Find the natural number `a` for which `sum_(k=1)^nf(a+k)=16(2^n-1),` where the function `f` satisfies the relation `f(x+y)=f(x)f(y)` for all natural number `x , ya n d ,fu r t h e r ,f(1)=2.`

The natural number a for which sum_(k=1,n) f(a+k)=16(2^n-1) where the function f satisfies the relation f(x+y)=f(x).f(y) for all natural numbers x,y and further f(1)=2 is:- A) 2 B) 3 C) 1 D) none of these

Let sum_(k=1)^(10)f(a+k)=16(2^(10)-1), where the function f satisfies f(x+y)=f(x)f(y) for all natural numbers x, y and f(1) = 2. Then, the natural number 'a' is

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If a function 'f' satisfies the relation f(x)f^('')(x)-f(x)f^(')(x) -f^(')(x)^(2)=0 AA x in R and f(0)=1=f^(')(0) . Then find f(x) .

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If a function f satisfies f (f(x))=x+1 for all real values of x and if f(0) = 1/2 then f(1) is equal to