Q. 64.0( 2 Votes )

Mark the correct alternative in the following:

Let f(x) = |x – 1|. Then,

A. f(x2) = [f(x)]2

B. f(x + y) = f(x) f(y)

C. f(|x|) = |f(x)|

D. None of these

Answer :

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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