Tuyen Quang Giai chi tiet de toan quoc gia 2017

BAI GIAI CHI TIET MOT SO CAU TRONG DE THI TOAN QUOC GIA 2017
MA DE 115
Cau 33 : Mot nguoi gui 50Trieu voi lai suat 6% neu khong rut thi sau moi nam so lai nhap vao
goc de tinh lai cho nam tiep theo . Hoi sau it nhat bao nhieu nam nguoi do nhan duoc so tien
hon 100Trieu
A. 12 nam

B. 14 nam

C. 13 nam

D. 11 nam

Ta da biet neu mot nguoi co so goc von ban dau la G gui vao ngan hang voi lai suat lar thi sau n nam nguoi

do duoc lanh ca goc va lai la TT. — G( +t)”

Goi n la so nam gui vao duoc lanh ra lon hon 100 trieu CA GOC VÀ LAI. Xay ra khi va chi

khi 100= 50(1+0,06)" =(I+0,06"=-°2=2=n=log,„2=— S8“ ~ 11,9
50
°
Jog(1,06)
vay yeu cau bai toan thoa so nam it nhat la 12 nen dap an chon

la A

Cau 37 (115) Mot chuyen dong trong 3 gio voi van toc v (km/h) phu
thuoc thoi gian t(h) co do thi van toc nhu hinh ve .Trong khoang thoi
gian | gio ke tu khi bat dau chuyen dong , do thi do la mot phan cua
parabon co dinh I(2; 9) khoang thoi gian con lai do thi chuyen dong

la mot doan thg song song voi truc Ox . Tinh quang dg cua chuyen
dong trong 3 gio:
A.s = 13,83(km)
B.s = 23,25(km)

7,1

4

C.s = 21,59(km)
D.s =15,50(km)
Trong mot gio dau vat chuyen dong co do thi van toc la parbol
co ptla y= 3x

+5x +4 tai x= Ì ta co v(l) = 7,75 = _
1

di duoc trong 3 gio la s = fart

r,
5
s= JC7x*+

1 2 3

Suy ra quang duong ma vat

3

5x+ 4)dx+ [ex

0

hay

oy

= 6,08(3)
+ 15.5 = 21,58

1

31
5x+4)dx + 47 ~ 21,58 tuc hai gio sau chuyen dong deu voi van

0

toc la “ (km/h)

Cau 40 (115)Cho ham so y = * —
X—
A.m>4

y=

ciom
(x—])

thoa man miny =3 Menh de nao duoi day dung
[2:4]

B.3
m< —Ithiy'>0=>

C.m<-l

D.1<

Miny [2:4]

1

4+m

m>—Ithiy’<0=> Miny =y@) =~
[2;4]

suy ra chon dap an A

m<3

a3 6 m=

(khong thoa m < —1)

=3 + m=5

(thoa dk m > —1)

Cau 42 (115) Cho F(x) = x’ la mot nguyen ham cua ham so f(x)e™* . Tim nguyen ham cua
ham so f '(x)e*

A. [f'\(x)e* dx =—2x?+2x+C

B.[f)e”“dx=2x?”—2x+C

C. f f'(x)e* dx =—x? +2x +C

D. [ f'(x)e™ dx =—x* +x+C

Cach 1:
4x

e
2x

J†@Œ)e”dx= fe

2x

2x

2x

e

4x

dx =2 (—2x)=2(x— x”)+c=—2x”+ 2x +c=

et

dap an A
Cach 2: dat

yee

du = 2e**dx
>

dv =f '(x)dx

v=f(x)=

2x
2x

=>

J†@œ)e”

2x

dx=e”.———

e

e*

— .2e°*dx

e

=2x

— 2x”

+C

Cau 46 (115) Cho hinh tu dien deu canh a. M, N lan luot la trung diem AB, BC, E la diem doi

xung voi B qua D khi do mp(MNE) chia tu dien thanh hai phan . Tinh the tich V cua khoi da
dien chua dinh A

ya va
216

3

By

228
18

3

cy — Uva

216

3

v_132a
216

E

B

NS
M
A

Giai tom tat
V acMNPO = Ve acmn — Vie ACPO

dt so ADO
sey, = 51 4(E(ABC)).
Ve scan = 51 4(E(ACMN)).t
|
7
Ve ace = 5| 4(E(ACPQ)).dt
sony = GE
(ACD)).
dt rg
V1iDIatrung diemcua BE=>

d(E, (ABC)) = 2d(D, (ABC))

⁄

3


= 2d(A,(BCD))TA CUNG CO d(E,(ACPQ)) = d(B, (ACPQ)) = d(B,(ACD)) = d(A, (BCD))
duuc _ BA BC _ 2BM 2BN _„

DO: MN//AC = ABAC~ ABMN =

——=
—_
BM'BN
BM’ BN

dt...

