Kien Truc He Thong File Ext2 Ext3 Ext4 Cua Cac He Dieu Hanh Ho Linux
Tinh Don Dieu Cua Ham So Hay
Transcript of Tinh Don Dieu Cua Ham So Hay
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CHUYN TON THPT V Trng SnCHUYN TON THPT V Trng Sn
CHUYN 1. TNH N IU CA HM S V CC NG DNG
VN 1: XT CHIU BIN THIN CA HM SQuy tc:
1. Tm TX ca hm s.2. Tnh o hm f(x). Tm cc im xi m ti o hm bng 0 hoc khng
xc nh.3. Sp xp cc im xi theo th t tng dn v lp BBT.4. Nu kt lun v cc khong ng bin, nghch bin ca hm s.
Bi 1. Xt chiu bin thin cc hm s sau:
2
3 2 4 2 3x 2 x 2x + 3a) y 2x + 3x + 1 b) y = x 2x 3 c) y d) yx 1 x 1
+ = + = =
+ +Bi 2. Xt tnh n iu ca cc hm s sau:
32
2 2
x x xa) y 25 x b) y c) y d) y
x 10016 x x 6
= = = =+
Bi 3. Chng minh rng:
a) Hm s 2y x 1 x= + ng bin trn khong1
1;2
v nghch bin trn khong
1;1
2
.
b) Hm s 2y x x 20= nghch bin trn khong ( ); 4 v ng bin trn
khong ( )5;+ .
Bi 4. Xt s ng bin, nghch bin ca cc hm s sau:
[ ]5
a) y x sin x, x 0;2 b) y x 2cos x, x ;6 6
= = +
Bi 4. Chng minh rng:
a) ( )f x cos2x 2x 3= + nghch bin trn R.
b) ( ) 2f x x cos x= + ng bin trn R.
Gii:
a) Ta c: f '(x) 2(sin 2x 1) 0, x R = + v f '(x) 0 sin 2x 1 x k , k Z4= = = +
Hm s f lin tc trn mi on ( )k ; k 14 4
+ + + v c o hm f(x) < 0 vi mi
( )x k ; k 1 , k Z4 4
+ + +
.
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Do , hm s nghch bin trn mi on ( )k ; k 1 , k Z4 4
+ + + .
Vy hm nghch bin trn R.
b) Ta c: f(x) = 1 sin2x; f '(x) 0 sin 2x 1 x k , k Z4
= = = +
NX: Hm s f lin tc trn mi on ( )k ; k 1
4 4
+ + + v c o hm f(x) > 0 vi mi
( )x k ; k 1 , k Z4 4
+ + +
.
Do hm s ng bin trn mi on ( )k ; k 1 , k Z4 4
+ + + .
Vy hm ng bin trn R.
VN 2: TM THAM S HM S N IU TRN MIN K
Phng php: S dng cc kin thc sau y:1. Cho hm s y = f(x) c o hm trn K.
Nu f '(x) 0, x K th f(x) ng bin trn K.Nu f '(x) 0, x K th f(x) nghch bin trn K.
2. Cho tam thc bc hai f(x) = ax2 + bx + c c bit thc 2b 4ac = . Ta c:
a 0f (x) 0, x R
0
>
a 0f (x) 0, x R
0
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Hm s nghch bin trn R khi v ch khi5
f '(x) 0, x R 0 a2
.
Bi 2
Vi gi tr no ca m, hm s ( )3 2f (x) mx 3x m 2 x 3= + + nghch bin trn R ?
Gii:TX: RTa c: 2f '(x) 3mx 6x m 2= +
Hm s nghch bin trn R khi v ch khi 2f '(x) 3mx 6x m 2 0, x R = +
m = 0, khi f(x) =1
6x 2 0 x3
: khng tha x R .
m 0 , khi m 0
f '(x) 0, x R 9 3m(m 2) 0
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o hm:( )
2
2
m 1y'
x m
=
+. Hm s ng bin trn tng khong xc nh khi
2y' 0, x m m 1 0 m 1 v m 1> > < >
Bi 5
Tm m hm s ( ) ( )3 21 1
y mx m 1 x 3 m 2 x
3 3
= + + ng bin trn [ )2;+ .
