题目

10、设函数f(x)=2x-cosx,｛an｝是公差为π/8的等差数列,f（a1）+f（a2）+…+f（a5）=5π
10、设函数f(x)=2x-cosx,｛an｝是公差为π/8的等差数列,f（a1）+f（a2）+…+f（a5）=5π,则[f(a3)]2-a1a5=______．
A 0 B 1/16*π2 C 1/8*π2 D 13/16*π2
选D
∵数列{an}是公差为π/8的等差数列,
且f(a1)+f(a2)+……+f(a5)=5π
2a1-cosa1+2a2-cosa2+2a3-cosa3+2a4-cosa4+2a5-cosa5=5π
∴2(a1+a2+……+a5)-(cosa1+cosa2+……+cosa5)=5π
∴(cosa1+cosa2+……+cosa5)=0
即2(a1+a2+……+a5)=2×5a3=5π,
a3=π/2,
a1=π/4
a5=3π/4
∴[f(a3)]²-a1a5
=(2a3-cosa3)²-a1a5
=(2*π/2-cosπ/2)²-π/4*3π/4
=π²-3π²/16
=13π²/16
我的疑问是这解题中的“∴(cosa1+cosa2+……+cosa5)=0“这步怎么来的?

提问时间：2021-03-12

答案

f(x)=2x-cosx所以f(a1）=2a1-cosa1f(a2)=2a2-cosa2.整理得因为f(a1)+f(a2)+……+f(a5)=5π2(a1+a2+……+a5)-(cosa1+cosa2+……+cosa5)=5π注意这里因为a5=a1+4*π/8=a1+π/2cosa1+cosa2+cosa3+cosa4+cosa5=[cos(a3-...