Question

Let $f\left(x\right)=|x-1|$. Then,

• A. $f\left(x^2\right)=\left[f\right(x){]}^2$
• B. $f(x+y)=f\left(x\right)f\left(y\right)$
• C. $f\left(|x|\right)=|f\left(x\right)|$
• D. None of these

Answer 1

The correct option is D

None of these

$f\left(x\right)=|x-1|$ Since, $|x^2-1|\ne |x-1{|}^2,f\left(x^2\right)\ne \left(f\right(x){)}^2$. Thus, (i) is wrong. Since, $|x+y-1|\ne |x-1||y-1|,f(x+y)\ne f\left(x\right)f\left(y\right)$. Thus, (ii) is wrong. Since $||x|-1|\ne ||x-1||=|x-1|,f\left(|x|\right)\ne |f\left(x\right)|$. Thus, (iii) is wrong. Hence, none of the given options is the answer.