Question

If $f\left(x\right)=\sqrt{3|x|-x-2},$ then which of the following is (are) CORRECT?

• A. Range of $f\left(x\right)$ is $[0,\infty )$
• B. Range of $f\left(x\right)$ is $[1,\infty )$
• C. Domain of $f\left(x\right)$ is $(-\infty ,-\frac{1}{2}]\cup [1,\infty )$
• D. Domain of $f\left(x\right)$ is $(-\infty ,\frac{1}{2}]\cup [1,\infty )$

Answer 1

Answer: (A), (C)

The correct options are
A Range of $f\left(x\right)$ is $[0,\infty )$
C Domain of $f\left(x\right)$ is $(-\infty ,-\frac{1}{2}]\cup [1,\infty )$

$f\left(x\right)=\sqrt{3|x|-x-2}$ is defined when $3|x|-x-2\ge 0$

Case $1:$ When $x<0$
$-3x-x-2\ge 0\Rightarrow x\le -\frac{1}{2}$
$\Rightarrow x\in (-\infty ,-\frac{1}{2}]$

Now, let $f\left(x\right)=y\Rightarrow y\ge 0$
$\Rightarrow y^2=-4x-2\Rightarrow y^2+2=-4x\because x\le \frac{1}{2}\Rightarrow -x\ge \frac{1}{2}\Rightarrow -4x\ge 2\Rightarrow y^2+2\ge 2\Rightarrow y\in [0,\infty )$

Case $2:$ When $x\ge 0$
$3x-x-2\ge 0\Rightarrow x\ge 1$
$\Rightarrow x\in [1,\infty )$

Now, let $f\left(x\right)=y\Rightarrow y\ge 0$
$\Rightarrow y^2=2x-2\Rightarrow y^2+2=2x\because x\ge 1\Rightarrow 2x\ge 2\Rightarrow y^2+2\ge 2\Rightarrow y\in [0,\infty )$

Hence, domain of $f\left(x\right)$ is $(-\infty ,-\frac{1}{2}]\cup [1,\infty )$ and range of $f\left(x\right)$ is $[0,\infty )$