Chng IHMS LNG GIC V PHNG TRNH LNG GICTit 1,2,3,4,5. HM S LNG
GICI. Mc tiu:Qua tit hc ny HS cn:1. V kin thc:- Hiu khi nim hm s
lng gic (ca bin s thc) v tnh tun hon ca cc hm s lng gic.2. V k
nng:-Xc nh c tp xc nh, tp gi tr, tnh cht chn, l; tnh tun hon; chu
k; sbin thin ca hm s y = sinx , y = cosx, y = tanx, y = cotx.- V c
th ca hm s v t suy ra th ca hm s y = sinx , y = cosx, y = tanx, y =
cotx.3. V t duy v thi :Tch cc hot ng, tr li cu hi. Bit quan st v
phn on chnh xc.II. Chun b ca GV v HS:GV: SGK, gio n,HS: Son bi trc
khi n lp, chun b bng ph, III. Phng php:Gi m, vn p, an xen hot ng
nhmIV. Tin trnh bi hc:Tit 1:1.n nh t chc lp2.Kim tra bi c3.Bi miHot
ng ca GV Hot ng ca HS Ni dungH1: Hnh thnh nh ngha hm s sin v
csinHTP 1(10): (Giibi tp ca hot ng 1 SGK)Yu cu HS xem ni dung hot
ng 1 trong SGK v tho lun theo nhm phn, bo co.Cu a)GV ghi li gii ca
cc HS tho lun theo nhm v c i din bo co.HS theo di bng nhn xt, sa
cha ghi chp.HS bm my cho kt qu:sin6=12, cos6=32, *S dng MTBT:sin6Th
thut tnh:chuyn qua n v rad:shift mode -4sin (shift - - -6- )- =Kt
qu:a)sin6=12, cos6=32nhm v cho HS nhn xt, b sung.-Vy vi x l cc s ty
(n v rad) ta c th s dng MTBT tnh c cc gi tr lng gic tng ng.GV v ng
trn lng gic ln bng v yu cu HS tho lun v bo co li gii cu b)Gi HS i
din nhm 1 ln bng trnh by li gii.GV gi HS nhn xt v b sung (nu cn).GV
chiu slide (sketpass) cho kt qu cu b).GV vi cch t tng ng mi s thc x
vi mt im M trn ng trn lng gic ta t tung v honh hon ton xc nh, vi
tung l sinx v honh l cosx, t y ta c khi nim hm s sin v csin.HTP2
(5):(Hm s sin v csin)GV nu khi nim hm s sin bng cch chiu slide.-Tng
t ta c khi nim hm s y = cosx.HS ch theo di ghi chp.HS tho lun theo
nhm v c i din bo co.HS nhn xt, b sung v ghi chp sa cha.HS trao i rt
ra kt qu t hnh v trc quan (ng trn lng gic)HS ch theo di trn bng v
ghi chp.HS ch theo di sin24 2 ; cos24 2sin(1,5)0,997;
cos(1,5)0,071xKHAOMsinx =OK;cosx =OH*Khi nim hm s sin:Quy tc t tng
ng mi s thc x vi s thc sinxsin:sin x y x ac gi l hm s sin, k hiu l:
y = sinxTp xc nh ca hm s sin l.*Khi nim hm s cos:Quy tc t tng ng mi
s thc x vi s thc cosxos:oscx y cx ac gi l hm s cos, k hiu l: y =
cosxTp xc nh ca hm s cos l.H2: Tnh tun hon ca hm s sinx v
cosxHTP1(10): V d v tnh tun hon ca hm Tm nhng s T sao cho f(x +T) =
f(x) vi mi x thuc tp xc nh ca cc hm s sau:a)f(x) =sinx; b)f(x) =
cosx.s y = sinx v y = cosxGV yu cu HS tho lun theo nhm v c i din bo
co.GV b sung (nu cn).GV ngi ta chng minh c rng T =2 l s dng nh nht
tha mn ng thc sin(x +T)= sinx v cos(x+T)=cosx.*Hm s y = sinx v y
=cosx tha mn ng thc trn c gi l hm s tun hon vi chu k2.HTP2: (5) (S
bin thin v th hm s lng gic y= sinx v y = cosx)-Hy cho bit tp xc nh,
tp gi tr, tnh chn l v chu k ca hm s y =sinx?GV cho HS tho lun theo
nhm v c i din ng ti ch bo co.GV ghi kt qu ca cc nhm v gi HS nhm khc
nhn xt b sung.GV ghi kt qu chnh xc ln bng.HTP3(10): (S bin thin ca
hm s y = sinx trn on [ ]0;)GV cho HS tho lun theo HS tho lun v c i
din bo co.HS nhm khc nhn xt b sung v ghi chp sa cha.HS ch theo di v
ghi nhHS tho lun theo nhm vo bo co.Nhn xt b sung v ghi chp sa
cha.HS da vo hnh v trao i v cho kt qu:-Xc nh vi mi x v 1 sinx 1 Tp
xc nh ; tp gi tr [ ]1;1 sin( ) sin x x nn l hm s l.Chu k 2.*T =2 l
s dng nh nht tha mn ng thc sin(x +T)= sinx v cos(x+T)=cosx.*Hm s y
= sinx v y = cosx tun hon vi chu k 2.*Hm s y = sinx:+Tp xc nh: ;+Tp
gi tr [ ]1;1 ;+L hm s l;+Chu k 2.*Hm s y = cosx:+Tp xc nh: ;+Tp gi
tr [ ] 1;1 ;+L hm s chn;+Chu k 2.nhm tm li gii v bo co.GV ghi kt qu
ca cc nhm v gi HS nhm khc nhn xt, b sung.Vy t s bin thin ca hm s y
= sinx ta c bng bin thin (GV chiu bng bin thin ca hm s y = sinx)GV
yu cu HS v th hm s y = sinx trn on[ ]0; v bng bin thin. Ly i xng th
qua gc ta (V y = sinx l hm s l )Vy v th ca hm s y=sinx ta lm nh th
no? Hy nu cch v v v th y = sinx trn tp xc nh ca n.GV gi HS nu cch v
v hnh v (trn bng ph).Cho HS nhm khc nhn xt, b sung. Tng t hy lm tng
t vi hm s y = cosx (GV yu cu HS t rt ra v xem nh bi tp nh)-HS ch
theo di hnh v v tho lun v bo co.-HS nhm khc nhn xt v b sung, ghi
chp sa cha.-HS trao i cho kt qu: x1, x20;21 1 ]v x1 HS ch theo di
trn bngHS ch theo di cc li gii HS xem ni dung H 3 v tho lun, trnh
by li gii cosinAO A B1 a >: phng trnh (1) v nghim.1 a : phng
trnh (1) c nghim:22 ,x kx k k + + ZNutha mn iu kin 2 2sinx =a ' th
ta vit =arcsina (c l ac-sin-a)Cc nghim ca phng trnh sinx = a cvit
l:arcsina 2arcsin 2 ,x kx ak k + + ZCh : (SGK)V d: Gii cc phng trnh
sau:a)sinx = 32;b)sinx = 23GV yu cu HS xem ni dung H 3 trong SGK v
tho lun tm li gii.GV gi 2 HS i din hai nhm trnh by li gii.GV hng dn
s dng my tnh b ti tm nghim gn ng.HS trao i v rt ra kt qu:a)x =
arcsin13+k2 x = -arcsin13+k2 , kZH 3: Gii cc phng trnh sau:a)sinx =
1;3b)sin(x +450)=22 .GV tng t vi vic gii phng trnh lng gic c bn
sinx = a ta cng c th gii c phng trnh cosx = a. y l ni dung ca tit
hc hm sau.*Cng c v hng dn hc nh:-Xem li v hc l thuyt theo SGK.-Xem
li cc v d gii v lm cc bi tp 1 SGK trang 28.II/ Rt kinh nghimTit
2:I.Tin trnh bi hc:*n nh lp, chia lp thnh 6 nhm.*Kim tra bi c: an
xen vi hot ng nhm*Bi mi:Hot ng ca GV Hot ng ca HS Ni dungH: (Phng
trnh cosx =a)HTP1( ): (Hnh thnh iu kin ca phng trnh cosx=a)Tp gi tr
ca hm s csin l g?By gi ta xt phng trnh: cosx = a (2)SGK v suy ngh
tr liV1 osx 1 c vi mi, nn tp gio tr ca hm s csin l on [ ]1;1 2.
