Sáng kiến kinh nghiệm NHỮNG KIẾN THỨC VÀ KĨ NĂNG cần KHẮC sâu TRONG mỗi CHƯƠNG KHI DẠY TOÁN 11, để HỌC SINH vận DỤNG được KHI HỌC 12

SÁNG KIẾN KINH NGHIỆM ĐỀ TÀI: "NH NG KIẾN TH M I H NG KHI V K N NG TOÁN KHI H N KH ỂH " S U TRONG SINH V N NG T V N Ề , , 12, - ôn g thi N I UNG Ề T I ề I .N , II  ề  ên III P    IV N I Đ : xác + ,  ,   (1-> 4) : lim f ( x); lim f ( x) x x  a f ( x)  x3  3x  c f(x) = - 3x2 – 2x4 2/ 3/ b f(x) = - 3x – x3 d f(x) = 2x4 - 3x – a Cho f ( x)  2x2  7x  x2 b Cho f ( x)  x  x  21 x  27 a Cho f ( x)  x  3x  x2 b Cho g ( x)  3x  2 x c Cho h( x )  x3 x  3x  2 lim f ( x); lim f ( x); lim f ( x) x 2 x x lim f ( x); lim f ( x); lim f ( x) x3 x x lim f ( x); lim  f ( x); lim f ( x); lim f ( x) x   x   x x lim g ( x); lim g ( x); lim g ( x); lim g ( x) x 2  x x 2 x lim h( x); lim h( x); lim h( x); lim h( x) x1 x x1 x lim h( x); lim h( x) x 2  4/ x 2 a Cho f ( x)  x   x  ) lim x f ( x)  a & lim  f ( x)  ax  b x x  ) lim f ( x)  a & lim  f ( x)  ax  b x   x  ) lim g ( x)  m & lim g ( x)  mx  n x   x x   b Cho g ( x)  x  x   ) lim x g ( x)  m & lim g ( x)  mx  n x x x   a.lim x 1 c.lim x 1 x2 1 3x   2  3x  x 0  x  b.lim x   3x  x2  5x  3 d.lim x2 x  17  x  x2  5x  lim    n    (3n  1) b lim n  n    n    3n 2n  2n  1  2n     4n c lim  a   a n ( a  1, b  1)  b   b n d lim    n    3n e  x   x 2009 lim x x  12009 f 2 x 3 - 2mx2 + m4 + 2m – x4 + 2mx2 – Đ  499 1 lim x 1  x    1  x 1000 :Đ ’ .C ,    , n a y  : sin x x  x cos x b y = c y = x x sin x  d y = cot2(1+x2) e y = h y = sin3(cos3x) i y = sin x  cos x  m.y = x sin x tan x q y = b c    a    (a, b, c  R) x x   l y = p.y = x a2  x2 sin x  cos6 x  sin x cos2 x k y = x x sin x  cos2 x sin x n y = ’ 2/ b Cho f(x) = g y = x.cot2x + sin2 x 1 tan x +2=0 ’ x2  2x  c Cho f(x) = cos4 - f(-1) =     f '    12 f    4 4     f '   f   6 4 e Cho f(x) = sin4x + cos4 f Cho y = x3 x4 cos x ’2 = (y- ’ ’’ ’ x  R g Cho y = 2x  x2 ’’ ’ - x2 ’’ 2 ’ ’ a y =  x 3x   2x  b.y = x4 – 2x2 +3 c.y = 4x3 – 3x4 d y = x2  2x x 1 e y = x3(1 – x)2 f y = 4x – 1+ g y = x x 1  x  2 h y = i y = x2 + j y = 2x  x2 m y = 4/ ).y x x 2 x  2 x x  3x  n y = 3x  10  x p y =    ;   ’ – c Cho f(x) = sin2x – x +5 d Cho f(x) = sin4x + cos2 – – l y = x  2  x k y = – x 1 2x  x x4      ;          ;  2     ;        ;   2 ’ ’ ’ ’ f Cho f(x) = sinx –      ;  2  g Cho f(x) = cos2x +cosx + h Cho