`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)`
`=g^(')(x)-3g(x)ge0` multiply `e^(-3x)`
`impliesd/(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),AAxge0,e^(-2x)`
`implies d/(dx)(f(x)e^(-2x)+2e^(x))ge0`
`impliesf(x)e^(-2x)+2e^(x)ge3`
`impliesf(x)ge3e^(2x)-2e^(3x),AAxge0`
comparing `ah(bx)-bh(ax)` with `3e^(2x)-2e^(3x)` we get `h(x)=e^(x), a=3,b=2`
`implies (a+b)h(0)=5`