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 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 duoc lanh ca goc va lai la TT — G( +t)” Goi n la so nam gui vao duoc lanh lon hon 100 trieu CA GOC VÀ LAI Xay va chi 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 gio voi van toc v (km/h) phu thuoc thoi gian t(h) co thi van toc nhu hinh ve Trong khoang thoi gian | gio ke tu bat dau chuyen dong , thi la mot phan cua parabon co dinh I(2; 9) khoang thoi gian lai thi chuyen dong la mot doan thg song song voi truc Ox Tinh quang dg cua chuyen dong gio: A.s = 13,83(km) B.s = 23,25(km) 7,1 C.s = 21,59(km) D.s =15,50(km) Trong mot gio dau vat chuyen dong co thi van toc la parbol co ptla y= 3x +5x +4 tai x= Ì ta co v(l) = 7,75 = _ di duoc gio la s = fart r, s= JC7x*+ Suy quang duong ma vat 5x+ 4)dx+ [ex hay oy = 6,08(3) + 15.5 = 21,58 31 5x+4)dx + 47 ~ 21,58 tuc hai gio sau chuyen dong deu voi van 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 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 mp(MNE) chia tu dien 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 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 | Ve ace = 5| 4(E(ACPQ)).dt sony = GE (ACD)) dt rg 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 _„ 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 tam cac tam giac dt BCE va ABE > DQ= =DA:DP = =bC — ADQP ~ ADAC > i _l => dt DAC oun = Tt, (BCD)).7 dtaco — VABCD 2a, (ABC)).dt = dt cpg = taco — tse sue, (ACD)).dt sono = sda, (CD) dt,.„ = Vane V acMNPO — YEACMN Vie ACPO — = 11 13a 183 CACH A V AcMNPO — VÀ MNPQ + cn SUY RA: YABCD ~ YABCD = ¬ 18 DAC 3 ] ] 8 Tát DQP + — 3a" = LIV2a 216 chon dap an C — Va CNP + VN Apo + VN AMQ =3 [dt ,cyp-d(A, (CNP)) + dt.ap ON, (APQ)) + dt ayo ACN, (AMQ))| dt ncnp = dtp amo a 9°93 2a N3 a2J3 12 A43 jlaa V3, 1, 2a v3, dt apo — dt scp - (dtApop + dt AACP) — 2332 _ a3 s 7a?AJ3 - a7v3 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 12 Vay Chon dap an C aJ6 SE a2J3 aJV6— 18 EE a2 V3 aV6 12 EN = Ila? VIS 216 = a?/2 216 Cau 47 (115): co? so phuc Z thoa |Z — 3i] =5va = la so thuan giasuZ=a+bi =|Z-3i|=5© 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 13a? —4bi (a-4) +bí —> 4a—6b—16=>b— a’ —4a +b’ =0(*)|a* —4a +b? =0 da? —32a „ 68a+64=086] 2â — a=4=> b= 0(oai) a= 16 g > b= (nhan) Vay co 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,) > => h(4) —h(-2) = h(4) > h(—2) ***) Tu (*), C**) va (***) suy 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 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) 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 cho AB nho nhat Biet d co mot vtcp u= (1;a;b) Tinh T=a- b Nhan xet : OM = V6 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 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—40 vab=(x + 2y) >0 > (*) S log,a+a=log,b+b 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 va chi (do x >Q) _ = s TT chan) 2-Jl1l x =———~— (loai) Suy ta co bbt cua P la x P pl SUY RA DAP AN CHON LA C x +2 `2 +0 + + ' „2+ DE 102 Cau 35 Cho ham so y = — A.mm—2 , oy a’ +4a+4=a°+1 aS =>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 mot nam la | ty dong Biet so tien phai tra sau moi nam tang them 15 % so voi nam truoc Hoi neu A lap cong ty tu nam 2016 thi nam dau tien nao A dung de tra luong cho cong nhan ca nam lon hon ty dong A Nam 2023 B Nam 2022 C Nam 