Mark the correct alternative in the following:
Let f(x) = |x – 1|. Then,
f(x)=|x-1|
f(x2 )=| x2-1|
[f(x)2=(x-1)2
= x2+1-2x
So, f(x2)≠[f(x)]2
f(x + y)=|x+y-1|
f(x)f(y)=(x-1)(y-1)
So, f(x + y) ≠ f(x)f(y)
f(|x|)=||x|-1|
Therefore, option D is correct.