=> dt un = " ape = dt oun = “ates . Tu gt thiet thi P, Q lan luot la trong tam cac tam giac

dt
BCE va ABE > DQ= =DA:DP = =bC — ADQP ~ ADAC > i. _l => dt
DAC

oun = 3 Tt, (BCD)).7 dtaco — 2 VABCD
2a, (ABC)).dt

= dt cpg = 9 taco — 9 tse

sue, (ACD)).dt sono = sda, (CD). dt,.„ = Vane

V acMNPO

—

YEACMN __ Vie ACPO —

3

8

= 11 13a
183
4

CACH 2

A

V AcMNPO — VÀ MNPQ +

cn

SUY RA:

2 YABCD ~ 9 YABCD
2

= ¬
18

DAC

3

3

]

]

8

8

9 Tát

DQP

2

3

+ là — 3a" = LIV2a
9
216

chon dap an C

— Va CNP + VN Apo + VN AMQ

=3 1 [dt ,cyp-d(A, (CNP)) + dt.ap ON, (APQ)) + dt ayo ACN, (AMQ))|

dt ncnp

=

dtp amo

a

9°93

2a N3
7

a2J3

2

12

5

A43 jlaa V3, 1, 2a v3,

dt apo — dt scp - (dtApop + dt AACP) —

4

2332

2 3 2

_ a3 s 7a?AJ3 - a7v3
4
36
18
a

d(A,(CNP)) = d(A,(BCD)) =, la — >

= ~,

d(N, (APQ)) = d(N, (ACD)) = s4, (ACD))

= sul, (BCD)) = = d(N,(AMQ)) =d(N,(ABD)) = sac, (ABD)) = sul, (BCD)) as
=
1.a7V¥3
= Vicyneo
= ae
3

12

Vay Chon dap an C

aJ6
3

SE

a2J3

aJV6—

18

6

EE

a2

V3 aV6.
12

EN

6

=. 1

3

Ila? VIS
216

=

1 a?/2

216


Cau 47 (115): co? so phuc Z thoa |Z — 3i] =5va =

la so thuan sao

giasuZ=a+bi

=|Z-3i|=5©
và

Z
Z—-4

—

mm...

atbi

(

nh.

_(a+bi( =4—bi)_ a -4a+b

a+bi-4

la so thuan ao <>

.=.

(a— 4)“ +b

a—4)/+bˆ=0_

(a—-4)/+bˆ

la +b-6b=16

)

+64

thay vao(*) > a?—4a +> —"*"# *
9

"9 6 13a?

—4bi
(a-4) +bí

—> 4a—6b—16=>b—

a’ —4a +b’ =0(*)|a* —4a +b? =0
da? —32a

„

68a+64=086]

2â

—

8

a=4=> b= 0(oai)
a=

16

g

> b= 9 (nhan)

Vay co duy nhat so phuc Z thoa dk nen chon dap an C
Cau 48 (115): Cho ham so y = f(x). Do thi cua ham so y = f(x) nhu hinh ve

Dat h(x) = 2f(x)T— x” . Menh de nao duoi day dung
A. h(4) = h(—2) < h(2)

B. h(2) > h(4) > h(—2)

C. h(2) > h(—2) > h(4)

D. h(4) = h(—2) > h(2)

Nhan xet : Cac diem A(-2;2), B(2;2), C(4; 4) cung nam tren dg thg y = x

Tu gt taco h’(x) = 2(f(x) - x) > h(x) = Jn@) dx = f (2e (x) — 2x]dx

Goi S,ladien tich hinh gioihan :

eS,

y=f'(x)y=x

X=—2;x=

&Š„ la dien tịch hình g1ol han :

y=f'(x)y=x

Xx=2;x=4

= [If(x)— x]dx => 2§, =2 [Tf &)— x]dx = [hŒ%)dx = g(x)”, = h(2)— h(—2) >0

= h(2) > h(—2) (*)
eS,= [| x—f) dx = 2S, =—2 [[f (x) —x]dx =— [h'(x)dx = —h(x)]}
=h(2) —h(4) >0=> h(2) > h(4) (**)
=> h(4) < h(2) = dap an D bi loai tiep theo ta so sanh h(4) voi h(- 2) ta co

h(4) —h(—2) = fh'(x)dx =2 | [f (@&) — x]dx = 2] [Tf (x) — x]dx + [Tf '(x) — x]dx
=2S, — 2S, = 2(S,—S,) > 0 => h(4) —h(-2) = h(4) > h(—2) ***)
Tu (*), C**) va (***) suy ra ta co:


Jlf'œ)— x|dx = []f'&)— x|dx + []f'Œ&)— x|dx= [Tf'Œ) — x]dxT— [Ix—f'%)dx =8, —S$, >0

S, la dien tich hinh phang gioi han boi doan CA voi do thi cua f(x) , C(-2; -2)
Osuy rataco h(2)>h(4) > h(-2). Vay dap an chon la dap an B

Cau 49 (115) trong he Oxyz cho (S): x* + y* +z? =9
diem M(1; 1; 2) va (P):x+y+z—-4=0.Goidla
dg thg di qua M , thuoc (P) cat (S) tai hai diem A, B
sao cho AB nho nhat . Biet d co mot vtcp u= (1;a;b)
Tinh T=a-

b

Nhan xet : OM = V6 <3=R; toa do M nghiem
dung pt (P) nen M nam trong mat cau va nam tren

P

(P) suy ra d qua M nam trong (P) cat (S) tai 2 diem A,B sao cho AB ngan nhat khi va chi

khid | HM (Hla hinh chieu cua
O tren (P)). Goi Us la vtep cua
d thi Us =[MH,np]
1
s—
3

