HDC MTCT cap tinh nam hoc 2012 2013 mon Toan THPT

dinh : Thi sinh trinh bay tom titt each giai, cong tlnrc ap dungjhoac quy trinh bam phim neu bai toan yeu cau va diSn ket qua vao 0 liSn kS.. Cac kSt qua tinh gftn dung, neu khong co chi[r]

so GIAO Dl)C vA DAo T ';'0 cuoc THI cAp TiNH GIAI ToAN TREN MAy TiNH CASIO, VINACAL NAM HQC 2012-2013 LAo CAl rHPn Mon fbi: Toan (Chuung trinh HUONG nAN CHAM DE THI cHiNH THUC Thai gian thi: 150 phut (kh6ng Ngay thi: 17/01/2013 Jed thoi gian giao thi) Huang ddn chdm g6m co 10 cdu, 07trang, m6i cdu 05 iliim (Chu y: Thi sinh lam bai thi vaa ban ild thi nay) DIEM B~ng eVA BAI s6 so PHAcH Giam khao (Ky va ghi ro ho ten) THI (Do Truong ban cham thi ghi) B~ng chii' * Quy dinh : Thi sinh trinh bay tom titt each giai, cong tlnrc ap dungjhoac quy trinh bam phim neu bai toan yeu cau) va diSn ket qua vao liSn kS Cac kSt qua tinh gftn dung, neu khong co chi dinh cu thS thi duoc lay chinh xac toi chtr s&thap phan Bai Cho ham s& f(x) =x - 3~ + Tim g~n dung tea dQ cac diem x+ C1,IC KETQUA CACHGIAI tri cua ham s& DIEM f'(x)=x +4x-7 (x+ 2)2 Taco: f'(x)=O<=>XJ ~1,3166;x2 (x;t:-2) ~-5,3166 1.0 1.0 Kit qua: A(l, 3166; -0,3668) B( -5,3 J 66; -13, 6332) Bili 2: Cho da thirc P(x) b~c 4, h~ so cua P(x)chia ~+l, x+2, x-I, x-2 du -2,4, (3.5 diemj'Iim da tlnrc P(x) CACHGIAI , ., .~ X4 1.5 1.5 bang -11, KETQUA DIEM , P(I)=-11,P(2)=6, Tu gia thiet co ( P -1)=-2,P(-2)=4 Xet da thtrc b~c b6n Q( x) thoa Q(I) = Q(2) Gia = Q(3) = Q( 4) = O su P(x) = Q(x) ta co h~ +a.x3 +bx2 + ex + d, 0.5 -ll=a+b+c+d = 8a + 4b + 2c + d 0.5 t -2=-a+b-c+d = -8a + 4b - 2c + d 2.0 ~276' a ~ ?_;c·~ ~ 37 d=_3}j' Kdt (lud: P ( x) = (x 23 +-x - 37 x-6 1)(x - 4) + % x 0.5 31 (1.5 diemj'I'inh : P(20) - P(1 0) ~ KETQUA CACHGIAI P(20) = 172737 P(10) = 11482 P(20) - P(10) = 161255 Bili 3: Giiii h~ phuong trinh : { :3 + ( - xy = x +y +x+y=5 KETQuA {(x+ y)[ (x+ y)' -3xy ] ~ 6+ xy (x + y)2 - 0.5 co h~ f'-3PS~6+P S2 +S-2P 0.5 =5 S3 + 4S2 - 14S + = • Giai phuong trinh t2 (SJ; Pi) tren ta DIEM 2xy + x + y::; f)~t S = x + y; p ::;xy ta ¢::> 0.5 0.5 0.5 (I) CACHGIAI (I)~ DIEM dUQ'C - 0.5 Sf + P = 'u'ng voi cac e~p SI ~ ~6,3702;~ ~ 14,6046 cac nghiem cua h~ S2 ~ 1,7379;Pz ~ -0,1209 1,5 S3 ~ 0, 6323; ~ ~ -1,9839 {x ~ 1,8049 {x ~ -0,0670 y ~ -0,0670' Y ~ 1,8049 ' 2.0 {X~-1'1274 {X~I,7597 y~I,7597 Ul BAi 4: Cho day s6: = 1;» = -2 _ { U + n ' y~-1,1274 - 2un+1 - 3un ViSt quy trinh bfim phim tinh V6i n>O;nEN Un; Sn (T6ng cua n s6 hang d~u cua day un) CACHGIAI DIEM Gan: -> A; -2 -> B ; -1 -> C (tong hai s5 hang dAu)2 - >D (bien dem) = 2B A: D = D + 1:A C = C + A: 1.0 D = D + 1: B = 2A - - B : C = C + B 2.0 ? KETQUA CACH GIAI Bai 5: , Tim :::: Sl5 = -12664,2542 , , I(x) f'(X):::: 5, x 5, + ~4- 3x2 KETQUA tren doan [- DIEM 0.5 X'et h'am so; T3 = 3x-2+ }4-3x' <=> X 1.0 1.0 = 3x - gia tri 1611nhat, gia tri nho nhat cua ham so : f(x) CACHGIAI BK : - Jj DIEM -5694,4751 Ul5 Jj;~] =1 Thuc hien tren may tinh f'(x)=O<=>x=l f(1) 0.5 1.0 = 2; 1.0 /(Jj) ~1,4641; f(- 1.0 v13) ~ -5,4641 Max/(x) = f(1):::: 1.0 [-}~J Min/(x) [_~;~] Bai : Cho tam giac MBC vuong tai A, AB = 13; ACE = 30° Cac duong phan giac BM, CN cua tam giac MBC e~t tai = /C-1):::: -5,4641 II (3di~m)Tinh dai BM, CN CACHGIAI BC= AB sin C = 13 KETQUA =213 sin 30° AC::;::3 BM =' BC AB+BC ~AB.BC.p(p , = 2.f3 AC=3 - AC) BM=2 CN= DIEM CN ~ 3,1058 ~AC.BC.p(p-AB) 0.5 0.5 1.0 1.0 AC+BC (2 diemj'Ifnh dien tich tam giac MMN CACHGIAI KETQuA DIEM Do duong phan giac BM, CN cat tai I nen l la tam duong ngoai tiep 2S r=IM=IN= r~ 0.6340 = 180°-15° -300 BIC=NIM V?y St;jMN 0.5 AB+BC+AC - - 0.5 = 135° ~ LO.63402.sin 1350 =~IM.JN.sin(MIN) ~ 0,1421 0.5 0.5 Bili 7: 1.