đề thi dap an hoc sinh gioi 11 2012-2013 THPT Diễn châu 2 nghe an

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Ih%'Yh&4`.c6 &4L%IKd%m<6K.'K&'(p&f(b%&'MI1J.GZ'5%'&'K(, 87 'K*GZI(FO%rO&4K%3&6O3(/.m,/.I1s%3&'k%3TM6*<K%3<K%3Y:(m> >&12%3R%3.X&./.Oj&0'k%3n7>nm7>nm7&'tK&'R&N&d(mu>u>u, 5O3(/&4hG:%%'V&.c6&`.'*mu,*u,*u>8(p&mv , ,x AC y AD z= = , Câu 8 (0,5 điểm),'K&6O3(/.m.E  , ,a BC b AC c AB= = = > { } 0 min , , 15A B C ≥ ,'R%3 O(%'4r%3 , ,ab bc ca .w%3GZIPfZ(86.d%'.c6OP&&6O3(/., """"""""""""""""""""""""""""""""""""""""""""""p&"""""""""""""""""""""""""""""""""""""""""""""""""""" Giám thị coi thi không giải thích gì thêm Họ và tên thí sinh……………………,,,,,,,,, Giámthị 1:………………………………… Số báo danh………………………x,,,,,,,,,, Giám thị 2:…………………xxxxx yz{     !"!# | yz{ $%&'()*+ Câu Ni dung Đim  67& ( ) ( ) 2 2 cos sin si n cos sin 0x x x x x⇔ − − − = ( ) ( ) cos sin cos sin sin 1 0x x x x x   ⇔ − + − =   ( ) 2 cos sin cos 0x x x⇔ − = cos sin 0 cos 0 x x x − =  ⇔  =  , ( ) 4 2 x k k Z x k π π π π  = +  ⇔ ∈   = +   , '12%3&45%'.E'6('l%3'(bOGZ ( ) , 4 2 x k x k k Z π π π π = + = + ∈ , !>@ !>@ !>@ 876.E ( ) ( ) 2 2 5 2 2 1 cos 2 16 sin 16sin sin 16 sin 4 sin cos 2 2 cos 2 1 2 x x x x x x x x −   = = = − +  ÷   v 1 cos 4 4 sin 8sin cos 2 4 sin . sin 5 5sin 3 10sin 2 x x x x x x x x + − + = − + , & 5 2 2 4 sin 5 0 2 3 5 2 4 x k x x k π π π π  = +  ⇔ − = ⇔   = +   ( ) 2 20 5 3 2 20 5 k x k Z k x π π π π  = +  ⇔ ∈   = +   , 9H^0'12%3&45%'.E'l%3'(bO ( ) 2 3 2 , 20 5 20 5 k k x x k Z π π π π = + = + ∈ , !>@ !>@ !>@  \!.';<=Ig.'K&6GH0I1J. 3 7 C 8P3=OA.';<=Q'/.%'6M>&4K%3 IEGM$%.EOj&./..';<=>>#, \O](8P%'1&'pGH0I1J. 4!3 ! <=.EA.';<=Q'/.%'6M&4K%3IE./. .';<=>>#GM$%IR%3QS%'6MnY:(TM^1:.&`%' ./.<=OZ.E.'; <=!IR%3IUM7,9H^.E 4!3 ! 3 7 C v@!?!n<=7, 4K%3@!?!<=I1J.&dK&'Z%'.E#}#} 2 6 C v@?!n<=73~OA.';<=Q'/. %'6MOZ.';<=!IR%3IUMYZ./..';<$>>#GM$%IR%3QS%'6M, 9H^.E@!?!•@?!v?@!!n<=.U%&5O7, !>@ !>@ !>@ !>@ # l(GZ8(p%.=TM-.UMGV^46&\'P0&'R%'V&GZOZMIW ⇒ n7v ! # l(mGZ8(p%.=TM-.UMGV^46&\'P0&'R'6(GZOZMIW ⇒ nm7v ! @ mGZ8(p%.=GV^46'6(TM-.UMOZMIW&\'6('P0€>mGZ'6(8(p%.= I$.GH0>/0f•%3.$%3&'R.%'‚%[/.<MV&&6.Enm7vn7,nm7v ?!! @ , 12%3&N&6&`%'I1J.[/.<MV&IFGV^I1J.'6(TM-.UMOZM&4X%3GZ ?!! ?D >[/.<MV&IFGV^I1J.'6(TM-.UMOZM[6%'GZ ?!! C! , 9H^[/.<MV&IFGV^I1J.'6(TM-.UM.a%3OZM&'W6Og%8Z(&K/%GZ ?!! @ ƒ ?!! ?D ƒ ?!! C! v @ D , !>@ !>@ !>@ !>@ ? 6.E ( ) 8 2 3 1 2x x+ − v ( ) [ ] ∑ = −=−+ C ! # C C # 7n k kk xxCxx v ∑ ∑ = = − − C ! ! # C 7,n7n k k i iiki k k xxCC v ∑∑ = = + − C ! !  C ,7n k k i ikii k k xCC , b<=.c6 C x R%3Y:(Q>(&'W6Og%      ≤ ∈ =+ ki Nik ik > C > 3(-('b%Z^&6I1J.nQ€(7vn#€7YZnQ€(7vn?€!7, 9H^'b<=.'R6 C x .c6Q'6(&4(F%GZ !! ? ? C  # # C  7,n,,7n −+− CCCC vB?, !>@ !>@  !>@ !