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Trang 1TIM HIEU KHA N A N G CUA HQC SINH L O P 12
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T H 6 N G Q U A M Q T THV'C NGHIf M SI/PHAM
NGUY£N HCru LQT T6M TAT
Bdi todn xet tinh dan diiu ciia mgt hdm si khd phi bien Irong chuang Irinh todnphi Ihdng Di gidi quyit bdi todn ndy cd nhiing cdng cu gidi khdc nhau: dimg dinh nghia, di/a vdo cdc yiu Id die trung cua hdm sS dugc cho, dua vdo dd Ihi cua hdm sd hay tinh iai hdm cip I cua hdm si dd Trong bdi bao ndy, chiing tdi thiit ki mgt tinh huing dgy hfc nhdm tim hieu khd ndng cua hgc sinh ldp 12 Irong viic vdn dung cdc cdng cu gidi bdi toan xit tinh dan diiu ciia hdm si mil Ddng thdi Ihdng qua dd phdt hiin nhiing sai lim hpc sinh mdc phdi khi gidi bdi lodn ndy
Tii khda: tinh don dieu, dao him, him sd mu
ABSTRACT
A research on twelfth graders' ability in solving the problem of examining the monotonicity of an exponential function through an educational experiment
The problem of examining Ihe monotonicity of an exponential fijnction is quite common in high school math curriculum To solve this problem, there are various tools such as: definition, characteristics of the given function, fitenlion graphs, or the first derivative of Ihe function In this article we designed a teaching scenario to examine the gbility oftwelflh groders in applying mathematical tools lo solve the problem of examining the monotonicity of an exponential function At the same time Me also wish to detect mistakes students often make when solving this type of problem
Keywords: monoticity, derivative, exponential hmction
1 D|t van de nhihig him so mu d dgng y=a'' hoac co
Chung tdi bit diu nghign cim tu thi dua dupc vl dgng y=a'' cd giip hgc viec phan tich sich giio khoa (SGK) sinh khai thic dupc hit cic cdng cy de Toin 12 (ning cao) Mpt dilu thii vj giii quylt bii toin xet tinh don difu ciia chung tdi cd dupc lign quan din bii toin him sd mu hay khdng? Nhihig sai lim xet tinh don difu cua him sd mu Cic hpc sinh mic phii khi giii quylt cic bai ham s i xgt tinh ddng biln, nghjch biln toin dgng nay Chung tdi thilt kl mot diu cd dang y=a'' hoic cd thi dua dupc tinh huong dgy hpc nhim tim hilu kha
vg dgng y=a'' Ldi giai mong dpi cua ning cua hpc sinh ttong vifc vgn d\iiig SGK cho thay hpc sinh chi cin dya vao cic cdng cy giai bii toin xet tinh don
CO sd cua him sd d i cho dl dua ra kit difu cua him so mu Ding thdi thdng qua lufn Lifu SGK d i gidi hgn viec khao sit dd phit hifn nhimg sai lim hpc sinh mic ThS, Sd Gigo dgc vS BSo tgo TPHCM phii khi giii cic bii toan nay
Trang 22 Thyc nghifm dii vdi hpc sinh
Thyc nghifm dupc tiln hinh trgn
hpc sinh ldp 12 ban khoa hpc ty nhign
vdi chuong trinh toan nang cao Thdi
dilm thyc hifn la sau khi hpc sinh da hpc
xong bii ham si mu Thdi gian thyc
nghifm dinh cho bai toin li 15 phut Hpc
sinh sg lim viec ci nhin
Hpc sinh se dupc phit giiy lim bii
fren dd cd in dg bii toin Giiy nhip cung
dupc phit cho hpc sinh vi thu lgi sau gid lim Dilu niy cho phgp chung tdi thu thip thgm diu vit thi hifn mil quan hf
ci nhin cua hpc sinh
Bii toin thyc nghifm:
Cd thg bigt dupe tinh ddng bign vi nghieh bign cua cic him sd cho trong bing sau diy hay khdng?
(Dinh diu X vio d mi em lya chpn
vi giii thich hoic cho ldi giai tuang irng)
Ham so
i)y = '1^
.2) '
V)y = K"
c ) y = 3 ' '
d ) r = 2>-'
Dirpc Khdng - Neu khdng, giii thich vi sao?
