Let f(x) =
Question:

Let f(x) = |x − 1|. Then,

(a) f(x2) = [f(x)]2

(b) f(x + y) = f(xf(y)

(c) f(|x| = |f(x)|

(d) None of these

Solution:

(d) None of these

$f(x)=|x-1|$

Since, $\left|x^{2}-1\right| \neq|x-1|^{2}$

$f\left(x^{2}\right) \neq(f(x))^{2}$

Thus, (i) is wrong.

Since, $|x+y-1| \neq|x-1||y-1|$

$f(x+y) \neq f(x) f(y)$

Thus, (ii) is wrong.

Since ||$x|-1| \neq|| x-1||=|x-1|$

$f(|x|) \neq|f(x)|$

Thus, (iii) is wrong.

Hence, none of the given options is the answer.

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