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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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CHUYN TON THPT V Trng SnCHUYN TON THPT V Trng Sn
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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CHUYN TON THPT V Trng SnCHUYN TON THPT V Trng Sn
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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CHUYN TON THPT V Trng SnCHUYN TON THPT V Trng Sn
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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CHUYN TON THPT V Trng SnCHUYN TON THPT V Trng Sn
(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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