Sáng kiến kinh nghiệm SKKN kiến thức và kỹ năng cần khắc sâu trong mỗi chương khi dạy bộ môn toán lớp 11

SÁNG KIẾN KINH NGHIỆM

ĐỀ TÀI:

"NH NG KIẾN TH V K N NG N KH S U TRONG

M I H NG KHI TOÁN Ể H SINH V N NG

KHI H "

T V N Ề

,

,

12,

-

ôn

g

thi

N I UNG Ề T I

I ề

N

,

II ề





ên III P







IV N

1 I Đ :

chính xác

+

,

 ,





(1-> 4) :

lim f(x); lim f(x) x x   

a 3 5 3 ) ( 3    x x x f b f(x) = 2 - 3x – x3 c f(x) = 5 - 3x2 – 2x4 d f(x) = 2x4 - 3x – 7 2/ a Cho 2 6 7 2 ) ( 2     x x x x f lim ( ); lim ( ); lim ( ) 2 f x f x f x x x x     b Cho 27 21 2 3 ) ( 3 2     x x x x f lim ( ); lim ( ); lim ( ) 3 f x f x f x x x x      3/ a Cho 2 1 3 2 ) ( 2     x x x x f

  ( ); lim  ( );lim ( );lim ( )

lim

2 2

x f x

f x

f x

f

x x

x

b Cho

x

x x g







2

1 3 ) ( lim ( ); lim ( ); lim ( ); lim ( )

2

x x

x

c Cho

2 3

3 )









x x

x x

h lim ( ); lim ( ); lim ( ); lim ( )

1

x x

x

lim ( ); lim ( ).

2

x

x   

4/ a Cho f(x)  x2  1 x

f x ax b a x x f x x        ( ) & lim ( ) lim )  a f x ax b x x f x x        ( ) & lim ( ) lim )  b Cho g(x)  2x 4x2  1

g x mx n m x x g x x        ( ) & lim ( ) lim )  m g x mx n x x g x x        ( ) & lim ( ) lim ) 

2 3 3 1 0 1 2 3 4 a.lim b.lim 3 5 2 1 1 x x x x x x          3 2 3 2 2 1 2 3 3 5 5 17 4 1 c.lim d.lim 2 5 3 5 6 x x x x x x x x x x            

) 1 3 (

4 1

2 1 lim        n n b      n   n  n n n n n n 4

2 2 1 2 2 3

4 2 lim            c ( 1 , 1 )

1

1 lim         b a b b a a n n d n n 3

3 1 2

2 1 lim       e  2009 2009 1

1 lim       x x x x f      1000 499 2 1

1 1 2 lim x x x x       

2 3 3 

4

- 2mx2 + m4 + 2m ’

– x4 + 2mx2 –

2 Đ : Đ

C

,





 ,

n :

x x x x y a cos sin   b y = x x 2 c y = 1 sin 3 2x d y = cot2(1+x2) e y = 1  2 tan 2x g y = x.cot2x + sin2 x h y = sin3(cos3x) i y = 1 cos 1 sin   x x k y = x x. 2 l y = 2 2 x a x  m.y = x x x tan sin n y = x x x 2 sin cos sin2  2 p.y = x x x x 2 2 6 6 cos sin 1 cos sin   q y = ( , , ) 4 2 a b c R x c x b a          2/ ’ 2

+ 2 = 0

b Cho f(x) = x2  2x 4 ’ - f(-1) = 0

c Cho f(x) = cos4 0

4

12 4























f f

























4 9

8 6

f f

e Cho f(x) = sin4x + cos4 cos 4x

4

1 ’ ’ xR

f Cho y =

4

3





x

x ’2

= (y- ’’

2xx 3 ’’

’

-2

1

x2 ’’ 2

).y ’ ’

2

3 3









x x x

b.y = x4 – 2x2 +3 c.y = 4x3 – 3x4

d y =

1

2 2





x

x x

e y = x3(1 – x)2 f y = 4x – 1+

1

1



x

g y =

1

2 

x

x

x

x 2 2

i y = x2 +

x

2

2xx k y = x2  3x 4 l y =   2

4

2 x

10

3x x p y =

4

1 2



x x x

4/ –   ;  ’

–    ;  

2 ’

c Cho f(x) = sin2x – x +5    

2

; 

 ’

d Cho f(x) = sin4x + cos2 –   ;  ’

–    

2

; 2





’

f Cho f(x) = sinx – 3    

2

; 

