函数y=log以三分之一为底(sin2x-cos2x)的单减区间.

问题描述:

函数y=log以三分之一为底(sin2x-cos2x)的单减区间.


y=log以三分之一为底(sin2x-cos2x)
y=[lg(sin2x-cos2x)]/[lg(1/3)]
y=-(lg3)lg(sin2x-cos2x)
y'=-2[(lg3)/(ln10)][(cos2x+sin2x)/(sin2x-cos2x)]
令:y'<0
有:-2[(lg3)/(ln10)][(cos2x+sin2x)/(sin2x-cos2x)]<0
即:(cos2x+sin2x)/(sin2x-cos2x)>0
(tan2x+1)/(tan2x-1)>0
有:tan2x>-1、tan2x>1………………(1)
或:tan2x<-1、tan2x<1………………(2)
由(1)得:tan2x>1,即:kπ/2+π/4>x>kπ/2+π/8
由(2)得:tan2x<1,即:kπ/2+π/8>x>kπ/2-π/4
即:函数的单调减区间是:x∈(kπ/2-π/4,kπ/2+π/8)∪(kπ/2+π/8,kπ/2+π/4)

求导得:y'=(2/ln3)*(1+sin4x)/cos4x
令y' 因此,原函数的单减区间为(π/8+kπ/2,3π/8+kπ/2),k∈Z。

复合函数
㏒1/3底t 它是减函数
∴t=sin2x-cos2x单调递增
t=√2sin﹙2x-π/4﹚
-π/2+2kπ ≦2x-π/4≦ π/2+2kπ
-π/8+kπ≦ x≦3π/8+kπ k∈z

y=log1/3 (sin2x-cos2x)
sin2x-cos2x=√2sin(2x- π/4)
kπ-π/8