40 Bài Hàm Số Chọn Lọc 2013

1. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 1 Bi 1. Cho hm s y = 2 5 3 2 2 4 x x 1. Kho st s bin thin v v thi (C) ca hm s. 2. Cho im M thuc (C) c honh xM = a. Vit phng trnh tip tuyn ca (C) ti M, vi gi tr no ca a th tip tuyn ca (C) ti M ct (C) ti hai im phn bit khc M. Gii. 2/ + V 2 5 3 2 ;)( 2 4 a a aMCM . Ta c: y = 2x3 6x aaay 62)(' 3 Vy tip tuyn ca (C) ti M c phng trnh : 2 5 3 2 ))(63( 2 4 3 a a axaay . + Xt pt : 0)632()( 2 5 3 2 ))(63( 2 5 3 2 2222 4 32 4 aaxxaxa a axaax x 0632)( 22 aaxxxg ax YCBT khi pt g(x) = 0 c 2 nghim phn bit khc a 1 3|| 1 03 0)( 0' 2 2 a a a a ag Bi 2. Cho hm s 1 x x y (C). 1. Kho st s bin thin v v thi (C) ca hm s. 2. Vit phng trnh tip tuyn vi th (C), bit rng khong cch t tm i xng ca th (C) n tip tuyn l ln nht. Gii. 2/ Gi s )() 1 ;( 0 0 0 C x x xM m tip tuyn vi th ti c khong cch t tm i xng n tip tuyn l ln nht. Phng trnh tip tuyn ti M c dng : 0 02 0 0 1 ( ) ( 1) 1 x y x x x x 2 0 2 2 0 0 1 0 ( 1) ( 1) x x y x x Ta c d(I ;tt) = 4 0 0 )1( 1 1 1 2 x x .t t = 1 1 0 x > 0 Xt hm s f(t) 4 2 ( 0) 1 t t t ta c f(t) = 2 4 4 (1 )(1 )(1 ) (1 ) 1 t t t t t t 0 1 f(t) = 0 khi t = 1 f(t) + 0 - Bng bin thin t bng bin thin ta c f(t) 2 d(I ;tt) ln nht khi v ch khi t = 1 hay ketnoitrithuc2013.blogspot.com - chia s kin thc n thi H 2. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 2 0 0 0 2 1 1 0 x x x + Vi x0 = 0 ta c tip tuyn l y = -x + Vi x0 = 2 ta c tip tuyn l y = -x+4 Bi 3. Cho hm s 2 4 1 x y x . 1. Kho st s bin thin v v th (C) ca hm s. 2. Tm trn th (C) hai im i xng nhau qua ng thng MN bit M(-3; 0) v N(-1; -1). Gii. 2. Gi 2 im cn tm l A, B c 6 6 ;2 ; ;2 ; , 1 1 1 A a B b a b a b Trung im I ca AB: I 2 2 ; 2 1 1 a b a b a b Pt ng thng MN: x + 2y +3= 0 C : . 0AB MN I MN => 0 (0; 4) 2 (2;0) a A b B Bi 4. Cho hm s 34 24 xxy . 1. Kho st s bin thin v v th )(C ca hm s cho. 2. Bin lun theo tham s k s nghim ca phng trnh k xx 334 24 . Gii. 2. th hm s 34 24 xxy gm phn nm pha trn Ox v i xng ca phn nm pha di Ox qua Ox ca th (C); k y 3 l ng thng song song vi Ox. T ta c kt qu: * 013 kk : phng trnh c 8 nghim, * 013 kk : phng trnh c 6 nghim, * 10331 kk : phng trnh c 4 nghim, * 133 kk : phng trnh c 3 nghim, * 133 kk : phng trnh c 2 nghim. Bi 5. Cho hm s 1 12 x x y 1. Kho st s bin thin v v th (C) ca hm s . 2. Tm ta im M sao cho khong cch t im )2;1(I ti tip tuyn ca (C) ti M l ln nht . Gii. 2. Nu )( 1 3 2; 0 0 C x xM th tip tuyn ti M c ph-ng trnh )( )1( 3 1 3 2 02 00 xx xx y hay 0)1(3)2()1()(3 0 2 00 xyxxx . Khong cch t )2;1(I ti tip tuyn l 2 02 0 4 0 0 4 0 00 )1( )1( 9 6 )1(9 16 19 )1(3)1(3 x x x x x xx d . Theo bt ng thc Csi 692)1( )1( 9 2 02 0 x x , vy 6d . Khong cch d ln nht bng 6 khi 3131)1( )1( 9 0 2 0 2 02 0 xxx x . x y O 3. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 3 Vy c hai im M : 32;31 M hoc 32;31 M Bi 6. Cho hm s 1x 2x y (C) 1. Kho st s bin thin v v th hm s (C). 