Đáp án chính thức môn Toán vào lớp 10_SGD&ĐT Bến Tre_2011-2012

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Đáp án chính thức môn Toán vào lớp 10_SGD&ĐT Bến Tre_2011-2012

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so GIA~ DVC vA I>Ao T~O BENTRE D.E THI TUY£N SINH LOP 10 TRUNG HQC PH6 THONG NAM HQC 2011·2012 Mon:TOAN ThOi gian: 120 phut (khong k8 pMt d~) ~ DE CHlNHTHUC cs« 2. (4,0 di€m) Cho phuong trinh X2 - 3x + m -1 = 0 (m lei tham s6) (1). a) Giai phirong trinh (1) khi m = 1. b) Tim cac gia tri cua tham s6 m d~ phuong trinh (1) c6 nghiem kep, c) Tim cac gia tri cua tham s6 m d~ phuong trinh (1) c6 hai nghiem XI; X 2 lei dQ dai cac canh cua mQt hlnh chtt nh~t c6 dien tich bing 2 (don vi dien tich). ca« 3. (6,0 dtJm) Cho cac ham s6 y = X2 c6 d6 thi lei (P) vei y = x+ 2 c6 d6 thi lei (d). a) Ve (P) va (d) tren cung mQt h~ true toa dQ vuong g6c (don vi tren cac true bing nhau). b) Xac dinh toa dQ cac giao di~m cua (P) va (d) bing phep tinh. c) Tim cac di~m thuoc (P) each d~u hai di~m A(.J3 + 1 ; 0) va B(O ; .Jj + 1). 2 2 Cdu 1. (4,0 di€m) Khong sir dung may tinh eArn tay: a) Tinh: P =Ji2 +5.J3 -~. b) Giai phuong trinh: X2 - 6x +8 = 0 . ) G ·,· h" h inh {X+2 Y =-3 c iai ~ p uong tn : . x-2y=5 Cdu 4. (6,0 di€m) Cho dirong tron tam 0 ban kinh R. Tir mQt di~m A nb ngoai duong tron ke cac ti~p tuy~n AM va AN voi duong tron (M, N lei cac tiSp di~m). a) Chung minh ring tU giac AMON nQi ti~p. b) Bi~t AM = R. Tinh QA theo R. - , . . . .t.' c) Duong thing d di qua di~m A, khong di qua di~m 0 va ~at duong tron tam 0 ~l h~l diem B, e. GQi I lei trung di~m cua Be. Chung to r~ng nam diem A, M, N, 0 va I cung nam tren mQt duong tron. ' H~t. so GIAo DVC VA DAO T~O BtNTRE ~" WONG DAN CHAM DE CHINHTmJf' rur TUYEN SINH LOP 10 TRUNG HQC PHO THONG NAM HQC 2011 -2012 Mon: ToAN Can Dap an Di~m Ghi chu Cau 1 4.00«1 a) P = v'12 +5.J3- if = 2.J3+ 5.J3 _!.J3 0,75 3 3 _ _ _ _- .• - •• -oo •••••• _ •• __ •• _._ •••••••• 1.J3 20.J3 =(2+5 ) 3 =- 3 0,50 3 3 b) Giai phuong trinh: X2 -6x+8=0. III = 9- 8= 1> 0 0,50 .•. - .• - • • -_ _ _ - =>R=l 0,25 _ _ -_ Phuong trinh (1) c6 nghiem: Xl =3-1=2; x 2 =3+1=4 0,50 c) Giai h~phuong trinh: { x+2y~-3 (I) . x-2y = 5 (2) 0,50 .(D.~{~).~ .~~~~: .2.~ .'7.~.:?.~.'7.!: • .•. - Thay X = 1vao (1) ta duoc: 1+2y = -3 ~ y =-2 0,50 -_._ _ __ ._ _ .•. _ • _ • _ _ _ .•. - .• - _ _ _ _ _- _._ •. _ - .• __ .•• _ _ _-_ V~y h~ c6 nghiem: { x~l y=-2 0,50 Cau2 4,00«1 Cho phuong trinh X2 -3x+m-l =0 a) Giai phuong trinh (1) khi m=l. Khi m=1 phuong trinh c6 dang: X2 -3x+l-1 =0 ~ X2 -3x = 0 0,75 _ •. _ _ _ _ __ _ _ .•. _ _ .• _ _ __ _ _ - .• - _ - _ [x~o ~ x(x-3)=O~ 0,50 x=3 _ -_ .• •. _ _ _ _ _ _ _- _ .•. _ _ _ _ _ _-_._- - .•. _- •. _ _ .•. -_ .•. V~y:Phuong trinh c6 nghiem la: X== 0; x=3 0,25 .!>l X~~~g.~~~~~~~.i~~ ~¢.P ~ A~.~ 0,50 _ _ _ .•. _ -_ • _ _ _ _ _ •.• •.• ~9-4m+4=0 0,50 •.• _-_ .•. _-_ .•. _ _ .•. - .• •. _ .• • •• _.__ ._._ __ _ _ - -_ __ . 