题规范与策略
规范解答 (Ⅰ)因为f(x)=excos x-x, 所以f '(x)=ex(cos x-sin x)-1,① 又f(0)=1, f '(0)=0,② 所以曲线y=f(x)在点(0, f(0))处的切线方程为y=1.③ (Ⅱ)由(Ⅰ)可设h(x)=ex(cos x-sin x)-1,则h'(x)=ex(cos x-sin x-sin x-cos x)= -2exsin x.④ 当x∈(0, π 2 )时,h'(x)<0,⑤