Phng trnh cosx = a:sinB
M
gii phng trnh ny ta phi lm g? V sao?Vy da vo iu kin:1 osx 1 c
gii phng trnh (2) ta xt hai trng hp sau (GV nu hai trng hp nh SGK v
v hnh hng dn rt ra cng thc nghim)1 a > khng tha mn iu kin1 cosx
1 (haycosx 1 ) phng trnh (2) v nghim.1 a cng thc nghim.GV nu ch nh
trong SGK c hai trnghp a) v b).c bit l phi nu cc trng khi a = 1, a
= -1, a = 0.(GV phn tch v nu cng thc nghim)HTP2( ): (V d p dng gii
phng trnh cosx = a)GV nu v d 1 v gi trnh by li gii.HTP3( ): (H cng
c kin thc)GV yu cu HS xem ni dung H 4 trong SGK v tho lun tm li
gii.HS do iu kin 1 sinx 1 nn ta xt 2 trng hp:1 1 a v a > HS ch
theo di trn bngHS ch theo di cc li gii HS xem ni dung H 4 v tho
lun, trnh by li giiHS trao i v rt ra kt qu: csinAO KA a
M B1 a >: phng trnh (2) v nghim.1 a : phng trnh (2) c
nghim:22 ,x kx k k + + ZNutha mn iu kin 0osx = c a ' th ta vit
=arccosa (c l ac-csin-a)Cc nghim ca phng trnh cosx = a cvit l:rccos
2r os 2 ,x a akx a ccak k + + ZCh : (SGK)V d: Gii cc phng trnh
sau:a)cosx = 32;b)cosx = 25GV gi 3 HS i din hai nhm trnh by li
gii.a)x =223k+ x= -223k+ ,kZb)x = arccos23+k2 x = -arccos23+k2,
kZc)x = 52 ,6k kt + ZH 3: Gii cc phng trnh sau:a)cosx = 1;2b)cosx =
23;c)cos(x +300)=32 .H2: (Bi tp p dng gii phng trnh cosx = a)GV yu
cu HS xem ni dung bi tp 3 d) v suy ngh tm li gii.GV gi 1 HS trnh by
li gii.Gi HS nhn xt, b sung (nu cn)GV nu li gii ng (nu cn)GV hng dn
s dng my tnh b ti tm nghim gn ng.HS theo di ni dung bi tp 3d) SGK v
suy ngh tm li gii.HS nhn xt, b sung v sa cha, ghi chp.HS trao i v
cho kt qu:cos2x = 141osx=2c tVy .Bi tp 3d) (SGK trang 28)*Cng c v
hng dn hc nh:-Xem li v hc l thuyt theo SGK.-Xem li cc v d gii v lm
cc bi tp 2,3 SGK trang 28.II/ Rt kinh ngimTit 8.I.Tin trnh bi hc:*n
nh lp, chia lp thnh 6 nhm.*Kim tra bi c: an xen vi hot ng nhm*Bi
mi:Hot ng ca GV Hot ng ca HS Ni dungH1: (Phng trnh tanx =a)HTP1( ):
(Hnh thnh iu kin ca phng trnh tanx=a)Tp gi tr ca hm s tangl g?Tp xc
nh ca hm s y = tanx?By gi ta xt phng trnh: tansx = a (3)GV yu cu HS
xem hnh 16 SGKVy da vo tp xc nh v da vo hnh 16 SGK ta rt ra cng thc
nghim (GV v hnh hng dn rt ra cng thc nghim) phng trnh (3) c cng thc
nghim.GV nu ch nh trong SGK c hai trnghp a) v b).(GV phn tch v nu
cng thc nghim)HTP2( ): (V d p dng gii phng trnh cosx = a)GV nu v d
1 v gi trnh by li gii.SGK v suy ngh tr liTp gi tr l khong (-; +)Tp
xc nh:\ , .2D k k + ' ) ZHS ch theo di trn bngHS ch theo di cc li
gii 1.Phng trnh tanx = a:sin B T
a csinAO A
M Biu kin ca phng trnh l: ,2x kk + ZNutha mn iu kin2 2tanx =a
< *n th cp s cng tng, ngc li cp s cng gim.Bi 2: HS suy ngh v tr
li tng t.HS cc nhm xem v tho lun theo nhm tm li gii.HS i din ln bng
trnh by li gii (c gii thch)HS nhn xt, b sung v sa cha ghi chp.Bi tp
1 n bi tp 4 (SGK)Bi tp 5a) (SGK)tp 5a) v tho lun suy ngh tr li.GV
gi HS i din ln bng trnh by li gii.GV gi HS nhn xt, b sung(nu cn)GV
nhn xt v nu li gii ng (nu HS khng trnh by ng li gii)HS trao i v rt
ra kt qu:t Bn = 13n-1Vi n = 1 th B1 = 131-1=12M6Gi s Bk = 13k-1M6Ta
phi chng minh Bk+1M6Tht vy, theo gi thit quy np ta
c:Bk+1=13k+1-1=13.13k-13+12=13(13k-1)+12=13.Bk+12V Bk M6 v 12M6 nn
Bk+1M6Vy Bn = 13n-1M6H2: HTP2: Xt tnh tng gim v b chn ca mt dy s.HS
cho HS cc nhm xem ni dung bi tp 7 v tho lun theo nhm tm li gii.GV
gi HS i din nhm ln bng trnh by li gii.Gi HS nhn xt, b sung (nu
cn)GV nhn xt v nu li gii ng (nu HS khng trnh by ng li gii)HTP2: Cc
bi tp v cp s cng v cp s nhn.GV yu cu HS cc nhm theo di bi tp 8 v 9
trong SGK v HS cc nhm xem v tho lun theo nhm tm li gii.HS i din nhm
ln bng trnh by li gii (c gii thch)HS cc nhm trao i v rt ra kt qu:Dy
(un) tng v b chn di bi 2.HS cc nhm tho lun v suy ngh tm li gii bi
tp 8 v 9, c i din ln bng trnh by li gii.HS nhn xt, b sung v sa Bi
tp 7 (SGK)Xt tnh tng, gim v b chn ca dy s (un), bit:1) na u nn +Bi
tp 8 v 9 (SGK)cho HS cc nhm tho lun tm li gii.Gi HS i din ln bng
trnh by li gii v gi HS nhn xt, b sung v GV nu li gii ng (nu HS khng
trnh by ng li gii)HTP3:GV cho HS cc nhm xem bi tp 10 v tho lun theo
nhm tm li gii.Gi HS i din nhm ln bng trnh by li gii.Gi HS nhn xt, b
sung (nu cn)GV nhn xt v nu li gii ng (nu HS khng trnh by ng kt
qu)cha ghi chp.HS cc nhm trao i v cho kt qu:8a)) u1=8; d = -3.8b)
u1=0, d = 3; u1=-12, d = 2159a)q = 2 v u1=69b) q = 2 v u1=12.HS cc
nhm tho lun tm li gii v c i din ln bng trnh by li gii c gii thch.HS
nhn xt, b sung v sa cha ghi chp.HS trao i v rt ra kt qu: + 4 .4 2 C
A B A A A = = =+ C2 = B.D nn 16A2 = 2A.D. suy ra: D = 8AA + B + C +
D = 3600 nn 15A = 3600Suy ra:A = 250,B = 480,C = 960, D = 1920 Bi
tp 10.Cho t gic ABCD c s o () ca cc gc lp thnh mt cp s nhn theo th
t A, B, C, D. Bit gc C gp bn ln gc A. Tnh cc gc ca t gic.- Cho cc
nhm cng tho lun gii bi ton- GV quan st v hng dn: Tnh cc gc B, C, D
theo ANhn v chnh xc kt qu nhm hon thnh sm nhtH3 : Cng c v hng dn hc
nh.*Cng c : -Gi HS nhn li khi nim cp s cng v cp s nhn, cng thc tnh
s hng tng qut, tnh cht ca cp s cng v cp s nhn v tng n s hng u ca mt
cp cp s cng v cp s nhn. -p dng gii bi tp 10 SGK trang 108.*Hng dn
hc nh : -Xem li l thuyt trong chng III.-Xem li cc bi tp gii v gii
cc bi tp cn li trong phn n tp chng
III.-----------------------------------
------------------------------------Tit 46. N TP HC K II.Mc tiu
:Qua bi hc HS cn :1)V kin thc :-HS h thng li kin thc hc t chng I n
chng III.2)V k nng :-Vn dng c cc pp hc v l thuyt hc vo gii c cc bi
tp- Hiu v nm c cch gii cc dng ton c bn.3)V t duy v thi :Pht trin t
duy tru tng, khi qut ha, t duy lgic,Hc sinh c thi nghim tc, say m
trong hc tp, bit quan st v phn on chnh xc, bit quy l v quen.II.Chun
b ca GV v HS:GV: Gio n, cc dng c hc tp,HS: Son bi trc khi n lp,
chun b bng ph (nu cn), III. Phng php: V c bn l gi m, vn p, an xen
hot ng nhm.*n nh lp, gii thiu, chia lp thnh 6 nhm.*Bi mi:Hot ng ca
GV D kin hot ng ca HS H1: n tp v h thng li kin thc hc trong chng I
n chng III.GV gi HS ng ti ch nu li cc kin thc c bn hc trong cc chng
I, II v III.-n tp li hm s lng gic, phng trnh lng gic, cng thc nghim
ca cc phng trnh lng gic c bn v thng gp.-n tp li cc quy tc m, ho v -
chnh hp- t hp, cng thc nh thc Niutn, php th v bin c, tnh xc sut ca
bin c.-n tp li dy s, cp s cng, cp s nhn t bit l cc cng thc trong dy
s, cp s cng v cp s nhn.HS ch theo di trn bng n tp kin thc v suy ngh
tr li HS ng ti ch tr li cc cu hi m GV t ra n tp kin thcH2: Gii mt s
kim tra tham kho:GV pht cho HS cc kim tra v hwongs dn gii.MT S KIM
TRA THAM KHA S 1I. TRC NGHIM Cu1: Biu thc no sau y cho gi tr ca
tng: S = 1 + 2 + 3 + + nA. n(n+1)B. ( 1)2n n +C. 12n +D. (2 1)2n n
+Cu 2: 1 1 1, ,2 4 6 l ba s hng u ca dy s (un) no sau yA. 12nnu =
B. 12nun=C. 1nun= D. 12 2nun=+Cu 3: Trong cc dy s (un) sau y, dy s
tngA. 211nun=+B.( 1) .nnu n = -C.12nnu = - D.2 12nnnu-=Cu 4: Trong
cc dy s (un) sau y, dy s no b chn trnA. 2 1nu n = + B. 21nu n = +
C. ( 1) 1nnu = - + D. 13nnu = - Cu 5: Trong cc dy s sau, dy s no
khng phi l cp s cngA. 2, 4, 8, 16, B. -1, -2, -3,- 4, C. 2, 2, 2,
2, D. 1, 2, 3, 4, Cu 6: Ba gc ca mt tam gic vung lp thnh mt cp s
cng. Gc nh nht ca tam gic bng bao nhiu ?A. 150B. 450 C. 300 D.
600Cu 7: Cho cp s nhn c u1 = 1, q = 2. S hng th 11ca cp s nhn l :A.
20 B. 2028 C. 22D. 1024Cu 8: Ba s to thnh mt cp s nhn, bit tng v
tch ca chng ln lt l 13 v 27. Tm s ln nht ?A. 3 B. 9C. 27D. 10II. T
LUNBi 1: Chng minh bng phng php qui np:*n , 3 n ta c 2n > 2n +
1Bi 2: Xc nh s hng u v cng sai ca cp s cng, bit 7 32 78. 75u uu u
'
P NI. TRC NGHIMCu 1: B Cu 2: B Cu 3: DCu 4: DCu 5: ACu 6: CCu
7:DCu 8:AII. T LUNBi 1: *n = 3 , bt : 23 > 2.3 + 1(ng)* Gi s bt
ng vi mt s t nhin bt k 3 n k , tc l 2k> 2k +1Ta chng minh:
2k+1> 2(k +1) +1Ta c 2k + 1 = 2k.2 > 2( 2k + 1) = 4k + 2 = 2k
+ (2k + 2) > 2k + 3 = 2(k+1) +1.Vy *n , 3 n ta c 2n > 2n +
1Bi 2:Dng cng thc: un = u1 + (n - 1).d1 11 121 11 16 ( 2 ) 8:( ).(
6 ) 75214 24 03 17hoc2 2u d u dTacu d u ddu uu ud d + + '+ + '+ + '
' S 2I. TRC NGHIMCu1: Biu thc no sau y cho gi tr ca tng: S = 1 2 +
3 4 + - 2n + (2n + 1)A. 1 B. 0 C. nD. n + 1Cu 2: Cho dy s (un) vi 1
( 1)nnun+ -= . Gi tr no sau y l s hng th 9 ca dy s (un) ?A. 19B.