1
—=-(];l;—2
)
3|

C2

zZ=t

= H(2;4;5)
= MA = (
3 3 3
——

Wo |

tu gt thi OH co pt; y=t> 3t—4=05St=

wu]&

x=t

= Us —[MH,n,]= (3; -3;0) vtcp cuad la (1; - 1; 0) = (1; a; b) =>a—b=—1—0=~-1
Suy ra dap an chon la B
Cau 50 (115) Xet cac so thục duong thoa log,
cuaP=x+y

-—

B we

X +

C

4y

=3xy +x+2y—4 . Tim gia tri nho nhat

tis

D HP

Giai tom tat

log, xi2y =3xy+xX+2y—4<©
log,(I— xy)+3(I— xy) +lI= log;(x +2y)+x+2y

< log, 3(1— xy) + 3(1 — xy) = log,(x + 2y)+ (x+2y) (*)

Dat a = 3(1 — 3xy) >0 vab=(x + 2y) >0

> (*) S log,a+a=log,b+b

1

Xet ham so f(t) = log,t+t>f'@Q=1+ tìn3 >0Vt>0 =>
3(l-xy)=x+2y

=>P=x+

—X +3

3x +2

= y(243x)=3-x

'

=> P'=1-

11

(3x+2)

y=

3x +2

_ 9x+l2x—7
=

(3x+2)

(*) xay ra khi va chi khi

(do
x >Q)

_

=

s.

TT
3

chan)

2-Jl1l.
x =———~— (loai)


Suy ra ta co bbt cua
P la

x

0
P
pl

SUY RA DAP AN CHON LA C

x

+2

`2

+0

+

+ 0

' „2+
3

DE 102
Cau 35

Cho ham so y =

—

A.m<0
l1—m

*

=

X

;

T1

=>m
[1:2]

[1:2

C.0
InY

|Miny

+

khong thoa dk m<1

= y(l) + y(2) =

Maxy = y+

.

(2)

2

m > Ithiy'<0 => Miny + Maxy = y(2) + y() = s1
[1:2]

16

thoa man miny + max y=

B.m>4

œx+DĐP

<>m=5

x+m

Il+m

=>

1

[1:2]

.

Menh de nao duoi day dung

+

D.22+m
5m+7

“es

5

3

=

a

7

=

16

6

=—

16

m=s

3

Dap an chon la B
Cau 37 Cho x.y la cac so thục duong thoa x° + 9y* =6xy

.Tinh

A.M=2

+
2

B.M=I

C. M=

Tu gt => x” +6xy +9y” =12xy ©(x+3y)” =—12xy
M=

1+log,, x +log,, y _

2log,,(x+3y)
Chon dap an B

log,, 12xy

_ log,, 12xy —Ị

log,,(x+3y)

log,,12xy

Cau 38 : Mot chuyen dong trong 3 gio voi van toc v (km/h)
phu thuoc thoi gian t(h) co do thi do la mot phan cua parabon
co ding I(2; 9) .Tinh quang dg cua chuyen dong trong 3 gio
A.s = 24,25(km)
B.s = 26,75(km)
C.s = 24,75 (km)

D.s = 25,25(km)

tu gt thi pt van toc cua chuyen dong duoc cho boi pt :

y= Wx

-+3x +6 suy ra quang duong vat di duoc trong 3 gio

la s= JCTx'+3x+60&

=1

= 24,75

0

Chon dap an C

Cau 39 Cho so phuc Z =a + bi thoa Z + 2 +¡ = |Z| Tỉnh
S = 4a + b

M= 1+ log,, x +log,, y
2log,,(x+3y)

D.M=-

]

3


Z+2+i=|Z)oatbi+2+i=
Va? +b? sa+2—ya’
+b’ +(b+)i=0

&

b+1=0

a+2— Ja’ +b? =0

5

=a+2=Nl+a

a>—2
,

oy

7

a’ +4a+4=a°+1

aS

3

4

=>S=4a+b=-3-1=-—4
Chon dap an D
Cau 40

Cho F(x) = (x—1)e* la mot nguyen ham cua ham so f(x)e™

. Tim nguyen ham cua

ham so f '(x)e*
A. [ £'(x)e* dx =(4—2x)e*+c

B.[f)e”

dx =

C. Jf'@)e”dx=(2—x)e*+c

D. | f'(x)e* dx =(x—2)e*+e

X

e

x

+c

tu et > (x-De* = J f(x)e“*dx > e* + (x—De* = xe* =e“f (x) > f(x) = —
e
dat

u=e* > du=2.e"*dx; v'=f(%)=>v=f4)=-~=
e

f'Œ&)e?* dx =xe*— [2e”*.-—dx

e

= xe*—2 | xe“dx = xe*— 2[xe" — Je*dx] = —xe* + 2e” +c=e”(2—x)+c

Dap an chon C
Cau 41 : Ong A lap cong ty tong so tien cong A phai tra cho cong nhan trong mot nam la | ty
dong . Biet so tien phai tra sau moi nam tang them 15 % so voi nam truoc do . Hoi neu A
thanh lap cong ty tu nam 2016 thi nam dau tien nao A dung de tra luong cho cong nhan trong ca
nam lon hon 2 ty dong
A. Nam 2023
B. Nam 2022
C. Nam 2021
D. Nam 2020
so tien phai tr sau nam thu 1 la N, =1+1.0,15 tiep theo sau nam thu 2 la
N, =N, +N,.0,15 =1+1.0,15 +(1+1.0,15)0,15 =(1 +1.0,15)[1 +0,15]=(1+1.0,15)°
nam thu n so tien phai tra la Ñ, =(1+ 0,15)” cho n lay gia tri tul,2,3... n taco
N,=LISty:N, =1,3225ty:N, =L520875ty: N, =1,74900625ty,N. =2,0113571S8ty
hay 2ty = (1+0,15)" >

so nam n=

log2
log1,15

~ 4.95948 ~5 Suy ra chon dap an C

Cau 42 (112) Cho ham so y= f(x) co bang bien thien nhu
hinh ve. Do thi ham so y = f (x)| co bao nhieu diem cuc tri
A.4

