262continuing from above post ,
f(z) = kz
putting z=1 ,
f(1)=k
but k = 1612-{f(1)}2 = f(1)
thus on solving , f(1) = 2,-4
thus f(x) =2x and f(x)=-4x are the solutions.
\hspace{-16}$If $\bf{x,y,z\in \mathbb{R}}$ and $\bf{f(x).f(y).f(z) = 12f(xyz) ô€€€ -16xyz}$\\\\ Then No. of function which satisy $\bf{f:\mathbb{R}\rightarrow \mathbb{R}}$ is
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4 Answers
21set x=y = 1
we have f(1)f(1)f(z) = 12 f(z) - 16 z
f(1)f(1) ≠12 because otherwise z = 0 for all z, contradiction
So now f(z) = 16z12 - f(1)f(1) = kz
So f is linear
262