DE Till TUYEN SINH LOP 10 TRUNG HQC PHO THONG Nam hoc 2010-2011 so GIAO DVC vA DAo T~o BENTRE ~ ~ , DE CHINHTlruC Mon:TOAN ThOi gian: 120phut, khong ki thili gian giao ad ca.u 1: (3,5 iliim) { 2x-3y =-4 a) Giai h~ phuong trinh x+3y=7 b) Giai phuong trinh: X4 -1 Ox 2 + 9 = 0 . bang phuong phap cong. ca.u 2: (3,5 iliim) Cho phuong trinh X2+2(m-l)x+m2 =0 (m leitham s6) (1). a) Giai phuong trinh (1) khi m = O. b) Tim cac gia tri cua tham s6 m de phirong trinh (1) co hai nghiem phan biet, c) Khi phuong trinh (1) co hai nghiem phan biet XI' X 2 • Tim cac gia tri cua tham s6 m sao cho XI +X2 +X I X 2 = 5. ca.u 3: (6,0 iliim) Cho cac ham s6 y = X2 co d6 thi lei(P) va y = -2x + 3 co d6 thi la (D). a) Ve (P) va (D) tren cling mot h~ true toa d9 vuong goc. b) Xac dinh toa d9 cac giao diem CUa (P) va (D) bang phuong phap dai s6. c) ViSt phuong trinh dirong thang tiSp xuc voi (P) va tao voi hai true toa d9 m9t tam giac co dien tich bang ± (don vi di~n tich). ca.u 4: (7,0 iliim) Cho mra dirong trim tam 0 dirong kinh AB = 2R. Tir A va B lfut hrot ke hai tiSp tuyen Ax va By voi mra dirong tron, Qua diem Mthu9c mra dirong trim (Mkhac A va B) ke tiSp tuyen thir ba c~t cac tiSp tuyen Ax va By l§n hrot tai C va D. a) Chirng minh r~ng: i) Tir giac AOMC n9i tiep. ii) CD = CA + DB va COD=900. iii) ACBD = R2. b) Khi RAM = 60°. Chung to tam giac BDMla tam giac dSu va tinh dien tich cua hinh quat tron ch~n cung MB cua mra dirong tron da cho theo R . HSt HUONG DAN CHAM CUA DE THI TUYEN SINH 10 THPT NAM HQC 2010 -2011 MON ToAN (Huang d~n cham g6m 04 trang) Cfiu Cau 1 3,50 d Cau 2 3,50 d Dap an Diim a) Giai h~ phuong trinh { 2x - 3 y = -4 bang phirong phap cong . x+3y=7 { 2X-3 Y =-4(1) ~ (1) cong (2) ta diroc 3x = 3 ~ x = 1 x+3y = 7 (2) _._ x = 1 => 1+ 3y = 7 ~ y = 2 0,50 _ _ _ _ _._ •. _ _. __ _._ _ _ _ _ Vay h~ phuong trinh co nghiem (x = l;y = 2). 0,50 0,50 b) Giai phuong trinh: X4 -1 Ox 2 + 9 = 0 . Dat: t = X2 (t ~ 0). Phuong trinh (1) co dang: (2 -lOt + 9 = 0 . 0,50 _ _ _._ _._ (0) [:: : ~ (thoa) 0,50 _._ _ _ _._ _._._ ._ _. Khi t = 1 ¢:::> X2 = 1 ¢:::> x = ±1 0,50 _._ _ .•. _.__ . __ __ . __ ._ _ _ _._ .•. _._ _._ _ _ __ ._ _._ _ • _ _. ~ Khi t = 9 ¢:::> X2 = 9 ¢:::> x = ±3 V~y: phirong trinh co bon nghiem x = ±I; x = ±3 0,50 Cho phuong trinh X2 + 2( m -I)x + m' = 0 a) Giai phirong trinh (1) khi m == 0 . Khi m = 0 phirong trinh co dang: X2 + 2.(0 -1)x + 0 2 = 0 0,50 - ¢:::> X2 - 2x = 0 ¢:::> x(x - 2) = 0 [ X = 0 V~ h inh cc hai hi ~ 0 50 ~ x = 2 ay: p uong tri co ai ng iem x = 0; x = 2 , b) Tim cac gia tri cua tham so m de phuong trinh (1) co hai nghiem pharr biet, Ta co: tl' = (m -1/ - m' = m2- 2m + 1- m' = -2m + 1 0,50 _._ _ _ _._ _._-_._ _ _-_._ _._-_._ Phuong trinh (1) co hai nghiem phan biet , 1 1 0,50 ~ tl > 0 ¢:::> -2m + 1> 0 <=> m <- Yay: m <- 2· 2 _ _ _. ._._ _ _ _._- _ __ ._ c) Khi phirong trinh (1) co hai nghiern phan biet Xl' x 2 Tim cac gia tri cua tham s6 m sao cho Xl + X 2 + X l X 2 = 5 . j Xl +X2 = -£. = -2(m -1) Khi m <.!. theo dinh 1y Viet ta co: a 0,50 2 c 2 X I .x 2 =-=m a 0,50 _ - _ _ _ •. __ _ _ _ __ .