- Neu c6, trinh biy ldi giii cua em
21 Phin tich mpt so yiu to tru&c khi
th^ nghidm
2.1.1 Cdc bien
Vifc chpn cic bai toin thyc nghifm
dupc dgt frgn CO sd lya chpn gid tti cdc
biln didactic sau diy
VI: "Him si li him si mu
hofc cd thi biin dii dupc vi ham si mii
bien x hay khdng?"
Hai gia tti cua biln:
- Hdm si Id hdm sd mU hogc cd thi
biin dii dugc vi him si mO biin x
- Hdm si khdng Id hdm sd mil hogc
khdng thi biin ddi dugc ve hdm sd mii
biin X
• V2: "Biiu thi:c mU li tuyin
tinh hay khdng tuyin tinh theo x?"
Biiu thiic mii Id tuyin tinh theo x
Biiu thirc mU khong Id tuyin tinh
theox
Ta bilt ring, him sd mu y = a" cd miln xic djnh li R Vi vay, him sd dang
y = a"''' chi la him sd mu cua biln t =
u(x) nlu nhu miln gii ttj cua u(x) la R Trudng hpp die biet: u(x) bilu diln tuyln tinh theo x thi a"'"' li him s i mu Ngupc lai, him si di cbo chi li
"mdt phin" ciia him sd mu (dd thi cua
nd chi l i mpt tfp con thyc sy cua bim sd mu), hoic khdng li ham sd mu
• V3: "Bi thi him si qua (0 ,1)
hogc (1, a) hay khong?"
Dd thi hdm si qua (0 ,1) hogc (I,
">•
Dd thi ham sd khdng qua (0 ,1) vd (1, a) Trong do a li co s i cda ham sd da
cho
2.1.2 Dgc trung cua bdi tadn dugc lua chon
Trang 3Bii niy dupc cho vdi nhilu him s6
khic nhau, trong dd cd nhttng him si
quen thupc (dupc cho trong SGK vi
SBT) vi khdng quen thufc Dieu niy cho
phgp chung tdi tim hieu ung xtt cua hpc
sinh trude nhttng him so khdng quen
thupc doi vdi kilu nhifm vy xgt tinh
ddng bien, nghjch bien cua him so Dfc
bift, gii trj cua biln VI dupc chpn trong
cau c) li hdm sd khdng Id hdm sd mu, sg
cho phgp lim rd moi quan bf ci nhan cua
hpc sinh dii vdi vifc xgt tinh don difu
him sd
Chdng tdi dy dodn ring cic ldi giai
cua hpc sinh sg su dyng kT thuft xgt co si
dg suy ra tinh don difu cua cic him so di
cho
2 1.3 Cdc chiin luge cd thi
Vdi cic him sd dupe lya chpn, bii
toin bao gdm nhimg dang him sd khic
nhau lien quan din ham si mu Tinh chit
cic him niy it nhigu dgu cd lign quan
din cic tinh chit cua ham sd mu Vi nhu
viy chung sg li ham sd mu hoic li mpt
phan cua him sd mu Cic chien lupc sau
diy dya ttgn co sd cac tinh chit cua him
simu
• ST„: "Chiin lupc ca si": doi
vdi him so y = a"'"', ip dyng ki thugt so
sinh ca sd vdi 1
Nlu a > 1 thi him sd ddng biln
Nlu a < 1 thi him sd nghjch biln
• STai: "Chiin lupc djnh li":
- Nlu (V X, > X2 => f(x,) > fl;x2)) thi
f dong biln
Nlu (V x, > X2 => f(x,) < fc)) thi
f nghjch biln
SThh: "Chien lupc him hpp":
Xgt tinh ddng biln, nghjch biln ciia cic him thinh phin
Xdt tinh ding biln, nghjch biln ciia him hpp di cho
• STai,: "Chiin lupc dpo him":
Xgt tinh dong biln, nghjch biln ciia ham
s i dya vio dau cua dgo him
2.1.4 Nhirng quan sdt cd thi
•"=©•
n Su lua chgn hdm sd
• Him sd niy tucmg ung vdi gia trj thu nhit cua tit ci cic bien VI, V2, V3