(2 di~m) Tinh gia tri eua bi~u thirc = ~ I + l)1 +.l + ]_ 1+.l + !_+ i A 2 3 ''-1+-_-1-+-.!_-+- -+-= -1 ~ 20 s: CACHGIAI Gan A ? KETQUA = 0, B ::;::0, C = Khaibao:A=A+ A K~t qua: 17667,9757 A • I Tinh tong : S=l C12013+22 C22013 +3 r 2013 C2013 -+ +2013 ( 2013 ) 2013()n-I1 C2013 -2013 CACHGIAI Ta co: (l+xY ¢::? ¢::? n(l +xr-I n.x(l+xy-l KETQUA +nC:x"-1 =xC~ +2C;X2 + +nC;xn 2c;xn-1 n(1 + xy-I + n(n -I)x.(l + xy-2 ",,}2C! + 22 C;x + + n ~ hex) ~ S=f = n(1 +xy-I (_1_) 2013 2013 +n(n-l)x.(l DIEM 0.5 0.5 =C~ +C~X+C;X2 + +C;xn = c~+2C;x+ 1.0 1.0 _!_ :C,=C.JB :B=B+ , ¢::? DIEM +xy-2 =2013(1 +_1_)2012 +2013.2012.-1-(1 2013 2013 0.5 0.5 +_1_)2011 2013 0.5 I Sd0930,2233 I Bai 8: Cho hinh ch6p SABCD co day ABCD la hinh thoi canh a = {j20 13; 0.5 BAD = 60° ; SO L (ABeD); SO = %a ; mat phang (a) chua AD va (a) vuong goc voi (SBC) s c 1.(1.5 di6m) Tinh VSABCD CACHGIAI KETQUA DIEM a J3 0.5 SAEeD =-.- VSABCD = -SO,SABCD 0.5 0.5 , (3.5 diem) Thill dien tich thiet dien tao boi (a) va hinh ch6p SABeD KETQuA CACHGIAI DIEM + dung d qua vuong goc voi BC cat BC, AD tai I;J + Til J ke JH lS1; tir H ke EF IIBC c~t SB, SC tai E; F Thi~t dien la tu giac AEFD + Xet MOl: 01 = BO.sin60o + IJ = 201 + SI = + 01 = aJ3 IJ = aJ3 = -.SO.lJ + HI = ~IJ2 ; ' 0.5 S/l.)'1J _HJ2 = -.JH.SI Suy JH = SO HI = aJ3 = suy EF a 3a + SAF.FD = 2·(a+2)·4 = -3a l S1 = 0.5 0.5 SI = aJ3 ~S02 +01 S/l.)lJ 1.0 -.BC= a - 2 0.5 = 9a = 9(\12013)2 ;;:;]1.7940 16 16 0.5 Bai 9: B6 ban A tang ban mot may tinh tri gia 5.000.000 d6ng bang each cho ban tien hang thang thea phirong thirc nhu sau: thang dftu tien ban nhan duoc 100.000 d6ng, cac thang ill thang thir hai tro di m6i thang nhan duoc 8<3tien han thang tnroc 20.000 d6ng NSu ban muon co may tinh d~ h9C bang phuong thirc tra gop hang thang bang s<3 tiSn b<3 cho voi 1ffi sudt 0,7%-1 thang, thi sau bao nhieu thang ban tra h~t n9'?(ViSt quy trinh b~m him) CACHGIAI KETQUA Thang thu nhat sau gop no: A = 5.000.000 - 100.000 = 4.900.000(d6ng) Gan: A = 4.900.000 ; B = 100.000 Thi thang sau gop: B = B + 20.000(gia tri o nho B cong them 20.000), no : A = A x 1.007 - B DIEM 0.5 0.5 0.5 2.0 0.5 20 thang S6 tiSn tra thang cu6i cung 1a 85392 dong 1.0 Bai 10 : Cho nr giac ABCD nQl tiep duong tron tarn ban kinh R= 4,20cm, AB = 7,69cm , BC = 6,96cm , CD = 3,85cm Tinh ti l~ phan tram dien tich giira phan to mau va phan khong to mau, (Xem hinh ve) c CACH GIAI AOB-2- Sin COB-2 - SIn CoD -I -I KETQUA AoB AB 2R - CB 2R CoB SIn -1 -CD 2R I ="2R 32°3i49" R: 111°54'16" - DoA = 360° - Doc' - CoB - AoB SABeD R: - - (sinAOB+sinBOC . - => Dien tich phdn to mau la S T'1 l~ s; y -Shin . +sinCOD+sin DOA) Im= lr.R2 - 25,8374 29,5803 -SABCIJ R: R: CoD R: 54°33'33" i50A R: 60°59'22" 29,5803 em' 25,8374 SABeD 100~ S S /mR: 1m s.; R: R: 29,5803 25,8374 DIEM 0.5 0.5 0.5 0.5 1.0 1.0 1.0 87 ,3466% -H1tll -