f(x) = 2x + sinx - ’      ;   2 ’    ;   ’ i Cho f(x) = cos2x – si j Cho f(x) = sin2x –      ;3   2 2 ’  0;2  ’ ’ 3x  b f(x) = sin 3x cos 3x    cos x  3 sin x   d 3   e f(x) = g f(x) 6/ c f(x) = x  3  x 60 64  5 x x3 a f(x) = cos x  sin x  x  = cos 3x  3 sin x cos x  7 5 a Cho f(x) = sin x  cos3 x  sin x cos x b Cho f(x) = x 1 cos2 x f(x) = sin2x+cosx+3x f f(x) = 3x  sin x  sin x h f(x) =3sinx - cos2x - 3cosx ’ – (x- ’ c Cho f(x) = - x3 – 3x2 ’ -x2) > ’ a f(x) = x4  x  2mx  4mx  b x x3   mx  mx  m 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) = * ’ x  x  mx  ≥ b Cho f(x) = * ’ xR  x  x  3m  2x   m  * ’ ,xR c Cho y = x  2mx  m  xm d Cho y =  x  mx  2m  x2 e Cho y = mx  3m  2x  m f Cho f(x) = sinx – m.sin2x - g.Cho f(x) = x(0; +) * ’ ’ x(-;0) < xR\{m} ’  0, xR\{-2} ’ xR\{  m } 3 x  mx  m  2x  ’ ’ ≥ xR  ’ 9/ * f(x) = x  mm  1x  m  xm * f(x) = b Cho f(x) =  c Cho f(x) = * f(x) = x  mx  m  x 1 x3  m  1x  mx  m   * f(x) =  x3  m  1x  2m  x  m  x  2mx  m  mx d Cho f(x) = mx  m  2x  m ’ ’ 11/ Cho y = xR\{m}, mR < xR\{  m }, mR - a b c - 17 = 3x  x 1 xR < a c < ’ -x3 b x  3x  m  m x2 d -4y + = e 12/ Cho y = x  3x  x 1 a -12 b -2 c - = d -5y + = - 3x2 a b c -2x4 + 4x2 a b 2x3 x  1 16/ Cho y = x  mx  2m  (Cm x 1 m -2x + 17/ Cho (C): y = 2x3  ax  b III H H Q - – ng + Tra  SG, SG mp(ABC) a b Cho SC = a  ’ ’ a  mp(SAC); BC  b  c ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’  SBC b  c  mp(SBC) C = 1200 d a =a , SA = SB = SC = SD SB a b ’ ’ ’ c ’ ’ ’ ’ ’  SC SC = a (ABCD); ACSK; CKSD a b   (SCD), SJ(SAB) AC  = 2a, SA (ABCD), SA = a a b c SD mp(BMH) d 5, a SA  SM b c  mp(ABC), DA = =a C, AB a BD a  mp(BCD) DH DC  mp(SAC) b ABCD) a  mp(ABCD) n mp(SBD) SA mp(ABCD), SA = (ABC), SA = m, AB = n  mp(SAB) mp(SOM)  mp(SBC) b.T d.( ) h/c ’ ’ ’ ’  a ’ () ’ ’ a ’ ’ ’ CD) ) ’ ’ ’ ’ ’ a ’ ’ ’ ’  ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ V Kế Qua 2009 - 2011 em 12C2 , KẾT U N c sinh nh ... mR - a b c - 17 = 3x  x 1 xR < a c < ’ -x3 b x  3x  m  m x2 d -4y + = e 12/ Cho y = x  3x  x 1 a -12 b -2 c - = d -5y + = - 3x2 a b c -2x4 + 4x2 a b 2x3 x  1 16/ Cho y = x  mx... ’ CD) ) ’ ’ ’ ’ ’ a ’ ’ ’ ’  ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ V Kế Qua 2009 - 2011 em 12C2 , KẾT U N c sinh nh ...T V N Ề , , 12, - ôn g thi N I UNG Ề T I ề I .N , II  ề  ên III P    IV N I Đ : xác + ,  ,   (1->