B. 2

hinh ve ben la do thi cua ham

so y = |f(x)| nen so diem cuc

trí cua do thi la 3

nen dap an chon C

C.3

xX

TP

0

+
+

0

PDN
= 0

D. 1
-

3

(

0

400

1

+

A

am

...


Cau 44 : cobao nhieu so phuc Z thoa |z +2 —]| = 2/2 va(z—1)° la so thuan ao

A.0
B.4
C. 3
D.2
Iz +2i]=2V2 â|a+2+(bD)i|=2v2 â(a+2)? +(b1) =8ôa? +4a +b? 2b=3
ơ
=[@D)+bi = (a — ĐẺ + 2(a— Đbi— bŸlasothuanao«> |8 — 9
2(a—])bz=0

=

ty

Ni.

=>

(a—1) =bˆ
(a—“ =b
Neu a-— 1 =b thi
ta co : a’ +4a+(a—l)’ —2(a—1l) =3 8 2a’ +3=3Sa=0;b=-1
Neu a-— 1 =b thi ta co : aˆ + 4a -+(a— Dˆ—2(—a)=3<>2a”

=

a=—1V3 >b=24+3
a=—14+¥3>b=2-V3

vay co ba so phuc thoa dk

+4a—I1=3<©>a” +2a—2=0

nen chon dap C

Cau 45 Tim cac gia tri thuc m de dg thg y = - mx cat do thi ham so y = x° —3x*—m+2
diem phan biet A, B, C sao cho AB = BC

_ A.m
€ (—co;3)

B.m € (—co;—1)

C.m € (—0o; +00)

tai 3

D.mc(l;-+e<)

pt hoanh do giao diem
x° —3x*-—-m+2+mx=06x*
>

—3x? +24+(x-lm=0
6 (x—1)(x*—2x+m-—2) =0

x=l
x—2x+m—2=0(*)

.

¬—

dk can la pt(**) co hai nghiemphan biet khac 1 =
khi do (*) co hai nghiem x,; =1*ƑV3—m

A'=3-m>0
I-2+m—2z0

Ầ©

m<3
m <3

om<3

suy ra do thi ham so co ba diem cuc trí

A(đ—43—m;—m(l—+/3— m));B(1;—m);C(I --2/3—m;—m(+4/3—m))
do

Xạ TẮc
—1.ŸA
Tửc
2
=k

— —m nen AB = BC. Vay đạp an chon laA

Chu y: voi m < 3 goix, =1—V3-

l;x,=l+3-m>=x,==A

S

xx,

lap

thanh cap so cong nen ba giao diem nam tren cung mot dg thg thoa man AB = BC

1—
Cau 46: Xet cac so thục duong x, y thoa log, =
= 2xy+x+y—3.
X
Ty
cua P=x+2y

1
log, —

Tim gia tri nho nhat

Giai tom tat

=
y

2xy+x+y—3< log,d—xy)+ 20

—xy)+1=log,(x+y)+x+y


& log, 2(1— xy) + 2 — xy) =log,(x + y)+(x+y) (*)

}=(*) S log,a+a=log,b+b
1

a = 2(1 — 3xy) >0 vab=(x + y)>0
Dat

Xet ham so f(t) = log,t+t>f'@Q=1+ tIa3 >0Vt>0 =>

2(1—xy)=x+y
pay

(*) xay ra khi va chi khi

=>y(+2x)=2—x=>y=—Š TZ (dox>0)
2x+Ì]

4 2X t4

pray

2x +1

10

(2x+1)

= 5%

2

T4

(2x+2)

v_—1+v10

2_

(nhan)

suy ra ta co bbt cua
P la

& 4x°+4x-9=06

-1-Lvio
_—I-ý10

=—z__

a

.

x |

(oat)

P

0

P

SUY RA DAP AN CHON LA A

mf(P): x + y +z=0. Xet dg the d thay doi thuoc (P)

¬

va di qua B, goi H la hinh chieu cua A tren d biet rang
khi thay doi H luon nam tren mot dg tron co dinh
Tinh R cua dg tron do

B.R=2
D.R=x2

0

\

Cau 47 : Trong he Oxyz chohai diem A(4;6;2), B(2; - 2;0)

A.R=x6
C.R=I

2

-

+

+ 0

Qi
2

\

P

Nhan xet : tu gt suy ra B€ (P)doAH L d>
EH L d,(E la hinh chieu cua
A tren(P)) —>khi d thay doi H luon nam tren đdg tron duong

kinh EB Toa do diem E la nghiem cua he
x=4+t

*

—6+t

z=2+t

=>

X+y+z=0

3t+12=0Sst=—4
=> E(0;2;—2)

EB =4/(0—2)?+(2+2)”+(~2—0)?
=26

>R=6

Chon dap an A.