2. Cho im A(0;a) .Xc nh a t A k -c hai tip tuyn ti (C) sao cho hai tip im t-ng ng nm v hai pha trc ox. Gii. 2. Ph-ng trnh tip tuyn qua A(0;a) c dng y=kx+a (1) iu kin c hai tip tuyn qua A: )3(k )1x( 3 )2(akx 1x 2x 2 c nghim 1x Thay (3) vo (2) v rt gn ta -c: )4(02ax)2a(2x)1a( 2 (4) c 2 nghim 1x l: 2a 1a 06a3' 03)1(f 1a Honh tip im 21 x;x l nghim ca (4) Tung tip im l 1x 2x y 1 1 1 , 1x 2x y 2 2 2 hai tip im nm v hai pha ca trc ox l: 0 )2x)(1x( )2x)(2x( 0y.y 21 21 21 3 2 a0 3 6a9 0 1)xx(xx 4)xx(2xx 2121 2121 Vy 1a 3 2 tho mn kin bi ton. Bi 7. Cho hm s 1 . 1 x y x 1.Kho st s bin thin v v th C ca hm s. 2.Bin lun theo m s nghim ca phng trnh 1 . 1 x m x Gii. 2. Hc sinh lp lun suy t th (C) sang th 1 ' 1 x y C x .Hc sinh t v hnh Suy ra p s 1; 1:m m phng trnh c 2 nghim 1:m phng trnh c 1 nghim 1 1:m phng trnh v nghim Bi 8. Cho hm s 2x 3 y x 2 c th (C). 1.Kho st s bin thin v v th ca hm s (C) 2.Tm trn (C) nhng im M sao cho tip tuyn ti M ca (C) ct hai tim cn ca (C) ti A, B sao cho AB ngn nht . Gii. 4. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 4 Vy im M cn tm c ta l : (2; 2) Bi 9. Cho hm s y = x3 3x2 +2 (1) 1. Kho st s bin thin v v th ca hm s (1). 2. Tm im M thuc ng thng y=3x-2 sao tng khong cch t M ti hai im cc tr nh nht. Gii. 2. Gi ta im cc i l A(0;2), im cc tiu B(2;-2) Xt biu thc P=3x-y-2 Thay ta im A(0;2)=>P=-4P=6>0 Vy 2 im cc i v cc tiu nm v hai pha ca ng thng y=3x-2, MA+MB nh nht => 3 im A, M, B thng hng Phng trnh ng thng AB: y= - 2x+2 Ta im M l nghim ca h: 4 3 2 5 2 2 2 5 x y x y x y => 4 2 ; 5 5 M Bi 10. Cho hm s 2 x xm y c th l )( mH , vi m l tham s thc. 1. Kho st s bin thin v v th ca hm s cho khi 1m . 2. Tm m ng thng 0122: yxd ct )( mH ti hai im cng vi gc ta to thnh mt tam gic c din tch l . 8 3 S Gii. 2. Honh giao im A, B ca d v )( mH l cc nghim ca phng trnh 2 1 2 x x mx 2,0)1(22 2 xmxx (1) Pt (1) c 2 nghim 21, xx phn bit khc 2 2 16 17 0)1(22)2.(2 01617 2 m m m m . Ta c .1617. 2 2 4)(.2)(.2)()( 21 2 12 2 12 2 12 2 12 mxxxxxxyyxxAB Khong cch t gc ta O n d l . 22 1 h 2. Ly im 1 M m;2 m 2 C . Ta c : 2 1 y' m m 2 . Tip tuyn (d) ti M c phng trnh : 2 1 1 y x m 2 m 2m 2 Giao im ca (d) vi tim cn ng l : 2 A 2;2 m 2 Giao im ca (d) vi tim cn ngang l : B(2m 2 ; 2) Ta c : 22 2 1 AB 4 m 2 8 m 2 . Du = xy ra khi m = 2 5. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 5 Suy ra , 2 1 8 3 1617. 2 2 . 22 1 . 2 1 .. 2 1 mmABhS OAB tha mn. Bi 11. Cho hm s 3 5 )23()1( 3 2 23 xmxmxy c th ),( mC m l tham s. 1. Kho st s bin thin v v th ca hm s cho khi .2m 2. Tm m trn )( mC c hai im phn bit );(),;( 222111 yxMyxM tha mn 0. 21 xx v tip tuyn ca )( mC ti mi im vung gc vi ng thng .013: yxd Gii. 2. Ta c h s gc ca 013: yxd l 3 1 dk . Do 21, xx l cc nghim ca phng trnh 3' y , hay 323)1(22 2 mxmx 013)1(22 2 mxmx (1) Yu cu bi ton phng trnh (1) c hai nghim 21, xx tha mn 0. 