Tran Sú Tuứng kim tra ẹaùi soỏ 10 CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 1 a) Trc nghim khỏch quan Cõu 1: (0,5) Tp xỏc nh ca hm s 3 1 y 1 x x 1 = + + l: a) D = (1; 1) b) D = (1; 1] c) D = (; 1] \ {1} d) D = (; 1] (1; + ) Cõu 2: (0,5) Cho hm s (P) : y = ax 2 + bx + c. Tỡm a, b, c bit (P) qua 3 im A(1; 0), B( 0; 1), C(1; 0). a) a = 1; b = 2; c = 1. b) a = 1; b = 2; c = 1. c) a = 1; b = 0; c = 1. d) a = 1; b = 0; c= 1. Cõu 3: (0,5) Cho hm s y = x 2 + mx + n cú th l parabol (P). Tỡm m, n parabol cú nh l S(1; 2). a) m = 2; n = 1. b) m = 2; n = 3. c) m = 2; n = 2. d) m= 2; n = 3. Cõu 4: (0,5) Cho hm s y = 2x 2 4x + 3 cú th l parabol (P). Mnh no sau õy sai? a) (P) i qua im M(1; 9). b) (P) cú nh l S(1; 1). c) (P) cú trc i xng l .thng y = 1. d) (P) khụng cú giao im vi trc honh. b) T lun Cõu 5: (8 im) Cho hm s y = (m 1)x 2 + 2x 3 (P m ) a) Kho sỏt v v th hm s vi m = 2 (tng ng l (P 2 )). Bng th, tỡm x y 0, y 0. b) Dựng th, hóy bin lun theo k s nghim ca phng trỡnh: 2 | x 2x 3 | 2k 1.+ = c) Vit phng trỡnh ng thng i qua nh ca (P 2 ) v giao im ca (P 2 ) vi trc tung. d) Xỏc nh m (P m ) l parabol. Tỡm to qu tớch nh ca parabol (P m ) khi m thay i. e) Chng minh rng (P m ) luụn i qua mt im c nh, tỡm to im c nh ú. ================ 1 kim tra ẹaùi soỏ 10 Tran Sú Tuứng CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 2 a) Trc nghim khỏch quan ( 3 ) : Cõu 1 : Tp xỏc nh ca hm s 1 y f(x) x 1 3 x = = + l: a) (1;3) , b) [1;3] , c) (1;3] , c) [1;3) Cõu 2: nh ca Parabol y = x 2 2x +2 l : a) I(1;1) b) I(1;1) c) I(1;1) c) I(1;2) Cõu 3 : Hm s y = 2x 2 4x + 1 a) ng bin trờn khong ( ; 1 ) b) ng bin trờn khong ( 1 ;+ ) c) Nghch bin trờn khong ( 1 ;+ ) d) ng bin trờn khong ( 4 ;2 ) b) T lun : ( 7 ) Cõu 5 ( 2 ) : Tỡm min xỏc nh v xột tớnh chn l ca hm s sau : 2 y x 1 x 1 = + + Cõu 6 ( 1,5 ): Xột s bin thiờn ca hm s : 3 y 2 x = trờn ( 2 ; + ) Cõu 7 : a) (1,5 ) Tỡm Parabol y = ax 2 + bx + 2 bit rng Parabol ú i qua im A(3 ; 4) v cú trc i xng 3 x 2 = . b) ( 2 ) Kho sỏt v v th hm s va tỡm c cõu a). ================= 2 Tran Sú Tuứng kim tra ẹaùi soỏ 10 CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 3 I. Phn trc nghim : ( 3 im ) Cõu 1: Hm s 2 4 x 1 y f(x) x . 