`f:RrarrR` be twice differentiable function satisfying
`f^(")(x)-5f^(')(x)+6f(x)ge0AAxge0` if `f(0)=1`
`f^(')(0)=0`. If `f(x)` satisfies `f(x),f(x)geae^(bx)-be^(ax), Aaxge0`, then find `(a+b)`

Answer

5

Explanation

Given inequality can be written as:
`f^(``)(x)-2f(x)ge3(f^(')(x)-2f(x)}`
let `f^(')(x)-2f(x)=g(x)`
`impliesg^(')(x)-3g(x)gt0` multiply `e^(-3x)`
`implies(d)/(dx)(g(x)e^(-3x))ge0impliesg(x)e^(-3x)` is non decreasing
Now `g(0)=f^(')(0)-2f(0)=-2`
`f^(')(x)-2f(x)ge-2e^(3x),AAxxge0,e^(-2x)`
`implies(d)/(dx)(f(x)e^(-2x))ge-2e^(x),AAxxge0`
`implies(d)/(dx)(f(x)e^(-2x)+2e^(x))ge0`
`impliesf(x)e^(-2x)+2e^(x)ge3`
`impliesf(x)ge3e^(2x)-2e^(3x),AAxxge0`
comparing `ah(bx)-hb(ax)` with `3e^(2x)-2e^(3x)` we get `h(x)=e^(x)`
`a=3,b=2`