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

) n Môn – 2017 : (5 1/ y = x3 – 3x2 – 2/ y = x4 – 2x2 : (1,5 y= 3x  2x  y= 4x  x2 y= x  0;3 x – 2sinx : y = f(x) = HD Đặ  x   x  (3  x)(6  x) = x   3;6 3 x  6 x ) Cho a  6, b  – 8, c  C x4  ax2 + bx + c ằ x  ta có: – = – 3x2 – D= ( ’=3 0,25 – 6x 0,5  x  0; y(0)  7  3x2 – 6x =    x  2; y(2)  11 ’= - 0,5 + ’ – + 0,75 y –  ;0), (2;+  H H H = H = = 2/ (2,5 = –7 0,25 = – 11 0,25 – 2x2 D= 0,25 ’=4 ’=  x  ; y(0)   4x – 4x =   x  ; y(1)  1  x  1; y(1)  1 – 4x 0,5 – –1 ’ – 0,5 + + – + 0,75 y –1;0), (1;+  H H H = H = y= (1,5 – 2017 3x  2x  ’= –  ;–1), (0;1) = =– 0,25 u y(–1) = y(1) = –1 0,25 x [0;3] (2x  1) ’> x [0;3] 0,5 + y(0) = –2 ; y(3) = + max y  y(3)  y  y(0)  2 x[0;3] x[0;3] 0,5 0,5 4x  x2 +y= D = \{2} + lim y  lim x 2  x 2  4x    x2 4x    , x 2 x  + lim y  lim x 2  V x=2 5 x(4  ) 4 x  lim x 4 + lim y  lim 2 x  x  x  x(1  ) 1 x x 5 x(4  ) 4 x  lim x  4, lim y  lim 2 x  x  x  x(1  ) 1 x x V 0,5 0,5 y=4 x – 2sinx +y= D= ’= – 2cosx  cosx = ’=   /2  x   k2 ; x    k2 6 0,25 0,25 ’’ = + ’’ 0,25   k2 ) = > H = + ’’    k2 ) = –1 <    =   k2 , y(   k2 )= + (   k2 ) 6 H y= Đặ     k2 , y(  k2 ) = (  k2) – 6 0,25  x   x  (3  x)(6  x) = 3 x  6 x + t2 = + (3  x)(6  x) 0,5 +  t   (3  x)(6  x)    18 nên t  3;3  +F(t) =  t  t  2 t  3;3  0,25 F’ = –t + 1; F’ + F(3) = ; F (3 2)   t=1 = 9 + max f (x)  max F(t)  F(3)   x   x =  x=–3, x = 3 x 6 3 t 3 0,25 + f (x)  F(t)  F(3 2)  3 x 6 3 t 3 Xé f’ f =4 = – 2ax – f’’ ó f’  f’ + Do a  b  –8 =  d 3 x  6 x =3  x =  – ax2 – bx – + Vì a  nên x  f’’ + x  9 2 – 2a  ó f’ 0,25 = – (2a + b) f’  x  d óf  + x  ta có f(x)  f(1) = – (a + b + c) + Vì a  6, b  –8 c  nên f(x)  0, x  + Hay x4  ax2 + bx + c, x  0,25 ... x  1; y(1)  1 – 4x 0,5 – –1 ’ – 0,5 + + – + 0,75 y –1;0), (1;+  H H H = H = y= (1,5 – 2017 3x  2x  ’= –  ;–1), (0;1) = =– 0,25 u y(–1) = y(1) = –1 0,25 x [0;3] (2x  1) ’> x [0;3]