Toán Lớp 11: Tìm min max y= sinx^6 + cosx^6

Question

Toán Lớp 11:  Tìm min max  y= sinx^6 + cosx^6

tuyen98 · Answer

$y= sin^6x+ cos^6x$   $= sin^4x+ cos^4x- sin^2x cos^2x$   $=( sin^2x+ cos^2x)^2-2 sin^2x cos^2x- sin^2x cos^2x$   $=1-3 sin^2x cos^2x$   $=1- dfrac{3}{4} sin^22x$   Ta c&oacute;: $0 le  sin^22x le 1$   $ to -1 le - sin^22x le 0$   $ to  dfrac{-3}{4} le  dfrac{-3}{4} sin^22x le 0$   $ to  dfrac{1}{4} le y le 1$   Vậy: $ min y= dfrac{1}{4};  max y=1$

buithaianh98 · Answer

Giải đáp:     Gi&aacute; trị lớn nhất: $1$   Gi&aacute; trị nhỏ nhất: $ dfrac{1}{4}$    Lời giải và giải thích chi tiết:    $A=sin^6x+cos^6x$   $=(sin^2x+cos^2x)(sin^4x-sin^2x.cos^2x+cos^4x)$   $=sin^4x-sin^2x.cos^2x+cos^4x$ (v&igrave; $sin^2x+cos^2x=1$)   $=(sin^2x+cos^2x)^2-3sin^2x.cos^2x$   $=1-3sin^2x.cos^2x$ (v&igrave; $sin^2x+cos^2x=1$)   $=1- dfrac{3}{4}.4.sin^2x.cos^2x$    $=1- dfrac{3}{4}.(2sinxcosx)^2$    $=1- dfrac{3}{4}.sin^22x$    Ta c&oacute;: $-1&le;sin2x&le;1$   $&rArr;0&le;sin^22x&le;1$   $&hArr;1&ge;1- dfrac{3}{4}.sin^22x&ge;1- dfrac{3}{4}$   $&hArr;1&ge;1- dfrac{3}{4}.sin^22x&ge; dfrac{1}{4}$ hay $1&ge;A&ge; dfrac{1}{4}$   +) $MaxA=1&hArr;1- dfrac{3}{4}.sin^22x=1$   $&hArr;sin^22x=0&hArr;sin2x=0$   $&hArr;x= dfrac{k pi}{2}$ $(k&isin; mathbb{Z})$   +)$MinA= dfrac{1}{4}&hArr;1- dfrac{3}{4}.sin^22x= dfrac{1}{4}$   $&hArr;sin^22x=1&hArr;1-cos^2x=1$   $&hArr;cos^22x=0$   $&hArr;cos2x=0$   $&hArr;x= dfrac{ pi}{4}+ dfrac{k pi}{2}$  $(k&isin; mathbb{Z})$