Question

已知函数y=f(x)所确定的曲线与x轴相切于原点且满足f(x)=2+sinx-f''(x),求f(

Answer 1

∵对任意x∈（-π/2，π/2）满足f′（x）cosx+f（x）sinx>0
∴[f(x)/cosx]'=[f′（x）cosx+f（x）sinx]/cos2x>0
∴f(x)/cosx是(-π/2,π/2)上的增函数
∴f(0)/cos0<f(π/4)/cosπ/4<f(π/3)/cosπ/3)
即f(0)< f(π/4)/(√2/2)<f(π/3)/(1/2)
∴f(0)<√2 f(π/4)<2f(π/3)
B,C,D均错

又f(-π/3)/cos(-π/3)<f(π/4)/cosπ/4
∴2f(-π/3)<√2f(π/4)
即.√2f（-π/3）<f（π/4）

选A