Question
If $f(x) \cdot f(y)=f(x)+f(y)+f(x y)-2 \forall x, y \in R$ and if $f(x)$is not a constant function, then the value of $f(1)$ is(A) 1(B) 2(C) 0(D) $-1$
Step 1
We get \[f(1) \cdot f(1)=f(1)+f(1)+f(1)-2\] which simplifies to \[f(1)^2=3f(1)-2\] or \[f(1)^2-3f(1)+2=0\] This is a quadratic equation in $f(1)$. Show more…
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