1 x + = = cú tp xỏc nh l : a) ( ] ;1 b) ( ) ;1 c) ( ] { } ;1 \ 0 d) ( ) { } ;1 \ 0 Cõu 2: Hm s no l hm s chn : a) 2 y 4x 2x= + b) y x 1 x 1= + c) ( ) 2 y x 1= d) y x 2 x 2= + + Cõu 3 : im ng qui ca 3 ng thng y 3 x; y = x+1; y = 2= l : a) ( 1; 2) b) ( 1; 2) c) (1; 2) d) (1; 2) Cõu 4 : th ca hm s no i qua im A ( 1; 3 ) v ct trc honh ti im cú x = 4 : a) 3 12 y x 5 5 = + b) 3 12 y x 5 5 = + c) 3 12 y x 5 5 = d) 3 12 y x 5 5 = Cõu 5 : Cho parabol ( P ) : 2 y x mx 2m= + .Giỏ tr ca m tung ca nh ( P ) bng 4 l : a) 3 b) 4 c) 5 d) 6 Cõu 6 : Hm s 2 y f(x) x 2x 5= = + : a) Gim trờn ( ) ; 1 b) Tng trờn ( ) 2;+ c) Gim trờn ( ) ;2 d) Tng trờn ( ) 1;+ II. Phn t lun : ( 7 im ) Bi 1 : ( 3 im ) a) V ba th ca ba hm s sau trờn cựng mt h trc ta Oxy : 1 (d ) : y 2x 2= + 2 (d ) : y x 2= + 3 (d ) : y x= b) Gi A,B,C l giao im cỏc th hm s ó cho. Chng t ABC vuụng. c) Vit ph.trỡnh .thng song song vi 1 (d ) v i qua giao im ca 2 3 (d ),(d ) Bi 2 : ( 2 im ) Lp bng bin thiờn v v th ca cỏc hm s sau : a) 2 x y 2 = b) 2 y 2x 4x 2= + Bi 3 : ( 2 im ) Xỏc nh a, b, c bit parabol 2 y ax bx c= + + a) i qua im A (8; 0) v cú nh I (6, 12 ) b) i qua A( 0 ; 1) , B(1 ; 1) , C (1 ; 1 ) . ================== 3 kim tra ẹaùi soỏ 10 Tran Sú Tuứng CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 4 I. PHN TRC NGHIM (3 im) Cõu 1 : Tp xỏc nh ca hm s y = x 5 4 2x+ l: a) D = ( ; 5] [2 ; ) + b) D = [5 ; 2] c) D = d) D = R Cõu 2 : Cho hm s f (x) = 2 16 x x 2 + . Kt qu no sau õy ỳng: a) f(0) = 2 ; f(1) = 15 3 b) f(1) = 15 ; f(0) = 8 c) f(3) = 0 ; f(1) = 8 d) f(2) = 14 4 ; f(3) = 7 Cõu 3 : Trong cỏc parabol sau õy, parabol no i qua gc ta : a) y = 3x 2 4x + 3 b) y = 2x 2 5x c) y = x 2 + 1 d) y = x 2 + 2x + 3 Cõu 4 : Hm s y = x 2 + 4x 3 a) ng bin trờn ( ; 2) b) ng bin trờn (2 ; )+ c) Nghch bin trờn ( ; 2) d) Nghch bin trờn (0 ; 3) Cõu 5 : Parabol y = 3x 2 2x + 1 cú trc i xng l: a) x = 1 3 b) x = 2 3 c) x = 1 3 d) y = 1 3 Cõu 6 : Ta giao im ca .thng y = x + 3 v parabol y = x 2 4x + 1 l: a) 1 ;1 3 ữ b) (0 ; 3) c) (1 ; 4) v (2 ; 5) d) (0 ; 1) v (2 ; 2) II. PHN T LUN (7 im) Bi 1: Vit phng trỡnh ng thng qua A(2 ; 3) v song song vi ng thng y = x + 1 Bi 2: Tỡm parabol y = ax 2 + bx + 1, bit parabol ú: a) i qua 2 im M(1 ; 5) v N(2 ; 1) b) i qua A(1 ; 3) v cú trc i xng x = 5 2 c) cú nh I(2 ; 3) d) i qua B(1 ; 6), nh cú tung l 3. =================== 4 Tran Sú Tuứng kim tra ẹaùi soỏ 10 CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 5 I. Phn trc nghim : Cõu 1 (0,5 im): Tp xỏc nh ca hm s 2 x 1 y x 1 + = l : a) R b) R\ {1; 1} c) R\ {1} c) (1; 1) Cõu 2 (0,5 im): Hm s y= ( 2 +m )x + 3m ng bin khi : a) m =2 b) m ? 