ẹE THI THệ SO 35 Thigian:180phỳt(khụngkthigiangiao) Cõu1(2,0im) Chohms y =- x3 + 3mx2 + 3(1- m2)x+ m3 - m (1) a)Khosỏtsbinthiờnvvthcahms (1)khim=1. b)Tỡmmthhms(1)cúhaiimcctrnm vcựngmtphớacangthng y=1 (khụngnmtrờnngthng). Cõu2(1,0im). Giiphngtrỡnh cos 2x+(1+ 2cosx)(sinx- cosx)= Cõu3(1,0im) Tỡmgiỏtrlnnhtvnhnhtcahms y =ex(x2 - x- 1) trờnon[02]. Cõu4(1,0im) Chonlstnhiờnthamón 2Cn2 + 3An2+2 = 326.Tỡmhsca x6 trongkhaitrinnhthc n ổ Niutnca ỗ 2x2 - ữ ,x> 0. xứ ố Cõu (1,0 im). Trong khụng gian vi h ta Oxyz, cho tam giỏc ABC vi A(1-12), B(ư113),C(021).TớnhdintớchtamgiỏcABCvtỡmtachõnngcaoktAcatam giỏcABC. Cõu6(1,0im).Chohỡnhchúp S.ABCcúỏy ABCltamgiỏcvuụngtiA,mtbờnSABltam giỏcuvnmtrongmtphngvuụnggúcvimtphng(ABC),giMlimthuccnhSC saocho MC =2 SM Bit AB =a , BC =a 3.TớnhthtớchcakhichúpS.ABCvkhongcỏch giahaingthng ACv BM. Cõu7(1,0im).TrongmtphngvihtoOxy,chotamgiỏcABCnitipngtrũn (T) cúphngtrỡnh (x-1)2 + (y- 2)2 = 25.CỏcimK(ư11),H(25)lnltlchõnngcaoh tA,BcatamgiỏcABC.Tỡmtacỏcnhcatamgiỏc ABCbitrngnhC cú honh dng. ỡù x2 + y + = y2 - 3x + Cõu8(1,0im) Giihphngtrỡnh ùợ y- 1+ 2y2 + 1= x + x2 + xy+ 3y Cõu9(1,0im). Cho x,y,zlcỏcsthcthamón x2 + y2 + z2 = 9, xyzÊ0 Chngminhrng 2(x+ y+ z)- xyzÊ 10. TRNGTHPTễNGSNI Kè THI KSCL TRCTUYNSINH NM 2015(LN1) PNTHANGIMMễNTON Cõu Nidung im 1a Khosỏthmsvvthhms 2,00 Khi m=1,tacúhms y =- x3 + 3x2 1)Tpxỏcnh: D = R 0,5 2)Sbinthiờn: *Giihn : lim y= lim(- x3 + 3x2)= +Ơ,lim y= lim(- x3 + 3x2)= -Ơ xđ-Ơ xđ-Ơ xđ+Ơ xđ+Ơ *ohm y=ư 3x2 +6x ,y=0 x=0, x =2. * Bngbinthiờn: x Ơ 2+ Ơ y' +0 Ơ + 0,5 y Ơ ưHmsnghch bintrờn cỏckhong (ư Ơ 0)v(2+ Ơ ),ng bintrờn khong (02) ưHmstcciti x=2,yC =4,tcctiuti x= 0, yCT =0. 3.th:thgiaovitrctungti y O(00),giaovitrchonhtiO(00) A(3 0), nhn im un I(12) lm tõm i xng *imun:y=ư 6x +6,y=0 x=1 thhmscú1imun I(12) O 1b A 2b 0,5 x Tỡmm thcú2cctr y'=-3x2 + 6mx+ 3(1- m2) 2,00 0,25 y'=0 -3x2 + 6mx+ 3(1- m2)= 0, y' cú D'= 9m2 + 9(1- m2)= 9> Suyra y' luụncúhainghimphõnbit x1 = m- 1, x2 =m+ Khiúhmscúhaicctrl y1 = y(x1)= 2(m- 1), y2 = y(x2)= 2(m+ 1) 0,5 Theobiratacú ( y1 - 1)( y2 - 1) > (2m - 3)(2 m + 1) > m > , m < - 2 1ử ổ ổ Vy