已知函数f(x)=x^2+ax+1,若存在t属于(π/4,π/2),f(sint)=f(cost),则实数a的取值范

f(x)=x^2+ax+1,
对称轴为x=-a/2
∵ t属于(π/4,π/2)
∴ sint≠cost
∴ 要满足f(sint)=f(cost),
则sint,cost关于-a/2对称
∴ sint+cost=-a
∴ -a=sint+cost
=√2*[sint*(√2/2)+cost*(√2/2)]
=√2*[sint*cos(π/4)+cost*sin(π/4)]
=√2sin(t+π/4)
∵ t属于(π/4,π/2)
∴ t+π/4∈(π/2,3π/4)
∴ sin(t+π/4)∈(√2/2,1)
∴ -a=√2sin(t+π/4)∈(1,√2）
∴ a∈(-√2,-1)