y=sin(2x+3/π）在区间[0,π]的一个单调递减区间是A[0,5π/12]B[π/12,7π/12]C[5π/12,11π/12]D[π/6,π/2]

Question

y=sin(2x+3/π）在区间[0,π]的一个单调递减区间是
A[0,5π/12]
B[π/12,7π/12]
C[5π/12,11π/12]
D[π/6,π/2]

Answer 1

正弦函数y=sinx的递减区间为[2kπ + π/2 ,2kπ + 3π/2] (k∈Z)
所以函数y=sin（2x+π/3）的递减区间为
{x|2kπ + π/2≤2x+π/3≤2kπ + 3π/2,k∈Z}
={x|kπ + π/12≤x≤kπ + 7π/12,k∈Z}
k = 0时,[π/12,7π/12] 刚好落在递减区间内.
所以选B.