Tuyen Quang Giai chi tiet de toan quoc gia 2017

Cau 37 115 Mot chuyen dong trong 3 gio voi van toc v km / h phu 7,75 thuoc thoi gian th co do thi van toc nhu hinh ve .Trong khoang thoi gian 1 gio ke tu khi bat dau chuyen dong , do thi[r]

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

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 4

dong trong 3 gio:

A.s = 13,83(km) B.s = 23,25(km)

C.s = 21,59(km) D.s =15,50(km)

Trong mot gio dau vat chuyen dong co do thi van toc la parbol oy 1 2 3

7,1

co ptla y= 3x +5x +4 tai x= Ì ta co v(l) = 7,75 = _ Suy ra quang duong ma vat

di duoc trong 3 gio la s = fart 5x+ 4)dx+ [ex = 6,08(3) + 15.5 = 21,58

hay s= JC7x*+ 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 = * — thoa man miny =3 Menh de nao duoi day dung

m< —Ithiy'>0=> Miny <y(4) =n a3 6 m= (khong thoa m < —1)

(x—]) m>—Ithiy’<0=> Miny =y@) =~ =3 + m=5 (thoa dk m > —1) 4+m

[2;4]

suy ra chon dap an A

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:

2x

dap an A

Cach 2: dat

> 2x => J†@œ)e” dx=e”.——— — 2e°*dx =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 By 228 18 cy — Uva 216 v_132a 216

E

M

A

Giai tom tat

V1iDIatrung diemcua BE=> d(E, (ABC)) = 2d(D, (ABC))

= 2d(A,(BCD))TA CUNG CO d(E,(ACPQ)) = d(B, (ACPQ)) = d(B,(ACD)) = d(A, (BCD))

duuc _ BA BC _ 2BM 2BN _„

=> 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

Tát

DQP 9

8 8 ] ] 3 3

= dt cpg = 9 taco — 9 tse 2a, (ABC)).dt oun = 3 Tt, (BCD)).7 dtaco — 2 VABCD

DAC

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

V acMNPO — YEACMN Vie ACPO — 2 YABCD ~ 9 YABCD

= ¬ = 11 13a + là — 3a" = LIV2a chon dap an C

V AcMNPO — VÀ MNPQ + cn — Va CNP + VN Apo + VN AMQ

1

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

2a N3 a2J3

a

dt ncnp = dtp amo 9°93 7 2 12 5

A43 jlaa V3, 1, 2a v3,

4 2332 2 3 2

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

_ a3 s 7a?AJ3 - a7v3

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 aJ6 a2J3 aJV6— a2 V3 aV6 1 Ila? VIS 1 a?/2

= Vicyneo = ae 3 12 3 SE 18 6 EE 12 EN = 6 3 216 = 216

Vay Chon dap an C

la so thuan sao

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

giasuZ=a+bi

=|Z-3i|=5© mm = nh

và Z — atbi _(a+bi( =4—bi)_ a -4a+b „ —4bi

la so thuan ao <> ( ) —> 4a—6b—16=>b— 2â 8

a’ —4a +b’ =0(*)|a* —4a +b? =0

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

y=f'(x)y=x

Xx=2;x=4

y=f'(x)y=x

X=—2;x = Goi S,ladien tich hinh gioihan : &Š„ la dien tịch hình g1ol han :

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

x=t

tu gt thi OH co pt; y=t> 3t—4=05St=

zZ=t

= Us —[MH,n,]= (3; -3;0) vtcp cua d la (1; - 1; 0) = (1; a; b) =>a—b=—1—0=~-1

Suy ra dap an chon la B

P

——

= H(2;4;5) 3 3 3 = MA = ( 1 C2 1

s— —=-(];l;—2

3 ) 3 |

wu]

Cau 50 (115) Xet cac so thục duong thoa log, =3xy +x+2y—4 Tim gia tri nho nhat

X + 4y cuaP=x+y

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 => (*) xay ra khi va chi khi

'

3(l-xy)=x+2y = y(243x)=3-x y= (do x >Q)

3x +2

TT chan)

Suy ra ta co bbt cua P la x 0 x +0

SUY RA DAP AN CHON LA C pl ' „2+

DE 102

Cau 35 Cho ham so y = T1 thoa man miny + max y= Menh de nao duoi day dung

<>m=5 khong thoa dk m<1

2 1 5 7 16

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

Dap an chon la B

1+ log,, x +log,, y 2log,,(x+3y) ]

+ 2 D.M=- 3

Cau 37 Cho x.y la cac so thục duong thoa x° + 9y* =6xy Tinh M=

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

M= 1+log,, x +log,, y _ log,, 12xy _ log,, 12xy —Ị

2log,,(x+3y) log,,(x+3y) log,,12xy

Chon dap an B

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

Chon dap an C

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

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

=>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*

e +c

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

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)=-~= f'Œ&)e?* dx =xe*— [2e”*.-—dx

= 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

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

log2 log1,15

hay 2ty = (1+0,15)" > so nam n= ~ 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

hinh ve ben la do thi cua ham xX 0 + 3 400 1

so y = |f(x)| nen so diem cuc TP + 0 - 0 ( +

= 0

nen dap an chon C

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

Iz +2—i]=2V2 ©|a+2+(b—D)i|=2v2 ©(a+2)? +(b—1)” =8«a? +4a +b? —2b=3

¬ Œ— ĐỸ =[@—D)+biƑ = (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” +4a—I1=3<©>a” +2a—2=0

a=—14+¥3>b=2- V3

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

_ A.m € (—co;3) B.m € (—co;—1) C.m € (—0o; +00) 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(*)

om<3

m <3

dk can la pt(**) co hai nghiemphan biet khac 1 = Ầ©

I-2+m—2z0 khi do (*) co hai nghiem x,; =1*ƑV3—m 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))

Xạ TẮc —1.ŸA Tửc —

do 2 =k —m nen AB = BC Vay đạp an chon la A

Chu y : voi m < 3 goi x, =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 Tim gia tri nho nhat

X Ty cua P=x+2y

Giai tom tat

1

log, — = 2xy+x+y—3< log,d—xy)+ 20 —xy)+1=log,(x+y)+x+y

y

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

Dat a = 2(1 — 3xy) >0 vab=(x + y)>0 }=(*) S log,a+a=log,b+b

1 Xet ham so f(t) = log,t+t>f'@Q=1+ tIa3 >0Vt>0 => (*) xay ra khi va chi khi

2(1—xy)=x+y =>y(+2x)=2—x=>y=—Š TZ (dox>0)

2x+Ì]

2

2x +1 (2x+1) (2x+2)

v_—1+v10 (nhan) suy ra ta co bbt cua P la

_—I-ý10 a x | 0 2 + 0

P \ Qi

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

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

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 3t+12=0Sst=—4 => E(0;2;—2)

X+y+z=0

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 gÖ3)>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

y=Í(%),y=-x

Tu gt ta co g’(x) = 2P (x) — 2(x +1) Goi S, ladthinh soihan: 3 \ va

x =—— ; x =

=f'(x),y=-x

S, ladt hinh gioi han: » Oy =>

X=l;x=3

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)<g(D C°) |

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

=2|ƒ fŒ&«)—(X+Ð dx+ [ fŒ)—(x+D) dx|=2S,—2S, =2(§,—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

Taco: V= 2 mm d(A, (BCD)) = — VoaBI = $A(C,(ABD).dt so

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

2J343 _ 3\J2 2

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

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

Goi r la ban kinh day cua (H) thi

2

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