Q6 of 118 Page 3

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.

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