C 42-C 43-D 44-C 45-A 46-A 47-D 48-A 49-B 50-A

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