Gii:Ta c: ( ) ( )2y ' mx 2 m 1 x 3 m 2= +
Hm s ng trn [ ) ( ) ( )22; y ' 0, x 2 mx 2 m 1 x 3 m 2 0, x 2+ +
( )2 26 2x
m x 2x 3 2x 6 0, x 2 m , x 2x 2x 3
+ +
+(v x2 2x + 3 > 0)
Bi ton tr thnh:
Tm m hm s ( ) 26 2xf x m, x 2x 2x 3= +
Ta c ( )( )
( )2
2
22
2x 12x 6f ' x , f ' x 0 2x 12x 6 0 x 3 6
x 2x 3
+= = + = =
+
BBT:x 2 3 6+ +
f(x) 0
f(x)
2
3
0
Ta cn c:[ )2;
2max f (x) m m
3+ . l cc gi tr cn tm ca tham s m.
Bi 6
Tm m hm s2mx 6x 2
yx 2
+ =
+nghch bin trn na khong [ )1;+ .
Gii:Ta c:
( )
2
2
mx 4mx 14y '
x 2
+ +=
+
Hm s nghch bin trn [ ) 21; y ' 0, x 1 mx 4mx 14 0, x 1+ + +
( )2 214
m x 4x 14, x 1 m , 1x 4x
+
+
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Bi ton tr thnh: Tm m hm s ( ) 214
f x m, x 1x 4x
=
+
Ta c:( )
22
14(2x 4)f '(x) 0, x 1
x 4x
+=
+
x 1 +f(x)
f(x)0
14
5
Ta cn c:[ )1;
14min f (x) m m
5+ . Vy
14m
5 l cc gi tr cn tm ca m.
Bi tp t gii:
Bi 1. Tm cc gi tr ca tham s a hm s ( ) 3 21
f x x ax 4x + 3
3
= + + ng bin trn R
Bi 2. Vi gi tr no ca m, hm sm
y x 2x 1
= + +
ng bin trn mi khong xc nh ?
Bi 3. nh a hm s ( ) ( )2 3 21
y a 1 x a 1 x 3x 53
= + + + + lun ng bin trn R ?
S: a 1 v a 2
Bi 4. Cho hm s( ) 2m 1 x 2x 1
yx 1
+ +=
+. Xc nh m hm s lun ng bin trn tng
khong xc nh ca n.
S: 1 m 2 Bi 5. Cho hm s ( ) ( )3 2 2y x m 1 x m 2 x m= + + + + . Chng minh rng hm s lun nghch
bin trn R vi mi m.Bi 6. Tm m hm s y = 3x3 2x2 + mx 4 ng bin trn khong ( )0;+ .
S:4
m9
.
Bi 7. Tm m hm s y = 4mx3 6x2 + (2m 1)x + 1 tng trn khong (0;2).
S:9
m
10
.
Bi 8. Cho hm s2x 2mx m 2
yx m
+ +=
.
a) Tm m hm s ng bin trn tng khong xc nh.b) Tm m hm s ng bin trn khong ( )1;+ .
VN 3:
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S SNG TNH N IU CA HM S CHNG MINH BT NG THC
Phng php: S dng kin thc sau: f(x) ng bin trn on [ ]a; b th ( ) ( ) ( ) [ ]f a f x f b , x a; b
f(x) nghch bin trn on [ ]a; b th ( ) ( ) ( ) [ ]f a f x f b , x a; b
Bi 1
Cho hm s ( )f x 2sin x tan x 3x= + .
a) Chng minh rng hm s ng bin trn na khong 0;2
.
b) Chng minh rng: 2sin x tan x 3x, x 0;2
+ >
.
Gii:
a) Hm s cho lin tc trn na khong 0; 2 v c
( ) ( )2
2 2
1 cos x 2cos x 11f '(x) 2cos x 3 0, 0;
cos x cos x 2
+ = + = >
. Do , hm s f ng
bin trn na khong 0;2
(pcm).
b) T cu a) suy ra f(x) > f(0) = 0, x 0; 2sin x tan x 3x, x 0;2 2
+ >
(pcm).
