MCQ
Let $f(x) = |x - 1|.$ Then:
  • A
    $(x^2) = [f(x)]^2$
  • B
    $f(x + y) = f(x)f(y)$
  • C
    $f(|x|) = |f(x)|$
  • None of these.

Answer

Correct option: D.
None of these.
$\text{f(x)}=|\text{x}-1|$
Since, $|\text{x}^2-1|\neq|\text{x}-1|^2$
$\text{f(x)}^2\neq(\text{f(x)})^2$
Thus, $(i)$ is wrong.
Since, $|\text{x}+\text{y}-1|\neq|\text{x}-1||\text{y}-1|$
$\text{f}(\text{x}+\text{y})\neq\text{f(x)}\text{f(y)}$
Thus, $(ii)$ is wrong.
Since, $|\text{|x|}-1\neq||\text{x}-1||=|\text{x}-1|$
$\text{f(|x|)}\neq|\text{f(x)}|$
Thus, $(iii)$ is wrong.
Hence, none of the given options is the answer.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free