Question

高中数学题。。请写出详细过程。谢谢

设函数f（x）＝2x-cosx,｛an｝是公差为8分之派的等差数列，f（a1）+f（a2）+……+f（a5）＝5派，则【f（a3）】的平方-a1×a3等于多少？

Answer 1

f(a1)+f(a2)+f(a3)+f(a4)+f(a5)=2(a1+a2+a3+a4+a5)-(cosa1+cosa2+cosa3+cosa4+cosa5)
=10a3-(cosa1+cosa2+cosa3+cosa4+cosa5)
=10a3-[cos(a3-π/4)+cos(a3-π/8)+cosa3+cos(a3+π/8)+cos(a3+π/4)]
=5π
∴10a3-5π=[cos(a3-π/4)+cos(a3-π/8)+cosa3+cos(a3+π/8)+cos(a3+π/4)]
=[cos(a3-π/4)+cos(a3+π/4)]+cosa3+[cos(a3-π/8)+cos(a3+π/8)]
=2cosa3cos(π/4)+cosa3+2cosa3cos(π/8)
=[1+2cos(π/4)+2cos(π/8)]cosa3
=[1+√2+√(2+√2)]cosa3
设g(x)=-[1+√2+√(2+√2)]cosx+10x-5π
则g'(x)=[1+√2+√(2+√2)]sinx+10＞0
∴g(x)严格单调递增，
∴g(x)=0最多有1个解。
显然x=π/2是g(x)=0的解
∴x=π/2是g(x)=0的唯一解
即a3=π/2
∴a1=π/4
∴f[(a3)]^2-a1a3=(2a3-cosa3)^2-a1a3
=﹙π-cosπ/2﹚²-π/4·π/2
=7π²/8