Bi 2
a) Chng minh rng hm s ( )f x tan x x= ng bin trn na khong 0;2
.
b) Chng minh rng3x
tan x x , x 0;3 2
> +
.
Gii:
a) Hm s cho lin tc trn na khong 0; 2
v c2
2
1
f '(x) 1 tan x 0,cos x= = >
x 0;2
. Do , hm s f ng bin trn na khong 0;2
.
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b) T cu a) suy ra f(x) > f(0) = 0, x 0; tan x x, x 0;2 2
>
.
Xt hm s3
xg(x) tan x x
3= trn na khong 0;
2
. Hm s ny lin tc trn na
khong 0;2
v c o hm2 2 2
2
1g '(x) 1 x tan x x 0, x 0;
cos x 2
= = >
, do
tan x x, x 0;2
>
.
Do , hm s g ng bin trn na khong 0;2
nn g(x) > g(0) = 0 x 0;2
3xtan x x , x 0;
3 2
> +
(pcm).
Bi 3
Chng minh rng : 2(x 1)ln x x 1> + , vi mi x > 1.
Gii:
Bt ng thc cho tng ng vi2(x 1)
ln x 0, x 1x 1
> >
+
Xt hm s ( )2(x 1)
f (x) ln x , x 0;x 1
= +
+. Ta c:
( )
( )
( )
( )2
2 2
x 11 4f '(x) 0, x 0;
xx 1 x x 1
= = +
+ +Suy ra hm s ng bin trn khong ( )0;+ nn cng ng bin trn khong ( )1;+ . Vy talun c f(x) > f(1) = 0 vi mi x > 1. cng l iu phi chng minh.Bi tp t gii:Bi 1.Chng minh cc bt ng thc sau:
a) sin x x, x 0< > v sin x 0, x 0<
c)
3x
sin x x , x 06> > v
3x
sin x x , x 06<
e)2x
sin x , x 0;2
>
f) tan x sin x> vi 0 x2
<
=Chng minh h trn c nghim duy nht.
Gii:Xt h:
x ye e ln(1 x) ln(1 y) (1)
y x a (2)
= + +
=vi iu kin xc nh x 1, y 1> >
T (1) y = x + a, th vo (1) ta c: x a xe e ln(1 x) ln(1 x a) 0+ + + + + = (3)Bi ton tr thnh chng minh (3) c nghim duy nht trn khong ( )1; + .t x a xf (x) e e ln(1 x) ln(1 x a)+= + + + + trn khong ( )1; +
Ta c f(x) l hm lin tc trn khong ( )1; + v c o hmx a x 1 1
f '(x) e e x 1 x a 1
+
= + + + +Do a > 0 nn vi mi x > -1, ta c:
x a xe e 0
1 10
x 1 x a 1
+ >
> + + +Nh vy f(x) > 0 vi mi x > -1 f(x) l hm s ng bin trn khong ( )1; +
Mt khc, ta c: x a1 x
f (x) e (e 1) ln1 a x
+= +
+ +
T ta tnh gii hn: x ax x x
1 xlim f ( x) lim e (e 1) lim ln
1 a x+ + ++= + = +
+ +v
x ( 1)lim f (x)
+ =
Vy, phng trnh (3) c nghim duy nht trn khong ( )1; + . T suy ra pcm.
Bi tp t luyn:Gii cc phng trnh sau:
a) 2 2x 15 3x 2 x 8+ = + + S: x = 1
b) ( ) ( ) ( ) ( )x 2 2x 1 3 x 6 4 x 6 2x 1 3 x 2+ + = + + + S: x = 7
VN 4:NG DNG CHIU BIN THIN CA HM S VO VIC BIN LUN
PHNG TRNH, H PHNG TRNH V BT PHNG TRNHCh . Cho f(x) l hm s lin tc trn T, th:
a) ( )f x a vi mi ( )x T a max f x
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b) ( )f x a vi mi ( )x T a min f x
c) ( )f x a c nghim ( )a min f x
d) ( )f x a c nghim ( )a max f x
Bi 1
Cho phng trnh
()2m x 2x 2 1 x(2 x) 0 + + + . Tm m phng trnh c nghim
x 0,1 3 + .