21 xx . 3 1 1 3 0 2 13 0)13(2)1(' 2 m m m mm Vy kt qu ca bi ton l 3m v . 3 1 1 m Bi 12. Cho hm s . 2 3 42 24 xxy 1. Kho st s bin thin v v th ca hm s cho. 2. Tm m phng trnh sau c ng 8 nghim thc phn bit . 2 1 | 2 3 42| 224 mmxx Gii. 2. Phng trnh 2 1 | 2 3 42| 224 mmxx c 8 nghim phn bit ng thng 2 12 mmy ct th hm s | 2 3 42| 24 xxy ti 8 im phn bit. th | 2 3 42| 24 xxy gm phn (C) pha trn trc Ox v i xng phn (C) pha di trc Ox qua Ox. T th suy ra yu cu bi ton 2 1 2 1 0 2 mm .1002 mmm Bi 13. Cho hm s mxxmxy 9)1(3 23 , vi m l tham s thc. 1. Kho st s bin thin v v th ca hm s cho ng vi 1m . 2. Xc nh m hm s cho t cc tr ti 21, xx sao cho 221 xx . Gii. 2. Ta c .9)1(63' 2 xmxy +) Hm s t cc i, cc tiu ti 21, xx ph-ng trnh 0'y c hai nghim pb l 21, xx Pt 03)1(22 xmx c hai nghim phn bit l 21, xx . O y x 6. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 6 31 31 03)1(' 2 m m m )1( +) Theo nh l Viet ta c .3);1(2 2121 xxmxx Khi 41214442 2 21 2 2121 mxxxxxx )2(134)1( 2 mm T (1) v (2) suy ra gi tr ca m l 313 m v .131 m Bi 14. Cho hm s 2)2()21( 23 mxmxmxy (1) m l tham s. 1. Kho st s bin thin v v th (C) ca hm s (1) vi m=2. 2. Tm tham s m th ca hm s (1) c tip tuyn to vi ng thng d: 07 yx gc , bit 26 1 cos . Gii. 2. Gi k l h s gc ca tip tuyn tip tuyn c vct php )1;(1 kn d: c vct php )1;1(2 n Ta c 3 2 2 3 0122612 12 1 26 1. cos 2 1 2 2 21 21 k k kk k k nn nn Yu cu ca bi ton tha mn t nht mt trong hai phng trnh: 1 / ky (1) v 2 / ky (2) c nghim x 3 2 2)21(23 2 3 2)21(23 2 2 mxmx mxmx 0 0 2 / 1 / 034 0128 2 2 mm mm 1; 4 3 2 1 ; 4 1 mm mm 4 1 m hoc 2 1 m Bi 15. Cho hm s y = 2 2 x x (C) 1. Kho st s bin thin v v th hm s (C). 2. Tm m ng thng (d ): y = x + m ct th (C) ti 2 im phn bit thuc 2 nhnh khc nhau ca th sao cho khong cch gia 2 im l nh nht. Tm gi tr nh nht . Gii. 2. (d) ct (C) ti 2 im phn bit th pt 2 2 x x m x hay x2 + (m - 4)x -2x = 0 (1) c 2 nghim phn bit khc 2. Phng trnh (1) c 2 nghim phn bit khc 2 khi v ch khi 2 16 4 0 m m (2). Gi s A(x1;y1), B(x2;y2) l 2 giao im khi x1, x2 l 2 nghim phng trnh (1). Theo nh l viet ta c 1 2 1 2 4 (3) 2 x x m x x m , y1=x1+m, y2=x2+m A, B thuc 2 nhnh khc nhau ca th th A, B nm khc pha i vi t x 2 = 0. A, B nm khc pha i vi t x 2 = 0 khi v ch khi (x1- 2)(x2 - 2) < 0 hay c nghim c nghim 7. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 7 x1x2 2(x1 + x2) +4 < 0 (4) thay (3) vo 4 ta c 4 < 0 lun ng (5) mt khc ta li c AB = 2 2 2 1 2 1 2 1 2 1 2( ) ( ) 2( ) 8x x y y x x x x (6) thay (3) vo (6) ta c AB = 2 2 32 32m vy AB = 32 nh nht khi m = 0 (7). T (1), (5), (7) ta c m = 0 tho mn . Bi 16. 