19- C. 0 D. 29Cu 3: Dy s no sau y khng phi l dy s tng ng thi cng
khng phi l dy s gim ?A. 1nnun=+ B. 13nnu = - C. 1nnun+= D. 2
12nn-Cu 4: Trong cc dy s sau, dy s no b chn ?A.1( 1)nun n=+ B. un =
2nC. un = 3n + 1D. (- 1)n .2nCu 5: Trong cc dy s sau, dy s no khng
phi l cp s cng ?A. un = 3n + 5B. un = 2n C. un = n2 D. 5 13nnu+=Cu
6: Tng 10 s hng u ca cp s cng bng bao nhiu nu bit u1 = 1 v u2 = 5
?A. 380 B. 190C. 95D. 195Cu 7: S hng th 11 ca cp s nhn: 2, - 4, 8,
. LA. 2048B. 1028 C. 1024 D. 2048Cu 8: Tm cng bi q ca cp s nhn, bit
u5 = 96 v u9 = 192A. q = 4 B. q = 3 C. q = 2D. q = 6II. T LUNBi 1:
Cho dy s (un), bit: 1113 i 1n nuu u v n+ =-= + a) Vit su s hng u ca
dy sb) D on cng thc s hng tng qut un v chng minh cng thc bng phng
php qui npBi 2: Xc nh cp s nhn (un), bit :356151350uuu ==
1.V d: Tm:( )2lim 3 2 n n +H2: HTP1:Bi tp ng dng thc t:GV gi HS
nu bi tp 1 trong SGK.GV cho HS cc nhm tho lun nhn xt tm li gii v gi
HS i din cc nhm ln bng trnh by li gii.Gi HS nhn xt, b sung (nu
cn).GV nhn xt v nu li gii ng (nu HS khng trnh by ng li gii).HS cc
nhm tho lun tm li gii v c i din ln bng trnh by li gii (c gii
thch).HS nhn xt, b sung v sa cha ghi chp.HS cc nhm trao i v a ra kt
qu:S:1 2 31 1 1) ; ; ;...2 4 8Bng quy np ta chng minh c:1.2n na u u
uu Bi tp 1: (SGK)HTP2: GV nu v chiu ln bng ni dung nh l 2.GV ly v d
minh ha(bi tp 8b) v cho HS cc nhm tho lun tm li gii, gi HS i din ln
bng trnh by li gii.GV gi HS nhn xt, b sung (nu cn)GV nhn xt, b sung
v nu li gii ng (nu HS khng trnh by ng li gii).HTP3: V d p dng:GV
cho HS cc nhm xem ni dung bi tp 8a) v cho HStho lun theo nhoma tm
li gii, gi HS i din ln bng trnh by li gii.Gi HS nhn xt, b sung (nu
cn)GV nhn xt, b sung v nu li gii ng (nu HS khng trnh by ng li
gii).( ) ( ) ( )6 6 3 91)lim lim 021 1 1 1) .10 10 10 10nnb uc g kg
kg _ , HS ch v theo di trn bngHS cc nhm tho lun tm li gii v c i din
ln bng trnh by (c gii thch)HS nhn xt, b sung v sa cha ghi chp.HS
trao i rt ra kt qu:2212lim lim1121 lim01lim limn nnnnnnnv
vvvvvvv+++ 3 1 3.lim 18 )lim 21 lim 1n nn nu uau u + +3)nh l:nh l
2: (SGK)a)Nu lim un = a v lim vn=t th lim 0nnuv.b)Nu lim un=a>0,
lim vn=0 v vn>0 vi mi n th limnnuv +c)Nu lim un=+ v lim
vn=a>0 th lim unvn=+V d: (Bi tp 8b SGK).Cho dy s (vn). Bit lim
vn=+Tnh gii hn:
22lim1nnvv +Bi tp 8a): (SGK)Cho dy s (un). Bit lim un=3.Tnh gii
hn:
3 1lim1nnuu+H3: Cng c v hng dn hc nh :*Cng c:-Nhc li cc nh l v
cc gii hn c bit.-p dng : Gii bi tp 7a) c) SGK trang 122.GV cho HS
tho lun theo nhm tm li gii v gi i din ln bng trnh by.GV gi HS nhn
xt, b sung v nu li gii ng (nu HS khng trnh by ng li gii).*Hng dn hc
nh:-Xem li v hc l thuyt theo SGK.-Xem li cc v d v bi tp gii.-lm thm
cc bi tp cn li trong SGK trang 121 v
122.-----------------------------------
------------------------------------ Tit 52I.Tin trnh bi hc : * n
nh lp : Chia lp thnh 6 nhm. *Kim tra bi c: Tnh : 33 1lim3 4++nn
*Bi mi : Hot ng hc sinh Hot ng gio vin Ni dungH1: Gii bi tp
2:GVcho HS cc nhm tho lun tm li gii bi tp 2 SGK v gi i din nhm ln
bng trnh by li gii.GV gi HS nhn xt, b sung (nu cn).GV nhn xt, b
sung v nu li gii ng (nu HS khng trnh by ng li gii ).HS cc nhm tho
lun tm li gii v c i din ln bng trnh by li gii (c gii thch)HS nhn
xt, b sung v sa cha ghi chp.HS trao i v rt ra kt qu:V 31lim 0nnn
31nc th nh hn mt s dng b ty , k t mt s hng no tr i, Bi tp 2:
(SGK)Bit dy s (un) tha mn 311nun 1 : f(x) = ax + 2 Hm s lin tc trn
(1 ; +) + x< 1: f(x) = x 22 +xHm s lin tc trn (-) 1 ; f(1) = a
+2 .2 ) 2 ( lim ) ( lim1 1+ + + + a ax x fx x.1 ) 1 ( lim ) ( lim21
1 + + x x x fx xa =-1th hm s lin tc Cc hm a thc c TX l g?Cc hm a
thc lin tc trn R.Tm TX? kt lun g v tnh lin tc ca hm s ?+ x > 1 :
f(x) = ?kt lun g v tnh lin tc ca hm s?+ x< 1 : f(x) = ?kt lun g
v tnh lin tc ca hm s?+ Xt tnh lin tc ca hm s ti x = 1?Tnh f(1)?? )
( lim1x fx? ) ( lim1x fx+ kt lun g v tnh lin lin tc trn [a ; b] nu
n lin tc trn (a ;b)v ) ( ) ( lim a f x fa x+ ) ( ) ( lim b f x fb x
Ch : th ca 1 hm s lin tc trn 1 khongl 1 ng lintrn khong .III,Mt s
nh l c bn.L 1: SGKL 2: SGK.V d: Xt tnh lin tc ca hm sy = 2cos tan )
1 ( +xx x xTX : D = R \{ 2; k +2,k Z }Vy hm s lin tc ti mi im x 2 v
x k +2( k) Z V d: Cho hm s f(x) = '< + +1 11 22khi x x xkhi x
axXt tnh lin tc ca hm s trn tontrc s. +x >1 : f(x) = ax + 2 nn
hm s lin tc.+x < 1: f(x) = x 12 +x nn hm s lin tc.+ti x = 1:f(1)
= a +2 .2 ) 2 ( lim ) ( lim1 1+ + + + a ax x fx x.1 ) 1 ( lim ) (
lim21 1 + + x x x fx xtrn R.a-1 th hm s lin tc trn ( - ) ; 1 ( ) 1
; + . GV treo bng ph hnh 59/ SGK v gii thch.GV nhn mnh L 3 c p dng
CM s tn ti nghim ca phng trnh trn 1khong.
a = -1 ; b = 1 hm s f(x) = x5 + x -1 lin tc trn R nn lin tc trn
on [-1;1] f(-1) = -3 f(1) = 1 f( -1) .f(1) = -3 < 0.tc ca hm s
trn ton trc s? HS quan st hnh v a = ?,b = ? hm s f(x) = x5 + x
-1lin tc ko? Tnh f (-1)? f(1) ?Kt lun g vduca f(-1)f(1)?a = -1 th )
1 ( ) ( lim ) ( lim1 1f x f x fx x + nn hm s lin tc ti x = 1.a1 hm
s gin on ti x = 1Vy:a = -1 th hm s lin tc trn R.a-1 th hm s lin tc
trn ( - ) ; 1 ( ) 1 ; + .L 3: Nu hm s y = f(x) lin tc trn on [ a;
b] v f(a).f(b) < 0 th tn ti t nht 1 im c ( a; b) sao cho f( c) =
0. Ni cch khc:Nu hm s y = f(x) lin tc trn [a ; b] v f(a).f(b) <
0 th phng trnh f(x) = 0 c t nht 1 nghim nm trong (a ; b).V d : Chng
minh rng phng trnh :x5 + x -1 c nghim trn(-1;1).Gii: Hm s f(x) = x5
+ x -1 lin tc trn R nn f(x) lin tc trn [-1; 1] . f(-1) = -3 f(1) =
1do f( -1) .f(1) = -3 < 0.Vy phng trnh c t nht 1 nghim thuc (
-1; 1). *Cng c v hng dn hc nh:Cng c:N hm s lin tc ti 1 im. N hm s
lin tc trn 1 khong. Mt s nh l c bn. BTVN: cc bi tp SGK.
-----------------------------------
------------------------------------Tit 59:I.Tin trnh bi hc:*n nh
lp, giithiu: Chia lp thnh 6 nhm* Kim tra bi c: Nu nh ngha, cc nh l
ca hm s lin tc ? Vn dng: Dng nh ngha xt tnh lin tc ca hm s:f(x) =
32 1 x x + ti03 x * Bi mi:Hot ng ca HS Hot ng ca GV Ni dungTXD: D =
R( )( )32 2228lim lim2lim 2 4 12x xxxg xxx x + + g(2) = 5 ( ) (
)2lim 2xg x g Hm s y = g(x) khng lin tc ti 02 x Hc sinh tr li- HS v
th- Da vo th nu cc khong hm s y = f(x) lin tc-Da vo nh l chng HD:
Tm tp xc nh?Tnh( )2limxgx v f ( 2) ri so snhHD: Thay s 5 bi s no hm
s lin tc ti 02 x tc l ( ) ( )x 2limg x 2 gHD: - V th y = 3x + 2 khi
x < - 1 ( l ng thng)- V thy = 21 x nu 1 x ( l ng parabol ) Bi tp
2: ( )38, 225 , 2xxg xxx 'a/ Xt tnh lin tc ca hm s y = g (x) ti02 x
KL: Hm s y = g(x) khng lin tc ti02 x b/ Thay s 5 bi s 12Bi tp 3:(
)23 2 , 11 , 1x xf xx x+ < ' a/ Hm s y = f(x) lin tc trn cc
khong( ) ; 1 v ( ) 1; +minh hm s lin tc trn cc khong ( ) ; 1 v ( )
1; +-Xt tnh lin tc ca hm s ti01 x -Tm tp xc nh ca cc hm s- Hm s y =
f(x) l hm a thc nn lin tc trn R - Chon a = 0, b = 1 - Chn c = -1, d
= -2-Hm s: f(x) = cosx x lin tc trn R - Chn a = 0, b = 1-Gi HS chng
minh khng nh cu a/ bng nh l- HD: Xt tnh lin tc ca hm s y = f(x) trn
TXD ca nHD: Tm TXD ca cc hm s , p dnh tnh cht ca hm s lin tcHD: Xt
tnh lin tc ca hm s ny v tm cc s a, b, c, d sao cho: f(a).f(b) <
0 v f(c).f(d) < 0Bin i pt: cosx = x tr thnhcosx x = 0t f (x) =
cosx x Gi HS lm tng t cu a/b/ -Hm s lin tc trn cc khong ( ) ; 1 v (
) 1; +- Ti01 x ( ) ( )1 1l imf x limx xf x + Hm s khng lin tc ti 01
x Bi tp 4: -Hm s y = f(x) lin tc trn cc khong ( ) ( ) ( ) ; 3 , 3;
2 , 2; +- Hm s y = g(x) lin tc trn cc khong ;2 2k k k Z _ + + ,Bi
tp 6: CMR phng trnh:a/32 6 1 0 x x + c t nht hai nghim b/ cosx = x
c nghim * Cng c:H thng l thuyt: nh ngha v tnh cht ca hm s lin tc*
Dn d: Xem li cc bi tp gii v chun b phn n tp chng
IV-----------------------------------
------------------------------------Tit 60,61:CU HI V BI TP N TP
CHNG IV I.MC TIU :Qua bi hc HS cn:1.Kin thc :bit cc nh ngha, nh l,
qui tc v cc gii hn dc bit.2.K nng: c kh nng p dng cc kin thc l
thuyt trn vo cc bi ton thuc cc dng cbn 3.T duy: tm cc phng php c th
cho tng dng ton. 4. Thi : Cn thn ,chnh xc. II.CHUN B CA GV V HS.GV:
gio n HS: n tp cc kin thc c v gii hn ca hm s. III.PHNG PHP DY HC:
phng php gi m ,vn p. IV.TIN TRNH BI HC:Tit 60:*n nh lp, gii thiu:
Chia lp thnh 6 nhm*Bi mi:Hot ng ca HS Hot ng ca GV Ni dungt n lm
nhn t c t v mu ri rt gn. lim21 3+nn= 3 Gi HS ln bng giiNu cch lm?Nu
kt qu?1. Tm cc gii hn sau:a, lim21 3+nn = lim )21 ()13 (nnnn+= lim
nn2113+ = 30 10 3+b,lim ( ) 22n n n += limnhn c t v mu cho lng lin
hip l n n n + +22) 2 )( 2 (2 2n n n n n n + + + = n2 22 n n + = 2n.
