A) \[f({{x}^{2}})={{(f(x))}^{2}}\]
B) \[f(x+y)=f(x)+f(y)\]
C) \[f(|x|)=|f(x)|\]
D) None of these
Correct Answer: D
Solution :
\[f(x)=|x-1|\left\{ \begin{matrix} -x+1, & x<1 \\ x-1, & x\ge 1 \\ \end{matrix} \right.\] Consider\[f({{x}^{2}})={{(f(x))}^{2}}\] If it is true it should be\[\forall \,\,x\] \[\therefore \]Put\[x=2\] \[LHS=f({{2}^{2}})=|4-1|=3\] \[RHS={{(f(2))}^{2}}=1\] \[\therefore \][a] is not correct Consider\[f(x+y)=f(x)+f(y)\] Put\[x=2,\,\,y=5\]we get \[f(7)=6;\,\,f(2)+f(5)=1+4=5\] \[\therefore \][b] is not correct Consider\[f(|x|)=|f(x)|\] Put\[x=-5\]then\[f(|-5|)=f(5)=4\] \[|f(-5)|\,\,=\,\,|-5-1|\,\,=6\] \[\therefore \][c] is not correct. Hence [d] is the correct alternative.
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