1. Kho st s bin thin v v th (C) ca hm s 2 1 1 x y x 2. Vit phng trnh tip tuyn ca (C), bit khong cch t im I(1;2) n tip tuyn bng 2 . Gii. 2. Tip tuyn ca (C) ti im 0 0( ; ( )) ( )M x f x C c phng trnh 0 0 0'( )( ) ( )y f x x x f x Hay 2 2 0 0 0( 1) 2 2 1 0x x y x x (*) *Khong cch t im I(1;2) n tip tuyn (*) bng 2 0 4 0 2 2 2 1 ( 1) x x gii c nghim 0 0x v 0 2x *Cc tip tuyn cn tm : 1 0x y v 5 0x y Bi 17. Cho hm s y = - x3 + 3mx2 -3m 1. 1. Kho st s bin thin v v th ca hm s khi m = 1. 2. Tm cc gi tr ca m hm s c cc i, cc tiu. Vi gi tr no ca m th th hm s c im cc i, im cc tiu i xng vi nhau qua ng thng d: x + 8y 74 = 0. Gii. 2. Ta c y = - 3x2 + 6mx ; y = 0 x = 0 v x = 2m. Hm s c cc i , cc tiu phng trnh y = 0 c hai nghim phn bit m 0. Hai im cc tr l A(0; - 3m - 1) ; B(2m; 4m3 3m 1) Trung im I ca on thng AB l I(m ; 2m3 3m 1) Vect 3 (2 ;4 )AB m m ; Mt vect ch phng ca ng thng d l (8; 1)u . Hai im cc i , cc tiu A v B i xng vi nhau qua ng thng d I d AB d 3 8(2 3 1) 74 0 . 0 m m m AB u m = 2 Bi 18. Cho hm s 133 xxy (1) 1. Kho st s bin thin v v th (C) ca hm s (1). 2. nh m phng trnh sau c 4 nghim thc phn bit: mmxx 33 33 Gii. 2. Phng trnh cho l phng trnh honh giao im gia th (C) ca hm s: 13 3 xxy v ng thng (d): 133 mmy ((d) cng phng vi trc honh) Xt hm s: 13 3 xxy , ta c: + Hm s l mt hm chn nn (C) nhn trc Oy lm trc i xng, ng thi 0x th 3 33 1 3 1y x x x x x y 0 1 2 1 2 1 1 3 (d) 8. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 8 + Da vo th (C) ta suy ra iu kin ca m phng trnh cho c 4 nghim phn bit l: 3 3 3 2 3 3 0 1 3 1 1 0 3 3 2 0 1 m m m m m m m m m Bi 19. Cho hm s 3 1 x y x c th l (C) 1) Kho st s bin thin v v th ca hm s. 2) Vit ph-ng trnh tip tuyn ca th hm s, bit tip tuyn ct trc honh ti A, ct trc tung ti B sao cho OA = 4OB Gii. 2. OA =4OB nn OAB c 1 tan 4 OB A OA Tip tuyn AB c h s gc k = 1 4 Phng trnh y = k 2 34 1 ... 5( 1) 4 x xx +) x = 3y=0, tip tuyn c ph-ng trnh 1 ( 3) 4 y x +) x= -5 y= 2, tip tuyn c ph-ng trnh 1 1 13 ( 5) 2 4 4 4 y x y x Bi 20. Cho ham so 1 1 x y x . 1) Khao sat s bien thien va ve o th (C) cua ham so. 2) Tm a va b e ng thang (d): y ax b cat (C) tai hai iem phan biet oi xng nhau qua ng thang ( ): 2 3 0x y . Gii. 2. Phng trnh cua ( ) c viet lai: 1 3 2 2 y x . e thoa e bai, trc het (d) vuong goc vi ( ) hay 2a Khi o phng trnh hoanh o giao iem gia (d) va (C): 1 2 1 x x b x 2 2 ( 3) ( 1) 0x b x b . (1) e (d) cat (C) tai hai iem phan biet A, B (1) co hai nghiem phan biet 0 2 2 17 0b b b tuy y. Goi I la trung iem cua AB, ta co 3 2 4 3 2 2 A B I I I x x b x b y x b . Vay e thoa yeu cau bai toan ton tai , ( ) ( ) A B AB I 2 2 3 0I I b a x y 2 3 ( 3) 3 0 4 a b b 2 1 a b . 9. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 9 Bi 21. Cho hm s 1 1 x y x ( 1 ) c th ( )C . 1. Kho st v v th ca hm s ( 1). 