t n lm nhn t chung cho c t v mu ri rt gn.lim) 121 (2+ +nnn = 1 0
12+ + = 1t n lm nhn t c t v mu ri rt gn.lim +7 32nnlim)73 ()2 1(nnn
nn+lim +7 32nn00 lim + nnq nu IqI :GV nu nh ngha v hng dn chng minh
(nh SGK)HTP2:GV yu cu HS cc nhm chng minh hai cng thc sau:HS ch
theo di trn bng lnh hi kin thcHS tho lun theo nhm tm li gii v c i
din ln bng trnh by li gii (c gii thch).I. o hm ca mt s hm s thng
gp:1)nh l 1: SGKHm s y = xn( ) , 1 n n > c o hm ti mi xv
(xn)=nxn-1 (c) = 0, vi c l hng s;(x) = 1GV gi HS i din cc nhm ln
bng trnh by li gii.Gi HS nhn xt, b sung (nu cn).GV nhn xt, b sung v
nu li gii ng (nu HS khng tnh by ng li gii)HS nhn xt, b sung v sa
cha ghi chpH2:HTP1: GV nu bi tp v cho HS cc nhm tho lun tm li
gii.GV gi HS nhn xt, b sung (nu cn).GV nhn xt, b sung v nu li gii
ng (nu HS khng trnh by ng)GV: Bi tp ta va chng minh chnh l ni dung
ca nh l 2.GV nu nh l 2 trong SGK.HTP2:GV: C th tr li ngay c khng,
nu yu cu tnh o hm ca hm s f(x) =x ti x = -3; x = 4?HS tho lun theo
nhm tm li gii v c i din ln bng trnh by (c gii thch)HS nhn xt, b
sung v sa cha ghi chp.HS trao i v chng minh tng t trang 158HS suy
ngh tr li:Ti x = -3 hm s khng c o hm.Ti x = 4 hm s c o hm bng ( )1
1' 442 4f V d: Cho hm sy x c o hm ti mi x dng. S dng nh ngha tnh o
hm ca hm sy x .H3: Tm hiu v o hm ca tng, hiu, tch, thng:HTP1: GV nu
nh l 3 v hng dn chng minh (nh SGK)HS ch theo di trn bng lnh hi kin
thcII. o hm ca tng, hiu, tch, thng:1)nh l:*nh l 3: SGKGi s u =
u(x), v = v(x) l cc hm s c o hm ti im x thuc khong xc HTP2: GV cho
HS tho lun theo nhm tm li gii v d H4, gi HS i din ln bng trnh by li
gii.Gi HS nhn xt, b sung (nu cn).HS tho lun theo nhm tm li gii v c
i din ln bng trnh by (c gii thch).HS nhn xt, b sung v sa cha ghi
chpHS trao i v rt ra kt qu:3 5 2 4 5 2 ' 15 10 y x x y x x 3 2 31yx
' 32x y x x xx nh. Ta c:(u + v) = u + v (1)(u - v) = u - v (2)(u.v)
= uv + vu (3)'2' ' ( ( ) 0)u u v v uv v xv v _ ,(4)V d H4: p dng
cng thc trong nh l 3, hy tnh o hm ca cc hm s:y = 5x3 2x5; y =
-x3x.H4: Cng c v hng dn hc nh:*Cng c: -Nhc li cc cng thc tnh o hm
ca hm s y = xn v y =x, cng thc tnh o hm tng, hiu, tch, thng.-p dng
gii cc bi tp sau:1)Tnh o hm ca hm s: 223 11x xyx x ++ +2) Tnh o hm
ca hm s: ( )102y x x +GV: Ch gi v hng dn v yu cu HS lm xem nh bi
tp.*Hng dn hc nh: -Xem li cc bi tp gii, xem li v hc l thuyt theo
SGK.- Son trc phn l thuyt cn li ca bi Quy tc tnh o hm.- Lm cc bi tp
1 v 2 trong SGK trang 162 v 163.-----------------------------------
------------------------------------Tit 67.I. Tin trnh bi hc:*n nh
lp, gii thiu- Chia lp thnh 6 nhm*Kim tra bi c: -Nu cc cng thc tnh o
hm ca tng, hiu, tch, thng.- p dng: Tnh o hm ca hm s: a) 54 3 y x x
;b)1 32 5xyx+*Bi mi:Hot ng ca GV Hot ng ca HS Ni dungH1: HTP1: GV
nu v d dn dc HS n vi h qu 1 v 2.GV: Nu ta t k = 6 v u=29 3 5 x x +
+ th ta c cng thc nh th no? (Ch n o hm ca u).y chnh l ni dung ca h
qu 1 trong SGK, Gv nu H qu 1.Tng t i vi H qu 2HTP2:GV yu cu HS cc
nhm suy ngh chng minh cc cng thc ca h qu 1 v 2.HS ch theo di trn
bng lnh hi kin thcNu k = 6 v u = 29 3 5 x x + + th ta c cng
thc:(ku) = k.uHS tho lun theo nhm chng minh cng thc o hm trong h qu
1 v 2HS nhn xt, b sung v sa cha ghi chpV d: Chng minh rng:a)( ) (
)'26 9 3 5 6 18 3 x x x 1+ + + ]2) H qu: *H qu 1: Nu k l mt hng s
th: (ku) = k.u*H qu 2: ( )'21 '( ) 0vv v xv v _ ,H2: Tm hiu v o hm
ca hm hp:HTP1: Tm hiu v hm hp:GV v hnh minh ha v phn tch ch ra khi
nim hm hpV d: Hm s 22 3 y x x + +l hm hp ca hm s : + +2 v i2 3 y u
u x xHTP2: p dng:GV cho HS cc nhm tho lun tm li gii v d sau:GV gi
HS nhm khc nhn xt, b sung HS ch theo di trn bng lnh hi kin thcHS
tho lun theo nhoma v ghi li gii vo bng ph, c i din ln bng trnh by
li gii HS nhn xt, b sung v II. o hm ca hm hp:1)Hm hp: (SGK)u= g(x)
l hm s ca x, xc nh trn khong (a; b) v ly gi tr trn khong (c; d); hm
s y = f(u) xc nh trn khong (c; d0 v ly gi tr trntheo quy tc sau:( )
( )x f g x aTa gi hm ( ) ( )y f g x l hm hp ca hm s y = f(u) vi
u=g(x).*V d: Hm s sau l hm hp ca hm no?+21)1a yx( ) +122 4) 1 2 b y
x xGV sa cha v ghi li gii ng (nu cn)HTP3: o hm ca hm hp:GV nu nh l
4 v ghi cng thc ln bngGV nu v d v ghi ln bng v cho HS cc nhm tho
lun tm li gii .GV gi HS nhn xt, b sung (nu cn)GV sa cha v b sung
(nu cn).GV yu cu HS c lp xem bng tm tt cc cng thc o hm trong SGK
trang 162sa cha ghi chp.HS ch theo di trn bngHS tho lun v ghi li
gii vo bng ph, c i din ln bng trnh by (c gii thch)HS nhn xt, b sung
v sa cha ghi chpnh l 4: SGKV d: Tnh o hm ca cc hm s sau:( )+ +
+262;6)4 5) 1 3 ;) 4 5c yxa y xb y x*H3: Cng c v hng dn hc nh:*Cng
c:- Nhc li cc cng thc tnh o hm ca tng, hiu, tch, thng; cng thc tnh
o hm ca hm hp.- p dng gi bi tp 2 d) v 3 a).*Hng dn hc nh:- Xem li v
hc l thuyt theo SGK, nm chc cc cng thc tnh o hm thng gp.- Lm cc bi
tp 1 n 5 trong SGK trang 162 v
163.-----------------------------------
-----------------------------------Tit 68.I. Tin trnh bi hc:*n nh
lp, gii thiu- Chia lp thnh 6 nhm*Kim tra bi c: Tnh o hm ca hm s: a)
221xyx;b) 22 5 y xx*Bi mi:Hot ng ca GV Hot ng ca HS Ni dungH1:
HTP1: GV nhc li 3 bc tnh o hm ti mt im HS tho lun theo nhm tm li
gii.Bi tp 1: SGKBng nh ngha, tm o hm ca cc hm s sau:bng nh ngha.GV
cho HS cc nhm tho lun tm li gii bi tp 1SGK trang 162Gi HS nhn khc
nhn xt, b sung (nu cn)GV chnh sa v b sungHTP2: S dng cc cng thc o
hm ca tng, hiu, tch, thng.GV nhc li cc cng thc tnh o hm ca tng,
hiu, tch, thng.GV cho HS cc nhm tha lun tm li gii bi tp 2a) d)Gi HS
i din ln bng trnh by.Gi HS nhm khc nhn xt, b sung (nu cn)GV chnh sa
v b sungi din nhm ln trnh by li gii HS nhn xt, b sung v sa cha ghi
chpHS trao i v rt ra kt qu:) 1; )10. a b HS tho lun theo nhm .i din
nhm ln bng trnh by li gii (c gii thch)HS nhn xt, b sung v sa cha
ghi chp.HS trao i v rt ra kt qu:2a) y =x4-12x2 +2;d)y =-63x6 +
120x4.2030) 7ti1;) 2 1 ti2.a y x x xb y x x x + + Bi tp 2: SGKTm o
hm ca cc hm s:( )5 35 2) 4 2 3;) 3 8 3a y x x xb y x x + H2: HTP1:
Tnh o hm ca cc hm s bng cch s dng cc cng thc v tng, hiu, tch,
thng:GV cho HS cc nhm tho lun tm li gii bi tp 3.Gi HS i din ln bng
trnh by li gii v gi HS nhn xt, b sung (nu cn)GV chnh sa v b sungHS
tho lun theo nhm, c i din ln bng trnh by.HS nhn xt, b sung v sa cha
ghi chpHS trao i v rt ra kt qu:Bi tp 3: SGKTnh o hm ca cc hm s
sau:( )( ) ( )37 22 2) 5 ;) 1 5 3 ;ayx xbyx x +22322) ;13 5) ;1)xc
yxxdx xne y mx + _ + ,HTP2: GV phn tch v hng dn gi bi tp 4 b),
4c).HTP3: GV cho HS tho lun theo nhm tm li gii bi tp 5 SGK.Gi HS
nhn xt, b sung (nu cn)GV chnh sa v b sung ( ) ( )( )( )( )( )25 5
5222222223 2) ' 3 5 7 10 ;) ' 4 3 1 ;2 1) ' ;15 6 2) ' ;16) 'a y x
x xb y x xxc yxx xd yx xn ne y mx x + + _ + ,HS ch theo di trn bng
lnh hi kin thcHS tho lun theo nhm v c i din ln bng trnh by li gii
(c gii thch)HS nhn xt, b sung v sa cha ghi chpHS trao i v rt ra kt
qu:a) x 2;b) 1 2 1 2. x < < +Bi tp 5: SGKCho hm s 3 23 2 y x
x +. Tm x :a)y > 0;b) y < 3.H3: Cng c v hng dn hc nh:*Cng
c:-Nhc li cc bc tnh o hm bng nh ngha, cc cng thc tnh o hm ca tng,
hiu, tch, thng, cc cng thc o hm thng gp.*Hng dn hc nh:- Xem li cc
bi tp hc, nm chc cc cng thc tnh o hm hc;- Son trc bi mi: o hm ca hm
s lng gic.-----------------------------------
------------------------------------Tit 69,70,71: O HM CA HM S LNG
GICI. Mc tiu:Qua tit hc ny HS cn:1.V kin thc:- Bit (khng chng
minh)0sinlim 1xxx- Bit o hm ca hm s lng gic.2.V k nng:-Tnh c o hm
ca cc ca mt s hm s lng gic.3. V t duy v thi :Tch cc hot ng, tr li
cu hi. Bit quan st v phn on chnh xc, bit quy l v quen.II. Chun b ca
GV v HS:GV: Gio n, phiu HT (nu cn),HS: Son bi trc khi n lp, chun b
bng ph, III. Phng php:Gi m, vn p, an xen hot ng nhm.Tit 69.IV. Tin
trnh bi hc:*n nh lp, gii thiu- Chia lp thnh 6 nhm*Kim tra bi c: -Nu