Diy la dgng him sd hoin toin quen thupc doi vdi hpc sinh ma SGK di dl cap Chung tdi chpn bii niy vdi myc dich lim CO sd dl so sinh iing xu cua hpc sinh ddi vdi nhihig dgng him sd khic Ttt do cung thiy dupc ring bufc cua thi chl len hpc sinh ttong vifc xgt tinh dan difu cua him so mu
• Cic chiin lupc cd thg:
STtj: "Chiin luge ca si": nhu tren
di ndi, diy li dgng him sd him co ban nhit cua him mu, rit sit vdi djnh nghia dupc ttinh biy ttong SGK, vi vfy mii chiin lupc CO sd chic chin se dupc hgc sinh lya chpn
STji; "Chiin luge dinh li"; STa,:
"Chiin luge dgo hdm "
Hai chiin lupc STj, vi STa, cd thi giii quylt bii toin, tuy nhign chung s5 khdng cd ca hfi dl xay ra vdi him si nay
vi chiin lupc ca sd di thdng Imh
D Cdi cd the quan sdt dugc ti hgc sinh:
Trang 4Ldi gidi tuang Ong vdi chiin luge
casdSTcs:
Vi — < 1 ngn y
bien
b)y =
7r' D Su lua chgn hdm sd
• Gii tri cua cie biln dupc
chpn:
Bii nay dupc xiy dyng dya trgn
biln V2 vdi gii ttj: Bieu thiic mu Id tuyin
tinh theo x
• Cic chiin lupc cd thi:
- ST„: "Chiin luge ca sd": diy li
chiin lupc dupc uu tign doi vdi him sd
dang mii Vi vgy, cd nhigu co hfi dg xiy
ra chien lupc niy cho du ham sd di cho
CO thda man digu kien cua ham sd mu
hay khdng
- STji: "Chiin luge dinh li": cung cd
I the xiy ra chiin lupc niy vi dang him sd
chua that sy dung vdi dang di djnh
nghia, bdi vi chung tdi di chpn gii tri thu
I nhat cua bign V2
I SThi,: "Chiin luge hdm hgp": ca
I hfi xiy ra chiin lupc niy cung bing nhu
i "chien luge dinh li"
I Him y = Jt^" li hpp cua hai him
thanh phin u(x) = 3x vi y = it" Him u(x)
I la dg dang bilt dupc tinh dan difu cua
nd Do dd tinh dom difu cua him Jt" cung
i dupe dl ding xic djnh
i! ST*: "Chiin luge daa hdm "
' D Cdi cd thi quan sdt dugc tir
hgc sinh:
t - Ldi gidi tucmg Ong vai chiin luge
easiSTr,:
Cd hai trudng hpp tuang img vdi chien lupc niy:
THI:
Tacd y = 7C^' = {7!^\
n^ > 1 ngn him so dong biln TH2: dong nhit tinh dan difu cda
him y = ir^" vdi him _y = ;r' Do dd, tinh don difu ciia him dupc cho xic djnh nhu sau:
Vi 71 > 1 ngn nen him sd da cho li ddng bign
Ldi gidi tuang iing vdi chiin luge dinh li STji:
V X|, X2: X] > X2 ta cd: 3xi > 3x2 =>
tdi gidi tuang ting vdi chien luge hdm hgp ST/,/,:
u(x) = 3x li ddng biln tgn R It" li ddng biln tren R ngn it""" li ddng biln fren R
Ldi gidi tuang ting vdi chien luge dgo hdm ST^/,:
y' = 3T^'\tm > 0 vdi mpi x thufc R
ngn him si y = it^" ddng biln ttgn R
e ) y = 3 ' '
D Su lua chgn hdm sd
• Gii trj cua cie biln dupc chpn:
Him s6 y = 3 thda cic gii trj cua
cic biln sau:
Gii tri tiiii h^i cua biln VI: Hdm sd khdng thi biin ddi dugc vi hdm sd mS biin x(y = a')
Gid ttj thd hai eua biln V2: Biiu thirc muld khdng tuyin tinh theo x
Trang 5Gia trj thd nhat cua bien V3: Dd thi