Gii:
Xt bt phng trnh : ( )2m x 2x 2 1 x(2 x) 0 (1) + + + t = + = 2 2 2t x 2x 2 x 2x t 2
Ta xc nh iu kin ca t :
Xt hm s = +2
t x 2x 2 vi x 0,1 3
+ Ta c: 2
x 1t ' , t ' 0 x 1
x 2x 2
= = =
+x 0 1 1 3+t 0 +
t2 2
1
Vy vi x0,1 3
+ th 1 t 2 .Khi :
(1) +
2t 2
mt 1
vi t [1;2]
Xt hm s
=+
2t 2
f(t)t 1
vi t [1;2] . Ta c:
f(t)+ +
= >
+
2
2
t 2t 20, x [1;2]
(t 1). Vy hm s f tng trn [1; 2].
Do , yu cu bi ton tr thnh tm m (1) c nghim t[1,2]
= =t 1;2
2m max f(t) f(2)
3.
l gi tr cn tm ca tham s.
Bi 2
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Tm m phng trnh 4 4x 13x m x 1 0 + + = c ng mt nghim.
Gii:
Ta c:4 4
x 13x m x 1 0 + + =4 4
x 13x m 1 x + =
( )4 3 24
x 1 x 1
4x 6x 9x 1 mx 13x m 1 x
= + =
Yu cu bi ton tr thnh tm m ng thng y = -m ct phn th f(x) = 4x 36x29x1ng vi x 1 ti mt im duy nht.Xt hm s f(x) = 4x3 6x2 9x 1 trn na khong ( ];1
Ta c: f'(x) = 12x2 12x 9 = 3(4x2 4x 3)
Cho f'(x) = 0 4x2
4x 3 = 0
1 3
x x2 2= =
x 1
2 1
f(x) + 0
f(x)
3
2
12
T bng bin thin ta thy:
Yu cu bi ton xy ra khi
3 3m m
2 2
m 12 m 12
= = < >
l cc gi tr cn tm ca tham s m.
Bi 3
Tm m h phng trnh ( )2x y m 0 Ix xy 1
=+ =
c nghim duy nht.
Gii:Ta c:
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(I)2x y m 0 2x y m 0
x xy 1 xy 1 x
= = + = =
Vi iu kin:xy 0
x 1
ta c:
(I) ( ) ( ) ( )22
y 2x m
y 2x m 1 xxy 1 x y x 1
x
= = = =
(Do x = 0 khng l nghim ca h)
( )2 21 x x 2x 1
2x m mx x
+ = = ()
Xt hm s2x 2x 1 1
f (x) x 2x x
+ = = + trn tp ( ] { }D ;1 \ 0=
Ta c hm s f(x) lin tc trn D v c o hm ( ) ( ]21
f '(x) 1 0, x ;0 0;1x
= + >
Gii hn :x x 0 x 0lim f (x) ; lim ; lim
+ = = + = v f(1) = 2
BBT :x 0 1
f(x) + +
f(x) + 2
T BBT ta thy :Yu cu bi ton xy ra khi m > 2. l cc gi tr cn tm ca tham s.
Bi 4 ( thi tuyn sinh i hc, Cao ng khi B 2004)
Tm m phng trnh 2 23 3log x log x 1 2m 1 0+ + = c t nht mt nghim thuc
31;3 .
Gii:
t2
3t log x 1= + . Vi x3
1;3
th t [1;2] .Khi phng trnh cho tng ng vi : 2t t 2 2m+ =Bi ton tr thnh tm m phng trnh 2t t 2 2m+ = c nghim t [1;2]Xt hm s f(t) = t2 + t 2 vi t [1;2] . Ta c : f(x) = 2t + 1 > 0, vi mi t [1;2]Vy yu cu bi ton xy ra khi :
x [1;2] x [1;2]min f (x) 2m max f (x) f (1) 2m f (2) 0 2m 4 0 m 2
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Bi 5 ( thi tuyn sinh i hc, Cao ng khi B 2004)
Tm m phng trnh ( )2 2 4 2 2m 1 x 1 x 2 2 1 x 1 x 1 x+ + = + + c nghim.