13 0,25 ~m=- 4 c) Hai nghiem Xl; x 2 cua phuong trinh (1) la dQdai cac canh cua mQthinh chu nh~t c6 dien tich bang 2 r"o f 3 - 4m <oo 1,00 ee S>O ~ 3>0 ~m=3 P=2 m-I=2 - .•. - .• .•. - .•. - .• ~ .•. ~ .•. - - .•.• _ - _ • - - - - V~y: m=3 0,25 - Clu3 6,00(1 1 ·(d) ·(i q~··(O;· iS~A· .(~2;0.) ······0,50······ . . ~ Do thi : b) Xac dinh toa dQgiao diem cua (P) va (d) bang phep tinh. Phuong trinh hoanh dQgiao di~m cua (P) va (d) la: ee X2 = X + 2 (:) X2 - X - 2 = 0 0,50 Ta co: a=b+c = 0 =>[XI = -1 050 ~=2 ' [ XI = -I => [YI = 1 x 2 = 2 Y2 = 4 0,50 ·V~yc~c·giaodi~m· iit:' ·A(~ i; i);" :8(2;4·)······· ······0,25······ . c) Tim cac di~m thuoc (P) each deu hai di~m A(fj +1;0) va B(O; fj +1) .Cac di~m each d8u hai 2 2 di~m A, B thuoc duong trung true cua doan thang AB va ~OAB vuong din tai 0 => cac di~m nay thuoc duong thang Y = x (duong phan giac AOB) 0,50 ······························l······················· • Do do: ~p hQ'Pcac diem nay la giao di~m cua (P) va duong thang Y = x . Phuong trinh hoanh dQ giao di~m cua (P) va duong thang Y = x la: (:) X2 = X (:) X2 - X = 0 (:) x( x-I) = 0 Cho cac ham s6 Y = X2 co db thi Ia (P) va Y = x + 2 co db thi la (d). a) Ve (P) va (d) tren cung mQt h~ true toa dQvuong goc, Bang mQt s6 gia tri: ~~ : ~-1 __-: 1·1 : ' ' y· l ysx2 Y"'x+2 ' ' - - _. - _. - - - - - - - - _. - - - - - - - - - - - - - - - - "" 4""""""" -xl I , , 1'" . =: 1 A, x ' , J (:) x=O; x=1 2 1,00 1,50 0,50 Cau4 6,00 (I 0,50 1) N8u hoc sinh lam bai khong theo huang dan chAm nhung dung van cho dli di€m theo timg cau. 2) HQc sinh co the dung may tinh cdm tay (cac loai may tinh duoc phep dem vao - phong thi) d~ lam bai n~u d~ bai khong yeu cAu giai theo phuong phap nao, 3 N a) Chung minh rang tU giac AMON nQi tiep, Ta co: AMO = ANa = 90° (tinh chAt ti8p tuyen) 0,50 + 0) ••• _ ._ •• __ •• _ •• __ •••••••••• __ •••••• _ •••••• w •• __ ••••••••••• w •• _ •••• _ ••••••••••• • ••••• __ •••••• _ •••••••• w_ •••••••••••••••••• _ ••••••• __ •••• ••••• •• _ ••••••••••••• '"' ••••• - •• - •• - •••• ~AMO+ANO=180o. 0,50 · 1 V~y: Trr giac AMON nQi tiep. 0,50 - b) Biet AM = R. Tinh OA theo R OA =.JAM2 +M0 2 =.JR2 +R2 =Rfi 0,75+q I) Chung to nam diem A, M, N, 0 va I cung nam tren mQt duong tron. Ta co: 01 1. Be (duong kinh di qua trung di€m cua mQt 0,50 f- O , _~Y.:_'?~£): - _ Cac tam giac AMO, ANO, AIO vuong ~ nam di~m A,M,l,O,N cung thuoc duong tron duong kinh OA 1,25 + 0, S . nam diem A, M, N, 0 va I cung nam tren mQt duong tron. ' H~t. so GIAo DVC VA DAO T~O BtNTRE ~" WONG DAN CHAM DE CHINHTmJf' rur TUYEN SINH LOP 10 TRUNG HQC PHO THONG NAM HQC 2011. (6,0 dtJm) Cho cac ham s6 y = X2 c6 d6 thi lei (P) vei y = x+ 2 c6 d6 thi lei (d). a) Ve (P) va (d) tren cung mQt h~ true toa dQ vuong g6c (don vi tren cac true bing nhau). b) Xac dinh toa dQ cac giao di~m cua (P) va (d) bing phep tinh. c). so GIA~ DVC vA I>Ao T~O BENTRE D.E THI TUY£N SINH LOP 10 TRUNG HQC PH6 THONG NAM HQC 2011·2012 Mon:TOAN ThOi gian: 120 phut (khong k8 pMt

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