Chuyn 1

Kho St S Bin Thin V V ThHm S

1. Tnh n iu Ca Hm SBi tp 1.1. Tm cc khong n iu ca cc hm s sau

a) y = 2x3 3x2 + 1. b) y = x3 3x+ 2. c) y = x3 + 3x2 + 3x.d) y = x4 2x2 + 3. e) y = x4 + 2x3 2x 1. f) y = x2 2x 3.g) y =

2x+ 3

x+ 2. h) y =

x+ 2

3x 1 . i) y =x2 4x+ 4

1 x .

Li gii.

a) Tp xc nh: D = R. o hm: y = 6x2 6x; y = 0[x = 0x = 1

. Bng bin thin:

x 0 1 + y + 0 0 +

y

1

0

+

Vy hm s ng bin trn cc khong (; 0), (1; +) v nghch bin trn (0; 1).b) Tp xc nh: D = R. o hm: y = 3x2 3 < 0,x R. Do hm s lun nghch bin trn R.c) Tp xc nh: D = R. o hm: y = 3x2 + 6x+ 3 0,x R. Do hm s lun ng bin trn R.d) Tp xc nh: D = R. o hm: y = 4x3 4x; y = 0

[x = 0x = 1 . Bng bin thin:

x 1 0 1 + y 0 + 0 0 +

y

+

2

3

2

+

Vy hm s ng bin trn cc khong (1; 0) , (1; +) v nghch bin trn cc khong (;1) , (0; 1).e) Tp xc nh: D = R. o hm: y = 4x3 + 6x2 2; y = 0

[x = 1x = 12

. Bng bin thin:

x 12 1 + y + 0 0

y

5162

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Vy hm s ng bin trn khong(; 12) v nghch bin trn khong ( 12 ; +).

f) Tp xc nh: D = (;1] [3; +). o hm: y = x 1x2 2x 3 ; y

= 0 x = 1. Bng bin thin:

x 1 3 + y +

y

+

0 0

+

Vy hm s ng bin trn khong (3; +) v nghch bin trn khong (;1).g) Tp xc nh: D = R\ {2}. o hm: y = 1

(x+ 2)2> 0,x D.

Do hm s ng bin trn cc khong (;2) v (2; +).h) Tp xc nh: D = R\{ 13}. o hm: y = 7(3x 1)2 < 0,x D.Do hm s nghch bin trn cc khong (; 13 ) v ( 13 ; +).i) Tp xc nh: D = R\ {1}. o hm: y = x

2 + 2x

(1 x)2 ; y = 0

[x = 0x = 2

. Bng bin thin:

x 0 1 2 + y 0 + + 0

y

+

2

+

0

Vy hm s ng bin trn cc khong (0; 1), (1; 2) v nghch bin trn cc khong (; 0), (2; +).Bi tp 1.2. Tm m hm s y = x3 + (m 1)x2 + (m2 4)x+ 9 lun ng bin trn R.Li gii. Tp xc nh: D = R. o hm: y = 3x2+2(m1)x+m24; = (m1)23(m24) = 2m22m+13.

Hm s lun ng bin trn R y 0,x R 0 2m2 2m+ 13 0

m 1 3

3

2

m 1 + 3

3

2

.

Vy vi m (; 1 3

3

2

][1 + 33

2; +

)th hm s cho lun ng bin trn R.

Bi tp 1.3. Tm m hm s y = mx3 + (3m)x2 2x+ 2 lun nghch bin trn R.Li gii. Tp xc nh: D = R. Vi m = 0, ta c: y = 3x2 2x+ 2 l mt parabol nn khng th nghch bin trn R. Vi m 6= 0, ta c: y = 3mx2 + 2(3m)x 2; = (3m)2 6m = m2 12m+ 9.Hm s lun nghch bin trn R y 0,x R

{m < 0m2 12m+ 9 0

{m < 0

6 33 m 6 + 33 m

Vy khng c gi tr no ca m hm s lun nghch bin trn R.

Bi tp 1.4. Tm m hm s y =mx 2m x lun ng bin trn mi khong xc nh.

Li gii. Tp xc nh: D = R\ {m}. o hm: y = m2 2

(m x)2 .

Hm s lun ng bin trn mi khong xc nh y > 0,x D m2 2 > 0[m 2m 2 .

Vy vi m (;2] [2; +) th hm s cho lun ng bin trn mi khong xc nh.Bi tp 1.5. Tm m hm s y =

mx 2x+m 3 lun nghch bin trn mi khong xc nh.

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Chuyn 1. Kho St S Bin Thin V V Th Hm S

Li gii. Tp xc nh: D = R\ {3m}. o hm: y = m2 3m+ 2

(x+m 3)2 .Hm s lun nghch bin trn mi khong xc nh y < 0,x D m2 3m+ 2 < 0 1 < m < 2.Vy vi m (1; 2) th hm s cho lun nghch bin trn mi khong xc nh.

Bi tp 1.6. Tm m hm s y = x+ 2 +m

x 1 lun ng bin trn mi khong xc nh.

Li gii. Tp xc nh: D = R\ {1}.o hm: y = 1 m

(x 1)2 =x2 2x+ 1m

(x 1)2 ; y = 0 x2 2x+ 1m = 0; = m.