2. Chng minh rng -ng thng ( ): 2d y x m lun ct (C) ti hai im phn bit A, B thuc hai nhnh khc nhau. Xc nh m on AB c di ngn nht. Gii. 2. Chng minh rng -ng thng ( ): 2d y x m lun ct (C) ti hai im phn bit A, B thuc hai nhnh khc nhau. Xc nh m on AB c di ngn nht . . -ng thng (d) lun ct ( C ) ti hai im phn bit th ph-ng trnh. 1 2 1 x x m x c hai nghim phn bit vi mi m v 1 21x x 1 ( 1)(2 ) 1 x x x m x c hai nghim phn bit 1 21x x 2 2 ( 3) 1 0 (*) 1 x m x m x c hai nghim phn bit 1 21x x 0 (1) 0f 2 ( 1) 16 0 (1) 2 ( 3) 1 2 0 m m f m m Vy vi mi gi tr ca m th-ng thng ( ): 2d y x m lun ct (C) ti hai im phn bit A, B thuc hai nhnh khc nhau. . Gi 1 1 2 2( ;2 ), ( ;2 )A x x m B x x m l hai im giao gia (d) v (C).( 1 2;x x l hai nghim ca ph-ng trnh (*)) Ta c 2 2 2 2 1 2 1 2 1 2 1 2 1( ;2( )) ( ) (2( )) 5( )AB x x x x AB x x x x x x Theo Vi t ta c 21 5 ( 1) 16 2 5 2 AB m m . 2 5 1AB m Vy vi m = -1 l gi tr cn tm. (R) Bi 22. Cho hm s 2 23 x x y c th (C) 1. Kho st s bin thin v v th (C) ca hm s. 2. Gi M l im bt k trn (C). Tip tuyn ca (C) ti M ct cc ng tim cn ca (C) ti A v B. Gi I l giao im ca cc ng tim cn. Tm ta M sao cho ng trn ngoi tip tam gic IAB c din tch nh nht. Gii. 2.Gi 2),() 2 23 ;( aC a a aM Phng trnh tip tuyn ca (C) ti M l: 2 23 )( )2( 4 2 a a ax a y () ng thng d1:x+2=0 v d2:y-3=0 l hai tim cn ca th d1=A(-2; ) 2 23 a a , d2=B(2a+2;3) Tam gic IAB vung ti I AB l ng knh ca ng trn ngoi tip tam gic IAB din tch hnh trn S= 8 )2( 64 )2(4 44 2 2 2 a a AB 10. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 10 Du bng xy ra khi v chi khi 4 0 )2( 16 )2( 2 2 a a a a Vy c hai im M tha mn bi ton M(0;1) v M(-4;5) Bi 23. Cho hm s 4 2 ( ) 8x 9x 1y f x 1. Kho st s bin thin v v th (C) ca hm s. 2. Da vo th (C) hy bin lun theo m s nghim ca phng trnh 4 2 8 os 9 os 0c x c x m vi [0; ]x . Gii. 2. Xt phng trnh 4 2 8 os 9 os 0c x c x m vi [0; ]x (1) t osxt c , phng trnh (1) tr thnh: 4 2 8 9 0 (2)t t m V [0; ]x nn [ 1;1]t , gia x v t c s tng ng mt i mt, do s nghim ca phng trnh (1) v (2) bng nhau. Ta c: 4 2 (2) 8 9 1 1 (3)t t m Gi (C1): 4 2 8 9 1y t t vi [ 1;1]t v (D): y = 1 m. Phng trnh (3) l phng trnh honh giao im ca (C1) v (D). Ch rng (C1) ging nh th (C) trong min 1 1t . Da vo th ta c kt lun sau: 81 32 m : Phng trnh cho v nghim. 81 32 m : Phng trnh cho c 2 nghim. 81 1 32 m : Phng trnh cho c 4 nghim. 0 1m : Phng trnh cho c 2 nghim. 0m : Phng trnh cho c 1 nghim. m < 0 : Phng trnh cho v nghim. Bi 24. Cho hm s: 1 2( 1) x y x 1. Kho st s bin thin v v th (C) ca hm s. 2. Tm nhng im M trn (C) sao cho tip tuyn vi (C) ti M to vi hai trc ta mt tam gic c trng tm nm trn ng thng 4x + y = 0. Gii. 2. Gi M( 0 0 0 1 ; 2( 1) x x x ) ( )C l im cn tm. Gi tip tuyn vi (C) ti M ta c phng trnh. : ' 0 0 0 0 1 ( )( ) 2( 1) x y f x x x x 0 02 00 11 ( ) 2( 1)1 x y x x xx Gi A = ox A( 2 0 02 1 2 x x ;0) 11. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 11 B = oy B(0; 2 0 0 2 0 2 1 2( 1) x x x ). Khi to vi hai trc ta OAB c trng tm l: G( 2 2 0 0 0 0 2 0 2 1 2 1 ; 6 6( 1) x x x x x . Do G ng thng:4x + y = 0 2 2 0 0 0 0 2 0 2 1 2 1 4. 