Functions 2 Question 11
12.
Let $f(x)=|x-1|$. Then,
(1983, 1M)
(a) $f\left(x^{2}\right)={f(x)}^{2}$
(b) $f(x+y)=f(x)+f(y)$
(c) $f(|x|)=|f(x)|$
(d) None of the above
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Answer:
Correct Answer: 12. (d)
Solution:
- Given, $f(x)=|x-1|$
$\therefore \quad f\left(x^{2}\right)=\left|x^{2}-1\right|$
and $\quad{f(x)}^{2}=(x-1)^{2}$
$\Rightarrow \quad f\left(x^{2}\right) \neq(f(x))^{2}$, hence (a) is false.
Also, $f(x+y)=|x+y-1|$
and $\quad f(x)=|x-1|$
$f(y)=|y-1|$
$\Rightarrow \quad f(x+y) \neq f(x)+f(y)$, hence (b) is false.
$ f(|x|)=|| x|-1| $
and $\quad|f(x)|=|| x-1||=|x-1|$
$\therefore \quad f(|x|) \neq|f(x)|$, hence (c) is false.
