已知函数f(x)=cosx4•cos(π2−x4)•cos(π−x2)
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已知函数f(x)=cos
•cos(
−
)•cos(π−
)
x |
4 |
π |
2 |
x |
4 |
x |
2 |
![已知函数f(x)=cosx4•cos(π2−x4)•cos(π−x2)](/uploads/image/z/18698100-60-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%EF%BC%9Dcosx4%E2%80%A2cos%28%CF%802%E2%88%92x4%29%E2%80%A2cos%28%CF%80%E2%88%92x2%29)
(1)f(x)=cos
x
4sin
x
4(-cos
x
2)=-
1
2sin
x
2cos
x
2=-
1
4sinx.
(2)由(1)知,f′(x)=-
1
4cosx,
令f′(x)=0得:cosx=0,
∴x=kπ+
π
2,k∈Z.
又x>0,
∴极值点从小到大排列依次为:
π
2,
3π
2,
5π
2,…
(2n−1)π
2,
故数列{an}的通项公式为:an=
(2n−1)π
2.
(3)由(2)知,bn=
1
(2n−1)π
2•
(2n+1)π
2=
4
π2•
1
(2n−1)(2n+1)=
2
π2(
1
2n−1-
1
2n+1),
∴Tn=
2
π2[(1-
1
3)+(
1
3-
1
5)+…+(
1
2n−1-
1
2n+1)]
=
2
π2(1-
1
2n+1)=
4n
π2(2n+1).
x
4sin
x
4(-cos
x
2)=-
1
2sin
x
2cos
x
2=-
1
4sinx.
(2)由(1)知,f′(x)=-
1
4cosx,
令f′(x)=0得:cosx=0,
∴x=kπ+
π
2,k∈Z.
又x>0,
∴极值点从小到大排列依次为:
π
2,
3π
2,
5π
2,…
(2n−1)π
2,
故数列{an}的通项公式为:an=
(2n−1)π
2.
(3)由(2)知,bn=
1
(2n−1)π
2•
(2n+1)π
2=
4
π2•
1
(2n−1)(2n+1)=
2
π2(
1
2n−1-
1
2n+1),
∴Tn=
2
π2[(1-
1
3)+(
1
3-
1
5)+…+(
1
2n−1-
1
2n+1)]
=
2
π2(1-
1
2n+1)=
4n
π2(2n+1).
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