Let `f: R->R` be a twice differentiable function such that `f(x+pi)=f(x)` and `f''(x)+f(x)geq0` for all `x in Rdot` Show that `f(x)geq0` for all `x in Rdot`

Options

A. `f'(1)lt0`
B. `f(2)lt0`
C. `f'(x)!=0` for an `x epsilon(1,3)`
D. `f'(x)=0` for some `x epsilon (1,3)`

Answer

A, B, C

Explanation

`f(x)=xF(x)`………….1
`:.f'(x)=xF'(x)+F(x)`…………….2
`impliesf'(1)=F'(1)+F(1)=F'(1)lt0`
`F(1)=0` and `F(3)=-4`
Also `F'(x)gt0` for all `xepsilon(1//2,3)`
So `F(x)` is decreasing and hence `F(2)lt0`.
`:.f(2)-2F(2)lt0`
Also for `x epsilon(1,3)`,
`f'(x)=xF'(x)+F(x)lt0`