Let the function defined as f : R → R be a twice differentiable function satisfying the relation f ′ ′ x - 5 f ′ x + 6 f ( x ) ≥ 0 ,   ∀   x ≥ 0 ,   f 0 = 1 and f ′ 0 = 0 . If f x satisfies the relation: f x ≥ a . h b x - b .   h a x , ∀ x ≥ 0 , then a + b equals:

Step-by-step solution

∵       f ′ ′ x - 5 f ′ x + 6 f x ≥ 0 Multiply both sides by e – 3 x , then arrange: d d x e - 3 x f ′ x - 2 e - 3 x f x ≥ 0 ∴         e - 3 x f ′ x - 2 e - 3 x f x ≥ - 2 ∴       f x . e - 2 x ≥ ∫ - 2 e x d x + c ⇒ f x ≥ - 2 e 3 x + c e 2 x , where c ≤   3 . ∴   f x ≥ 3 e 2 x - 2 e 3 x . ∴   a = 3 ,   b = 2 ∴   a + b