Hm s ng bin trn mi khong xc nh khi v ch khi

y 0,x D x2 2x+ 1m 0,x D 0 m 0Vy vi m 0 th hm s cho lun ng bin trn mi khong xc nh.

Bi tp 1.7. Tm m hm s y =mx+ 4

x+mnghch bin trn (; 1).

Li gii. Tp xc nh: D = R\ {m}. o hm: y = m2 4

(x+m)2.

Hm s nghch bin trn (; 1) khi v ch khi y < 0,x (; 1)

{ m / (; 1)m2 4 < 0

{ m 12 < m < 2 2 < m 1

Vy vi m (2;1] th hm s cho lun ng bin trn mi khong xc nh.

Bi tp 1.8. Tm m hm s y =mx 2x+m 3 nghch bin trn (1; +).

Li gii. Tp xc nh: D = R\ {3m}. o hm: y = m2 3m+ 2

(x+m 3)2 .Hm s nghch bin trn (1; +) y < 0,x (1; +)

{

3m / (1; +)m2 3m+ 2 < 0

{3m 11 < m < 2

m

Vy khng c gi tr no ca m hm s nghch bin trn (1; +).Bi tp 1.9. Tm a hm s y = x3 + 3x2 + ax+ a nghch bin trn on c di bng 1.

Li gii. Tp xc nh: D = R. o hm: y = 3x2 + 6x+ a; = 9 3a. 0 a 3 y 0,x R: Hm s ng bin trn R nn khng tha mn yu cu bi ton. > 0 a < 0, y c hai nghim x1, x2 (x1 < x2). Theo nh l vi-t c x1 + x2 = 2;x1x2 = a3 .Bng bin thin:

x x1 x2 + y + 0 0 +

y

y(x1)

y(x2)

+

T bng bin thin ta c hm s nghch bin trn [x1;x2].Do hm s nghch bin trn on c di bng 1 khi v ch khi

|x1 x2| = 1 (x1 x2)2 = 1 (x1 + x2)2 4x1x2 = 1 4 4a3

= 1 a = 94(tha mn)

Vy vi a = 94 th hm s cho nghch bin trn on c di bng 1.

Bi tp 1.10. Tm m hm s y = x3 + 3x2 +mx+ 2 ng bin trn on c di bng 3.Li gii. Tp xc nh: D = R. o hm: y = 3x2 + 6x+m; = 9 + 3m. 0 m 3 y 0,x R: Hm s nghch bin trn R nn khng tha mn yu cu bi ton. > 0 m < 3, y c hai nghim x1, x2 (x1 < x2). Theo nh l vi-t c x1 + x2 = 2;x1x2 = m3 .Bng bin thin:

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x x1 x2 + y 0 + 0

y

+

y(x1)

y(x2)

T bng bin thin ta c hm s ng bin trn [x1;x2].Do hm s ng bin trn on c di bng 3 khi v ch khi

|x1 x2| = 3 (x1 x2)2 = 9 (x1 + x2)2 4x1x2 = 9 4 + 4m3

= 9 m = 154

(tha mn)

Vy vi m = 154 th hm s cho ng bin trn on c di bng 3.

2. Cc Tr Ca Hm SBi tp 1.11. Tm cc tr ca cc hm s sau

a) y = 2x3 3x2 + 1. b) y = x3 3x+ 2. c) y = x3 + 3x2 + 3x.d) y = x4 2x2 + 3. e) y = x4 + 2x3 2x 1. f) y = x2 2x 3.g) y =

2x+ 3

x+ 2. h) y =

x+ 2

3x 1 . i) y =x2 4x+ 4

1 x .

Li gii.

a) Tp xc nh: D = R. o hm: y = 6x2 6x; y = 0[x = 0x = 1

. Bng bin thin:

x 0 1 + y + 0 0 +

y

1

0

+

Vy hm s t cc i ti x = 0; yC = 1 v t cc tiu ti x = 1; yCT = 0.b) Tp xc nh: D = R. o hm: y = 3x2 3 < 0,x R. Do hm s khng c cc tr.c) Tp xc nh: D = R. o hm: y = 3x2 + 6x+ 3 0,x R. Do hm s khng c cc tr.d) Tp xc nh: D = R. o hm: y = 4x3 4x; y = 0

[x = 0x = 1 . Bng bin thin:

x 1 0 1 + y 0 + 0 0 +

y

+

2

3

2

+

Vy hm s t cc i ti x = 0; yC = 3 v t cc tiu ti x = 1; yCT = 2.e) Tp xc nh: D = R. o hm: y = 4x3 + 6x2 2; y = 0

[x = 1x = 12

. Bng bin thin:

x 12 1 + y + 0 0

y

5162

Vy hm s t cc i ti x = 12 ; yC = 516 .f) Tp xc nh: D = (;1] [3; +). o hm: y = x 1

x2 2x 3 ; y = 0 x = 1. Bng bin thin:

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Chuyn 1. Kho St S Bin Thin V V Th Hm S

x 1 3 + y +

y

+

0 0

+

Vy hm s khng c cc tr.

g) Tp xc nh: D = R\ {2}. o hm: y = 1(x+ 2)2

> 0,x D. Do hm s khng c cc tr.

h) Tp xc nh: D = R\{ 13}. o hm: y = 7(3x 1)2 < 0,x D. Do hm s khng c cc tr.i) Tp xc nh: D = R\ {1}. o hm: y = x

2 + 2x

(1 x)2 ; y = 0

[x = 0x = 2

. Bng bin thin:

x 0 1 2 + y 0 + + 0

y

+

2

+

0

Vy hm s t cc i ti x = 2; yC = 0 v t cc tiu ti x = 0; yCT = 2.

