MCQ
If $f : [-2, 2] \rightarrow R$ is defined by $\text{f(x)}=\begin{cases}-1,&\text{for}-2\leq\text{x}\leq0\\\text{x}-1,&\text{for }0\leq\text{x}\leq2\end{cases}$ then $\{\text{x}\in[-2,2]:\text{x}\leq0\text{ and }\text{f}(\text{|x|})=\text{x}\}=$
  • A
    $\{-1\}$ 
  • B
    $\{0\}$
  • $\Big\{-\frac{1}{2}\Big\}$
  • D
    $\phi$

Answer

Correct option: C.
$\Big\{-\frac{1}{2}\Big\}$
Given,
$\text{f(x)}=\begin{cases}-1,&\text{for}-2\leq\text{x}\leq0\\\text{x}-1,&\text{for }0\leq\text{x}\leq2\end{cases}$
We know,
$|\text{x}|\geq0$
$\Rightarrow\text{f(|x|)}=|\text{x}|-1\ ...(\text{i})$
Also,
If $\text{x}\leq0,$ then $|\text{x}|=-\text{x}\ ...(\text{ii})$
$\{\text{x}\in[-2,2]:\text{x}\leq0$ and $\text{f}(\text{|x|})=\text{x}\}$
$=\{\text{x}:|\text{x}|-1=\text{x}\} \ [$Using $(i)]$
$=\{\text{x}:-\text{x}-1=\text{x}\}\ [$Using $(ii)]$
$=\Big\{\text{x}:2\text{x}=\frac{-1}{2}\Big\}$
$=\Big\{\text{x}:\text{x}=\frac{-1}{2}\Big\}$
$=\Big\{-\frac{1}{2}\Big\}$

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