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Question

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

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

The correct option is C -f(x)

We have, f(xy)=f(x)f(y)f(ax)f(a+y)
Putting x=a & y=xa, we get
f(a(xa))=f(a)f(xa)f(0)f(x)(i)
f(2ax)=f(a)f(xa)f(x)[f(0)=1]
Now putting x=0,y=0, we get
f(0)=(f(0))2[f(a)]2f(a)=0[f(0)=1](ii)
Using (i) & (ii), we get
f(2ax)=f(x)


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