de thi kscl dau nam mon toan lop 12 truong thpt nguyen hue nam 2016 2017

n

)

Môn - – 2017 : (5

1/ y = x3 – 3x2 – 7 2/ y = x4 – 2x2

: (1,5

y = 3x 2 2x 1



 x 0;3

y = 4x 5

x 2





y = 3 x – 2sinx

:

y = f(x) = 3 x  6 x  (3 x)(6 x)  x  3;6

HD Đặ = 3 x  6 x )

5 Cho a 6, b– 8, c 3 C ằ x 1 ta có:

x4  ax2 + bx + c

– – 2017

(

5 = 3

– 3x2 – 7 D = ’ = 3 2 – 6x ’ = 3x2 – 6x = 0  x 0; y(0) 7 x 2; y(2) 11          - 0 2 +

’ – 0 +

y

H –;0), (2;+  H

H = = –7

H = = – 11

0,25 0,5 0,5 0,75 0,25 0,25 2/ (2,5 = 4 – 2x2 D = ’ = 4 3 – 4x ’ = 4x3 – 4x = 0  x 0 ; y(0) 0 x 1 ; y(1) 1 x 1; y( 1) 1                – –1 0 1 +

’ – 0 + 0 – 0 +

y H –1;0), (1;+  H –;–1), (0;1) H = =

H = = – u y(–1) = y(1) = –1 0,25 0,5 0,5 0,75 0,25 0,25

(1,5

y = 3x 2

2x 1



 x[0;3]

’ = 7 2

(2x 1) ’ >  x [0;3]

+ y(0) = –2 ; y(3) = 1

+

x [0;3]

max y y(3) 1

x [0;3]min y y(0) 2

0,5

0,5 0,5

+ y =

4x 5

x 2



 D = \{2}

+

x 2 x 2

4x 5 lim y lim

x 2





+

x 2 x 2

4x 5 lim y lim

x 2



V x = 2

+

x(4 ) 4

lim y lim lim 4

x(1 ) 1

x(4 ) 4

lim y lim lim 4

x(1 ) 1

,

V y = 4

0,5

0,5

+ y = 3 x – 2sinx

D =

’ = 3 – 2cosx

’ = cosx = 3/2  x k2 ; x k2

’’ =

+ ’’ k2

6

) = 1 > 0

H = k2

6

, y( k2 6

) = ( k2 ) 3 6

– 1

+ ’’ k2

6



  ) = –1 < 0

H = k2

6



6



  )= 1 + ( k2

6



0,25 0,25

0,25

0,25

y = 3 x  6 x  (3 x)(6 x) 

Đặ = 3 x  6 x

+ t2 = 9 + 2 (3 x)(6 x) 

+ 9t2  9 2 (3 x)(6 x)    9 9 18 nên t 3;3 2

+F(t) = 1t2 t 9

   t 3;3 2

0,5

0,25

F’ = –t + 1; F’ =  t = 1

+ F(3) = 3 ; F(3 2) 6 2 9

2



+

3 x 6 3 t 3 2

max f (x) max F(t) F(3) 3

        khi 3 x  6 x = 3x=–3, x = 6

+

3 x 6 3 t 3 2

6 2 9 min f (x) min F(t) F(3 2)

2

    



   khi 3 x  6 x = 3 2 x = 3

2

0,25

Xé f = 4

– ax2 – bx –  1

f’ = 4 3

– 2ax – f’’ = 2 – 2a + Vì a 6 nên  x 1 f’’  d ó f’ 1

+ x 1 ó f’ f’ = 4 – (2a + b)

+ Do a 6 và b –8 f’  0 x 1 d ó f 1

+ x 1 ta có f(x) f(1) = 1 – (a + b + c)

+ Vì a 6, b –8 và c 3 nên f(x)  0, x 1

+ Hay x4 ax2 + bx + c,  x 1

0,25

0,25