• _ _ _ _. __ ._ _ __ _ _ _ _._ _ _ __ __ _ _ _ _ _ _ ~ [ml = -I 0,25 m 2 =3 ~ _ _._ _._-_._ _._ _._ _ - _._ _ _._._._._ _._ V~y: K~t hQ'Pv61 diSu kien m < i, ta dircc m = -1 0,25 _ _._ _._ _._ _._._ _._-_ _._ _ _-_. '-_ __ ._ _ 1 1 1 ~ _._ _._ _._ _ _ D6 thi: b) Xac dinh toa dQ giao di~m cua (P) va (D) bang phirong phap dai s6. Phuong trinh hoanh dQ giao diSm cua (P) va (D) la: <=> X2 = -2x + 3 <=> X2 + 2x - 3 = 0 Ta c6: a+b+c=O::::>[X I =1 x 2 =-3 + 1 [ X = 1 [y = 1 I ::::> I x 2 =-3 Y2 =9 V~y cac giao diSm la: M(l;l); N(-3;9) c) Tim phirong trinh dirong thang tiep xuc voi (P) va tao voi hai true toa dQmot tam giac, c6 d~~ntich bang 1. b;fo G9i (d) la duong thang can tim. Phuong trinh (d): Y = ax + b (a:t:. 0) Phuong trinh hoanh dQ giao diSm cua (P) va (d) la: X2 - ax - b = o. _(~)t~~E_~~~_ v6i (~L~=_~ = !!_2 + ~~_~_~_i2 . ~5~ _ Giao diSm cua (d) voi cac true toa dQ 1a A(-!! ;O), B(O;b). a D ' A , h ' ., s Ilbll bl b 2 ien tic cua tam giac OAB ="2 ;; = 2I a /' Cau 3 Cho cac ham 56 Y = X2 c6 d6 thi la (P) va y = -2x + 3 c6 d6 thi la (D). 6,00 d a) Ve (P) va (D) tren cling met h~ true toa dQvuong g6c. Bang mot 56 gia tri: ._ ~ -2 -1 0 1 2 4 1 0 1 4 (D) di qua (0;3)va (1;0) 2 y 3 _____________ _!t _ -~ -2 o 1 -1 x 2 1,00 0,50 1,50 0,50 0,50 0,50 Chirng minh rang: iii) AC.BD = R2. Ta co: {CM = CA ~ AC.BD = CM.DM DM=DB 0,50 Ma CM.DM =OM 2 =R 2 ~ Ae. [SO = R L 0,50 b) Khi iiAM = 60°. Chirng to tam giac BDMla tam giac dSu Ta co: iiAM = MiD = 60° (cling chan cung MB) L _~._._ __._ .__. ._ _.__.___ __ _'_"" ._ __ ._ .__ __ __ __ _ . _ _ _ __ _ _._-_._ Q!.?Q Cau4 7,00 c1 0,50 1 b 2 1 2 SOAB = - <=> -1-1 = - <;:> 2b = lal (2) 4 2 a 4 0,50 [ b =0 Tir (1) va (2) suy ra 4b 4 + 4b = 0 <=> b =-1 V6i b=O=>a=O (loai) V oi b = -1 => a = ±2 V~y: (d) c6 phirong trinh y = ±2x-l 0,50 0,50 a) Chtrng minh rang: i) Tir giac AOMC noi tiep. { - ° CAO=90 Ta co: _ (tinh chat tiep tuyen) CMO=900 ._ _._-_. __ _ _ _ _ _._ _ _ _._-_ _ _._ _._ _ _._ _ _ _. -_._-_._ , _. => CAO + CMO = 180° Vay: Tir giac AOMC n9i tiep dirong tron duong kinh oc. 0,50 0,25 ii) CD = CA + DB Ta co : {CM = CA (tinh ch~t hai tiep tuyen dt nhau tai mo.t diem) 0 50 DM=DB ' V~y CD=CM+DM=CA+DBo Chirng minh CoD = 90° : T . {AOC=Coii (inh hI. h 0.;' I.':' nh 0 ~ diem) 050 a co : _ _ t c at ai tiep tuyen cat au tai mot iern ' DOB=DOM _._ ._ Ma ADM + JiOiJ = 180° (kS bu) 0,50 _._ _._ _._ ._ j - - - 180° Yay: COD = COM + MOD = = 90° o 2 0,50 3 ~ _ _._ -_._ ' _- Ma MlMD din tai D (DM=DB) V~y MlMD d~u. 0,50 Tinh dien tich cua hinh quat tron chan cung MB cua mra dirong tron da cho theo R. Ta co: BAM = 60 0 => J.iCiB = 120 0 0,50 f S = 7rR 2 n = 7rR 2 120 = 7rR2 (dvdt) 0,75 q 360 360 3 1) N~u hoc sinh him bai khong theo huong d~n chsm nhtrng dung v~n cho du di€m theo tung cau. 2) H9C sinh co th€ dung may tinh dm tay (cac loai may tinh diroc phep dem vao phong thi) d€ him bai n~u d€ bai khong yeu du giai theo phuong phap nao. HET ~. ~ , - . DE Till TUYEN SINH LOP 10 TRUNG HQC PHO THONG Nam hoc 2010-2011 so GIAO DVC vA DAo T~o BENTRE ~ ~ , DE CHINHTlruC Mon:TOAN ThOi gian: 120phut, khong ki