Cau 48: Cho ham so y = f(x)co do thi cua y = f(x) nhu hinh ve . Dat g(x) = 2f(x) -(x+1)° .
Menh de nao duoi day dung

A. g3)>g(3)>g)

B.g()>g(—3)>g)

C. gƯ3)>g(—3)>g0)

D.g()>g(3)> g(—3)

Tu hình ve ta thay cac diem

A(-3; -2), B(1;2), C(3;4) nam tren dg thg d: y=x + 1

Tu gt ta co g’(x) = 2P (x) —

2(x +1)

S, ladt hinh gioi han:

»

. Goi S, ladthinh soihan:

=f'(x),y=-x
Oy
=>

X=l;x=3

y=Í(%),y=-x
x

=——

3

; x

=

\

va


es = fi f(x) —(x +1) dx +25, = 2 f f(x) —(x +1) dx = ƒs'x)dx =[g(x)]|',
= œ() —8(—=3))>0>g)> a3)

(*)

:

°S,= [lot 1) —f '(x)]dx > 28, = -2[Ifœ —(x+ I)]dx = -2ƒ g{x)dx
= “21803

—ø()]>0> g(3)— gq) < 0>gG)

|

e Taco: g(3)— g(—3) =g@)|Ì, = [ g6) dx = =2 [Tf @đ)@+D]dx
1

3

3

1

=2| f&ô)(X+é dx+ [

f)(x+D) dx|=2S,2S, =2(Đ,S,) >0

= 2[g(3) g(—3)] > 0 = g(3) > g(—3) @**)
Tom laitaco:

g(3)>g(—3)>g()

. Suy ra dap an chon D

Cau 49 : Xet khoi tu dien ABCD co canh AB = x va cac canh con lai deu bang 2V3 .
Tim x de the tich V cua khoi tu dien nay lon nhat
A.x=x6
Taco:

V=

B.x=4A14
2

mm

CxS?

d(A, (BCD))

=
—

VoaBI

= $A(C,(ABD).dt

_223 = ALBLsi n(AIB) = v3 38)
3

2

D.x — 2/3
so

sin(AIB)
= 3V3.sin(AIB)

3

=> Vlon nhat khi sin(AIB) lon nhat © ⁄AIB =90” khi do

AIB la tam giac vuong can nen AB = x = BI N2 = V2.
Chon dap an C

2J343 _ 3\J2

2

Cau 50: Cho mat cau (S) co R = 4, , hinh tru (H) co chieu cao h = 4 va hai dg tron day nam tren
(S) . Goi V la the tich cua (H) va V’ la the tich cua (S) . Tinh ti so vị

` —^

B.

=1

cv

V'
«16
V'
3
Goi r la ban kinh day cua (H) thi

r=A42—2? =2\32=>V=x 2x5

V'

2

3

1ó

4 = 4805 V's ond =

p.~-=2
V'

3

CHON A

DE 103
Cau 35 : mot vat chuyen dong trong 4 gio voi van toc v (km/h) phu thuoc thoi gian t (h) co do
thi van toc nhu hinh ve . Trong thoi gian 3 gio tu khi bat dau chuyen dong , do thi la mot phan
Prabol co dinh I(2; 9) khoang thoi gian con lai do thi la mot dg thg song song voi truc Ox . Tinh
quang dg vat di duoc trong 4 gio


A. s = 26,5(km)

B.s = 28,5(km)

C.s =27,0km)

Tu gt do thi van toc
v la do thi ham so y = -ox? 9x

D. s = 24,0(km)

=> v)=“

trong khoang thoi gian tu 3h den 4h vat chuyen dong deu voi van
27
.
toc v= T suy ra quang dg vat di duoc trong
3

4 gio bang: [(—2x?+9x)dx +”.
»
4
4

l=—8 4 ot
4
4

8
4

=

4

9

e

97

Vay chon dap an C

2

1
Cau 37: Cho F(x)= —a*
X
ham so f*(x)Ìnx

Jf@)

A. | f(x

Inxdx=—==

Inx
+

1

FC

SE

A} eo,

3x

x

Ox

— 10). 1 1,
x

x

xX

f(x y=

x

x

u=Inx

dat;

du —

41>

— —

—Inx

Joa

1

—=|

3x°

J

¢dx

Inx

‘

x?

Ì=

-—3

1

3.

—3x?

+—.

+

3x

poginx = 3

x°

X

4

x’
1

sa

X

;
3
1
ff (x)Inxdx = { ——-Inxdx =—3 J —lInxdx

—3} | —Inxdx|=—3

t

1
—y +C

D. ff Goinxdx =-"* 4

3x

2

Inx
Inxdx =———

|ft(x
Jf@)

B.

`...
Tu gt

4

10) . Tim nguyen ham cua
la mot nguyen ham cua ham so ——
X

sẻ

X

3

x°

nx =

x"

nx

=>

vn
Inx

1

x?