>@ @ 6.E 7n  −+= + nn uu  ⇔ 7n  −+= + nn uu ⇔ 7n  +=+ + nn uu , j& += nn uv > ≥n >&6.Efg^ { } n v GZOP&.V0<=%'‚%Y:( !?  =+= uv >.$%38P(Tv, 6.E 7n,,,,7n7n,,, !!!! −++−+−=+++= vvvuuuS n v !!7,n!?!!   ,!!7,,,,n !! !! !! −−=− − − =−+++ q q vvvv !>@ !>@ !>@ A l(GZI(FOI=([R%3.c6 TM6 GZI(FOI=([R%3.c6 TM6m, 'MY(&6O3(/. v ƒƒ  vƒƒ ≥ Q'$%3Ie(, VM8r%3[-^46Q'(>>>&'k%3'Z%3, 9H^.'MY(&6O3(/. %'W%'V&v '(>>>&'k%3'Z%3, 5O>, l(GZ'5%'.'(pMYM$%33E..c6 &4L% ⇒ n€7fKIEn#€7, l(„GZ'5%'.'(pMYM$%33E..c6 &4L%m ⇒ „n"€7fKIEn"@€7, '12%3&45%'./.I1s%3&'k%3m) !B# =+− yx €) !# =−+ yx € )^•v!,v…>v…m, KIE&l6IP./.I(FO>.U%&5OGZ)n€7>n 7€ # @ − , !>@ !>@ !>@ mn"B†#€!7 n"€?7 n#€!7 n€!7 n"@€7 n#€7 B 67 Kn7TM6 <K%3<K%3Y:(%L% n7…nm7v ‡> ‡††,Kn7<K%3<K%3Y:(m%L% n7…nm7v > ††m€ n7…nm7v‡>‡††m, =( >>‡>‡ &6I1J.&'(p&f(b%GZ &R3(/. ‡,'(p&f(b%GZ'5%'85%''Z%', ƒ7R3(/. ‡GZ'5%'&'K( ⇔ v ‡, 6.E) AD MQ AB BM BC MN AB AM == >       +=+= + ⇒ ADBC MN AD MQ BC MN AB MBAM  ADBC ADBC MN + =⇒ , ADAB ADAB BC MNAB AM + ==⇒ ,, , 9H^ &4L%.d%'m<6K.'K ADAB ADAB AM + = , &'5&'(p&f(b%&'MI1J.GZ '5%'&'K(, !>@ !>B@ !>@ !>@ 87ƒ79ˆu>u>mu, 4K%3O0nm7%=(*.X&m&d(„,4K%3O0n„7 ‰I1s%3&'k%3TM6*<K%3<K%3Y:(.X&„&d( u,/.'[/.Ih%'&12%3&N.'K./.I(FOmuYZu, 6.E) BCD OBC S S JD OJ AD OD == Š € BCD ODC S S AB OB = Š > BCD OBD S S AC OC = Š 9H^  ŠŠŠ =++=++ BCD OBD BCD OBC BCD OBC S S S S S S AC OC AB OB AD OD +0f•%3m6M.'^&6I1J. # ,, ŠŠ,Š, ,# ŠŠŠ ACABAD OCOBOD AC OC AB OB AD OD ≥++  ADACABODOCOB ,, B  ŠŠ,Š, ≤⇒ , VM‹vŒ[-^46 OBDOCDOBC SSS ==⇔ ⇔ *GZ&4l%3&‚O&6O3(/.m, 9H^n*mu,*u,*u7 O6[ v xyz B  Q'(*GZ&4l%3&‚O&6O3(/.m, !>@ !>@ C KO(% { } { } !! @<(%<(%><(%><(%O(%@>> ≥⇒≥ CBACBA , 6.E)<(%@ ! v ≈ − = −  #  #!.K<  !>@CC, (-<•%31J.Gd(4r%3 , ,ab bc ca Q'$%3GZIPfZ(86.d%'.c6OP& &6O3(/.>&6.E&'F3(-<• cabcab +≥ ≤+⇔ b c a c ⇒ &4K%3'6(<= b c a c > .EOP&<=Q'$%3G:%'2%   ,(-<•   ≤ a c @>! ?  <(% ?  <(% ?  <(% <(% ?  =≤≤⇔≤⇔≤⇔ AC A C a c nO‚M&'M_%7,n 7, !>@ !>@ nĐáp án gồm 3 trang. Học sinh giải theo cách khác đng vẫn cho đim tối đa7 """""""""""""""""""""""""""""""""""""""""""""""""""""p&""""""""""""""""""""""""""""""""""""""""""""""  muu u  m  * „    ‡  m   . 4 sin cos 2 2 cos 2 1 2 x x x x x x x x −   = = = − +  ÷   v 1 cos 4 4 sin 8sin cos 2 4 sin . sin 5 5sin 3 10sin 2 x x x x x x x x + − + = − + , & 5 2 2 4 sin 5 0 2 3 5 2 4 x k x x. +  ⇔ − = ⇔   = +   ( ) 2 20 5 3 2 20 5 k x k Z k x π π π π  = +  ⇔ ∈   = +   , 9H^0' 12% 3&45%'.E'l%3'(bO ( ) 2 3 2 , 20 5 20 5 k k x x k Z π π π π =. $%&'()*+ (Thi gian 150 pht không k thi gian giao đ) Câu 1 (2 điểm),(-(./.0' 12% 3&45%') 67 cos 2 sin cos sinx x x x= − 87 5 2 16 sin 10 sin 5 sin 3 2 x x x− = − , Câu 2 (1 điểm),9:(./..';<=!>>>#>?>@>A>B>C>D.E&'FGH0I1J.86K%'(LM<=&N%'(L%.E</M .';<=I=(OP&Q'/.%'6M<6K.'K./..';<=>>#IR%3QS%'6M, Câu