# Quantitative Aptitude – Algebra – Functions

## Slot – 1 – Quantitative Aptitude – Algebra – Functions – Consider a function f satisfying

Q. Consider a function f satisfying f(x+y) = f(x) f(y) where x, y are positive integers, and f(1) = 2 if f(a+1) + f(a+2)+…+f(a+n)+16 (2^n – 1) then a is equal to?

Solution: Given, f(x+y) = f(x)*f(y), and f(1) =2
f(a+1) + f(a+2) + f(a+3) + …..+f(a+n) = 16 (2^n -1)————–1)
putting n =1
f(a+1) = 16
f(a)*f(1) = 16
or f(a) = 8 as f(1) =2
Putting n =2 in eq 1)
F(a+1) + f(a+2) =16*(2^2 -1)
16 + f(a)*f(2) = 16*3 = 48
8*(f2) = 48 -16 = 32
Or f(2) = 4
By putting n = 3 , in eq 1 we get,
f(a+1) + f(a+2) + f(a+3) = 16*(2^3 -1)
16 + 32 + f(a)*f(3) = 16*7 = 112
F(a)*f(3) = 112 – 16 – 32 = 64
8*f(3) = 64
F(3) = 8
As we know f(a) = 8 so a = 3

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