 ’

g Cho f(x) =

2

1

cos2x +cosx +3

2    

2 3 ; 2   ’

h Cho f(x) = 2x + 2 1    

2 ; 2   ’

i Cho f(x) = cos2x – si   ;  ’

j Cho f(x) = sin2x – 2sinx - 3 0 ; 2  ’

’

a f(x) = 3 60643  5 x x x c f(x) =   2 9 3 x x  b f(x) =          3 3 cos sin 3 cos 3 3 sin x x x x d f(x) = sin2x+cosx+3x e f(x) = 3 cos 2x sin 2x 2x 1 f f(x) = x x sin 2x 4 4 sin 3   g f(x) = 7 5 5 cos 5 5 sin 3 3 3 cos 2 x x  x h f(x) =3sinx - cos2x - 3cosx 6/ a Cho f(x) = x x x x cos sin 1 cos sin3 3   ’

b Cho f(x) = x 2x cos 2 1  – (x- ’

c Cho f(x) = - x3 – 3x2 ’ -x2 ) > 0 ’

a f(x) = 2 4 1 3 4 3 2 4  x  mx  mx x b x  x  mx2 mxm 3 4 2 1 3 4 c f(x) = mx4 + (m2-9)x2 + 10m -7 d f(x) = (m -1)x4 + (4- m2)x2 + 2m -3 8/ a Cho f(x) = 2 2 3 1x3  x2 mx

* ’ ≥ xR * ’ x(0; +) b Cho f(x) = x3  6x2 3m 2x 2 m

* ’  0 ,xR * ’ < x(-;0)

c Cho y =

m x

m mx x







2

’ xR\{m}

d Cho y =

2

4 2 2











x

m mx

x ’ 0, xR\{-2}

e Cho y =

m x

m mx







2

4

3 ’ xR\{

2

m

 }

f Cho f(x) = sinx – m.sin2x -

3

1 ’ ≥ xR

3

1x3 mx2  m x ’

9/ ’ 

* f(x) =   m x m x m m x      1 3 1 2 * f(x) = 1 2 2 2     x m mx x * f(x) =  1 1 3 2 3      m x mx m x * f(x) = 2 3 2 2     x m m x x b Cho f(x) =  1 2 3 1 3 2 2 3       x m x m x m ’ < xR c Cho f(x) = x m m mx x     2 2 2 ’ < xR\{m}, mR d Cho f(x) = m x m mx    2 3 ’ < xR\{ 2 m  }, mR -x3

a

b - 9

c - 17 = 0 11/ Cho y = 1 1 3   x x

a

b

c

d -4y + 8 = 0 e

12/ Cho y =

1

2 3 2







x

x

x

a -12

b -2

c - 1 = 0 d -5y + 7 = 0 3

- 3x2

a

b

c

-2x4 + 4x2

a

b

1

2 3 2x3  x2 

0

16/ Cho y = 1 1 2 2     x m mx x (Cm m

-2x + 1 17/ Cho (C): y = x ax2 b 3 3 2

3 III H H Q

- – ng

+ Tra

nhau a  SG, SG mp(ABC) b Cho SC = a 3



’ ’

a  mp(SAC); BC  ’

b  ’ ’ ’ ’

c ’ ’ ’ ’ ’ ’



SBC b

c   mp(SBC) d C = 1200

2

6

a

, SA = SB = SC = SD

=a 3

a SB

b

c ’ ’ ’ ’ ’ ’ ’ ’

 SC

SC =

a 2

a (ABCD); ACSK; CKSD

b  

(SCD), SJ(SAB)

AC



= 2a, SA (ABCD), SA = a 2

a

b

c SD

d mp(BMH)

5, SA

a  SM

b

c

 mp(ABC), DA = a 3 C, AB = a 2

BD

DC DH

a  mp(BCD)

b  mp(SAC) ABCD)

0

SA mp(ABCD), SA = 2 3 a 0

 mp(ABCD)

n mp(SBD)

0

(ABC), SA = m, AB = n

 mp(SAB)

0

mp(SOM)  mp(SBC) b.T

d.( ) h/c )

’ ’ ’ ’

a  ’

()

’ ’ ’ ’ ’

a CD)

’ ’ ’ ’ ’

a 3 ’ ’ ’

’

 ’ ’ ’ ’

’ ’ ’

0 ’ ’ ’

’ ’

’

’ ’

V Kế

Qua

2009 - 2011 2 và

các em 12C2

,

KẾT U N c sinh nh