2 c) m > 2 c) m < 2 Cõu 3 (0,5 im): Hm s y = f(x) = x ( x4 +3x2 + 5) l : a) Hm s chn b) Hm s l c) Hm s khụng chn, khụng l c) C 3 kt lun trờn u sai Cõu 4 (0,5 im): Cho hm s 2x 1 ;x 1 y x 7 ;x 1 2 + = + > Bit f(x 0 ) = 5. thỡ x 0 khụng õm tng ng l: a) 2 b) 0 c) 1 c) 3 Cõu 5 (0,5 im): nh ca parabol y = ax 2 + bx + c l a) b ; a 4a ữ b) b ; a 4a ữ c) b ; 2a 4a ữ c) b ; a 4a ữ Cõu 6 (0,5 im): th ca hm s y = 3x 2 +2 c suy ra t th ca hm s y = 3x 2 nh phộp tnh tin song song vi trc Oy a) lờn trờn 3 n v b) lờn trờn 2 n v c) xung di 3 n v c) xung di 2 n v II : T LUN Cõu 1 (2 im): Tỡm tp xỏc nh cỏc hm s sau : a) 2 x 1 y x 5x 6 = + + b) 1 y 2 3x x 1 = + + Cõu 2 (3 im): Lp bng bin thiờn v v th hm s y = x 2 + x + 2 Cõu 3 (2 im): Xỏc nh hm s bc hai bit th ca nú l mt parabol cú tung nh l 13 4 , trc i xng l ng thng x = 3 2 , i qua im M (1 ; 3) ============= 5 kim tra ẹaùi soỏ 10 Tran Sú Tuứng CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 6 Phn 1:Trc nghim khỏch quan (3 im) Cõu 1: Tp xỏc nh ca hm s 2 x 2 y x 4x 3 = + l: a) { } D \ 1; 2; 3= Ă b) { } D \ 1; 3= Ă c) { } D \ 2= Ă d) ] [ D ( ; 1 3; )= + Cõu 2: Hm s y = x 2 4x + 1 a) ng bin trờn khong (; 0) v nghch bin trờn khong (0; + ). b) Nghch bin trờn khong (; 0) v ng bin trờn khong (0; + ). c) ng bin trờn khong (; 2) v nghch bin trờn khong (2; + ). c) Nghch bin trờn khong (; 2) v ng bin trờn khong (2; + ). Cõu 3: Tp xỏc nh v tớnh chn, l ca hm s 2 2 x y x 1 = l: a) D = Ă ; hm s chn. b) { } D \ 1= Ă ; hm s chn. c) { } D \ 1= Ă ; hm s chn. c) { } D \ 1= Ă ; hm s khụng chn, khụng l. Cõu 4: Cho hm s f(x) = 3x cú tp xỏc nh l tp Q . Tỡm x f(x) = 1. a) x = 1 b) x = 3 c) x = 1 3 c) Tt c u sai. Cõu 5: Giao im ca th hai hm s y = x + 3 v y = x 2 4x + 1 l: a) (4; 1) v (5; 2) b) (1; 4) v (2; 5) c) (1; 4) v (2; 5) c) (4; 1) v (5; 2) Cõu 6: Ph.trỡnh .thng i qua A(0; 2) v song song vi ng thng y = x l: a) y = x + 2 b) y = 2x c) y = 1 x 2 c) y = 2x + 2 Phn II: T lun (7 im) Cõu 7: (2 im) Tỡm tp xỏc nh ca cỏc hm s sau: a) 1 y x 4 2 x = + + b) 2 y (x 2) x 1 = + + Cõu 8: (1 im) Xột tớnh chn, l ca hm s f(x) = 3x.x Cõu 9: (2 im) Lp bng bin thiờn v v th hm s y = x 2 + 2x + 3 Cõu 10:(2 im) Xỏc nh hm s y = ax 2 + bx + c (a 0), bit th hm s i qua cỏc im: A(0; 3); B(1; 4); C(1; 6). ================ 6 Tran Sú Tuứng kim tra ẹaùi soỏ 10 CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 7 I/ Phn trc nghim (4 im) Bi 1: Hm s y= x x 1 l: a) hm s chn b) hm s l c) hm s khụng chn, khụng l d) hm s va chn, va l Bi 2: Hm s y= x 2 +2x +1 ng bin trong khong : a) ( ;1) b. ( ;1) c. (1;+ ) d. 1 kt qu khỏc Bi 3: Tp xỏc nh ca hm s y= 6 3x+ l : a) ( ;2) b. ( ;2) c. (2;+ ) d. [2;+ ) Bi 4 : th hm s :y= x 2 +2x+3 cú nh l : a) I(1;4) b. I(1;3) c. (1;4) d. 1 kt qu khỏc II/ Phn t lun (6im) Bi1: Tỡm tp xỏc nh ca hm s : y= 1 2 x 2x 1 + Bi 2: Xột tớnh chn l ca hm s : y= x 2 2x +3 Bi 3: Xột s bin thiờn v v th ca hm s : y=x 2 +4x+3 ================= 7 kim tra ẹaùi soỏ 10 Tran Sú Tuứng CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 8 I/ Phn trc nghim (4 im) Bi 1: Hm s y= 2 2 x x 1+ l: a) hm s chn b) hm s l c) hm s khụng chn khụng l d) hm s va chn, va l Bi 2: hm s y= x 2 2x +1 ng bin trong khong : a) ( ;1) b. ( ;1) c. (1;+ ) d. 1 kt qu khỏc Bi 3: Tp xỏc nh ca hm s y= x 2 x 3x 4 + l : a) R b. R\ } { 1,4 c. R\ } { 2 d. 1 kt qu khỏc Bi 4 : th hm s :y= x 2 6x+1 cú nh l : a) I(3;4) b. I(3;8) c. (3;8) d. 1 kt qu khỏc II/ Phn t lun (6im) Bi1: Tỡm tp xỏc nh ca hm s : y= 2x 4 Bi 2: Xột tớnh chn l ca hm s : y= 2x 7 Bi 3: Xột s bin thiờn v v th ca hm s : y=x 2 2x+1 ================= 8 Tran Sú Tuứng kim tra ẹaùi soỏ 10 CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 9 I/ Phn trc nghim (4 im) Bi 1: Hm s y= x 3 +x+1 l: a) hm s chn b) hm s l c) hm s khụng chn khụng l d) hm s va chn, va l Bi 2: Hm s y= x 2 +2x +1 nghch bin trong khong : a) ( ;1) b. ( ;1) c. (1;+ ) d. 1 kt qu khỏc Bi 3: Tp xỏc nh ca hm s y= 2 x 1 x 4x 3 + + l : a)R b. R\ } { 1,3 c. R\ } { 2 d. 1 kt qu khỏc Bi 4 : th hm s y= x 2 +4x+1 cú nh l : a) I(2;4) b. I(2;8) c. (2;3) d. 1 kt qu khỏc II/ Phn t lun (6im) Bi1: Tỡm tp xỏc nh ca hm s : y= 3x 4+ bi 2: Xột tớnh chn l ca hm s : y= x 2 + 2x +4 Bi 3 : Xột s bin thiờn v v th ca hm s : y=x 2 2x+2 =================== 9 kim tra ẹaùi soỏ 10 Tran Sú Tuứng CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 10 Phn I: T lun (7 im) Cõu 1 (2 im): Vit phng trỡnh dng y = ax + b ca cỏc ng thng: a) i qua hai im A(2;1) v B(5;2). b) i qua im C(2;3) v song song vi ng thng y = 1 2 x Cõu 2 (3 im): Cho hm s y = 3x 2 2x + 1 a) Lp bng bin thiờn v v th c) ca hm s. b) Tỡm ta giao im ca th c) v ng thng (d): y = 3x 1. Cõu 3 (2 im): Xột tớnh chn, l ca cỏc hm s sau: a) y = 3x + 5 b) y = 2x 2 + 1 c) y = 1 x d) y = x Phn II: Trc nghim khỏch quan (3 im) Cõu 1 (0,5 im): Chn mnh ỳng trong cỏc mnh sau: a) th ca hm s chn nhn trc honh lm trc i xng. b) th ca hm s l nhn trc tung lm trc i xng. c) th ca hm s chn nhn trc tung lm trc i xng. c) th ca hm s l nhn trc honh lm trc i xng. Cõu 2 (0,5 im): Cho hm s y = 2 x 1 (x 2) x 2 (x 2) + < Giỏ tr ca hm