Bi tp 1.12. Tm m hm s y = x3 3mx2 + 3 (2m 1)x 2a) C cc tr. b) t cc tr ti x = 0. c) t cc i ti x = 1.

Li gii. Tp xc nh: D = R. o hm: y = 3x2 6mx+ 3(2m 1); = 9m2 18m+ 9.a) Hm s c cc tr y c hai nghim phn bit 9m2 18m+ 9 > 0 m 6= 1.b) Hm s t cc tr ti x = 0 y(0) = 0 3(2m 1) = 0 m = 12 .Vi m = 12 y = 3x2 3x; y = 6x 3; y(0) = 3 < 0 hm s t cc i ti x = 0 m = 12 tha mn.Vy vi m = 12 th hm s cho t cc tr ti x = 0.c) Hm s t cc i ti x = 1 y(1) = 0 3 6m+ 3(2m 1) = 0 0 = 0 (ng m R).Li c: y = 6x 6m; y(1) = 6 6m.Vi y(1) > 0 m < 1 hm s t cc tiu ti x = 1 m < 1 khng tha mn.Vi y(1) < 0 m > 1 hm s t cc i ti x = 1 m > 1 tha mn.Vi y(1) = 0 m = 1, ta c y = 3x2 6x+ 3 = 3(x 1)2 0,x R hm s khng c cc tr.Vy vi m > 1 th hm s cho t cc i ti x = 1.

Bi tp 1.13. Cho hm s y = 13x3 mx2 + (m2 m+ 1)x+ 1. Vi gi tr no ca m th hm s

a) t cc i ti x = 1. b) C cc i, cc tiu. c) Khng c cc tr.

Li gii. Tp xc nh: D = R. o hm: y = x2 2mx+m2 m+ 1; = m 1.a) Hm s t cc i ti x = 1 y(1) = 0 1 2m+m2 m+ 1 = 0 m2 3m+ 2 = 0

[m = 1m = 2

.

Vi m = 1 y = x2 2x+ 1 = (x 1)2 0,x R hm s khng c cc tr. Vi m = 2 y = x2 4x+ 3; y = 2x 4; y(1) = 2 < 0 hm s t cc i ti x = 1.Vy vi m = 2 th hm s cho t cc i ti x = 1.b) Hm s c cc i, cc tiu y c hai nghim phn bit > 0 m 1 > 0 m > 1.c) Hm s khng c cc tr y khng c hai nghim phn bit 0 m 1 0 m 1.

Bi tp 1.14. Cho hm s y = x4 2 (m+ 1)x2 + 2m+ 1. Vi gi tr no ca m th hm sa) C ba im cc tr. b) t cc tiu ti x = 0. c) t cc tr ti x = 1.

Li gii. Tp xc nh: D = R. o hm: y = 4x3 4(m+ 1)x.a) y = 0

[x = 0x2 = m+ 1

. Hm s c ba im cc tr y c ba nghim phn bit m+ 1 > 0 m > 1.b) Hm s t cc tiu ti x = 0 y(0) = 0 0 = 0 (ng m R).Li c: y = 12x2 4(m+ 1); y(0) = 4(m+ 1).Vi y(0) > 0 m < 1 hm s t cc tiu ti x = 0 m < 1 tha mn.Vi y(0) < 0 m > 1 hm s t cc i ti x = 0 m > 1 khng tha mn.Vi y(0) = 0 m = 1, ta c y = 4x3; y = 0 x = 0.

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x 0 + y 0 +

y

+

1

+

Suy ra hm s t cc tiu ti x = 0. Vy vi m 1 th hm s cho t cc tiu ti x = 0.c) Hm s t cc tr ti x = 1 y(1) = 0 4 4(m+ 1) = 0 m = 0.Vi m = 0 y = 4x3 4x; y = 12x2 4; y(1) = 8 > 0 hm s t cc tiu ti x = 1 m = 0 tha mn.Vy vi m = 0 th hm s cho t cc tr ti x = 1.

Bi tp 1.15. Tm m hm s y = x4 + 2 (2m 1)x2 + 3 c ng mt cc tr.

Li gii. Tp xc nh: D = R. o hm: y = 4x3 + 4(2m 1)x = 4x(x2 + 2m 1). y = 0[x = 0x2 = 2m 1 .

Hm s c ng mt cc tr y c ng mt nghim 2m 1 0 m 12 .Bi tp 1.16. (B-02) Tm m hm s y = mx4 +

(m2 9)x2 + 10 c ba im cc tr.

Li gii. Tp xc nh: D = R. o hm: y = 4mx3 + 2(m2 9)x = 2x(2mx2 +m2 9). Vi m = 0, ta c y = 18x c mt nghim nn hm s khng th c ba cc tr. Vi m 6= 0, ta c y = 0

[x = 0

x2 = 9m2

2m

.

