Toán Lớp 11: f(x)= 1/tan^4x +1/(2-tan^4x.) x ∈(0 ; π/4)

Question

Toán Lớp 11:  f(x)= 1/tan^4x +1/(2-tan^4x.)   x ∈(0 ; π/4)

ngoclanhoa · Answer

Giải đáp:   Ta c&oacute;      t=2tan1&minus;tan2(x)&rArr;2t=1tan(x)&minus;tan(x)&rArr;4t2=1tan2(x)+tan2(x)&minus;2 t=2tan1-tan2x&rArr;2t=1tanx-tanx&rArr;4t2=1tan2x+tan2x-2    Từ đ&oacute;      (4t2+2)2=(1tan2(x)+tan2(x))2&rArr;1tan4(x)+tan4(x)=16t4+16t2+2 4t2+22=1tan2x+tan2x2&rArr;1tan4x+tan4x=16t4+16t2+2    L&uacute;c đ&oacute;     f(t)=16t4+16t2+2 ft=16t4+16t2+2  với t = tan(2x)   Khi     x&isin;(0;&pi;4) x&isin;0;&pi;4  th&igrave; t = tan(2x) v&agrave; li&ecirc;n tục tr&ecirc;n miền đ&oacute; n&ecirc;n ta c&oacute;:   f(t)=16t4+16t2+2 ft=16t4+16t2+2    Bắt đầu từ đ&acirc;y ta c&oacute;:      f(sin(x))+(cos(x))=16sin4(x)+16sin2(x)+2+16cos4(x)+16cos2(x)+2=16(1sin4(x)+1cos4(x))+16(1sin2(x)+1cos2(x))+4 fsinx+cosx=16sin4x+16sin2x+2+16cos4x+16cos2x+2=161sin4x+1cos4x+161sin2x+1cos2x+4    &Aacute;p dụng bất đẳng thức Cauchy, ta c&oacute;:     1sin4(x)+1cos4(x)&ge;2sin2(x)cos2(x)=8sin2(2x)&ge;8&forall;x&isin;(0;&pi;2)1sin2(x)+1cos2(x)&ge;2sin(x)cos(x)=4sin(2x)&ge;4&forall;x&isin;(0;&pi;2) 1sin4x+1cos4x&ge;2sin2xcos2x=8sin22x&ge;8&forall;x&isin;0;&pi;21sin2x+1cos2x&ge;2sinxcosx=4sin2x&ge;4&forall;x&isin;0;&pi;2    Cuối c&ugrave;ng ta thu được f(sinx) + f(cosx)  &ge;196 &ge;196    Dấu bằng xảy ra khi v&agrave; chỉ khi     x=&pi;4 x=&pi;4