s ó cho ti x = 1 l: a) 3 b) 2 c) 1 c) 0 Cõu 3 (0,5 im): Giao im ca parabol (P): y = 3x 2 + x + 3 v ng thng (d): y = 3x 2 cú ta l: a) (1;1) ; ( 5 3 ;7) b) (1;1); ( 5 3 ;7) c) (1;1) ; ( 5 3 ;7) c)(1;1) ; ( 5 3 ;7) Cõu 4 (0,5 im): Hm s y = x 2 + 2x + 1 : a) ng bin trờn khong ( ;1). b) Nghch bin trờn khong ( ;2). c) ng bin trờn khong (2;+ ). c) Nghch bin trờn khong (1;+ ). Cõu 5 (0,5 im): Parabol (P): y = x 2 4x + 3 cú nh l: a) I(2;1) b) I(2;1) c) I(2;1) c) I(2;1) Cõu 6 (0,5 im): Tp xỏc nh ca hm s y = 1 2x 3 1 2x + l: a) 1 3 ; 2 2 b) 3 ; 2 + ữ c) c) 1 ; 2 ữ . ================ 10 [...]... R\ { 3} 4+x 1 1 D I( ;1) 3 2 4 Cõu 4 .thng no sau õy l trc i xng ca th hm s y = 2x x + 3 5 1 1 A x = B y = C x = 5 D y = 5 5 5 Cõu 5 ng thng song song vi ng thng y = 3 x l : 3 3 A y = x + 5 B y = x C 3 x + y = 0 D x = 3 y 3 3 Cõu 6 Ta giao im ca hai .thng d1: y = 3x + 5 v d2: 2x + 3y 1 = 0 l: 14 13 13 14 14 13 13 14 A ( ; ) B ( ; ) C ( ; ) D ( ; ) 11 11 11 11 11 11 11 11 Cõu 7 Hm s y = x2 6x + 5... im : 2 2 1 1 , , (a) ( ) (b) ( ) m 1 m 1 1 m 1 m 2 2 1 2 , , (C ) ( ) (d) ( ) 1 m 1 m 1 m 1 m PHN 2 :T LUN ( 7 im ) Cõu1 (1 ) Cho hm s y = x2 + bx + c Tớnh b v c bit rng hm s t giỏ tr nh nht bng 1 khi x = 1 Cõu2 (1, 5 ) V th , lp bng bin thiờn v xột tớnh chn l ca hm s sau õy : y = x ( x 2) Cõu3 (2 ) Cho hm s y = x2 mx + m 2 cú th l parabol (Pm) a) Xỏc nh giỏ tr ca m sao cho (Pm) i qua im A (2 ;1) ... x +1 (x > 2) ta sau õy, cú bao nhiờu im thuc th ca hm s f ? M (0 ;1) , N( 2; 3), E (1; 2) , F( 3;8) , K( 3;8 ) (a) 1 (b) 2 (c) 3 (d) ỏp s khỏc x2 + 1 (x 2) Cõu 5 ( 0,5 im) Cho hm s f(x) = 2 Hi cú bao nhiờu x 8x + 17 (x > 2) im thuc th ca hm s f cú tung bng 2 ? (a) 2 (b) 3 (c) 1 (d) 4 kim tra ẹaùi soỏ 10 12 Tran Sú Tuứng Cõu 6 Ta nh ca parabol (P) : y = (m 2 1) x2 + 2( m + 1 )x + 1 (m 1) ... + 2 + 3 x + m = 0 cú 2 nghim phõn bit khi : A m > 5 B m < 5 C m 5 D m > 5 mx + 1 3 Cõu 9 Phng trỡnh = 2 cú nghim x = khi : x 1 m 2 m 1 m 1 A m 1 B m 2 C D m2 m2 4 2 Cõu 10 Phng trỡnh 2x 7x + 6 = 0 cú s nghim l : A 2 B 1 C 3 D 4 kim tra ẹaùi soỏ 10 16 Tran Sú Tuứng B / Phn t lun : (6 im) Cõu 1 (3 im) Cho phng trỡnh (m 1) x 2 + 2x m +1 = 0 a) Chng minh rng vi m 1 , phng trỡnh luụn cú 2. .. + 2 + 3 x + 1 = m cú 2 nghim phõn bit khi a) m > 6 b) m< 6 c) m 6 d) m 5 Cõu 8 Ba ng thng d1 y= ( m 2 ) x + 2m + 3, d2 : y = 2x +1 d3 : y = 3x + 6 ng qui vi giỏ tr ca m l: a) 2 b) 3 c) 2 d) 1 Cõu 9 Parabol ( P) y= ax2 +bx + 2 i qua 2 im A ( 1 ; 5 ) , B ( 2 ; 