Question

求f(x)=(sinxcosx)/(1+sinx+cosx)的递增区间?

Answer 1

令sinx+cosx=-1

x=kπ/2

先考虑一个周期内的情形

x的定义域为[π/4,9π/4)-{kπ/2}

令t=sinx+cosx=根号2 sin(x+π/4)

因此

当x属于[π/4,5π/4)-{kπ/2}时,t单调减

当x属于[5π/4,5π/4)-{kπ/2}时,t单调增

因为t^2=1+2sinxcosx,所以

f(x)=(t^2-1)/(2+2t)=(t-1)/2

关于f(x)关于t单调增

因此在[π/4,9π/4)-{kπ/2}上,f(x)的单调增区间为

(5π/4,3π/2),(3π/2,2π),(2π,9π/4)

因此f(x)在定义域上的单调增区间为

(5π/4+2kπ,3π/2+2kπ),(3π/2+2kπ,2π+2kπ),(2π+2kπ,9π/4+2kπ),k为任意整数