3x°

+C=—-+
2 54+C

Cach khac dat :
dx

u=Inx
vi=f (x)

du = —
v=f{(x%)=

Inx

]
X

3

=> [f'@Inxdx =—>x”

dx

Inx

X

X

]

[[=—>+—5+4+C
3x

Chon dap an C

Cau 4Ï: Mot vat chuyen dong theo qui luat S = _. + 6t* voi t (giay ) la khoang
thoi gian tinh tu khi vat bat dau chuyen dong va s (metf) la quang duong di duoc cua cua vat
trong khoang thoi gian do . Hoi trong khoang thoi gian 6 giay ke tu khi bat dau chuyen dong
van toc lon nhat cua vat dat duoc bang nhieu
A. 24 (m/s)
B. 105 (m/s)

C. 18 (m/s)
D. 64 (m/s)


Ta co v toc cua chuyen dong trong khoang thoi gian 6 giay duoc cho boi pt :

V —-

+12t
=> v' =—3t +12=0<>t=4=
van toc lon nhatcua

t

chuyen dong trong khoang thoi gian 6g1ay la tai thoi diem t = 4giay > V_,. = V(4) = 24(m/s)
Chon dap an A.
Cau 44 : Xet khoi chop S.ABC day la tam gic vuong can tai A, SA vuong goc day khoang
cach tu A den mf(SBC) bang 3 . Goi a la goc giua hai mat phang (SBC) & (ABC),
tinh cosa sao cho V cua khoi chop do nho nhat
A.cosa =4

Beose— X3
3

3

Tu gt thi AI =

AH

SIn œ

V=2SAd,u„e=2—

f(t)

_

t—t

ta co f()=

nho nhonhat
nhat

=> BC=

5

SInœ

&

D. cosa =<

2

2cosa

V nho nhat khi

voI voi mol moi t €

sina

sina

2sin°

acosa

433

_t | 0

(0;1)
(0;l)vat

3

SA=Altana

—-BCAI=——_”“——==—ˆ

cosa

(Ì— cos“ œ)€osœ
=

3

Sin @

=cosœ

tạ
f(t)

—

f(

1

l

3t? —1
1
VB
N Su”
=> —=U©t=-==—— suy ra bbt cua f(t) tu bbt suy ra V nho
(t—)
W3
3

nhat khi t =cosa =

= |Én

=——2

3

=

Coosa — x2
2

Nen ta chon dap an B

Cau 45: Tim cac gia tri m de do thi ham so y = x* —2mx’ co ba diem cuc tri tao

thanh tam giac co dt nho hon 1

A.m>0

B.m<1

C.0
¥4

D.0
y'=4x° —4mx = 4x(x*— m) > dkcanla y'=0 phaicoba nghiem phan biet > m > 0
Khi do do thi ham so co ba diem cuc tri A(-Nm;—n? );O(0;0);B(Vm;—m?)

= dt ron = = ABOH = 22mm

=mVm <1e0

Nen ta chon dap an D
Cau 46 : Cho ham so y = f(x) co do thi cua y = f’(x) nhu hinh ve Dat g(x) = 2f(x) + x’ thi

menh de nao duoi day dung

A. 83)
B.g)
C.g)
Nhan xet: A(-3; 3), BCI; - 1), C(3; -3) cung nam tren dg thg y = - x

Tu gt ta co gˆ(x) = 2F(x) † 2x

D.g(—3)

= ƒø)dx=g(@)|'¿=g)—gC-3)
S„la dt hinh phang gioi han:

Goi |

y=f(x),y=-x

x=-—3;x=1

S,la dt hinh phang gioi han

y=f(x)y=_—x

X=lÌl;x=3

© s,= f (—x)—f (x) dx= 25, =—2 Í f'%)+x dx =—2 [g'%)dx =—2[g(@)l|';
= —2(g(1) — g(—3)) > 0 = g() eS, = [If'«)—(—x)]dx = 2S; =2 [[f'&)—(—x)]dx =2

[ g'(x)dx

= 2[gG) — g()] > 0= g(3) > g() (**).

e Ta co :[g(3)—g(—3)]
= g(x)|?, = f g'@dx ==2 If œ)
+ x]dx
=2)[ f&«)—(—x) dx+[ f(x) —(—x) dx} =—28, +28, =—2(8,-S,)
<0
= g(3)— g(—3) <0 = g(3)
Tom laitaco:

g(1) < g(3) < g(-3) .Suy ra dap an chon B
Z

Cau 48 : Co bao nhieu so phuc Z tho |z + 3i| = x13 va
A. Vo so

B. 2

la so thuan ao

z+2
D. 1

C. 0

e |z-E3i|= V13 <>|a-+(b+3)i|=A13 ©a?+(b+3)?=13a?+6b-+b?=4
7

e

z+2_

Ta

—

a +bi

(a+2)~+bi

a +b+6b=4
a

+b’

+ 2a=0

_=—

a(a+ 2) + b”-+ (2ab + 2b)i

5
5
(a+ 2)° +b

›

=3b—a=2=a=3b—2=
9b“ —-12b+4+b“ˆ

10b* — 6b =0 S b=0>a =~—2(loai) hay b =šÐ

a?”-++-2a + bˆ=0

la so thuan ao <©>

(a+ 2)

›

3

+b

3

z0

+6b=4

a ——s (nhan)

Vay chon dap an D.
Cau 49 Trong he toa do Oxyz Cho hai diem A(3; - 2 ; 6), B(Q; 1; 0) va mat cau (S):

(x—1)* + (y—2)° + (z—3)° =25 . Mat phang (P): ax + by +cz — 2 = 0 di qua A, B va cat (S)
theo giao tuyen la dg tron co ban kinh nho nhat .Tinh T=a+b+c