Hm s c ba cc tr y c ba nghim phn bit 9m2

2m> 0. Bng xt du:

m 3 0 3 +9m2 0 + | + 0

2m | 0 + | +VT + 0 || + 0

T bng xt du ta c m (;3) (0; 3).

Bi tp 1.17. Xc nh gi tr ca m hm s y =x2 +mx+ 1

x+ma) Khng c cc tr. b) t cc tiu ti x = 1. c) t cc i ti x = 2.

Li gii. Tp xc nh: D = R\{m}.a) o hm: y =

x2 + 2mx+m2 1(x+m)2

; y = 0 x = m 1 hm s lun c cc tr.Vy khng c gi tr no ca m hm s khng c cc tr.

b) Hm s t cc tiu ti x = 1 y(1) = 0 m2 + 2m

(m+ 1)2= 0

[m = 0m = 2 .

Vi m = 0 y = x2 1x2

; y = 0 x = 1.

x 1 0 1 + y + 0 0 +

y

2

+

2

+

T bng bin thin suy ra hm s t cc tiu ti x = 1 m = 0 tha mn. Vi m = 2 y = x

2 4x+ 3(x 2)2 ; y

= 0 x = 1 hoc x = 3.

x 1 2 3 + y + 0 0 +

y

0

+

4

+

6

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Chuyn 1. Kho St S Bin Thin V V Th Hm S

T bng bin thin suy ra hm s t cc i ti x = 1 m = 2 khng tha mn.Vy vi m = 0 th hm s cho t cc tiu ti x = 1.

c) Hm s t cc i ti x = 2 y(2) = 0 m2 + 4m+ 3

(m+ 2)2= 0

[m = 1m = 3 .

Vi m = 1 y = x2 2x

(x 1)2 ; y = 0 x = 0 hoc x = 2.

x 0 1 2 + y + 0 0 +

y

1

+

3

+

T bng bin thin suy ra hm s t cc tiu ti x = 2 m = 1 khng tha mn. Vi m = 3 y = x

2 6x+ 8(x 3)2 ; y

= 0 x = 2 hoc x = 4.

x 2 3 4 + y + 0 0 +

y

1

+

5

+

T bng bin thin suy ra hm s t cc i ti x = 2 m = 3 tha mn.Vy vi m = 3 th hm s cho t cc i ti x = 2.

3. Gi Tr Ln Nht V Gi Tr Nh Nht Ca Hm SBi tp 1.18. Tm gi tr ln nht v gi tr nh nht (nu c) ca cc hm s sau:

a) y = 1 + 8x 2x2 trn [1; 3]. b) y = x3 3x2 + 1 trn [2; 3]. c) y = 1 + 4x3 3x4 trn [2; 1].d) y = x3 3x2 + 1 trn (1; 4). e) y = x 5 + 1x trn (0; +). f) y = x 1x trn (0; 2].g) y =

4

1 + x2. h) y = x4 + 2x2 1. i) y = x+4 x2.

Li gii.a) Ta c: y = 8 4x; y = 0 x = 2; y(1) = 9, y(2) = 9, y(3) = 7.Vy max

[1;3]y = y(2) = 9; min

[1;3]y = y(1) = 9.

b) Ta c: y = 3x2 6x; y = 0 x = 0 hoc x = 2; y(2) = 19, y(0) = 1, y(2) = 3, y(3) = 1.Vy max

[2;3]y = y(0) = y(3) = 1; min

[2;3]y = y(2) = 19.

c) Ta c: y = 12x2 12x3; y = 0 x = 0 hoc x = 1; y(2) = 79, y(0) = 1, y(1) = 2.Vy max

[2;1]y = y(1) = 2; min

[2;1]y = y(2) = 79.

d) Ta c: y = 3x2 6x; y = 0 x = 0 hoc x = 2.x 1 2 4

y 0 +

y

1

3

17

Vy min(1;4)

y = y(2) = 3; hm s khng c gi tr ln nht.e) Ta c: y = 1 1x2 ; y = 0 x = 1.

x 0 1 + y 0 +

y

+

3

+

7

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Vy min(0;+)

y = y(1) = 3; hm s khng c gi tr ln nht.f) Ta c: y = 1 + 1x2 > 0,x (0; 2].

x 0 2

y +

y

32

Vy max(0;2]

y = y(2) = 32 ; hm s khng c gi tr nh nht.

g) Tp xc nh: D = R. Ta c: y = 8x(1 + x2)2

; y = 0 x = 0.

x 0 + y + 0

y

0

4

0

Vy maxR

y = y(0) = 4; hm s khng c gi tr nh nht.

h) Tp xc nh: D = R. Ta c: y = 4x3 + 4x; y = 0 x = 0.

x 0 + y 0 +

y

+

1

+

Vy minRy = y(0) = 1; hm s khng c gi tr ln nht.

i) Tp xc nh: D = [2; 2]. Ta c: y = 1 x4 x2 ; y

= 0 x = 2; y(2) = 0, y(2) = 22, y(2) = 0.Vy max

[2;2]y = y(

2) = 2

2; min

[2;2]y = y(2) = 0.