2021 D Nam 2020 so tien phai tr sau nam thu la N, =1+1.0,15 tiep theo sau nam thu 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 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 hinh ve ben la thi cua ham so y = |f(x)| nen so diem cuc trí cua thi la nen dap an chon C C.3 xX TP + + PDN = D - ( 400 + 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 D.2 Iz +2—i]=2V2 ©|a+2+(b—D)i|=2v2 â(a+2)? +(b1) =8ôa? +4a +b? 2b=3 =[@D)+bi = (a — ĐẺ + 2(a— Đbi— bŸlasothuanao«> |8 — 2(a—])bz=0 = ty Ni => (a—1) =bˆ (a—“ =b Neu a-— =b thi ta co : a’ +4a+(a—l)’ —2(a—1l) =3 2a’ +3=3Sa=0;b=-1 Neu a-— =b thi ta co : aˆ + 4a -+(a— Dˆ—2(—a)=32a” = a=—1V3 >b=24+3 a=14+Ơ3>b=2-V3 vay co ba so phuc thoa dk +4aI1=3a +2a—2=0 nen chon dap C Cau 45 Tim cac gia tri thuc m de dg thg y = - mx cat thi ham so y = x° —3x*—m+2 diem phan biet A, B, C cho AB = BC _ A.m € (—co;3) B.m € (—co;—1) C.m € (—0o; +00) tai D.mc(l;-+e —3x? +24+(x-lm=0 (x—1)(x*—2x+m-—2) =0 x=l x—2x+m—2=0(*) ¬— dk can la pt(**) co hai nghiemphan biet khac = (*) co hai nghiem x,; =1*ƑV3—m A'=3-m>0 I-2+m—2z0 Ầ© m0 Dat Xet ham so f(t) = log,t+t>f'@Q=1+ tIa3 >0Vt>0 => 2(1—xy)=x+y pay (*) xay va chi =>y(+2x)=2—x=>y=—Š TZ (dox>0) 2x+Ì] 2X t4 pray 2x +1 10 (2x+1) = 5% T4 (2x+2) v_—1+v10 2_ (nhan) suy ta co bbt cua P la & 4x°+4x-9=06 -1-Lvio _—I-ý10 =—z a x | (oat) P 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 thay doi H luon nam tren mot dg tron co dinh Tinh R cua dg tron B.R=2 D.R=x2 \ Cau 47 : Trong he Oxyz chohai diem A(4;6;2), B(2; - 2;0) A.R=x6 C.R=I - + + Qi \ P Nhan xet : tu gt suy 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 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 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 + 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 =—— ; x = \ va es = fi f(x) —(x +1) dx +25, = 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)0 = 2[g(3) — g(—3)] > = g(3) > g(—3) @**) Tom laitaco: g(3)>g(—3)>g() Suy dap an chon D Cau 49 : Xet khoi tu dien ABCD co canh AB = x va cac canh lai deu bang 2V3 Tim x de the tich V cua khoi tu dien lon nhat A.x=x6 Taco: V= B.x=4A14 mm CxS? d(A, (BCD)) = — VoaBI = $A(C,(ABD).dt _223 = ALBLsi n(AIB) = v3 38) D.x — 2/3 so sin(AIB) = 3V3.sin(AIB) => Vlon nhat sin(AIB) lon nhat © ⁄AIB =90” AIB la tam giac vuong can nen AB = x = BI N2 = V2 Chon dap an C 2J343 _ 3\J2 Cau 50: Cho mat cau (S) co R = 4, , hinh tru (H) co chieu cao h = 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' Goi r la ban kinh day cua (H) thi r=A42—2? =2\32=>V=x 2x5 V' 1ó = 4805 V's ond = p.~-=2 V' CHON A DE 103 Cau 35 : mot vat chuyen dong gio voi van toc v (km/h) phu thuoc thoi gian t (h) co thi van toc nhu hinh ve Trong thoi gian gio tu bat dau chuyen dong , thi la mot phan Prabol co dinh I(2; 9) khoang thoi gian lai thi la mot dg thg song song voi truc Ox Tinh quang dg vat di duoc gio A s = 26,5(km) B.s = 28,5(km) C.s =27,0km) Tu gt thi van toc v la thi ham so y = -ox? 9x D s = 24,0(km) => v)=“ khoang thoi gian tu 3h den 4h vat chuyen dong deu voi van 27 toc v= T suy quang dg vat di duoc gio bang: [(—2x?