0 6 6( 1) x x x x x 2 0 1 4 1x (v A, B O nn 2 0 02 1 0x x ) 0 0 0 0 1 1 1 2 2 1 3 1 2 2 x x x x Vi 0 1 1 3 ( ; ) 2 2 2 x M ; vi 0 3 3 5 ( ; ) 2 2 2 x M . Bi 25. Cho hm s y = x3 3x2 + mx + 4, trong m l tham s thc. 1. Kho st s bin thin v v th ca hm s cho, vi m = 0. 2. Tm tt c cc gi tr ca tham s m hm s cho nghch bin trn khong (0 ; + ). Gii. 2. Hm s cho nghch bin trn khong (0 ; + ) y = 3x2 6x + m 0, x > 0 3x2 + 6x m, x > 0 (*) Ta c bng bin thin ca hm s y = 3x2 + 6x trn (0 ; + ) T ta c : (*) m 0. Bi 26. Cho hm s 2 12 x x y c th l (C) 1.Kho st s bin thin v v th ca hm s 2.Chng minh -ng thng d: y = -x + m lun lun ct th (C) ti hai im phn bit A, B. Tm m on AB c di nh nht. Gii. 2. Honh giao im ca th (C ) v -ng thng d l nghim ca ph-ng trnh )1(021)4( 2 2 12 2 mxmx x mx x x Do (1) c mmmvam 0321)2).(4()2(01 22 nn -ng thng d lun lun ct th (C ) ti hai im phn bit A, B. Ta c yA = m xA; yB = m xB nn AB2 = (xA xB)2 + (yA yB)2 = 2(m2 + 12) suy ra AB ngn nht AB2 nh nht m = 0. Khi 24AB Bi 27. Cho hm s y = 1 12 x x (1) 1/ Kho st s bin thin v v th ca hm s (1) 2/ nh k ng thng d: y = kx + 3 ct th hm s (1) ti hai im M, N sao cho tam gic OMN vung gc ti O. ( O l gc ta ) Gii. x y 0 0 12. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 12 2. / Xt pt: )(04)1()1(3 1 12 2 xgxkkxxkx x x d ct th hs (1) ti M, N 347347 0 0)1( 0 0 kk k g k k xx k k xx kkk xxkxxkkxkxxxONOMONOM NM NM NMNMNMNM 4 . 1 53046 09)(3).)(1(0)3)(3(.0. 2 2 Bi 28. Cho hm s y = x3 + mx + 2 (1) 1. Kho st s bin thin v v th ca hm s (1) khi m = -3. 2. Tm m th hm s (1) ct trc hanh ti mt im duy nht. Gii. . 2.Pt : x3 + mx + 2 = 0 x xm 22 ( x )0 Xt f(x) = 2 2 2 2)(' 2 x xxf x x = 2 3 22 x x Ta c x - 0 1 + f(x) + + 0 - f(x) + -3 - - - th hm s (1) ct trc hanh ti mt im duy nht 3 m . Bi 29. Cho hm s y = x3 3x + 1 c th (C) v ng thng (d): y = mx + m + 3. 1/ Kho st s bin thin v v th (C) ca hm s. 2/ Tm m (d) ct (C) ti M(-1; 3), N, P sao cho tip tuyn ca (C) ti N v P vung gc nhau. Gii. 2. Phng trnh hanh giao im ca (C) v (d): x3 (m + 3)x m 2 = 0 Hay : (x + 1)(x2 x m 2) = 0 (*)02 3,1 2 mxx yx (*) phi c hai nghim phn bit ( m > ) 4 9 , xN v xP l nghim ca (*) Theo gi thit: 133 22 PN xx 3 223 3 223 01189 2 m m mm Bi 30. Cho hm s 2 4 1 x y x . 1) Kho st v v th C ca hm s trn. 2) Gi (d) l ng thng qua A( 1; 1 ) v c h s gc k. Tm k sao cho (d) ct ( C ) ti hai im M, N v 3 10MN . Gii. 13. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 13 2. T gi thit ta c: ( ): ( 1) 1.d y k x Bi ton tr thnh: Tm k h phng trnh sau c hai nghim 1 1 2 2( ; ), ( ; )x y x y phn bit sao cho 2 2 2 1 2 1 90(*)x x y y 2 4 ( 1) 1 ( )1 ( 1) 1 x k x Ix y k x . Ta c: 2 (2 3) 3 0 ( ) ( 1) 1 kx k x k I y k x D c (I) c hai nghim phn bit khi v ch khi phng trnh 2 (2 3) 3 0(**)kx k x k c hai nghim phn bit. Khi d c c 3 0, . 