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Q.

f is a function defined as ∑k=1nf(a+k)=162n−1 and f(x+y)=f(x).f(y) and f(1) = 2 then integral value of a

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a

3

b

0

c

2

d

1

answer is A.

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Detailed Solution

fx+y=fxf(y) x=y=1⇒f2=f1f1=4 x=1,y=2⇒f3=f1f2=8⇒fx=2x ∑k=1   n2a+k=16(2n-1) 2a(2+22+23+...........+2n)=16(2n-1) use G.P formula
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