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Q. A real valued function f(x) satisfies the functional equation f(xy)=f(x)f(y)f(ax)f(a+y), where a is any constant and f(0)=1, then f(2ax) is equal to

Relations and Functions

Solution:

Given f(xy)=f(x)f(y)f(ax)f(a+y)
Let x=0=y
f(0)=(f(0))2(f(a))2
1=1(f(a))2
f(a)=0
f(2ax)=f(a(xa))
=f(a)f(xa)f(a+xa)f(0)
=0f(x)(1)=f(x)(f(a)=0,f(0)=1)