hdm sd qua (0,1) hodc (I,a)
Him so niy thda hai digm dfc bift
cua him so mu li (0,1) vi (l,fl) (trudng
hpp niy a bing 3) Tuy nhien, him niy
chi li "mgt phdn" cua him so mu bdi
ring tfp gii trj cua nd li [l.+co)
Him y = 3' khdng cd tinh chit
ludn ting hoic ludn gidm nhu tinh dan
difu cua him so mu Do dd, khdng thi
khio sit tinh chit niy bing ki thuit so
sinh CO so cua nd vdi 1
Cic chiin lupc cd the:
Ngu hpc sinh cho ring cd the biet
dupc tinh ding biln, nghjch biln cua
him sd y = 3^ thi cic chign lupc sau
diy cd thg xiy ra:
STcs: "Chiin luge ca sd ": mac du
y = 3' khdng phii li him sd mQ nhung
him niy cd hai digm die bift (0,1) va
(l,a) vi cd dang a""" ngn cd nhilu co hpi
xuit hien chign lupc niy
STjih: "Chiin luge hdm hgp ": chign
lupc nay cd thg xiy ra trong trudng hpp
hpc sinh nhin dang dupc him Tuy nhign
theo chung tdi, thg chl da khdng tgo co
hdi cho chiin lupc nay xiy ra
STah: "Chiin luge dgo hdm": mfc
du cd SGK cd gidi thifu phuong phip xet
tinh bign thign cua mpt him sd bing dgo
him, tuy nhign ddi vdi bai niy phuang
phip dgo him khdng li ttpng tim, do do
chung tdi nghi ring cd rit it co hfi xiy ra
chiin lupc niy
D Cdi cd thi quan sdt dugc tii
hgc sinh:
Ldi gidi tuang img vai chiin luge
ca si ST„:
2 ,
VI 3 > 1 ngn y = y ddng bign fren
R
Lai gidi tuang img vol chiin luge hdm hgp ST/,/,:
Him so x^ dong biln trgn [0, +»)
vi nghjch bien trgn (-oo, 0]
Do do him sd y = y ddng bign
trgn [0, +M) vi nghjch biln tren (-«, 0]
Lai gidi tuang ling vol chiin luge dga hdm STj/,:
y' = 2x3''
Khi X > 0 thi y' > 0 ngn ham s6 ddng biln
Khi X < 0 thi y' < 0 ngn him so nghjch bign
d)y = 2 ' '
D Su lua chgn hdm sd
• Gii trj cua cic biln dupc chpn:
- Gii trj thu nhit cua biln VI: Ham
sd Id hdm sd mi hogc cd thi biin ddi ve hdm sd mi biin x (y = cf)
Gii trj thu nhit cua biln V2: Bien thirc mi la tuyin tinh theo x
Him nay dupc cho vdi myc dich tim higu moi quan hf ca nhin cua hoc sinh vl ham sd mu Mpt cich rd rang hon chung tdi mudn kiim chung ring cd phai hpc sinh di thgt sy gin lign hay dong nhit him si mu vdi mft bilu dien bao gdm ca s i a vi mft bilu thdc mu Him s6 y = 2 ' " li ham sd mu,my nhign nd li him mu vdi ca so i Vi nhir
Trang 6Nguyen Hiiu Lpi
vfy tinh don difu cua nd dupc xet theo
bieu thuc him >' = 2
• Cic chiin lupc cd thi:
STcs: "Chien luge ca si"', 57^:
"Chiin luge dinh li"; SThi,: "Chiin luge
hdm hgp "; STji,: "Chiin luge dgo hdm "
D Cdi cd thi qugn sdt dugc tir
hgc sinh:
Ldi^idi tuang ting vdi chiin luge
easdSTcs:
Cd hai trudng hop xiy ra vdi chiin
lupc niy:
THI: y = 2'"" dupc biln ddi thinh
dgng y = 2 — , Idi giii nhu sau:
Vi — <1 ngn him da cho nghieh
2
77/2: y = 2 ' " , Idi giii nhu sau:
Vi 2 > 1 ngn him di eho ddng biln
Ldi gidi tuang iing vdi chiin luge dinhliSTj,:
Vdi mpixi >X2tacd:
-X| < -X2 => 1 - X| < 1 - X2 => 2'-'i < 2'-'2
Viy him sd di cho nghjch biln
Ldi gidi tuang img vdi chiin luge hdm hgp ST/,/,:
Him u(x)=l-x li nghjch biln trgn R ngn him 2'"" la nghjch biln trgn R
Ldi gidi tuang irng vai chiin luge dgo hdm ST^/,:
y' = -2'"''ln2 < O V x e R n g n y = 2'"" nghjch bien
2.2 Phan tich chl tiit kit qui th{rc nghiim
bien
Bdng thdng ke cdc ldi gidt cda hgc sinh
Cau
a
b
c
d
Chien lirffc
Cor so
E
|3g (11)1,22(11^)
M
33(l/2),|l4(2i
Dinh li
1
1
Ham hpp
3
1
Dao ham
12
14
14
13
Khdng biet
1
M
g
Bd trdng
N
S
Tdng si
74
74
74
74 Trong bing, chung tdi die bift chu
S den cic sd cd ddng khung Nhttng so
nay phan anh quan he ca nhin cua hpc
sinh ddi vdi him si mil
Dung nhu dy doan, da sd hpc sinh
sir dung chiin lupc co sd dg xet tinh ddng
bien, nghieh bign cua him sd niy Die
bift luu y him si d ciu c) Diy khdng
phai li ham s i mu nhu dinh nghTa d
SGK, tuy nhign cd 31/74 (41,8%) hpc
sinh ip dyng ki thuft xet tinh don difu ciia him sd mu dg giii Ldi giii dign hinh
ttong trudng hpp niy li: "a=3>l=> hdm
sd ddng biin "
Ngoii ra cd 21/74 (28,3%) hpc sinh
cho ring khdng thi bilt dupc tinh don difu cua hmn niy vi sd mu li x^ Cdc giii
thich tucmg dng li: "i^i chua biit gid tri
Trang 7cua A nin khdng xdc dinh dugc ddng
biin, nghieh biin": "khdng thi biit tinh
ddng biin, nghieh biin vi khdng cd dgng
y=d'": "VI khdng biit dugc gid tri cua x
thugc khodng ndg nin khdng xit dugc
tinh ddng biin, nghieh biin " Mpt si Idi
giii thich khic tuy cd dung dgo him dl
khio sit nhung di din kit lufn li: "Viy'
cdn chiro tham sd x nin dau cua y' chua
xdc dinh Vl vgy khdng thi xdc dinh hdm
sd ddng biin hay nghieh bien " RO ring
Ii hpc sinh di khdng kllm tra him di cho
cd la him so mu hay khdng, tuy nhign hp
di ip dyng tinh chat cua him so mu (dgo
him ludn khdng chua tham so) dl giii
Mpt trudng hpp tuang ty cd thi
thiy d ciu b) nhu sau:
Theo SGK thi y=7i''' li him s i mu
CO sd It', tuy nhign cd din 38/74 (51,3%)
hpc sinh giai him niy vdi ca sd it Ddi
vdi bam sd d ciu d) y=2'"'' cung cd den
14/74 (18,9%) hpc sinh giii him niy vdi
CO sd li 2 Dilu nay cho thiy hpc sinh da
khdng kigm tra sy thda ding cua him so
3 Ket luan
Thyc nghifm dua din mft sd kgt
qua sau:
Ttt cic kgt qui cd dupc, chung tdi
nhgn thiy da sd hpc sinh cua ldp dupc
thyc nghifm tip trung cich giai cua minh
vio vifc xgt cdc ca sd: a > 1 thi him so dong biln, 0 < o < 1 thi ham si nghjch bien Cic em khdng cd nhifm vy di kiim tra him si dupc cho cd la him sd mu hay khdng? Do dd, cic em di khdng thi giii quylt dupc cau c vi d Ngoai ra ta cdn thiy, khdng nhilu hpc sinh su dyng dao him dl xgt tinh dan difu cua him si Nlu cd, cic em cung khdng thinh cdng
dl dua din kit qui sau cung Dilu niy cd thi dupc giii thich li SGK chi dua ra nhihig dgng toin chi cin xgt dgn co sd la
