DE23 THPT chuyên vĩnh phúc (l3,a) _ ĐỀ THI THỬ THPT QUỐC GIA TOÁN NĂM 2015

SGD&T TRNGTHPT THI TH CHUYấNVNHPHC KHOSTCHTLNG CCMễNTHITHPTQUCGIALN3ưNMHC2014 K THI THPT QUC GIA 2015 - S- 2015 23 MễN:TONưKHI12A+B Thi gian lm bi 180 phỳt Thigian180phỳt(Khụngkthigiangiao) oOo -thigm01trang Cõu1(2,0im).Chohms y = x + x + mx + m -2 ( m lthams)cúthl ( Cm ). a)Khosỏtsbinthiờnvvthcahmskhi m =0 b)Xỏcnh m ( Cm)cúcỏcimccivcctiunmvhaiphớatrchonh Cõu2(1,0im). Giiphngtrỡnh: 2cos x + 2cos x - cos x = sin x + Cõu3(1,0im). x + x ex Tớnh: I = ũ dx x + e - x Cõu4(1,0im). a) Giiphngtrỡnh: log x + log x + log x =log36 x ( ) n 2ử ổ b) Tỡmshngkhụngphthucvo x trongkhaitrinnhthcNiutn ỗ x2 + ữ x ứ ố n+2 n+1 * (vi x ạ0), bitrng n ẻ Ơ v Cn+5 - Cn+ = ( n +4) Cõu5(1,0im). Cho hỡnh chúp S ABCD cú ỏy ABCD l hỡnh ch nht vi AB = 3a AD =2a Hỡnh chiu vuụnggúcca S lờnmtphng ( ABCD) lim H thuccnh AB saochoAH =2HB Gúc gia mt phng ( SCD) v mt phng ( ABCD) bng 600.Tớnh theo a th tớch chúp S ABCD vtớnhkhongcỏchgiahaingthng SC v AD Cõu6(1,0im). Trong mt phng ta Oxy cho tam giỏc cõn ABC cú ỏy BC nm trờn ng thng d :2 x - y + =0, cnh AB nm trờn ng thng d  :12 x - y - 23 =0. Vit phng trỡnh ngthng AC bitnúiquaim M( 31). Cõu7(1,0im). Trong khụng gian Oxyz , cho A (100 ) , B ( 20 ) , C( 003).Vit phng trỡnh mt phng ( P)iqua O,C saochokhong cỏcht A n ( P) bngkhong cỏcht B n ( P). Cõu8(1,0im). ỡù x + xy + y + x + xy + y = 3( x + y) Gii h phngtrỡnh: ùợ x + y + + 12 x + y + = xy + x + Cõu9(1,0im). Chobasthcdng a, b,c thamón a + b + c =3. ổ 1 1ử Tỡmgiỏtrinh nhtcabiuthc S = ( a + b + c)+ 5ỗ + + ữ ố a b c ứ ưưưưưưưưưưưưưHtưưưưưưưưưưưư Thớsinhkhụngcsdngtiliu.Cỏnbcoithikhụnggiithớchgỡthờm. Hvtờnthớsinh:Sbỏodanh: CmnthyNguynDuyLiờn(lientoancvp@vinhphuc.edu.vn)ógiti www.laisac.page.tl 133 TRNGTHPTCHUYấNVNHPHC. HNGDNCHMKSCL LN3 NM2015 (Hngdnchmcú 5trang) Mụn:TON ư12AB I.LUíCHUNG: 1)Nuthớsinhlmbikhụngtheocỏchnờutrongỏpỏnnhngvnỳngthỡchosim tngphnnhthangimquynh. 2)Vicchitithoỏthangim(nucú)tronghngdnchmphimbokhụnglmsai lchhngdnchmvphicthngnhtthchintrongcỏcgiỏoviờnchmthihhosỏt. 3)imtonbitớnhn0,25im.Saukhicngimtonbi,ginguyờnktqu. II.PN: Cõu í Nidungtrỡnhby a Chohms y = x + x + mx + m -2 ( m lthams)cúth l ( C ). m im ồ1, a)Khosỏtsbinthiờnvvthcahmskhi m =0 Khi m =0 hmstrthnh y = x + x - ã TX: D = R ã Sbinthiờn: ộ x= +)Chiubinthiờn: y = x + x, y' = ởx = -2 Hmsngbintrờncỏckhong ( -Ơ -2 ) , ( +Ơ),nghchbintrờn ( -20 ) 0.25 +)Cctr :Hmstcciti xCD = -2 yCD = y ( -2) =2 Hmstcctiuti xCT = yCT = y (0) = -2 +)Giihn: lim y = -Ơ lim y = +Ơ x đ-Ơ 0.25 xđ+Ơ Bngbinthiờn: x ư2 -Ơ +0 y' y +Ơ 0+ +Ơ 0.25 ư2 -Ơ ( )( ã th:ct Ox ti ( -1 ) , -1 + 30 , -1 - ) thnhnimunU(-10 )ltõmixng. ( Giỏmkhotv) b b)Xỏcnh m ( C )cúcỏc imccivcctiunmvhaiphớatrchonh m Phngtrỡnh honh giaoim ca ( Cm)vtrchonh l x + x + mx + m - = ( x - 1) ( x + x + m - ) =0 (1) 0.25 ồ1, 0.25 ộ x= -1 (1) ởg ( x ) = x + x + m - = ( 2) 134 0.25 ( Cm) cúhaiimcctrnmvhaiphớaivitrc Ox PT (1)cúba nghim ỡù D = - m> phõnbit ( ) cú hainghimphõnbit khỏc -1 m< ùợg ( -1)= m - Vykhi m 0).Ta cú ã ABC = ã ACB < 900 