8 k k Ta bin i (*) tr thnh: 2 22 2 2 1 2 1 2 1(1 ) 90 (1 )[ 4 ] 90(***)k x x k x x x x Theo nh l Viet cho (**) ta c: 1 2 1 2 2 3 3 , , k k x x x x k k th vo (***) ta c phng trnh: 3 2 2 8 27 8 3 0 ( 3)(8 3 1) 0k k k k k k 16 413 16 413 3 kkk . KL: Vy c 3 gi tr ca k tho mn nh trn. Bi 31. Cho hm s 12 2 x x y 1. Kho st s bin thin v v th (C) ca hm s cho. 2. Tm nhng im trn th (C) cch u hai im A(2 , 0) v B(0 , 2) Gii. 2. Pt ng trung trc an AB : y = x Nhng im thuc th cch u A v B c hong l nghim ca pt : x x x 12 2 2 51 2 51 012 x x xx Hai im trn th tha ycbt : 2 51 , 2 51 ; 2 51 , 2 51 Bi 32. Cho hm s 2 32 x x y 1. Kho st s bin thin v v th (C) ca hm s. 2. Cho M l im bt k trn (C). Tip tuyn ca (C) ti M ct cc ng tim cn ca (C) ti A v B. Gi I l giao im ca cc ng tim cn. Tm to im M sao cho ng trn ngoi tip tam gic IAB c din tch nh nht. Gii. 2. Ta c: 2x, 2x 3x2 ;xM 0 0 0 0 , 2 0 0 2x 1 )x('y Phng trnh tip tuyn vi ( C) ti M c dng: 2x 3x2 )xx( 2x 1 y: 0 0 02 0 To giao im A, B ca v hai tim cn l: 2;2x2B; 2x 2x2 ;2A 0 0 0 14. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 14 Ta thy M0 0BA xx 2 2x22 2 xx , M 0 0BA y 2x 3x2 2 yy suy ra M l trung im ca AB. Mt khc I = (2; 2) v tam gic IAB vung ti I nn ng trn ngoi tip tam gic IAB c din tch S = 2 )2x( 1 )2x(2 2x 3x2 )2x(IM 2 0 2 0 2 0 02 0 2 Du = xy ra khi 3x 1x )2x( 1 )2x( 0 0 2 0 2 0 Do c hai im M cn tm l M(1; 1) v M(3; 3) Bi 33. Cho hm s 2 2 1 x y x (C) 1. Kho st hm s. 2. Tm m ng thng d: y = 2x + m ct th (C) ti 2 im phn bit A, B sao cho AB = 5 . Gii. 2. Phng trnh honh giao im: 2x2 + mx + m + 2 = 0 , (x - 1) (1) d ct (C) ti 2 im phn bit PT(1) c 2 nghim phn bit khc -1 m2 - 8m - 16 > 0 (2) Gi A(x1; 2x1 + m) , B(x2; 2x2 + m. Ta c x1, x2 l 2 nghim ca PT(1). Theo L Vit ta c 1 2 1 2 2 2 2 m x x m x x . AB2 = 5 2 2 1 2 1 2( ) 4( ) 5x x x x 2 1 2 1 2( ) 4 1xx x x m2 - 8m - 20 = 0 m = 10 , m = - 2 ( Tha mn (2)) Bi 34. Cho hm s 3 2 2 3 3 3( 1)y x mx m x m m (1) 1.Kho st s bin thin v v th ca hm s (1) ng vi m=1 2.Tm m hm s (1) c cc tr ng thi khong cch t im cc i ca th hm s n gc ta O bng 2 ln khong cch t im cc tiu ca th hm s n gc ta O. Gii. 2. Ta c , 2 2 3 6 3( 1)y x mx m hm s c cc tr th PT , 0y c 2 nghim phn bit 2 2 2 1 0x mx m c 2 nhim phn bit 1 0, m Cc i ca th hm s l A(m-1;2-2m) v cc tiu ca th hm s l B(m+1;-2-2m) Theo gi thit ta c 2 3 2 2 2 6 1 0 3 2 2 m OA OB m m m Vy c 2 gi tr ca m l 3 2 2m v 3 2 2m . Bi 35. 1) Kho st s bin thin v v th (C) ca hm s : y = x3 3x2 + 2 2) Bin lun theo m s nghim ca phng trnh : 2 2 2 1 m x x x Gii. 2. Ta c 2 2 2 2 2 2 1 1 1 m x x x x x m,x . x Do s nghim ca phng trnh bng s giao im ca 2 2 2 1y x x x , C' v ng thng 1y m,x . 15. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 15 V 2 1 2 2 1 1 f x khi x y x x x f x khi x nn C' bao gm: + Gi nguyn th (C) bn phi ng thng 1x . + Ly i xng th (C) bn tri ng thng 1x qua Ox. Da vo th ta c: + 2m : Phng trnh v nghim; + 2m : Phng trnh c 2 nghim kp; + 2 0m : Phng trnh c 4 nghim phn bit; + 0m : Phng trnh c 2 nghim phn bit. Bi 36. 