Let f : 0,1 → R be such that f x y = f x . f y , for all x , y ∈ 0,1 , and f ( 0 ) ≠ 0 . If y = y ( x ) satisfies the differential equation, d y d x = f x with y 0 = 1 then y 1 4 + y 3 4 is equal to:

Step-by-step solution

Given relation is f x y = f x .   f y   ∀   x ,   y ∈ R   .....(i) On putting x =   y = 0 , we get f 0 = f 2 0 ⇒ f 0 = 0 ,   1 but f 0 ≠ 0 ⇒ f 0 = 1 , Now if we put y = 0 in (i), we get f x = 1 Hence d y d x = 1 ⇒ y = x + c ⇒ y = x + 1 s i n c e   y 0 = 1 ⇒ y 1 4 + y 3 4 = 1 4 + 1 + 3 4 + 1 = 3 .