cc bc tnh o hm bng nh ngha ca mt hm s y = f(x) ti x ty .*Bi mi:Hot
ng ca GV Hot ng ca HS Ni dungH1: HTP1: GV cho HS tho lun theo nhm
tm li gii v d H1 SGK/163.GV: Ta c nh l quan trng sau (tha nhn khng
chng minh) (GV nu nh lv ghi ln bng)HTP2:GV ly v d v cho HS tho lun
theo nhm tm li gii.Gi HS i din ln bng trnh by li gii ca nhm.GV chnh
sa v b sung...HS tho lun theo nhm v bm my tnh tm li gii.Kt qa:
sin0,010,9999833334;0,01sin0,0010,99999983330,001HS tho lun theo
nhm tm li gii v ca i din trnh by....HS nhn xt, b sung v sa cha ghi
chp...HS trao i rt ra kt qu :a) 1;b)5; c) 1.1. Gii hn ca sinxx:nh l
1: 0sinlim 1xxxV d: Tnh:0tan)lim ;xxax0sin5)lim ;1sin) lim
.1xxxbxxcx+H2: Tm hiu v o hm ca hm s y = sinx:2.Hm ca hm s y =
sinx:nh l 2: SGK.HTP1: GV nu nh l v hng dn chng minh tng t SGK.GV:
Da vo nh l 2 v da vo cng thc tnh o hm ca hm hp hy suy ra cng thc
tnh o hm ca hm s y = sinu vi u = u(x).GV ly v d minh ha v hng dn
gii.HTP2: GV nu v d p dng v yu cu HS cc nhm tho lun tm li gii.Gi HS
i din ln bng trnh by. GV chnh sa v b sung ...HS ch theo di lnh hi
kin thc...HS: Da vo nh l 2 v cng thc tnh o hm ca hm hp ta c:( )sin
' '.cos u u u HS ch theo di lnh hi phng php gii...HS tho lun theo
nhm tm li gii v c i din ln bng trnh by...HS nhn xt, b sung v sa cha
ghi chp...Hm s y = sinx c o hm ti mix v ( )sin ' os x cx Chng minh:
SGKCh : Nu y = sinu v u = u(x) th:( )sin ' '.cos u u u V d p
dng:Tnh o hm ca cc hm s sau:( )23) sin3 ;) sin 2a y xb y x H3: Tm
hiu v o hm ca hm s y = cosx:GV cho HS cc nhm tho lun tm li gii v d
H2.GV nu nh l v hng dn chng minh tng t SGK.GV: Da vo nh l 3 v da vo
cng thc tnh o hm ca hm hp hy suy ra cng thc tnh o hm ca hm s y =
cosu vi u = u(x).GV ly v d minh ha v hng dn gii.HS tho lun theo nhm
tm li gii ca v d H 2 v c i din ln bng trnh by...HS nhn xt, b sung v
sa cha ghi chp...HS: Da vo nh l 3 v cng thc tnh o hm ca hm hp ta
c:( )os ' '.sin cu u u HS ch theo di lnh hi phng php gii...HS tho
lun theo nhm tm li gii v c i din ln bng trnh by...2.Hm ca hm s y =
sinx:V d H2: SGKnh l 3: SGK.Hm s y = cosx c o hmti mix v ( )os '
sin cx x Ch : Nu y = cosu v u = u(x) th:( )os ' '.sin cu u u V d p
dng:Tnh o hm ca cc hm s sau:HTP2: GV nu v d p dng v yu cu HS cc nhm
tho lun tm li gii.Gi HS i din ln bng trnh by. GV chnh sa v b sung
...HS nhn xt, b sung v sa cha ghi chp...( )23) os3 ;) cos 2a y c xb
y x H4: Cng c v hng dn hc nh:*Cng c:- Nhc li cc cng thc tnh o hm ca
cc hm s sinx v cosx.- p dng gi bi tp 3a) SGK:*Tnh o hm ca hm s sau:
a) y = 5sinx 3 cosx.*Hng dn hc nh:- Nm chc cc cng thc v o hm hc;-
Xem li cc v d gi;- Son phn cn li ca bi v lm cc bi tp sau: 1; 2;
3b), 3d), 4a), b) c) v e).-----------------------------------
-----------------------------------Tit 70.I. Tin trnh bi hc:*n nh
lp, gii thiu- Chia lp thnh 6 nhm*Kim tra bi c: -Nu cc cc cng thc
tnh o hm ca cc hm s y = sinx v y = cosx, y = sinu v y = cosu-p dng
: Tnh o hm ca hm s sau:sin,os 2xy x k kcx _ + ,*Bi mi:Hot ng ca GV
Hot ng ca HS Ni dungH1: HTP1: GV: da vo vi d trn ta c nh l sau: (GV
nu nh l 4)GV da vo cng thc tnh o hm ca hm hp hy suy ra o hm ca hm s
y =tanu vi u = HS ch theo di lnh hi kin thc...HS suy ngh nu cng
thc...( )2'tan 'cosuuuHS ch theo di lnh 4. o hm ca hm s y = tanx:nh
l 4: (SGK)( )21tan ',os 2x x k kc x _ + ,Ch : SGK( )2'tan 'cosuuuV
d: Tnh o hm ca hm u(x).HTP2:GV nu v d minh ha v hng dn
gii...HTP3:GV nu v d (hoc pht phiu HT) v cho HS cc nhm tha lun tm
li gii Gi HS i din ln bng trnh by, gi HS nhn xt, b sung (nu cn).GV
chnh sa, b sung ...hi kin thc...HS tho lun theo nhm tm li gii ...HS
i din trnh by li gii (c gii thch)HS trao i rt ra kt qu :...s:y=
tan(4x3 7x +1)Phiu HT 1:Tnh o hm ca cc hm s:a) y = tan(3 4x4);b) y
= tan(5x3 + 2).H2:HTP1:GV cho HS cc nhm tho lun tm li gii v d H 4 v
gi HS i din ln bng trnh by li gii.Gi HS nhn xt, b sung (nu cn)GV
chnh sa v b sung ...GV ta c: tan ?2x _ ,GV nu nh l 5(SGK)GV da vo
cng thc tnh o hm ca hm hp hy suy ra o hm ca hm s y =cotu vi u =
u(x).HTP2:GV nu v d minh ha v hng dn gii...HTP3:HS tho lun theo nhm
tm li gii v d H4 ...HS i din ln bng trnh by (c gii thch)HS nhn xt,
b sung ...HS ch theo di trn bng lnh hi kin thc...HS suy ngh nu cng
thc...( )2'cot 'sinuuuHS ch theo di lnh hi kin thc...HS tho lun
theo nhm tm li gii ...5. o hm ca hm s y = cotx:V d H5: SGKnh l 5:
(SGK)( ) ( )21cot ',sinx x k kx Ch : SGK( )2'cot 'sinuuuV d: Tnh o
hm ca hm s:y= cot(3x3 7)Phiu HT 1:Tnh o hm ca cc hm s:GV nu v d
(hoc pht phiu HT) v cho HS cc nhm tha lun tm li gii Gi HS i din ln
bng trnh by, gi HS nhn xt, b sung (nu cn).GV chnh sa, b sung ...HS
i din trnh by li gii (c gii thch)HS trao i rt ra kt qu :...a) y =
cot(5 4x4);b) y = cot(x3 + 2).H3: Cng c v hng dn hc nh:*Cng c: GV:
Gi HS nhc li cc cng thc tnh o hm hc;GV: Ghi l bng cc cng thc o hm
nh bng o hm trang 168 SGK.p dng: Gii bi tp 1a) 2b) v 3c) SGK.*Hng
dn hc nh:- Xem li v hc l thuyt theo SGK;- Xem li cc v d v bi tp
gii;- Lm cc bi tp t bi 1 n bi 8 SGK trang 168 v
169.-----------------------------------
------------------------------------ Tit 71.I. Tin trnh bi hc:*n nh
lp, gii thiu- Chia lp thnh 6 nhm*Kim tra bi c: -Nu cc cng thc tnh o
hm m em hc.-p dng cng thc tnh o hm hy gii bi tp 1b)*Bi mi:Hot ng ca
GV Hot ng ca HS Ni dungH1: HTP1:GV cho HS cc nhm tho lun tm li gii
cc bi tp 1c) v 1d).Gi HS i din trnh by li gii.GV gi HS nhn xt, chnh
sa v b sung...HS tho lun theo nhm tm li gii v c i din ln bng trnh
by...HS nhn xt, b sung v sa cha ghi chp...HS trao i v rt ra kt
qu:Bi tp 1: SGK trang 168 v169HTP2:GV phn tch v hng dn gii bi tp
2a) v yu cu HS lm bi tp 2c) tng t....GV cho HS tho lun theo nhm v
giHS ln bng trnh by li gii ...Gi HS nhn xt, b sung (nu cn)GV chnh
sa v b sung ...( )( )( )222222 2 3 91 ) ' ;3 410 6 91 ) ' .3x xc
yxx xd yx x +HS tho lun thoe nhm v c i din ln bng...HS nhn xt, b
sung ...HS trao i v rt ra kt qu:Tp nghim:( ) ( ) 1;1 1;3 S Bi tp 2:
SGKH2: HTP1:GV cho HS 6 nhm tho lun tm li gii bi tp 3 v gi HS i din
ln bng trnh by li gii ...Gi HS nhn xt, b sung ...GV chnh sa, b sung
v nu li gii ng...HTP2:GV hng dn v gi bi tp 5 SGKHS tho lun v c i
din trnh by li gii...HS nhn xt, b sung ...HS trao i v rt ra kt
qu:....HS ch theo di lnh hi kin thc...HS ch theo di trn bng lnh hi
kin thc...Bi tp 3: SGKTm o hm ca cc hm s:) 5sin 3 os ;sin os) ;sin
os) .cot ;) 1 2tan .a y x cxxcxb yxcxc y x xe y x + +Bi tp 5:
SGK...H3: HTP1:GV cho HS tho lun theo nhm tm li gii bi tp 6 v gi HS
i din ln bng trnh by.Gi HS nhn xt, b sung (nu cn)GV chnh sa v b
sung ...(GV gi :a) Dng hng ng thc:HS tho lun theo nhm tm li
gii...HS nhn xt, b sung v sa cha ghi chp...HS trao i rt ra kt
qu:...a) y = 0b) y = 0Bi tp 6: SGK( ) ( )3 3 2 2a b aba abb + +
+b)S dng cng thc cung gc b nhau: 2 2 v;v 3 3 3 3x x x x + +)HTP2:
GV phn tch v hng dn gii bi tp 7 v 8 (nu cn thi gian)HS ch thoe di
trn bng lnh hi kin thc...H4: Cng c v hwngs dn hc nh:*Cng c:Nhc li
cc cng thc tnh o hama v cc cng thc o hm ca mt s hm s c bit.*Hng dn
hc nh:-Xem li cc bi tp gii;- Xem v son trc bi: . Vi Phn
-----------------------------------
------------------------------------Tit 72: Kim tra mt titTit 73:VI
PHNI. Mc tiu:Qua bi hc HS cn:1)V kin thc v k nng:Bit v nm vng nh
ngha vi phn ca mt hm s:( ) ( ) 'hay' dy f x x dy f x dx - p dng gii
c cc bi tp c bn trong SGK;- ng dng c vi phn vo php tnh gn ng.2. V t
duy v thi :Tch cc hot ng, tr li cu hi. Bit quan st v phn on chnh
xc, bit quy l v quen.II. Chun b ca GV v HS:GV: Gio n, phiu HT (nu
cn),HS: Son bi trc khi n lp, chun b bng ph, III. Phng php:Gi m, vn
p, an xen hot ng nhm.IV. Tin trnh bi hc:*n nh lp, gii thiu: Chia lp
thnh 6 nhm.*Bi mi:Hot ng ca GV Hot ng ca HS Ni dungH1:HTP1: V d dn
n nh ngha vi phn.GV cho HS cc nhm tho lun tm li gii v d H 1 trong