Gii:iu kin xc nh ca phng trnh : x [ 1;1]
t 2 2t 1 x 1 x= + . Vi x [ 1;1] , ta xc nh iu kin ca t nh sau :
Xt hm s 2 2t 1 x 1 x= + vi x [ 1;1]
Ta c :
( )2 22 2 4
x 1 x 1 xx xt '
1 x 1 x 1 x
+ += + =
+ , cho t ' 0 x 0= =
x 1 0 1t 0 +
t2 2
0
Vy vi x [ 1;1] th t 0; 2 T 2 2 4 2t 1 x 1 x 2 1 x 2 t= + = . Khi , phng trnh cho tng ng vi :
( )2
2 t t 2m t 2 t t 2 mt 2
+ ++ = + + =
+
Bi ton tr thnh tm m phng trnh2t t 2
mt 2
+ + =+
c nghim t 0; 2
Xt hm s2
t t 2f(t)
t 2
+ +=
+vi t 0; 2 . Ta c : ( )
2
2
t 4tf '(t) 0, t 0; 2
t 2
= < +
Suy ra : ( )t 0; 2t 0; 2
max f (t) f (0) 1, min f (t) f 2 2 1
= = = =
By gi, yu cu bi ton xy ra khit 0; 2 t 0; 2
min f (t) m max f (t) 2 1 m 1
. y l cc
gi tr cn tm ca tham s.
Bi 6 ( thi tuyn sinh i hc, Cao ng khi B 2006)
Tm m phng trnh 2x mx 2 2x 1+ + = + c nghim thc phn bit.
Gii:
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Ta c: ( )( )
2
2
1x
2x mx 2 2x 1 1
3x 4x 1 mx 2
+ + = + + =
(*)
NX : x = 0 khng phi l nghim ca (2). Do vy, ta tip tc bin i :
( )2
1x2
(*)3x 4x 1
m 3x
+ =
Bi ton tr thnh tm m (3) c nghim thc phn bit { }1
x ; \ 02
+
Xt hm s23x 4x 1
f(x)x
+ = vi { }
1x ; \ 0
2
+ . Ta c :
{ }2
2
3x 1 1f '(x) 0, x ; \ 0x 2
+ = > + BBT :
x 0 1f(x) + +
f(x)
+ +
9
2
T BBT, ta thy : Yu cu bi ton xy ra khi9
m2
.
Vy vi9
m2
th phng trnh cho c nghim thc phn bit.
Bi 7 ( thi tuyn sinh i hc, Cao ng khi A 2007)
Tm m phng trnh ( )243 x 1 m x 1 2 x 1 1 + + = c nghim.
Gii:iu kin xc nh ca phng trnh : x 1Khi :
( )( )
( )2
44 2
x 1 x 1 x 1 x 11 3 m 2 3 m 2 2
x 1 x 1 x 1x 1
+ = + =
+ + ++
t 4x 1
tx 1
=
+( t 0 ). V 4 4x 1 21 1
x 1 x 1
= 0, phng trnh 2x 2x 8 m(x 2)+ = lun c hai nghimthc.
Gii:
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Bi 9 ( thi tuyn sinh i hc, Cao ng khi D 2007)
Gii:
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Bi 10 ( thi tuyn sinh i hc, Cao ng khi A 2008)
Gii:
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Bi tp t gii
Bi 1. Tm m bt phng trnh ( ) ( ) 2x 4 6 x x 2x m+ + ng vi mi x [ 4; 6] .
S : m 6Bi 2. Tm m bt phng trnh x 1 4 x m+ c nghim.
S : m 5
Bi 3. Tm m phng trnh 22 x 2 x 4 x m + + = c nghim.S : 2 2 2 m 2
Bi 4. Tm m h phng trnhx y 1
x x y y 1 3m
+ =
+ = c nghim.
S:1
0 m4
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