0.25 r r r r ị cos ã ABC = cos ã ACB cos ( n AB , nBC ) = cos ( nBC ,nCA) 136 r r r r n AB nBC nCA nBC 145 2a - 5b r = a - 100ab - 96b2 = r = r r 2 n AB nBC nCA nBC a +b 0.25 a + 12b = a - 8b =0 r Vi a + 12b =0 Chn a = 12, b = -1 thỡ nCA = (12 -1)ị AB AC (loi) 0.25 Vi 9a - 8b =0 Chn a = 8, b =9 nờn AC : ( x - ) + ( y - 1)=0 0.25 ị AC : x + y - 33 =0 Trongkhụng gian Oxyz ,cho A (100 ) , B ( ) , C( 03).Vitphngtrỡnhmt phng ( P)iqua O,C saochokhongcỏcht A n ( P) bngkhongcỏcht B ồ1,0 n ( P). Do ( P)cỏchu A v B nờnhoc ( P ) AB hoc ( P) iquatrungim AB. 0.25 ỡ qua O( 00) ù Khi ( P ) AB ị ( P ) ị ( P ): x - y= r uuur uuur r ùợvtpt n ộở AB, OC ựỷ = ( -63 ) ị n = ( -1 0) 0.25 ổ Khi ( P) iquatrungim I ỗ 0ữ ca AB. Ta cú : ố ứ ỡ qua O( 00 0) ù ị (P)ớ ị ( P ): x + y = r uur uuur r ổ ùvtpt n0 ộở IC , OC ựỷ = ỗ 20 ữ ị n0 = ( 210) ố ứ ợ 0.25 Vyphngtrỡnhmtphng ( P ) : x - y = 0, 0,25 ( P ): x + y =0 ỡù x + xy + y + x + xy + y = ( x + y) Giih phngtrỡnh: ùợ x + y + + 12 x + y + = xy + x + (1) ( 2) ỡ5 x + xy + y2 ù iukin: x + xy + y x + y+ 0. ù x + y + ợ ồ1,0 0.25 ( ) Khihcú nghim ( x y ) ắắ đ x + y Tathy x + xy + y x + y ( *)dubngkhi x = y thtvy (*) x + xy + y ( x + y ) Tngt 2 ( x - y ) luụnỳngvimi x,y ẻ Ă x + xy + y x +2 y (**)dubngkhi x = y 0.25 T (*) & (**) ị VT (1) = x + xy + y + x + xy + y 3( x + y ) =VP (1) Dungthcxy rakhi x = y ( 3) 137 Th ( 3) vo ( 2) tac: ( ) ( x - x ) + ( x + ( x - x) + 3x + + 19 x + = x + x + ( 4) iukin x - ) ( ) 3x + + x + - 19 x + = 0.25 x2 - x x - x + 2ì = 2 x + + x+ ( x + ) + ( x + ) 19 x + + (19 x +8) ộ ự ỳ 1 ỳ = ( x - x) + + 2ì 2 ỳ 3 x + + x+ + ( x + ) 19 x + + (19 x + 8) ỳ ( x + 24 ) 2444444444444 144444444444 > ởờ ỷỳ 0,25 ( 3) ộ x = ắắ đ y= x - x= Tha móniukin ( 3) ờởx = ắắđ y = Vyh phng trỡnh cúhainghim ( x y ) = ( 00 ) & ( x y ) =(11) Chobasthcdng a, b,c thamón a + b + c =3. ổ 1 1ử Tỡmgiỏtrinhnhõtcabiuthc S = ( a + b + c)+ 5ỗ + + ữ ố a b c ứ Nhn xột: 8a+ 8a + ồ1,0 3a2 + 23 , (1) vimi < a < dubngkhi a =1 thtvy a 3a2 + 23 3a - 16a + 23a - 10 Ê ( a - 1) ( 3a- 10 )Ê0 luụnỳng a 0.25 vimi < a < dubngkhi a =1 Tngt 8b+ 3b2 + 23 , ( 2) dubngkhi b = b 3c2 + 23 8c+ , ( 3) dubngkhi c =1 c 2 2 ổ 1 1ử ( a + b + c )+ 69 T ắắắắđ S = ( a + b + c) + ỗ + + ữ = 39 ố a b c ứ 0.25 (1),( ) &( 3) 0.25 Dubngxyrakhi a = b = c =1 Vygiỏtrnhnht ca S =39 tckhivchkhi a = b = c =1 0,25 Chỳý: tỡmravphica(1)tasdngphng phỏptiptuyn CmnthyNguynDuyLiờn(lientoancvp@vinhphuc.edu.vn)ógiti www.laisac.page.tl 138 ...TRNGTHPTCHUYấNVNHPHC. HNGDNCHMKSCL LN3 NM2015 (Hngdnchmcú 5trang) Mụn:TON ư12AB I.LUíCHUNG: 1)Nuthớsinhlmbikhụngtheocỏchnờutrongỏpỏnnhngvnỳngthỡchosim... lchhngdnchmvphicthngnhtthchintrongcỏcgiỏoviờnchmthihhosỏt. 3)imtonbitớnhn0,25im.Saukhicngimtonbi,ginguyờnktqu. II.PN: Cõu í Nidungtrỡnhby a Chohms y = x + x + mx + m -2 ( m lthams)cúth l ( C ). m im ồ1, a)Khosỏtsbinthiờnvvthcahmskhi... ồ1, a)Khosỏtsbinthiờnvvthcahmskhi m =0 Khi m =0 hmstrthnh y = x + x - ã TX: D = R ã Sbinthiờn: ộ x= +)Chiubinthiờn: y = x + x, y' = ởx = -2 Hmsngbintrờncỏckhong ( -Ơ -2 ) , ( +Ơ),nghchbintrờn (