SGK.GV:Hy p dng nh ngha trn vo hm s y = x ?GV : Do dx =x nn vi hm s
y = f(x) ta c:dy = df(x) = f(x) x =f(x)dxHTP2:GV nu v d p dng v gi
HS ln bng trnh by...Gi HS nhn xt, b sung (nu cn)GV nhn xt, chnh sa
v b sung....HS tho lun thoe nhm tm li gii.C di din ln bng trnh byHS
nhn xt, b sung v sa cha ghi chp...HS trao i rt ra kt qu:...HS suy
ngh trnh by:dx = d(x)=(x) x =x HS tho lun theo nhm v c i din ln bng
trnh by.HS nhn xt, b sung v sa cha ghi chp. HS ch trn bng lnh hi
kin thc.1. nh ngha: (Xem SGK)Cho hm s y= f(x) xc nh trn khong (a;b)
v c o hm ti ( ) ; x a b . Gi sx l s gia ca x.Ta gi f(x) x l vi phn
ca hm s y = f(x) ti x ng vi s giax K hiu: df(x) hoc dy, tc l:dy =
df(x) = f(x) x .V d: Tm vi phn ca cc hm s sau:a) y = x4- 2x2 +1b) y
= cos2xH2: HTP1:GV nu v phn tch 2. ng dng o hm vo php tnh gn
ng:Theo nh ngha o hm, ta c:tm cng thc tnh gn ng.HTP2:GV nu v d v
cho HS tho lun theo nhm.Gi HS i din ln bng trnh by li gii.Gi HS nhn
xt, b sung v sa cha ghi chp.GV nhn xt, chnh sa v b sung...HS ch
theo di lnh hi kin thc....HS tho lun theo nhm tm li gii v c i din
ln bng trnh by....HS nhn xt, b sung v sa cha ghi chp.0'( ) limxyf
xx( ) ( )( ) ( ) ( )0 0 0 nh th' ' .'(1)xyf x y f x xxf x x f x f x
x + + (1) l cng thc gn ng n gin nht.V d: Tnh gi tr gn ng ca:3,99Li
gii:t ( ) ( )( ) ( ) ( ) ( ) ( )( )1'23, 99 4 0.01 4 ' 4 0, 0113,
99 4 0,01 4 . 0,01 1, 99752 4f x x f xxf f f f + + H3: Bi tp p
dng:GV cho HS tho lun theo nhm tm li gii bi tp 1 v 2 SGK trang
171.Gi Hs i din cc nhm ln bng trnh by li gii.Gi HS nhn xt, b sung
(nu cn).GV nhn xt, chnh sa v b sung ...HS tho lun theo nhm tm li
gii v c i din ln bng trnh by (c gii thch)HS nhn xt, b sung v sa cha
ghi chp.Ch theo di lnh hi kin thc.Bi tp:1)Tnh vi phn ca cc hm s
sau:( ) ( )2 2) ( ,l hng s);) 4 1 .xa y a babb y x x x x+ + + 2) Tm
dy, bit:a) y = tan2x;b) 2cos.1xyx H4: Cng c v hng dn hc nh:*Cng c:-
Nhc li cng thc tnh vi phn ca mt hm s, cng thc tnh gn ng.*Hng dn hc
nh:-Xem li v hc l thuyt theo SGK, cc bi tp gii.- Xem v son trc bi:
5. o hmcp 2. -----------------------------------
------------------------------------Tit 74: O HM CP HAII. MC TIU:
Qua bi hc gip hc sinh: 1.V kin thc:- Nm c cng thc tnh o hm cp 2 ca
hm s y = f(x) - Nm c ngha hnh hoc; ngha vt lo hm cp mt v y ngha c
hc ca o hm cp hai tm gia tc tc thi ti thi im t ca chuyn ng.- Bc u
vn dng c cng thc tnh o hm cp cao tnh cc o hm n gin- Nm c nh ngha o
hm cp hai; o hm cp n ca hm s y = f(x)-Hiu c ngha c hc ca o hm cp
hai -Nm vng cc cng thc tm o hm cc hm s lng gic. 2.V k nng: - Gip hc
sinh c k nng thnh tho trong vic tnh o hm cp hu hn ca mt s hm s thng
gp - Bit cch tnh o hm cp n ca mt s hm n gin nh hm a thc , hm 1a x +
by v cc hm s y = sinax ; y = cosax ( a l hng s )3.V t duy v thi : -
Tch cc tham gia cc hot ng xy dng ni dung bi hc - Bit quan st v phn
on chnh xc cc ni dung v kin thc lin quan n ni dung ca bi hc , bo m
tnh nghim tc khoa hc.II.CHUN B: - Gio vin: Son bi, dng c ging dy ,
my chiu - Hc sinh: Son bi, nm vng cc kin thc hc v cch xc nh o hm
bng nh ngha v cng thc tnh o hm ca hm s y = sinx, lm bi tp nh, chun
b cc dng c hc tp. III.PHNG PHP DY HC : - Gi m vn p thng qua cc hot
ng iu khin t duy , an xen hot ng nhm . - Pht hin v gii guyt vn .IV.
TIN TRNH BI HC : Kim tra bi c : Cho hm s f(x) = x3 x2 + 1- Tnh
f/(x)- Tnh [f/(x)]/ Bi mi :Hot ng ca GV Hot ng ca HS Ghi bng- Gi
thiu bi hc , t vn vo bi thng qua phn kim tra bi cH1: .- Gi thiu o
hm cp hai ca hm s y = f(x) da trn phn kim tra bi c - Cng c nh ngha
trn c s cho hc sinh gii cc v d v H1 : sgk.V d1: Gai bi tp 42/218sgk
f(x) = x4 cos2x f(x) = (x +10)6V d2: Gai H1 sgk- Tr li cc cu hi kim
tra f(x) = x3 x2 + 1 f/(x) = 3x2 2x[f/(x)]/ = 6x- 4- Theo di, ghi
nhn ni dung Tham gia tr li cc cu hi - Rt ra qui tc tnh o hm cp hai
ca hm s y = f(x)- Tin hnh gii bi tp sgk f(x) = x4 cos2xf/(x) = 4x3+
2sin2xf//(x) = 12x2+ 2cos2xf///(x) = 24x- 4sin2x f(x) = (x
+10)6f/(x) = 6(x +10)5f//(x) = 30(x +10)4f///(x) = 120(x
+10)3f(4)(x) = 360(x +10)2f(5)(x) = 720(x +10)f(6)(x) = 7201. o hm
cp hai : a. nh ngha: (Sgk)f/(x) gi l o hm cp mt ca y = f(x)f//(x)
gi l o hm cp hai ca y = f(x) f(n)(x) gi l o hm cp n ca y = f(x)b. V
d1: Tm o hm ca mi hm s sau n cp c cho km theo f(x) = x4
cos2xf(4)(x) = 48- 8cos2x f(x) = (x +10)6f(6)(x) = 720 Cho hm s y
=x5.Tnh y(1); y(2); y(5) ; y(n) y/ = 5x4;y// =20x3 . y(5) = 120Vy
y(n)(x) = 0 (vi n >5)c. V d 2: H1 : sgk.Hot ng ca GV Hot ng ca
HS Ghi bngH2: Gi thiu ngha c hc ca o hm cp 2- Cho hs nhc li ngha o
hm cp mtGii thiu nghao hm cp hai- Gi thiu gia tc tc thi ti thi im
t0 ca chuyn ng - Gi thiu cng thc tnh gia tc tc thi ti thi im t0 ca
chuyn - Theo di, ghi nhn ni dung - Tham gia tr li cc cu hi - Rt ra
qui tc tnh gia tc tc thi ti thi im t0 ca chuyn ng- Tin hnh gii bi
tp sgk a(t) = v/(t) = 8 + 6t v(t) = 11m/s2. ngha c hc ca o hm cp
2a. Gia tc tc thiXt chuyn ng s = s(t) ( )00limtvatt l gia tc tc thi
ti thi im t0 ca chuyn ng ( ) ( )/0 0at s t b. V d1:Gai bi tp
44/218sgkng - Cng c ngha c hc ca o hm cp 2 trn c s cho hs gii cc v
d v H2 : sgk.V d1: Gai bi tp 44/218sgk v(t) = 8t + 3t2 V d 2: H1 :
sgk218 3 1111/ 3tt tt +
- Tin hnh suy lun nu kt qu v gii thch - Theo di, ghi nhn ni dung
cc cu hi cng c ca GV - - Tham gia tr li cc cu hi a(4) = v/(4) =
32m/s2 t = 1s th a(1) = 14m/s2c. V d 2: H1 : sgk.Hot ng ca GV Hot
ng ca HS Ghi bngH3: .- Gi thiu o hm cp cao ca hm s y = f(x) trn c s
o hm cp hai Lu : Cc bc khi tnh o hm cp n ca hm s y = f(x)Tnh f/(x)
; f//(x) ; f///(x)Tm qui lut v du , h s v bin s tm ra o hm cp n-
Cng c o hm cp cao trn c s cho hc sinh gii cc v d v H3 : sgk.V d1:
Gai bi tp 42/218sgk f(x) = (x +10)6V d2: Gai H3 sgkH4 : Cng c l
thuyt- Hc sinh nhc li cc cng thc tnh o hm cp hai v o hm cp n ca hm
s y = f(x)- Theo di, ghi nhn ni dung Tham gia tr li cc cu hi - Rt
ra qui tc tnh o hm cp o hm cp n ca hm s y = f(x)- Tin hnh gii bi tp
sgk f(x) = (x +10)6f(6)(x) = 720f(n)(x) = [f(n-1)(x)]/3. o hm cp
cao : a. nh ngha: (Sgk)f(n)(x) gi l o hm cp n ca y = f(x)f(n)(x) =
[f(n-1)(x)]/b. V d1: Tm o hm cp n ca cc hm s sau f(x) = (x
+10)6f(n)(x) = 0 f(x) =cosxc. V d 2: H3 : sgk. f(x) =sinx( )( )
sin2nnf x x _ + ,H5 : Luyn tp thng qua cc cu hi trc nghim khch quan
v t lun theo nhm- Cu hi t lun theo nhmHot ng ca GV Hot ng ca HS-
Chia hc sinh thnh cc nhm nh. mi nhm gm 4 hc sinh - Phn chia thnh
hai nhm chnh nhm trao i gii cng mt lc hai bi tp sgk- Giao nhim v
cho mi nhm gii mt bi tp Bi tp 43/219sgk : Chng minh vi mi 1 n ta c
: a. y = ( )( )( )( )11 . !1 nnnnf x thf xx x+ b. y =( )( )( )4 4s
sinn nf x inax th f x a ax Lu : ( ) ( )/21 1f x f xx x v o hm cc hm
s y = sin u(x) v y = cosu(x) lm bi- Yu cu cc nhm tin hnh trao i v
trnh by bi gii vo bng ph - Chn mt s nhm c ni dung hay d sai hay ng
ln trnh by - Cho hc sinh tham gia ng gp kin v cc bi lm ca cc
nhm---- Nhn xt kt qu bi lm ca cc nhm , pht hin cc li gii hay v nhn
mnh cc im sai ca hs khi lm bi - Ty theo ni dung bi lm ca hc sinh,
GV hon chnh ni dung bi gii . Nu ni dung trnh by kh v cha p mt GV
trnh chiu kt qu chun b .- Ch cch phn chia nhm v ni dung cu hi ca
nhm do Gv phn cng - c hiu yu cu bi ton.- Theo di, ghi nhn cc kin
thc gi ca Gv - Tho lun nhm tm kt qu-Tin hnh lm bi theo nhm - i din
nhm trnh by kt qu bi lm ca nhm - Nhn xt kt qu bi lm ca cc nhm v gp
nhm hon thin ni dung ca bi gii- Theo di v ghi nhn cc phn tch ca cc
bn v ca thy gio -* Cu hi Trc nghim khch quanCu 1 :o hm cp n ca hm s
11yx+ l:A.( )( )11 . !( 1)nnnnyx++B.( )1. !( 1)nnnyx++C.( )( )11 .(
1)nnnyx++D.( )1( )11 . !( 1)nnnnyx+++Cu 2 :o hm cp n ca hm s ( ) ln
1 y x + l:A.( ) ( )1( )1 . 1 !( 1)nnnnyx +B.( ) ( )1( )11 . 1 !(
1)nnnnyx +C.( ) ( )1( )1 . 1 !( 1)nnnnyx+ ++D.( ) ( )1( )11 . 1 !(
1)nnnnyx++ ++Cu 3 :o hm cp n ca hm s ( )11yx x l:A.( )( )1 11 . !.
