问题 解答题
已经函数f(x)=2sinxcosx+sin2x-cos2x.
(1)求f(x)递增区间;
(2)求f(x)当x∈[0,
π
2
]时的值域.
答案

∵f(x)=2sinxcosx+sin2x-cos2x

=sin2x-cos2x

=

2
(cos(-
π
4
)sin2x+sin(-
π
4
)cos2x)

=

2
sin(2x-
π
4

(1)f(x)递增区间为2x-

π
4
∈[-
π
2
+2kπ,
π
2
+2kπ]
  k∈Z

即递增区间为x∈[-

π
8
+kπ,
8
+kπ]k∈Z)

(2)当x∈[0,

π
2
]

即2x-

π
4
∈[-
π
4
4
]

∴f(x)min=

2
sin(-
π
4
)=-1

f(x)max=

2
sin(
π
2
)=
2

即f(x)当x∈[0,

π
2
]时的值域为[-1,
2
]

选择题
选择题