Bi tp 1.19. Tm gi tr ln nht v gi tr nh nht (nu c) ca cc hm s saua) y = x+

2 cosx trn

[0; pi2

]. b) y = 2 sinx 43 sin3x trn [0;pi]. c) y = sin4x 4sin2x+ 5.

d) y = sin4x+ cos4x. e) y = 5 sinx 12 cosx 5. f) y = sin2x+ sin 2x+ 2cos2x.Li gii.

a) Ta c: y = 12 sinx; y = 0 sinx = 12 x = pi4 ; y(0) =

2, y(pi4 ) =

pi4 + 1, y(

pi2 ) =

pi2 .

Vy max[0;pi2 ]

y = y(pi4 ) =pi4 + 1; min

[0;pi2 ]y = y(0) =

2.

b) t sinx = t, t [0; 1]. Hm s tr thnh y = f(t) = 2t 43 t3.Ta c: f (t) = 2 4t2; f (t) = 0 t = 1

2; f(0) = 0, f( 1

2) = 2

2

3 , y(1) =23 .

Vy max[0;pi]

y = max[0;1]

f(t) = f( 12) = 2

2

3 ; min[0;pi]y = min

[0;1]f(t) = f(0) = 0.

c) Tp xc nh: D = R. t sinx = t, t [1; 1]. Hm s tr thnh y = f(t) = t4 4t2 + 5.Ta c: f (t) = 4t3 8t; f (t) = 0 t = 0 hoc t = 2 (loi); f(0) = 5, f(1) = 2.Vy max

Ry = max

[1;1]f(t) = f(0) = 5; min

Ry = min

[1;1]f(t) = f(1) = 2.

d) Tp xc nh: D = R.Ta c: y = sin4 x+ cos4 x = 1 12 sin2 2x. t sin 2x = t, t [1; 1]. Hm s tr thnh y = f(t) = 1 12 t2.o hm: f (t) = t; f (t) = 0 t = 0; f(1) = 12 , f(0) = 1.Vy max

Ry = max

[1;1]f(t) = f(0) = 1; min

Ry = min

[1;1]f(t) = f(1) = 12 .

e) Ta c: y = 5 sinx 12 cosx 5 5 sinx 12 cosx = y + 5.Phng trnh c nghim 52 + 122 (y + 5)2 y2 + 10y 144 0 18 y 8.

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Chuyn 1. Kho St S Bin Thin V V Th Hm S

Vy maxR

y = 8; minRy = 18.

f) Ta c: y =1 cos 2x

2+ sin 2x+ 1 + cos 2x = sin 2x+

1

2cos 2x+

3

2 2 sin 2x+ cos 2x = 2y 3.

Phng trnh c nghim 22 + 12 (2y 3)2 4y2 12y + 4 0 3

52 y 3+

5

2 .Vy max

Ry = 3+

5

2 ; minRy = 3

5

2 .

Bi tp 1.20. Cho parabol (P ) : y = x2 v im A (3; 0). Tm im M (P ) sao cho khong cch AM ngnnht v tnh khong cch .

Li gii. Ta c: M (P )M(t; t2) AM = (t+ 3; t2) AM = (t+ 3)2 + t4 = t4 + t2 + 6t+ 9.Xt hm s f(t) = t4 + t2 + 6t+ 9 trn R; f (t) = 4t3 + 2t+ 6; f (t) = 0 t = 1. Bng bin thin:

t 1 + f (t) 0 +

f(t)

+

5

+

T bng bin thin ta c minRf(t) = f(1) = 5.

Suy ra AM t gi tr nh nht bng

5 khi t = 1M(1; 1). Vy M(1; 1).Bi tp 1.21. Tm m hm s y = x3 + 3x2 mx 4 ng bin trn (; 0).Li gii. Ta c: y = 3x2 + 6xm.

Hm s ng bin trn (; 0) 3x2 + 6xm 0,x (; 0) m 3x2 + 6x, x (; 0) (1)Xt hm s f(x) = 3x2 + 6x trn (; 0] c f (x) = 6x+ 6; f (x) = 0 x = 1. Bng bin thin:

x 1 0f (x) 0 +

f(x)

+

3

0

T bng bin thin ta c min(;0]

f(x) = f(1) = 3. Do (1) m min(;0]

f(x) m 3.

Bi tp 1.22. (BT-79) Tm m hm s y = 13x3 + (m 1)x2 + (m+ 3)x 4 ng bin trn (0; 3).Li gii. Ta c: y = x2 + 2(m 1)x+m+ 3 = m(2x+ 1) x2 2x+ 3.

Hm s ng bin trn (0; 3) m(2x+ 1) x2 2x+ 3 0,x (0; 3) m x2 + 2x 32x+ 1

,x (0; 3) (2).

Xt hm s f(x) =x2 + 2x 3

2x+ 1trn [0; 3] c f (x) =

2x2 + 2x+ 8

(2x+ 1)2> 0,x [0; 3]. Bng bin thin:

x 0 3

f (x) +

f(x)

3

127

T bng bin thin ta c max[0;3]

f(x) = f(3) =12

7. Do (2) m max

[0;3]f(x) m 12

7.