!1nn nnnxx+ ++( )( )1 11 . !. !1nn nnnxx+ +( )( )1 11 . !. !1nn
nnnxx+ ++Kt qu khcCu 4 :o hm cp n ca hm s y = cosx l:A.( )cos( .
)2ny x n +B.( )cos( . )ny x n + C.( )sinny x D.( )cosny x Cu 5 : o
hm cp n ca hm s y = sin3x l y(n) bng ::A.3 sin(3 . )2nx n +B.3
cos(3 . )2nx n +C.3 sin(3 . )2nx n +D.3 cos(3 . )2nx n +Cu 6 : o hm
cp n ca hm s y = sinax lA.sin( . )2na ax n +B.cos( . )2na ax n +C.
-sin( . )2na ax n +C. -cos( . )2na ax n +Cu 7 : o hm cp 2010 ca hm
s y = cosx l :A. sinx B. cosx C. -cosx D. -sinxCu 8 : o hm cp 2007
ca hm s y = cosx l :A. -cosx B. -sinx C. cosx D. sinxCu 10 :o hm cp
n ca hm s y = cos2x l:A.x cos y) n (B.( )cos( . )ny x n +C.( )sinny
x D.( )2 cos( . )2n ny x n +H6 : Hng dn v dn d bi tp chun b cho tit
hc sau -Nm vng cc cng thc tm o hm cc hm s thng gp , cc hm s lng gic
v o hm cp cao. - Gii cc bi tp n tp chng V.
-----------------------------------
------------------------------------Tit 75: CU HI V BI TP N TP CHNG
V.I. MC TIU: Qua bi hc , hc sinh cn nm c:1)V kin thc: - Nm vng cc
cng thc tm o hm cc thng gp, o hm cc hm s lng gic v o hm cp cao. -
Nm vng cc ngha hnh hc v ngha c hc ca o hm2)V k nng: - Gip hc sinh
vn dng thnh tho cng thc tm o hm v ngha ca o hm vo vic gii cc bi ton
lin quan n o hm 3)V t duy v thi : - Tch cc tham gia cc hot ng xy
dng ni dung bi hc - Bit quan st v phn on chnh xc cc ni dung v kin
thc lin quan n ni dung ca bi hc , bo m tnh nghim tc khoa hc.II.
CHUN B : - Gio vin: Son bi, dng c ging dy , my chiu - Hc sinh: Nm
vng cc kin thc hc trong chng o hm v vn dng cc kin thc gii cc bi tp
n tp chng III. PHNG PHP DY HC : - Thng qua hot ng kim tra cc kin
thc hc gii v sa cc bi tp sgk. - Pht hin v gii guyt vn sai ca hc
sinh nhm khc phc cc im yu ca hc sinh khi tin hnh gii bi tp.Tit
74:IV. TIN TRNH BI DY:*n nh lp, gii thiu-Chia lp thnh 6 nhm.*Kin
tra bi c: Kt hp vi iu khin cc hot ng nhm.*Bi mi:Hot ng ca GV v HS
Ni dung H1: Kim tra v n luyn kin thc v o hm s hc - Nu cng thc tnh o
hm hm s thng gp v o hm cc hm s lng gic - Trnh chiu cc cng thc tnh o
hm ca cc hm s hc v hm s hp ca chngI. n luyn l thuyt v cng thc tnh o
hm ca cc hm s : 1. Cc qui tc tnh o hm : ( )// /u v u v t t ( ) ( )/
// / /. ku u v u v v u v ku + // // u u v v uv v _ , ' ' ' . x u xy
y u2. o hm ca cc hm s thng gp : (u = u(x)) ( C )/ = 0 ( C l hng s )
( x )/ = 1 (xn)/ = nxn - 1 (n2 ;nN)/21 1x x _ , vi0 x ( )/12xxvi (x
> 0) (un)/ = nun 1u/ //21 uu u _ , vi0 x ( )//2uuu = x 21
vi (x > 0)3. o hm ca cc hm slng gic : (u = u(x)) (sinx)= cosx
(cosx)= -sinx xx2/cos1) (tan xx2/sin1) (cot (sinu)= cosu.u/ (cosu)/
= - sinu. u/uuu2//cos) (tan uuu2//sin) (cot II. n luyn bi tp v cng
thc tnh o hm ca H2:Vn dng cc kin thc v o hm gii cc bi tp n tp chng
o hm Gi nhiu HS gii nhanh Bi tp - HS tin hnh gii cc bi tp- GV kim
tra bi tp HS - HS theo di v gp di s dn dt ca GV hon thnh ni dung bi
tp- GV rt ra nhn xt v cch gii ca hs v nu cc cch gii hay v nhanh Hng
dn hs cch tm o hm cp cao ca hm s y = f(x) Lu : Cc bc khi tnh o hm
cp n ca hm s y = f(x)Tnh f/(x) ; f//(x) ; f///(x)Tm qui lut v du ,
h s v bin s tm ra o hm cp n- Gi nhiu hs gii Bi tp - Cng c o hm cp
cao trn c s sa bi tp ca HS. Gup hs tm c qui lut khi tnh o hm cp cao
H3 : Kim tra v n luyn kin thc v ngha ca o hm - Nu ngha hnh hc ca o
hm - Nu phng trnh tip tuyn ca th hm s y = f(x) ti im M0(x0; y0)- p
dng gii Bi tp - HS tin hnh gii cc bi tp- HS theo di v gp di s dn dt
ca GV hon thnh cc hm s : 1. Tnh o hm ca cc hm s sau : a.4 3/ 3 25
12 1: 2 54 3 2 + + + x xy x KQ y x xx
b. ( )2 2 2 2/23 2 3:11+ + x x a x x ay KQ yxx c.( )2 / 22 cos 2
sin: sin + y x x x x KQ y x xd. 2 2 /3 2 2sintan tan: 2cos cos _ +
+ ,x xy x x KQ yx x 2. Tnh o hm cp caoca cc hm s sau : a.// / / /
/sin cos: sin cos y x y xKQ y x y x b.( )1sin sin5 cos 4 cos 62y x
x x x ( )( )4: 128cos 4 648cos 6 KQ y x x xc.( )( )( ) ( )54: 0 n 6
ny x KQ y xe. ( )( )( )( )11 . ! 1 :2 1 2 1+ + +nnnny KQ y xx xIII.
n luyn v ngha ca o hm : 1. Phng trnh tip tuyn ca th hm s y = f(x)
ti im M0(x0; y0) l : ( ) ( )/0 0 0y f x x x y + 2. p dng gii bi tp
7 SGK trang 176.ni dung bi tp. Hot ng ca GV Hot ng ca HS Ghi
bngH1:GV cho HS cc nhm tho lun tm li gii bi tp 2 trong SGK trang
176. Gi HS a din ln bng trnh by.Gi HS nhn xt, b sung (nu cn)GV nhn
xt, chnh sa v b sung ...HS tho lun theo nhm tm li gii v c i din ln
bng trnh by (c gii thch)HS nhn xt, b sung v sa cha ghi chp.HS ch
theo di lnh hi kin thc.Bi tp 2: SGKTnh o hm ca cc hm s sau:os) 2
sin ;3 os) ;2 12 os -sin) .3sin oscxa y x xxcxb yxcd yc ++H2: Gii
bi tp 5SGKGV cho HS tho lun theo nhm v gi HS i din ln bng trnh
by.Gi HS nhn xt, b sung (nu cn)GV nhn xt, chnh sa v b sung...HS tho
lun v c i din ln bng trnh by.HS nhn xt, b sung v sa cha ghi chp.HS
trao i v rt ra kt qu:K0 xTa c:( )22 64 244 260 192'( ) 33 60 192='
0 3 60 192 02; 4xf xx xx xxf x x xx x + + + t tVy tp nghim:{ } 4;
2;2;4 S Bi tp 5:Gii phng trnh f(x) = 0, bit rng:( )360 643 5 f x xx
x + +H3: Gi bi tp 9 SGK.GV cho HS tho lun theo nhm v gi HS i din ln
bng trnh by.Gi HS nhn xt, b sung (nu cn)GV nhn xt, chnh sa HS tho
lun v c i din ln bng trnh by li gii.HS nhn xt, b sung v sa cha ghi
chp.HS trao i rt ra kt qu:Bi tp 9: SGK.Cho hai hm s:21 v y=2
2xyxVit phng trnh tip tuyn vi th ca mi hm s cho ti giao v b
sung.012;222 ;21 . 2 1 nn hai 2tip tuyn vung gc i nhauVy gc gia hai
ng thng l:90y xy xVv + im ca chng. Tnh gc gia hai tip tuyn k trn.H
4 : Cng c v hng dn hc nh:*Cng c: Nhc li cc cng thc tnh o hm hc;
Phng trnh tip tuyn ca mt ng cong ti mt im, song song, vung gc vi mt
ng thng, vi phn, o hm cp hai,...*Hng dn hc nh:- Xem li cc bi tp
gii, hc v nm chc cng thc o hm, o hm cp hai, vi phn v phng trnh tip
tuyn.- Lm trc cc bi tp cn li trong phn n tp cui
nm.-----------------------------------
------------------------------------Tit 76. CU HI V BI TP N TP CUI
NM..Mc tiu :Qua bi hc HS cn :1)V kin thc :-HS h thng li kin thc hc
c nm, khc su khi nim cng thc cn nh.2)V k nng :-Vn dng c cc pp hc v
l thuyt hc vo gii c cc bi tp- Hiu v nm c cch gii cc dng ton c
bn.3)V t duy v thi :Pht trin t duy tru tng, khi qut ha, t duy
lgic,Hc sinh c thi nghim tc, say m trong hc tp, bit quan st v phn
on chnh xc, bit quy l v quen.II.Chun b ca GV v HS:GV: Gio n, cc dng
c hc tp,HS: Son bi trc khi n lp, chun b bng ph (nu cn), III. Phng
php: V c bn l gi m, vn p, an xen hot ng nhm.*n nh lp, gii thiu,
chia lp thnh 6 nhm.*Bi mi:Hot ng ca GV D kin hot ng ca HS H1 : n tp
kin thc :GV cho HS tho lun theo nhm tm li gii cc bi tp t bi 1 n bi
18 trong phn cu hi.GV gi HS ng ti ch trnh by.Gi HS nhn xt, b sung
(nu cn).HS ch theo di lnh hi kin thc.HS tho lun theo nhm tm li gii
v c i din ng tich trnh by.HS nhn xt, b sung v sa cha ghi chp.H2 :GV
cho HS tho lun v gii bi tp 1 trong SGK.Gi HS i din trnh by li
gii.Gi HS nhn xt, b sung (nu cn)GV nhn xt, chnh sa v b sung..LG
:a)cos2(x+ k ) = cos(2x + 2k) = cos2x.b)y = -2sin2x3 1' 2. 3;3 2 3
2y y _ _ , ,Phng trnh tip tuyn ca (C) ti 3xl :3 133 2y x + Bi tp 1:
SGKCho hm s : y = cos2x.a) Chng minh rng cos2(x + k) = cos2x vi mi
s nguyn k. T v th (C) ca hm s y = cos2x.b) Vit phng trnh tip tuyn
ca th (C) ti im c honh 3x.c) Tm tp xc nh ca hm s :21 os21 os 2c xyc
x+H3 : GV cho HS tho lun tm li gii bi tp 13 SGK trang 180. Gi HS i
din ln bng trnh by li gii.Gi HS nhn xt, b sung (nu cn)GV nhn xt,
chnh sa v b sungHS tho lun theo nhm tm li gii v c i din ln bng trnh
by.HS nhn xt, b sung v sa cha ghi chp.HS trao i v rt ra kt qu :a) 4
; b)116 ; c)- ; d)- ; e) 2 ; f)13 ;g)+.H3: Cng c v hng dn hc
nh:-Xem li cc bi tp gii v h thng li kin th c bn trong phn n tp cui
nm.- Lm tip cc bi tp 3, 10, 14, 15, 17 v 19 SGK trang 179, 180 v
181.