mẻỗ - Ơ- ữ ẩ ỗ +Ơ ữ 2ứ ố ố ứ 2a 0,5 0,5 0,5 0,25 Giiphngtrỡnhlogarit iukin: 0 nờn C(5-1) ịớ ớ 2 ợ(x- 1) + (y- 2) = 25 ợ y= -1 ợ y= ngthngACiquaCvcúvectchphng l CH =(-36) nờnACcúphng trỡnh x+y- 9= 0. DoAlgiaoca ACv(T)nờntaimAlnghimcah ỡ2x + y- 9= ỡ x= ỡ x= ịớ (loi).Doú A(17) 2 ợ(x- 1) + (y- 2) = 25 ợ y= ợ y= -1 0,25 ngthngBCiquaCvcúvectchphng l CK =(-62) nờnBCcúphng trỡnh x+3 y- 2= 0. DoBlgiaoca BCv(T)nờntaimBlnghimcah ỡ x +3y- 2= ỡ x= -4 ỡ x= (loi).Doú B(-42) ịớ ,ớ 2 ợ(x- 1) + (y- 2) = 25 ợ y= ợ y= -1 Vy A(17) B(-42) C(5-1). Giihphngtrỡnh 0,25 0,25 0,25 0,25 0,25 0,25 2,00 ỡù x2 + y + 3= y2 - 3x + (1) Tacúhphngtrỡnh ùợ y- 1+ 2y2 + 1= x + x2 + xy+ 3y (2) iukin: y1,x 0,y2 3x. 0,25 (2) y- 1- x + (y2 - 2y+ 1)- x2 + (y2 - xy- y)= y- 1- x + (y- 1)2 - x2 + y(y- x- 1)= y- 1+ x ổ ( y- x- 1)ỗỗ ố + 2y - 1+ xữữ = y - 1+ x ứ ổ y = x+ ỗ Do ỗ ố 0,5 + 2y- 1+ x> 0,"y 1,"x 0ữ ữ y-1+ x ứ +)Thyvo(1)tac x2 + x+ 1- x2 - x+ 1= - (3) Xột f(x)= x2 + x+ 1- x2 - x+ 1, 2x+ 2x- 2x+ 2x- f'(x)= = 2 2 x + x+ x - x+ (2x+ 1) + (2x- 1)2 + t Xột g (t ) = t +3 , g '(t ) = (t + 3)3 > 0,"tẻ R suyrag(t)ngbintrờn R Do x+1> 2x- nờn g(2x+1)> g(2x- 1) suyra f '( x) = g (2 x + 1) - g (2 x - 1) > 0,"x ẻR Doú f(x) ngbintrờn R ,nờn (3) f(x)= f(2) x= 2ị y= Vyhóchocúnghim ( xy)=(23) Chngminhbtngthc Gis x Ê yÊ z,do xyzÊ0 nờn xÊ0 y2 + z2 2(x + y+ z)- xyzÊ 2x+ 2(y2 + z2)- x. 2 x(9- x ) x 5x =2x + 2(9- x2)= + 2(9- x2) 2 0,25 0,25 0,5 x3 5x 2x + 2(9- x2) vi xẻ[-30] ị f '(x)= x2 - 2 2 9- x2 f'(x)=0 2x x - = 9- x2(5- 3x2)= -4 2x 2 9- x 0,5 (9- x 2)(5- 3x2)2 = 32x2 (iukin 5-3x2 0) 9x6 - 111x4 + 327x2 - 225= x2 = 1,x2 = 3,x2 = Do x2 Ê 0,5 2,00 y2 + z2 ổ y+ zử Do x + y + z = ị x Ê ị x ẻ [ -3 0]. Tacú yz Êỗ ,doú ữ Ê ố ứ Xột f (x)= 0,5 25 nờn x2 =1 x= -1,x= 1(loi). f( -3)= -6, f(-1)= 10, f(0)= suyra max f(x)= f(-1)= 10 [ -30] 0,25 Nhvy 2(x+ y+ z)- xyzÊ f(x)Ê 10 ỡ x= -1 ùù ỡ x= -1 Dubngxyrakhi y = z ợ y = z= ù 2 ùợy + z = 2( y + z ) = Vy 2(x+ y+ z)- xyzÊ 10.ngthcxyrakhi(xyz)lmthoỏnvca(ư122) ưưưưưưưưưưưưưưưư***Ht***ưưưưưưưưưưưưưưưư Cm nthyoTrngXuõn(trongxuanht@gmail.com)ógitiwww.laisac.page.tl 0,5