8 ) vi a) a= 2 v b = 1 b) a= 2 v b = 8 c) a = 1 v b = 3 d) a = 2 v b= 0 mx + 1 3 Cõu 10 Phng trỡnh = 2 cú nghim x = khi x 1 m 2 a) m 1. .. x4 + (m1)x3 +mx2 1 ================ Tran Sú Tuứng 15 kim tra ẹaùi soỏ 10 CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 15 A / Phn trc nghim : (4im) Cõu 1 Tp xỏc nh ca hm s y = x + 3 l : A D = ( ; 3] B D = (3;+ ) C D = R Cõu 2 Trong cỏc hm s sau õy , hm s no l hm s l : A y = 1 x + 1 + x B y = 4 x x 1 C y = 2 D y = 3x2 + 2 x x x Cõu 3 Parabol y = 3x2 2x + 1 cú nh l : 1 1 2 1 2 A I( ;2) B I(... l hm s l : a y = 1 x + 1 + x b y = 4 x 4 + x x 1 c y = 2 d y = 3x2 + 2 x 1 x x Cõu 3 Parabol (P) y= x2 4x +5 cú ta nh l : a ( 2 ; 1) b ( 2 ; 1 ) c ( 2 ; 1 ) d ( 2 ; 1 ) Cõu 4 ng thng song song vi ng thng 3 x + y = 2 l a) y = 3 x + 1 b) x + 3 y + 2 = 0 c) x 3 y = 2 d) 3 x + 2y = 0 Cõu 5 Parabol (P) y= x2 4x +5 cú trc i xng l ng thng a) x= 2 b) y = 2 c) y= 2 d) x= 2 2 Cõu 6 Hm s y= x 6x + 5... +mx ============= kim tra ẹaùi soỏ 10 Tran Sú Tuứng 14 CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 14 BI 1: Tỡm TX ca cỏc hm s sau: 2 a) y = 2x 4 2 x 3x + 2 b) y = x +2 x 2 + 3x 1 BI 2: Xột tớnh chnl ca hm s: a) y = f(x) = x2 + x4 + 5 b) y = f(x) = x3 + BI 3: Xột tớnh bin thiờn ca hm s: a) y = x2 2x + 3 trong ( ; 1) b) y = 5x 1 x 3 trong (2 ; +) x2 BI 4: Cho hm s y=x2 2x + 3 (P) a) Kho sỏt... 1 = 3x + m 1 (m l Cõu 2 (2im) Gii v bin lun phng trỡnh tham s) x Cõu 3 (1im) Tỡm m hm s y = x m + 1 xỏc nh trờn 2m + 1 x [ 2; 3] ======================== Tran Sú Tuứng 17 kim tra ẹaùi soỏ 10 CHNG II: HM S BC NHT BC HAI KIM TRA 1 TIT S 16 TRC NGHIM : (4 im) Cõu 1 Tp xỏc nh ca hm s y= x + 1 l : a) (1; + ) b) ( ; 1 ] c) R d) [ 1 ; 1 ] Cõu 2 Trong cỏc hm s sau õy , hm s no l hm s l : a y = 1. .. sau: 2x a) y = x + 1 2 x 3x + 2 BI 2: Xột tớnh chnl ca hm s: a) y = f(x) = x4 + x2 + 1 b) y = b) y = f(x) x3 BI 3: Xột tớnh bin thiờn ca hm s: a) y = x2 2x + 1 trong (1 ; +) 5x b) y = x2 x x +2 1 x 3 trong ( ; 2) x2 BI 4: Cho hm s y=x2 2x + 1 (P) a) Kho sỏt v v th ca hm s trờn b) Tỡm giao im ca (P) v ng thng (d): y=x +1 (Bng pp i s v bng th) BI 5: Tỡm m hm s sau l hm s l: y = f(x) = x3 + (m1)x2 . .thng d 1 : y = 3x + 5 v d 2 : 2x + 3y 1 = 0 l: A. ( 14 11 ; 13 11 ) B. ( 13 11 ; 14 11 ) C. ( 14 11 ; 13 11 ) D. ( 13 11 ; 14 11 ) Cõu 7. Hm s y = x 2 6x. (a). ( 2 2 , m 1 m 1 ) (b). ( 1 1 , 1 m 1 m ) (C ). ( 2 2 , 1 m 1 m ) (d). ( 1 2 , 1 m 1 m ) PHN 2 :T LUN ( 7 im ) Cõu1. (1 ) Cho hm s y = x 2 + bx