A.T=3
C.T=2

B. T=5
D. T=4

Tu gt suy ra A nam ngoai , B nam trong mat cau nen AB
cat mat cau tai hai diem co dinh M, N .Goi H la tam dg
tron giao tuyen thi H nam tren trung truc cua MN va H

Zs
[Pe
|

,


la hinh chieu cua I tren (P). Goi K la hinh chieu cua I tren

AB thi HK

AB khi do ban kinh r=

HK? + KN’ =>

r nho nhat khi HK= 0 xay khi HI cung vuong goc AB do mat
phang (P) nhan IK lam vtpt
BA= (3;—3;6) > n= (1;—1;2) la mot vtpt cua mf(Q) qua I

vuong goac AB co pt laco pt: x y + 2z— 5=0
AB co pt x =t; y=1—t; z=2t suy ra toa do K la nghiem
cuahe x=t; y=1—t;z=2t va pt:x —y+2z—5=0=>6t—6=05t=1>H=K(;0;2)

— JH =(0;2;1) la mot vtpt cua (P) nen (P) co pt
la Ox + 2y+z+2=0 >T=a+b+c=3.
Chon dap anA
t

Cau 50: Xet ham so f(t) = 9 am

(m la so thuc ) .Goi S la tap hop tat ca cac gia tri cua m sao

m

cho f(x) + f(y) = 1 voi moi so thuc x, y thoa man e*””
A. 0

B. 1

f(x)
+ f(y) =— 2

x

O*+m*

+

T

C. Vo so

y

x+y

__

4

>m

2 _

=—3°”

¬ax†y

D. 2

—12,9Y2 + (0* +99) m2 0X + (9 +9) m2-+ mê

9'-+mÏ

>
=>0”'”>m

. tim so phan tu cua S

.

=> tatimm
sao cho

m

e*”

ty

axty

x+y>0

=3


xet hamso f(t)=e'—e.t >f'(t)h=e'-e=0St=1=>

y

y)

=>im"

2

=3

ơax+y

e*
t>l=f'đ>0
t<l>f'(t)<0O

=> bbt cua y =f (t) tu bbt ta thay voi moi 0
voi moi 1 0 tuc la
e' >e.t voi moi t>0,tl=>

(j0

e*'”
khi x+y = 1 khi do taco m’ =3' =3 6m=4v3

—

f0

-

0

+

N77

nen chi co hai gia tri cua m thoa man . Vay chon dap an D
DE 104
1
Cau 34; Mot vat chuyen dong theo qui luat S = 3° +6tˆ voi t (giay)la khong thoi gian tu khi
vat chuyen dong va S la (m) la quang dg vat di duoc trong khoang thoi gian do . hoi trong 9
giay van toc lon nhat cua vat dat duoc bang bao nhieu
A. 144(m/s)

S'=_—

B. 36(m/s)

C. 243(m/s)

D. 27(m/s)

+12t=—(— 6Ý +36 <36 vay van toc lon nhat la 36(m/s)

suy ra dap an B
Cau 35: Mot nguoi chay trong thoi gian t gio van toc (km/h) phu thuoc thoi gian t(h)

co do thi la mot phan cua Pra bol dinh I(

8)

)..., nhu hinh ve .Tinh quang dg S nguoi do

chay trong khoang thoi gian 45 phut
A. S= 4, 0 (km)
B. S=2,3(km)

C.S =4,5 (km)

D. 5,3 (km)

Vi do thi cua chuyen dong la mot phan cua do thi ham so y = y = —32x* + 32x
Nen quang duong di trong khoang thoi gian 45 phut la :
0,75

S= | (—32x”+32x)dx=4,5(km)

Dapan C

0

Cau 42: Ham so F(x)=

A. ƒf@0dx=—|

x

—, la mot nguyan ham cua 10) Tim
X

X

+

Ị |+‹

B.ƒf@)dx=X +

2x7

x”

c feax =-[BE +4] +
xổ

f '(x)dx = _nx
Œ)
x?

+

x?

D. Ġ)dx= "+

x?

+c
| +e

x”

2x7

du — 2%

Tutti | XLp f X
J

J f '(x)dx

—
u=Inx
oy
he
=—Laat|
=x
px
X
X
vi=f'x

X

x°

v=f(x)=-

— —| RE—_—+
x?

! |+‹ Dap an chon A.

2x?

Cau 46 : Xet cac so nguyen duong a, b sao cho pt aln* x + bInx +5 co hai nghiem phan biet
X,,X, va pt 5log’ x + blogx +a co hai nghiem phan biet x,,x, thoamanx,.x, >x,.x, . Tim
gia tri nho nhat cua S = 2a + 3b
A. 30
B.25
C. 33
D. 17

aln* x +blInx +5=0 va: 5log’ x + blogx +a =0 cohainghiemphan
biet = b* — 20a > 0
.

Khi do t, =Inx,

tee

> x, =e';t, =Inx,

ot

> x, =e*

__

> x,.x, =e

t', =logx, > x, =e"';t', =logx, > x, =e”? >x,.x, =e"
=b

>e'?

=b

>10°

b? — 20a > 0

=b

=Ine?