+9x)dx +” » 4 l=—8 ot 4 = e 97 Vay chon dap an C Cau 37: Cho F(x)= —a* X ham so f*(x)Ìnx Jf@) A | f(x Inxdx=—== Inx + FC SE A} eo, 3x x Ox — 10) 1, x x xX f(x y= x x u=Inx dat; du — 41> — — —Inx Joa —=| 3x° J ¢dx Inx ‘ x? Ì= -—3 —3x? +— + 3x poginx = x° X x’ sa X ; ff (x)Inxdx = { ——-Inxdx =—3 J —lInxdx —3} | —Inxdx|=—3 t —y +C D ff Goinxdx =-"* 3x Inx Inxdx =——— |ft(x Jf@) B ` Tu gt 10) Tim nguyen ham cua la mot nguyen ham cua ham so —— X sẻ X x° nx = x" nx => Inx x? 3x° +C=—-+ 54+C Cach khac dat : dx u=Inx vi=f (x) du = — v=f{(x%)= Inx ] X => [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 vat bat dau chuyen dong va s (metf) la quang duong di duoc cua cua vat khoang thoi gian Hoi khoang thoi gian giay ke tu 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 khoang thoi gian giay duoc cho boi pt : V —- +12t => v' =—3t +12=0t=4= van toc lon nhatcua t chuyen dong 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 Goi a la goc giua hai mat phang (SBC) & (ABC), tinh cosa cho V cua khoi chop nho nhat A.cosa =4 Beose— X3 3 Tu gt thi AI = AH SIn œ V=2SAd,u„e=2— f(t) _ t—t ta co f()= nho nhonhat nhat => BC= SInœ & D cosa =< 2cosa V nho nhat voI voi mol moi t € sina sina 2sin° acosa 433 _t | (0;1) (0;l)vat SA=Altana —-BCAI=——_”“——==—ˆ cosa (Ì— cos“ œ)€osœ = Sin @ =cosœ tạ f(t) — f( l 3t? —1 VB N Su” => —=U©t=-==—— suy bbt cua f(t) tu bbt suy V nho (t—) W3 nhat t =cosa = = |Én =——2 = Coosa — x2 Nen ta chon dap an B Cau 45: Tim cac gia tri m de thi ham so y = x* —2mx’ co ba diem cuc tri tao tam giac co dt nho hon A.m>0 B.m|a-+(b+3)i|=A13 ©a?+(b+3)?=13a?+6b-+b?=4 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 (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) › +b z0 +6b=4 a ——s (nhan) Vay chon dap an D Cau 49 Trong he toa Oxyz Cho hai diem A(3; - ; 6), B(Q; 1; 0) va mat cau (S): (x—1)* + (y—2)° + (z—3)° =25 Mat phang (P): ax + by +cz — = 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 A nam ngoai , B nam 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 ban kinh r= HK? + KN’ => r nho nhat HK= xay HI cung vuong goc AB 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 toa 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) = am (m la so thuc ) Goi S la tap hop tat ca cac gia tri cua m m cho f(x) + f(y) = voi moi so thuc x, y thoa man e*”” m _ =—3°” ¬ax†y D —12,9Y2 + (0* +99) m2 0X + (9 +9) m2-+ mê 9'-+mÏ > =>0”'”>m tim so phan tu cua S => tatimm cho m e*” ty axty x+y>0 =3 0 tf'(t) bbt cua y =f (t) tu bbt ta thay voi moi e.t voi moi t>0,tl=> (j0 e*'” x, =e* > x,.x, =e t', =logx, > x, =e"';t', =logx, > x, =e”? >x,.x, =e" =b >e'? =b >10° b? — 20a > =b =Ine? =b >Inl0° b —b _aa =e =e> —b b +—=>—lnl0=>5a> a b> 2V5a Tom lai ta phai co dk: | a> 2.2 b> 25a n T02 2200 đk na anguyen duong nen a nho nhat bang b> 2/5a => b> 215 ~7,75 vay b nho nhat bang suy 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 cho DA, DB, DC doi mot vuong goc voi va f(a, b, c) la tam mat cau ngoai tiep tu dien ABCD Tinh S=a+b+c A.S=-4 B.S=- C.S=-2 D.S=-3 Tu gt suy A, B, C lan luot nam tren cac truc Ox, Oy, Oz va OA = OB = OC = suy ABC la tam giac deu canh co dai 2/2 suy DA= DB =DC nen O, D cung nam tren dg thg A qua O vuong goc voi mf(ABC) dg thg 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 => D(-$3-5:-5).Goi Ilatam mat cau ngoai tiep > Ice A> doaiviD +O) (G60 (ot; = 0,0 1 1 2 = 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 thi cua y = Ÿ(x) nhu hinh ve Dat g(x) = 2f(x) + (x+ thi menh de nao sau day dung A 3) 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 + Jư (x) + (x+ DJdx| = —2S + 2S' = —2(S—S') g(3) — g(—3)