函数y=lg(sinxcosx)的单调递减区间是?

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函数y=lg(sinxcosx)的单调递减区间是?
A.（kπ-π/4,kπ+π/4) B.(kπ+π/4,kπ+π/2)
C.(2kπ-π/4,2kπ+π/2) D.(kπ-π/4,kπ) (k∈Z)

复合函数问题.
设 f(x) = lg x,g(x) = sin x cos x
则y = f(g(x))
f(x)本身是单调递增函数,且定义域为x > 0.所以若要使f(g(x))单调递减,g(x)必须递减,且g(x) > 0.
g(x) = sin x cos x = (1/2) sin 2x
g(x)的递减区间为(kπ+π/4,kπ+3π/4); g(x) > 0的区间为(kπ,kπ+π/2)
二者相交,得到(kπ+π/4,kπ+π/2),
即B选项.