Bi tp 1.23. Tm m hm s y = mx3 3 (m 1)x2 + 9 (m 2)x+ 1 ng bin trn [2; +).Li gii. Ta c: y = 3mx2 6(m 1)x+ 9(m 2) = 3m(x2 2x+ 3) + 6x 18.

Hm s ng bin trn [2; +) khi v ch khi

3m(x2 2x+ 3) + 6x 18 0,x [2; +) m 6 2xx2 2x+ 3 ,x [2; +) (3)

Xt hm s f(x) =6 2x

x2 2x+ 3 trn [2; +) c f(x) =

2x2 12x+ 6(x2 2x+ 3)2 ; f

(x) = 0 x = 3

6.

Bng bin thin:

9

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x 2 3 +

6 + f (x) 0 +

f(x)

23

f(3 +

6)

0

T bng bin thin ta c max[2;+)

f(x) = f(2) =2

3. Do (3) m max

[2;+)f(x) m 2

3.

Bi tp 1.24. Tm m hm s y = x3 + 3x2 + (m+ 1)x+ 4m ng bin trn (;2) v (2; +).Li gii. Ta c: y = 3x2 + 6x+m+ 1.

Hm s ng bin trn (;2) v (2; +) khi v ch khi

3x2 + 6x+m+ 1 0,x (;2) (2; +) m 3x2 6x 1,x (;2) (2; +) (4)

Xt hm s f(x) = 3x2 6x 1 trn (;2] [2; +) c f (x) = 6x 6; f (x) = 0 x = 1.Bng bin thin:

x 2 2 + f (x) +

f(x)

1 25

T bng bin thin ta c max(;2][2;+)

f(x) = f(2) = 1. Do (4) m max(;2][2;+)

f(x) m 1.

Bi tp 1.25. (BT-50) Tm m hm s y =mx2 + 6x 2

x+ 2nghch bin trn [1; +).

Li gii. Hm s xc nh trn [1; +). o hm: y = mx2 + 4mx+ 14

(x+ 2)2=m(x2 + 4x) + 14

(x+ 2)2.

Hm s nghch bin trn [1; +) m(x2 + 4x) + 14

(x+ 2)2 0,x [1; +) m 14

x2 + 4x,x [1; +) (5).

Xt hm s f(x) =14

x2 + 4xtrn [1; +) c f (x) = 28x+ 56

(x2 + 4x)2> 0,x [1; +). Bng bin thin:

x 1 + f (x) +

f(x)

145

0

T bng bin thin ta c min[1;+)

f(x) = f(1) = 145. Do (5) m min

[1;+)f(x) m 14

5.

Bi tp 1.26. Tm m hm s y =x2 2mx+ 2m2 2

xm ng bin trn (1; +).

Li gii. Tp xc nh: D = R\ {m}. o hm: y = x2 2mx+ 2(xm)2 .

Hm s ng bin trn (1; +) y 0,x (1; +)

{m / (1; +)x2 2mx+ 2 0,x (1; +)

{m 1m x2+22x ,x (1; +)

(1)

Xt hm s f(x) =x2 + 2

2xtrn (1; +) c f (x) = x

2 22x2

; f (x) = 0 x =

2. Bng bin thin:

10

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Chuyn 1. Kho St S Bin Thin V V Th Hm S

x 1

2 + f (x) 0 +

f(x)

32

2

+

Do (1){m 1m 2 m 1. Vy vi m 1 th hm s ng bin trn (1; +).

Bi tp 1.27. Tm a hm s y =x2 2ax+ 4a2

x 2a ng bin trn (2; +).

Li gii. Tp xc nh: D = R\ {2a}. o hm: y = x2 4ax

(x 2a)2 .Hm s ng bin trn (2; +) y 0,x (2; +)

{

2a / (2; +)x2 4ax 0,x (2; +)

{2a 2a x4 ,x (2; +)

{a 1a 12

a 12

Vy vi m 12 th hm s ng bin trn (2; +).

4. ng Tim Cn Ca Th Hm SBi tp 1.28. Tm tim cn (nu c) ca cc hm s sau

a) y =2x 1x 2 . b) y =

x 3x+ 2 . c) y =

3 4xx+ 1

.

d) y =x2 + x

x 1 . e) y =x+ 3

x+ 1. f) y = 2x 1 + 1

x.

g) y =x2 4x+ 4

1 x . h) y =x2 + x 1. i) y = x+

x2 + 2x.

Li gii.a) Tp xc nh: D = R\ {2}.Ta c lim

x y = 2 TCN l y = 2; limx2+ y = +, limx2 y = TC l x = 2.Vy hm s c tim cn ngang y = 2 v tim cn ng x = 2.b) Tp xc nh: D = R\ {2}.Ta c lim

x y = 1 TCN l y = 1; limx2+ y = +, limx2 y = TC l x = 2.Vy hm s c tim cn ngang y = 1 v tim cn ng x = 2.c) Tp xc nh: D = R\ {1}.Ta c lim

x y = 4 TCN l y = 4; limx1+ y = +, limx1 y = TC l x = 1.Vy hm s c tim cn ngang y = 4 v tim cn ng x = 1.d) Tp xc nh: D = R\ {1}.Ta c lim

x y = 1 TCN l y = 1; limx1+ y = +, limx1 y = TC l x = 1.Vy hm s c hai tim cn ngang y = 1 v tim cn ng x = 1.e) Tp xc nh: D = R\ {1}.Ta c lim

x y = 0 TCN l y = 0; limx1+ y = +, limx1 y = TC l x = 1.Vy hm s c tim cn ngang y = 0 v tim cn ng x = 1.f) Tp xc nh: D = R\ {0}. Hm s vit thnh y = 2x

2 x+ 1x

.

