Question:
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
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.