A) 1
B) 2
C) 0
D) \[-\,1\]
Correct Answer: B
Solution :
We have, |
\[f(x)\cdot f(y)=f(x)+f(y)+f(xy)-2\] |
Put \[x=y=1,\]we get |
\[{{(f(1))}^{2}}=3f(1)-2\] \[\Rightarrow \]\[f(1)=1\]or 2 |
Let \[f(1)=1,\]then put \[y=1\] |
\[\therefore \]\[f(x)\cdot f(1)=f(x)+f(1)+f(x)-2\] \[\Rightarrow \]\[f(x)=2f(x)+1-2\] \[\Rightarrow \]\[f(x)=1\] constant function, then \[f(1)=2,f(1)\ne 1\] |
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