Question

已知函数f(x)=sinx+ sin^2(x/2)-con^2(x/2)+ a的最小值为根号2 ；(1)求a (2)求函

Answer 1

解: (1).f(x)=sinx+(1-cosx)/2-(1+cosx)/2+a.

=sinx+cosx+a.

∴fx)=√2sin(x-π/4)+a.

当sin(x-π/4)=-1时,f(x)min=√2.

√2*(-1)+a=√2.

∴a=2√2.

(2). ∵ sinx的x∈(2kπ-π/2,2kπ,+ρ/2)为增函数, x∈(2kπ+π/2,2kπ+3π/2)为减函数.

f(x)=sin(x-π/4)+2√2的单调递增区间为:x∈(2kπ-π/4,2kπ+π/4);

f(x)的单调递减区间为:x∈(2kπ+π/4,2kπ+5π/4), k∈Z.