(1)由f(x)<0,得a<(x-cosx)•ex, 记g(x)=(x-cosx)•ex, 则g′(x)=(1+sinx)•ex+(x-cosx)•ex =(1+sinx-cosx+x)•ex, ∵0<x<1, ∴sinx>0,1-cosx>0,ex>0,∴g′(x)>0, ∴g(x)在(0,1)上为增函数. ∴-1<g(x)<(1-cos1)•e,故a≤-1.
(2)构造函数h(x)=e-x+sinx-1-(0<x<1),且h(0)=0, 则h′(x)=-e-x+cosx-x, 由(1)知:当a=-1时,f(x)=-e-x+cosx-x<0(0<x<1), ∴h(x)在(0,1)单调递减,∴h(x)<h(0)=0, 即e-x+sinx<1+(0<x<1). |