1. kho st s bin thin v v th ( C) ca hm s: 2 32 x x y 2. Tm m ng thng (d): y = 2x + m ct th (C ) ti hai im phn bit sao cho tip tuyn ca (C ) ti hai im song song vi nhau. Gii. 2. Phng trnh honh giao im ca (d) v (C) l: 032)6(22 2 32 2 mxmxmx x x (x = 2 khng l nghim ca p trnh) (d) ct (C ) ti hai im phn bit m tip tuyn ti song song vi nhau (1) c hai nghim phn bit x1; x2 tho mn: y(x1) = y(x2) hay x1+x2= 4 2 4 2 6 0)32(8)6( 2 mm mm Bi 37. Cho hm s : 3 3y x m x( ) (1) 1) Kho st s bin thin v v th (C) ca hm s (1) khi m = 1. 2) Tm k h bt phng trnh sau c nghim: 3 2 3 2 2 1 3 0 1 1 log log ( 1) 1 2 3 x x k x x Gii. 2. Ta c : x k x x 3 2 3 2 2 3 3x 0 (1) 1 1 log log ( 1) 1 (2) 2 3 . iu kin (2) c ngha: x > 1. T (2) x(x 1) 2 1 < x 2. H PT c nghim (1) c nghim tho 1 < x 2 x k x k x x 3 3 ( 1) 3x 0 ( 1) 3x < 1 2 1 2 t: f(x) = (x 1)3 3x v g(x) = k (d). Da vo th (C) (1) c nghim x (1;2] 1;2 min ( ) (2) 5k f x f . Vy h c nghim k > 5 Bi 38. Cho hm s 3 2 2 3( 1) 2y x mx m x (1), m l tham s thc 1. Kho st s bin thin v v th hm s khi 0m . 2. Tm m th hm s ct ng thng : 2y x ti 3 im phn bit (0;2)A ; B; C sao cho tam gic MBC c din tch 2 2 , vi (3;1).M Gii. 1+1- - 2 m 1 2 16. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 16 2. Phng trnh honh giao im ca th vi ( ) l: 3 2 2 3( 1) 2 2x mx m x x 2 0 2 ( ) 2 3 2 0(2) x y g x x mx m ng thng ( ) ct d th hm s (1) ti ba im phn bit A(0;2), B, C Phng trnh (2) c hai nghim phn bit khc 0 2 2 1 ' 0 3 2 0 2 (0) 0 3 2 0 3 m hoacm m m g m m Gi 1 1;B x y v 2 2;C x y , trong 1 2,x x l nghim ca (2); 1 1 2y x v 1 2 2y x Ta c 3 1 2 ;( ) 2 h d M 2 2.2 2 4 2 MBCS BC h M 2 2 2 2 2 1 2 1 2 1 1 2( ) ( ) 2 ( ) 4BC x x y y x x x x = 2 8( 3 2)m m Suy ra 2 8( 3 2)m m =16 0m (tho mn) hoc 3m (tho mn) Bi 39. Cho hm s 3 22 3(2 1) 6 ( 1) 1y x m x m m x c th (Cm). 1. Kho st s bin thin v v th ca hm s khi m = 0. 2. Tm m hm s ng bin trn khong ;2 Gii. 2. 3 22 3(2 1) 6 ( 1) 1y x m x m m x )1(6)12(66' 2 mmxmxy y c 01)(4)12( 22 mmm 1 0' mx mx y Hm s ng bin trn ;2 0'y 2x 21m 1m Bi 40. Cho hm s y = 1 x x 1. Kho st s bin thin v v th (C) ca hm s. 2. Tm ta im M thuc (C), bit rng tip tuyn ca (C) ti M vung gc vi ng thng i qua im M v im I(1; 1). (M(0 ; 0) ; M(2 ; 2) ) Gii. 2. Vi 0 1x , tip tuyn (d) vi (C) ti M(x0 ; 0 0 1 x x ) c phng trnh : 0 02 0 0 1 ( ) ( 1) 1 x y x x x x 2 0 2 2 0 0 1 0 ( 1) ( 1) x x y x x (d) c vec t ch phng 2 0 1 ( 1; ) ( 1) u x , 0 0 1 ( 1; ) 1 IM x x (d) vung gc IM iu kin l : 0 0 2 00 0 01 1 . 0 1.( 1) 0 2( 1) 1 x u IM x xx x + Vi x0 = 0 ta c M(0,0) + Vi x0 = 2 ta c M(2, 2) VNMATH.COM VNMATH.COM VNMATH.COM VNMATH.COM VNMATH.COM VNMATH.COM 17. Cc bi tp d v c bn v KS hm s Trong n thi i Hc nm 2012 -2013 VNMATH.COM 17