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Question

2f(x)=f(xy)+f(xy), x,yR+
If f(1)=0 and f(2)=256, then f(256) is equal to

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Solution

2f(x)=f(xy)+f(xy)
f(xy)=2f(x)f(xy).......(i)
Put x=1
f(y)=2f(1)f(1y)f(y)=f(1y)........(ii)(f(1)=0)
Put y=x
f(x2)=2f(x)f(1)f(x2)=2f(x)......(iii)
Put y=x2
f(x3)=2f(x2)f(1x)f(x3)=2f(x)+f(x) ( from (ii))f(x3)=3f(x)
f(xn)=nf(x)
Differentiating both the sides w.r.t. x,
nxn1f(xn)=nf(x)
f(xn)=f(x)xn1

Now, 256=28=xnx=2, n=8

f(256)=f(2)27 =256128 =2

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