# Mark the correct alternative in the following:Let f(x) = |x – 1|. Then,A. f(x2) = [f(x)]2B. f(x + y) = f(x) f(y)C. f(|x|) = |f(x)|D. None of these

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.

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