Ta c limx [y (2x 1)] = 0 TCX l y = 2x 1; limx0+ y = +, limx0 y = TC l x = 0.

Vy hm s c tim cn xin y = 2x 1 v tim cn ng x = 0.g) Tp xc nh: D = R\ {0}. Hm s vit thnh y = x+ 3 + 1

1 x .Ta c lim

x [y (x+ 3)] = 0 TCX l y = x+ 3; limx1+ y = , limx1 y = + TC l x = 1.Vy hm s c tim cn xin y = x+ 3 v tim cn ng x = 1.h) Tp xc nh: D =

(; 1

5

2

][1 +5

2; +

). Ta c

11

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limx+

x2 + x 1

x= 1; lim

x+

(x2 + x 1 x

)=

1

2 TCX l y = x+ 1

2.

limx

x2 + x 1

x= 1; lim

x

(x2 + x 1 + x

)= 1

2 TCX l y = x 1

2.

Vy hm s c hai tim cn xin y = x+ 12 v y = x 12 .i) Tp xc nh: D = (;2] [0; +). Ta c limx+

x+x2 + 2

x= 2; lim

x+

(x+

x2 + 2 2x

)=

1

2 TCX l y = 2x+ 1

2.

limx

(x+

x2 + 2x

)= limx

x2 (x2 + 2x)xx2 + 2x = 1 TCN l y = 1.

Vy hm s c tim cn xin y = 2x+ 12 v tim cn ngang y = 1.

Bi tp 1.29. Tmm th hm s y =mx2 2m (m 1)x 3m2 +m 2

x+ 2c tim cn xin i qua A (1;3).

Li gii. Tp xc nh: D = R\ {2}. Hm s vit thnh y = mx 2m2 + m2 +m 2x+ 2

.

Do vi m 6= 0,m 6= 1,m 6= 2 hm s c tim cn xin y = mx 2m2.Khi tim cn xin qua A(1;3) 3 = m 2m2

[m = 1 (loi)m = 32

.

Vy vi m = 32 th tim cn xin ca hm s cho qua A(1;3).

Bi tp 1.30. Tmm hm s y =2x2 + (m+ 1)x 3

x+mc giao hai tim cn nm trn parabol (P ) : y = x2+2x1.

Li gii. Tp xc nh: D = R\ {m}. Hm s vit thnh y = 2xm+ 1 + m2 m 3x+m

.

Do vi m 6= 1

52 hm s c tim cn xin y = 2xm+ 1 v tim cn ng x = m

Suy ra giao hai tim cn l I(m; 1 3m).Khi I (P ) 1 3m = m2 2m 1

[m = 1m = 2 (tha mn).

Vy vi m = 1 v m = 2 th hm s c giao hai tim cn thuc (P ).

Bi tp 1.31. (A-08) Tm m gc gia hai tim cn ca hm s y =mx2 +

(3m2 2)x 2x+ 3m

bng 450.

Li gii. Tp xc nh: D = R\ {3m}. Hm s vit thnh y = mx 2 + 6m 2x+ 3m

.

Vi m = 13 hm s khng c tim cn nn khng tha mn yu cu bi ton. Vi m = 0 hm s c tim cn ngang y = 2 v tim cn ng x = 3m.Khi gc gia hai tim cn bng 900 nn khng tha mn yu cu bi ton. Vi m 6= 13 ,m 6= 0 hm s c tim cn xin y = mx 2 v tim cn ng x = 3m.Khi gc gia hai tim cn bng 450 gc gia tim cn xin v tia Ox bng 450 hoc bng 1350 m = 1.Vy vi m = 1 th hm s c gc gia hai tim cn bng 450.

Bi tp 1.32. Tm m th hm s y =x2 +mx 1

x 1 c tim cn xin to vi cc trc to mt tam gic cdin tch bng 4.

Li gii. Tp xc nh: D = R\ {1}. Hm s vit thnh y = x+m+ 1 + mx 1 .

Do vi m 6= 0 hm s c tim cn xin y = x+m+ 1.Tim cn xin ct Ox ti A(m 1; 0) OA = |m+ 1| v ct Oy ti B(0;m+ 1) OB = |m+ 1|.Khi SOAB = 12OA.OB =

12 (m+ 1)

2 12 (m+ 1)2 = 4 m = 1 2

2.Vy vi m = 1 22 th tim cn xin to vi hai trc ta mt tam gic c din tch bng 4.

Bi tp 1.33. Tm m th hm s y =2x2 (5m 1)x+ 4m2 m 1

xm c tim cn xin to vi cc trc to mt tam gic c din tch bng 4.

Li gii. Tp xc nh: D = R\ {m}. Hm s vit thnh y = 2x 3m+ 1 + m2 1x 1 .