=b

>Inl0°

b

—b

_aa

=e

=e>

—b

5

b

+—=>—lnl0=>5<aln0=>a>
a

b> 2V5a Tom lai ta phai co dk: |

a> 2.2

b> 25a

n

T02 2200 đk
na

do anguyen duong nen a

nho nhat bang 3 khi do b> 2/5a => b> 215 ~7,75 vay b nho nhat bang 8 suy ra
S =2a+ 3b =2.3 + 3.8 = 30. Vay S nho nhat bang 30.
NenchondapanA
Cau 47: Trong he Oxyzcho ba diem A(-2;0;0), B(O;- 2;0), C(O; 0; - ;-2) . Goi D la diem khac O

sao cho DA, DB, DC doi mot vuong goc voi nhau va f(a, b, c) la tam mat cau ngoai tiep tu dien
ABCD. Tinh S=a+b+c
A.S=-4
B.S=- 1
C.S=-2
D.S=-3


Tu gt suy ra A, B, C lan luot nam tren cac truc Ox, Oy, Oz va OA = OB = OC = 2

suy ra ABC

la tam giac deu canh co do dai 2/2 suy ra DA= DB =DC nen O, D cung nam tren dg thg A
qua O vuong goc voi mf(ABC) dg thg do co mot vtcp bang
x=t

[AB, AC] = (4;4;4) => u= (1;1;1)la
mot vtepcuaA => Acopt:;y=t>
z=t

D(;t;Ð > DA = (t+ 2:t;);DB = (t;t+ 2;);doDA | DB = t(t+ 2) +(t+2)t+
=>

t=——
t=0

—0

4
3

=> D(-$3-5:-5).Goi

Ilatam mat cau ngoai tiep > Ice A>

doaiviD +O)

(G60 (ot; = 0,0

1
1
1
1
2
2
4
= 3(t+ 2)4, =(t+ 2) 4+ 2t st=— a l(-33-35-9)
—2),doID=IA

suyraat+b+c=—1.

Vay chon
dạp an B

Cau 48: Cho ham so y = f(x) co do thi cua y = Ÿ(x) nhu hinh ve. Dat g(x) = 2f(x) + (x+
thi menh de nao sau day dung

A. 3)
B.gq)

C. g()<g(—3)>g()

D.g) = g(—3) >g)

Tu gt >gx)=2f {x) +2(«k+1) Goi S, S’ lan luot la dt hinh phang gioi han boi

y=f(x4;y=-x-l
x=—3;x=1

va

|y=f(x);y=-x-—l

x=3;x=1

eS= [rot 1) —f '(x)]dx > 2S = -2ƒ[I&+D +f '(x)]dx = _[Ƒgtx)dx = —gŒ)|';

= “(at —gø(—3)) >0= g()— sí-3) <0>g0) eS'= fee (x)—(—x—1)]dx > 28' =2/F (x) +x+D]dx = [etnydx — 9(3)—g(1) >0
> 203) > g(1) (**)

|

|

e (3) —g(—3) = [ g%)dx =2 [[f'œ) +(x+1)]dx =2| [[f'œ) +(xX+1)]dx
—3

+ 2 Jư (x) + (x+ DJdx| = —2S + 2S' = —2(S—S') <0 => g(3) — g(—3) <0

= g(3) < g(—3) Œ**#*). Tu (5), CF"), (**) suy ra g(1) < g(3) < g(-3)
Chon dapana .

Cau 49: Trong tat ca hinh chop tu giac deu noi tiep mat cau co ban kinh bang 9 .Tinh the tich V
cua khoi chop co the tich lon nhat
A. V= 144

B. V= 576

C.V =576V2

Goi H la hinh chieu cua tam [ mat cau (S) datTH =x

D. V = 1446
O
°

=> AHE=V8I—x” =dt,we› =5-AC.BD=-2AH.2AH
— 2 AH’ = 2(81— x”) vaSH=9+x
]
2
= V =~ ] dt.) SH = —.2(81
— x”).(x +9) = £(x
+ 9)(81 — x”)

3

2

3

›

3

l Cà

> Wa Bll x )+ (K+ 9)(—2x)]
= 2-3-1814

811=0]

X

x=3

9

0

g 7 bt cua ham so V(x)

x=

8

4

3

7

¬

Vay V lon nhat bang 576 . Chon dap an B
Cau 50: Goi S la tap hop cac gia tri thuc cua m de ton tai duy nhat so phuc Z thoa man
Z.Z=l va Z—v3+Ì =m .Tim so phan tu cua S
A.2

B.4

C.1

D.3

(a-bi)(a— bi)—1<>a?+b =I; la + bi —^A/3-Lil—m>0<>(a43)? +(-LD
& a2 +b? —2aJ3 +2b+4=—m? = taco

—24-3+2b+5—m” ¬p—
esate mast)

2

a+b? =1
a +b? —2aV3 +2b+4=m’

d+ ANB _— 2

=>

=]

2

=1

4a? +12a? + 4aA3(m°—5)-+(m?—5)” =4

16a? + 4./3(m?—5)a+(m?—5)* —4 =0 (*)=> yebtthoamankhi

nhat nghiem a

=m’

A'= [2/3 (m?— 5)

—=-12(mˆ— 5)“ — 16(m“—
5)” +60

2

(*)co

duy

—16[(m”— 5)”—4]=0
©4(m—5)“

=—61 ©(mˆ—5)” =l6

© m —5=4;hay m—5—=—4—>>m
=3 hay m=I (do m>0) Vay chon dap an A.