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Question

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

A
f(x)
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B
f(x)
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C
f(x)
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D
f(a)+f(ax)
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Solution

The correct option is B f(x)
Given that :
f(xy)=f(x)f(y)f(ax)f(a+y) x,y R
Putting y=0, we get
f(x)=f(x)f(ax)f(a)
f(ax)f(a)=0
f(a)=0 (As f(ax)0)

Now, f(2ax)=f(a(xa))
=f(a)f(ax)f(aa)f(a+xa)
=0f(x)
Therefore, f(2ax)=f(x)

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