Do vi m 6= 1 hm s c tim cn xin y = 2x 3m+ 1.Tim cn xin ct Ox ti A( 3m12 ; 0) OA = |3m1|2 v ct Oy ti B(0;3m+ 1) OB = |3m 1|.Khi SOAB = 12OA.OB =

14 (3m 1)2 14 (3m 1)2 = 4

[m = 1 (loi)m = 53

.

Vy vi m = 53 th tim cn xin to vi hai trc ta mt tam gic c din tch bng 4.

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Chuyn 1. Kho St S Bin Thin V V Th Hm S

Bi tp 1.34. Cho hm s y =3x 1x 2 . Chng minh tch cc khong cch t im M nm trn th hm s n

hai tim cn khng i.

Li gii. Tp xc nh:D = R\ {2}. Hm s c tim cn ngang y = 3 y3 = 0 v tim cn ng x = 2 x2 = 0.Ly M(x0; 3x01x02 ) thuc th ta c

d (M,TCN) =5

|x0 2| ; d (M,TC) = |x0 2| d (M,TCN) .d (M,TC) = 5

Vy tch cc khong cch t M n hai tim cn l mt hng s (pcm).

Bi tp 1.35. Cho hm s y =x2 + 4x 3

x 2 . Chng minh tch cc khong cch t im M nm trn th hms n hai tim cn l mt hng s.

Li gii. Tp xc nh: D = R\ {2}. Hm s vit thnh y = x+ 2 + 1x 2 .

Do hm s c tim cn xin y = x+ 2 x+ y 2 = 0 v tim cn ng x = 2 x 2 = 0.Ly M(x0;x0 + 2 + 1x02 ) thuc th ta c

d (M,TCN) =1

|x0 2| ; d (M,TC) = |x0 2| d (M,TCN) .d (M,TC) = 1

Vy tch cc khong cch t M n hai tim cn l mt hng s (pcm).

Bi tp 1.36. Tm M thuc th hm s y =3x 5x 2 tng khong cch t M n hai tim cn l nh nht.

Li gii. Tp xc nh:D = R\ {2}. Hm s c tim cn ngang y = 3 y3 = 0 v tim cn ng x = 2 x2 = 0.Ly M(x0; 3x01x02 ) thuc th ta c d (M,TCN) =

1|x02| ; d (M,TC) = |x0 2|.

Khi d (M,TCN) + d (M,TC) = 1|x02| + |x0 2| 2. Du bng xy ra 1|x02| = |x0 2| [x0 = 1x0 = 3

.

Vy tng cc khong cch t M n hai tim cn nh nht bng 2 khi M(1; 2) v M(3; 4).

Bi tp 1.37. Tm im M thuc th hm s y =x2 + 2x 2

x 1 tng khong cch t M n hai tim cn lnh nht.

Li gii. Tp xc nh: D = R\ {1}. Hm s vit thnh y = x+ 3 + 1x 1

Do hm s c tim cn xin y = x+ 3 x y + 3 = 0 v tim cn ng x = 1 x 1 = 0.Ly M(x0;x0 + 3 + 1x01 ) thuc th ta c d (M,TCN) =

1|x01| ; d (M,TC) = |x0 1|.

Khi d (M,TCN) + d (M,TC) = 1|x01| + |x0 1| 2. Du bng xy ra 1|x01| = |x0 1| [x0 = 0x0 = 2

.

Vy tng cc khong cch t M n hai tim cn nh nht bng 2 khi M(0; 2) v M(2; 6).

5. Kho St S Bin Thin V V Th Hm SBi tp 1.38. Kho st s bin thin v v th ca cc hm s sau

a) y = x3 + 3x2 4. b) y = x3 + 3x 2. c) y = x3 + 1. d) y = x3 + 3x2 + 3x+ 1.e) y = x3 + x 2. f) y = 2x3 x 3. g) y = x3 + 3x2 1. h) y = 13x3 x2 3x 53 .

Bi tp 1.39. Kho st s bin thin v v th ca cc hm s saua) y = x4 2x2 3. b) y = x4 + 2x2 1. c) y = 12x4 + x2 32 . d) y = 3 2x2 x4.e) y = x4 + 2x2 2. f) y = 2x4 4x2 + 1. g) y = 2x4 4x2 + 1. h) y = x4 4x2 + 3.

Bi tp 1.40. Kho st s bin thin v v th ca cc hm s sau

a) y =4

2 x . b) y =x 32 x . c) y =

x+ 3

x 1 . d) y =x+ 22x+ 1

.

e) y =x 2x+ 1

. f) y =x+ 2

x 1 . g) y =2 xx+ 1

. h) y =x+ 3

x 2 .

Bi tp 1.41. Kho st s bin thin v v th ca cc hm s sau

a) y =x2 + 2x+ 2

x+ 1. b) y =

x2 2x 3x 2 . c) y =

2x2 + 5x+ 4

x+ 2. d) y =

x2 2xx+ 1

.

e) y =x2 2xx 1 . f) y =

2x2 x+ 11 x . g) y = x+ 2 +

1

x 1 . h) y = x 1 +1

x+ 1.

Hc sinh t gii.

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