试题:
已知函数f(x)=
sin2x+cos2x+1
2cosx

(1)求f(x)的定义域和值域;
(2)若x∈(-
π
4
π
4
),且f(x)=
3
2
5
,求cos2x
的值.
(3)若曲线f(x)在点P(x0,f(x0))(-
π
2
x0
π
2
)
处的切线平行直线y=
6
2
x
,求x0的值.

答案:

我来补答
解(1)f(x)=
2sinxcosx+2cos2x-1+1
2cosx
=sinx+cosx=
2
sin(x+
π
4
)
(2分)
2cosx≠0,得x≠kπ+
π
2
(k∈Z)

x+
π
4
≠kπ+
4
(k∈Z)
(4分)
f(x)的值域为{y|-
2
≤y≤
2
}
(6分)
(2)∵f(x)=
3
2
5
,∴
2
sin(x+
π
4
)=
3
2
5

sin(x+
π
4
)=
3
5
(7分)
-
π
4
<x<
π
4
,∴0<x+
π
4
π
2

cos(x+
π
4
)=
4
5
(8分)
cos2x=sin(2x+
π
2
)
=2sin(x+
π
4
)cos(x+
π
4
)=
24
25
(10分)
(3)f/(x)=cosx-sinx
由题意得f/(x0)=cosx0-sinx0=
2
cos(x0+
π
4
)
=
6
2
(12分)
cos(x0+
π
4
)=
3
2
又∵-
π
4
x0+
π
4
4

x0+
π
4
=
π
6
,-
π
6
x0=-
π
12
,-
12
(14分)
 
 
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