cd thi kit luin dupc tinh dan difu ciia him so mu
Thyc nghifm cung md ra hudng xiy dyng bii tip cho hpc sinh ma gido vign cin phii cin nhic Khdng phii liic nio cung dg xuit cho hpc sinh cic bai tap quen thupc Do dd, giao vien cd thi tgo ra cic tinh hudng hpc tfp nhim giup hpc sinh cd the van dyng cic kign thuc co lign quan Ching ban, ddi vdi bii him so
mu, giio vign cin xiy dyng hf thdng bai tgp sao cho bufc cdc em phii van dyng nhihig phuang phip giii khdc nhau de giii quylt hit hf thdng bii tgp dd Tu do gdp phin khic phyc cac ldi mic phii cua hpc sinh nhu da chi ra trong thyc nghifm tten
TAI L i f u THAM KHAO
Cue Nhi gido vi Cin bp quin 11 Giio dye (2008), Hudng ddn thuc hiin chuang trinh
vd sdch gido khoa ldp 12 THPT, Nxb Giio dye
Ngd Vilt Diln (2001), Phuang phdp chgn lgc gidi Todn hdm si mi vd logarit, Nxb
Dai bpc Qudc gia Hi Npi
Tran Vin Hao, Phan Truang Din (1991), Sdch gido vien Dgi si vd gidi tich II, Nxb
Giio dye
Lg Thj Thign Huong, Nguyen Anh Tuin, Lg Anh Vii (2000), Todn cao cdp, Nxb
Giio dye
Trang 85 NguySn HOu Lpi (2008), Khdi niem hdm sd mii d trudng Trung hgc phd thdng, Lufn
van Thac si chuyen nginh Phuang phip giang day Toin, Trudng Dgi hoc Su pham TPHCM
6 Doin Quynh, Nguyen Huy Doan (2007), Dgi si vd Gidi tich 11 ndng eao, Nxb Giio
due
7 Doan Quynh, Nguyen Huy Doan (2008), Gidi tieh 12 ndng cao Nxb Giao due
8 Doan Quynh, Nguyen Huy Doan (2008), Sich Gido vien Giai tich 12 ndng eao Nxb
Giao dye
9 Le van Tiln (2005), Phuangphdp dgy hgc mdn Todn d trudng phd thdng, Nxb Dai
hpc Quic gia TPHCM
(Ngiy Tda soan nh$n dugc bii 20-02-2012: ngiy chip nh$n ding: 19-6-2012)
SV" CAN THIET CUA MO HINH HOA
(Tiep theo trang 121)
TAI LIEU THAM KHAO
1 Nguygn Thj Tan An, Tran Ding (2009), "Su dung md hinh hoa toan hpc ttong vifc
day hpc loan", Tgp chi Gido due, (219)
2 Gabriele Kaiser, Werner Blum, Rita Borromeo Ferri, Gloria Stillman (2011), Trends
in Teaching and Learning of Mathematical Modelling Springer
3 Gabriele Kaiser, Bharath Sriraman (2006), A Global Survey of International Perspectives on Modelling in Mathematics Eduacation, ZDM Vol 38(3)
4 Hans-Stefan Siller, Modelling in Classroom 'Classical Models' (in Mathematics Education) and recent developments
www.algebra.tuwien.ac.at/kronfellner/ ESU-6/ /l-13-SilIer.pdf
5 OECD (2003), The Pisa 2003 - Assessment Framework - Mathematics, Reading, Science and Problem Solving Knowledge and Skills, OECD, Paris, France
6 Rita Borromeo Ferri (2006) Theoretical and Empirical Differentiations of Phases in the Modelling Process ZDM Vol.38(2)
7 Werner Blum, Peter L Galbraith, Hans-Wolfgang Henn, Mogens Niss (2007),
Modelling and Applications in Mathematics Education Springer
(Hgiy Tda soan nhin dui?c bii: 01-02-2012: ngiy chip nhin ding: 19-6-2012)