-----------------------------------------------------------------------Tit
77.KIM TRA HC K III.Mc tiu:1)V kin thc:-Cng c li kin thc c bn ca nm
hc. 2)V k nng:-Lm c cc bi tp ra trong kim tra.-Vn dng linh hot l
thuyt vo gii bi tp3)V t duy v thi :Pht trin t duy tru tng, khi qut
ha, t duy lgic,Hc sinh c thi nghim tc, tp trung suy ngh tm li gii,
bit quy l v quen.II.Chun b ca GV v HS:GV: Gio n, cc kim tra, gm 4 m
khc nhau.HS: i s: n tp k kin thc trong chng IV v V. HH: n tp k kin
thc trong chng II v III.IV.Tin trnh gi kim tra:*n nh lp.*Pht bi kim
tra: Bi kim tra gm 2 phn:Trc nghim gm 16 cu (4 im)T lun gm 3 cu (6
im)*Ni dung kim tra:S GIO DC V O TOTRNG THPT VINH LC THI HC K II -
MN TONLP 11 C BNNm hc: 2007 - 2008Thi gian lm bi: 90 pht; (16 cu
trc nghim)H, tn th
sinh:.............................................................Lp
11B....I. Phn trc nghim: (4 im)Cu 1: Gii hn sau bng bao nhiu: 3lim2
nA. 3 B. 32C. 0 D. +Cu 2: 12 1lim1xxx bng:A. -2 B. 0 C. +D. Cu 3:
Cho hm s:( )4 nu4 2 1 3 nu4xxf xxm x + 'Hm s cho lin tc ti x = 4
khi m bng:A. 3 B. -2 C. 2 D. -3Cu 4: Gii hn sau bng bao nhiu:
223lim1 2xxx++A. 3 B. 0 C. 12D. -2Cu 5: Cho hm s( ) 2 f x x Chn mnh
ng trong cc mnh sau:A. ( )4lim 2xf xB. ( )0lim 2xf x C. ( )1lim 1xf
xD. ( ) limxf x +Cu 6: Cho hm s( )1.1f xx Chn kt qu sai:A. Hm s lin
tc ti mi1 x B. Hm s lin tc ti mi( ) 1; x +C. ( )2lim 1xf xD. (
)32lim2xf xCu 7: Cho hm s( )203; 1; f x x x x Chn s gia tng ng y di
y cho thch hp:A.( )210 y x B.( )21 2 y x + +C.( )21 10 y x + D.(
)21 1 y x + Cu 8: Cho hm s( )25 f x x +. Phng trnh tip tuyn vi th
ca hm s ti im M0 c honh x0 = -1 l:A.( ) 2 1 6 y x +B.( ) 2 1 6 y x
+ C.( ) 2 1 6 y x + +D.( ) 2 1 6 y x +Cu 9: Vi( )21 f x x th( ) ' 2
f l kt qu no sau y:A. Khng tn ti B. ( )2' 23f C. ( )2' 23f D. ( )2'
23f Cu 10: Hm s 22cos y x c o hm l:A. 2' 2sin y x B. 2' 4 . os y
xcx C. 2' 2 .sin y x x D. 2' 4 .sin y x x Cu 11: Cho ng thng ( ) a
v ng thng( ) b .Mnh no sau y ng?A. Nu( ) //( ) th a//b B. Nu( ) //
( ) th a// ( ) v b// ( ) C. Nu a//b th( ) // ( ) D. a v b cho
nhau.Cu 12: Cho hnh lp phng ABCD.ABCD, tuuur rDA a ,uuur rBA b, '
AA c uuur r.Khng nh no sau y ng ?A.' + +uuuur r r rAC a b c B.'
+uuuur r r rAC a b c C.' uuuur r r rAC a b c D.' +uuuur r r rAC a b
cCu 13: Cho hai ng thng a , b v mp( ) . Khng nh no sau y ng ?A. Nu
( ) / / b th tn ti '( ) bv '/ / b bB. Nu ( ) / / b v( ) ct a th b
ctaC. Nu ( ) a v ( ) / / bth/ / a bD. Nu a v b cng song song vi( )
th a v b song song vi nhau.Cu 14: Cho a,b nm trong ( ) v a,b nm
trong ( ) .Mnh no sau y ng ?A. Nu a//b v a//b th ( ) //( ) B. Nu( )
//( ) th a//a v b//bC. Nu a//a v b//b th ( ) //( ) D. Nu a ct b, ng
thi a//a v b//b th ( ) //( ) Cu 15:Cho hnh chp S.ABCD c y ABCD l
hnh vung tmO .Bit SA=SB=SC=SD. Khng nh no sau y sai ?A. ( ) SO ABCD
B. ( ) AB SAC C. ( ) AC SBD D.SD AC Cu 16: Cho mt phng () v hai ng
thng a v b. Khng nh no sau y ng ?A. Nu/ /( ), a b ath ( ) b B. Nu/
/( ), / /( ) a bth/ / b aC. Nu/ /( ), ( ) a bth a b D. Nu( ), a b
ath / /( ) bII. Phn t lun: (6 im)*i s:Cu 1: (2 im)a) Tnh gii hn: 22
3lim5 1xx x xx+ + +b) Tnh '''(2) f bit:( )5( ) 2 3 f x x . Cu 2: (2
im)Cho ng cong (C) c phng trnh: 32 5 y x x + .a) Chng minh rng phng
trnh 0 yc t nht mt nghim thuc khong (0;2);b) Vitphng trnh tiptuyn
ca ngcong (C). Bit rngh s gccatip tuyn bng 5.*Hnh hc: (2 im)Cho hnh
chp S.ABCD c ( ) SA ABCD , y ABCD l hnh thang vung ti A v D viSA 3
a , ABAD =DC ==2a. Gi I l trung im ca AB.a) Chng minh rng:( ) DI
SAC ;b) Tnh gc to bi gia hai mt phng (ABCD) v (SCD);c) Tnh khong cc
gia hai ng thng cho nhau AB v
SC.---------------------------------------------------------- HT
----------I. p n trc nghim: (4 im)1. abCd2. abCd3. Abcd4. abCd5.
Abcd6. Abcd7. abcD8. abCd9. Abcd10. abcD11. aBcd12. abcD13. Abcd14.
abcD15. aBcd16. abCdII. Phn T Lun: (6 im)p n im*i s:Cu 1: (2 im)22x
x2 31 1x 2x 3 x 2x xa) lim lim15x 1 55x+ + + + + + ( )( ) ( )( )( )
( ) ( )( )( ) ( ) ( ) ( )( ) ( )4325'5 4 4' '4 4 3 3' '23 3 2 210 2
380 2 3y: ''' 2 480. 2.2 3 480.1 480) ( ) 2 3 t 2 3 ' 2' 5. . '
10.'' 10. 10. 10.4. . ' 80.''' 80. 80. 80.3. . ' 480 =480. 2 3xxV
fb f x xu x uf x u u u uf x u u u u uf x u u u u u x =1 0,25
0,250,250,25Cu 2: (2 im)a) Xt hm s f(x) = x3 + 2x 5Ta c: f(0) = -5
v f(2) = 7. Do f(0).f(2) < 0.(Cch 2: f(1).f(2) = -14 < 0)y =
f(x) l mt hm s a thc nn lin tc trn . Do n lin tc trn on [0;2]. Suy
ra phng trnh f(x) = 0 c t nht mt nghim x0( ) 0;2 .b)Do phng trnh
tip tuyn vi ng cong (C) c h s gc k = 5, nn ta c: f(x0) = 5(vi x0 l
honh tip im)
320x+ 2 = 520x= 1 00x 1x 1 *Khi x0 = 1 y0 = -2, ta c phng trnh
tip tuyn l: y + 2 = 5(x 1) y = 5x -70,50,250,25 0,25 *Khi x0 = -1
y0 = -8, ta c phng trnh tip tuyn l: y + 8 = 5(x + 1) y = 5x -3Vy c
hai phng trnh tip tuyn vi ng cong (C) c h s gc bng 5 l:( )1y = 5x
-7 v( )2y = 5x -30,25 0,25 0,25 *Hnh hc: (2 im)a)Chng minh( ) DI
SAC :ABCD l hnh thang vung ti A v D vI l trung im ca AB, ABAD DC2
nn t gic AICD l hnh vung.( ) DI AC SAC ( ) 1Theo ra, ta c:
( )( )SA ABCDSA DIDI ABCD ) Hay( ) ( ) DI SA SAC 2 T (1) v (2)
ta c:( ) DI SAC (pcm)AD CBISb) Tnh gc to bi gia hai mt phng (ABCD)
v (SCD):Ta c: 0,25 0,25 0,25 0,25 ( ) ( )( )( )DC ABCD SDCDC AD
ABCDDC SD SCD gc to bi gia hai mt phng (ABCD) v (SCD) l gc: SDAXt
tam gic SAD vung ti A, ta c:0SA a 3tanSDA 3AD aSDA 60 Vy gc gia hai
mt phng (ABCD) v (SCD) bng 600.c)Tnh khong cc gia hai ng thng cho
nhau AB v SC:( )( )Ta c:AB//DCAB/ / SDCDC SDC)Mt khc, ta c:( ) SC
SCD nn khong cch gia hai ng thng AB v SC chnh bng khong cch t mt im
bt k nm trn ng thng AB n mt phng (SCD).Trong tam gic vung SAD vung
ti A, gi H l hnh chiu vung gc ca A ln cnh SD, khi ta c: ( ) ( )d
AB; SCD AH p dng h thc lng vo tam gic vung SAD vung ti A ta
c:SA.ADAH.SD SA.AD AHSD (*)Ta c: SD2 = SA2 + AD2 2 2 2 2SD 3a a 4a
+ SD 2a (3)Thay (3) vo (*) ta c:2a 3 a 3AH2a 2 Vy khong cch gia hai
ng thng cho nhau AB v SC bng a 32.0,25 0,25 0,25 0.25
---------HT---------
