Question

Consider the following lists.

List-I		List-II
(A)	$f(x) = \frac{|x+2|}{x+2} , x \ne -2 $	(I)	$[\frac{1}{3} , 1 ]$
(B)	$(x)=|[x]|,x \in [R$	(II)	Z
(C)	$h(x) = |x - [x]| , x \in [R$	(III)	W
(D)	$f(x) = \frac{1}{2 - \sin 3x} , x \in [R$	(IV)	[0, 1)
		(V)	{ -1, 1}

• A. (A)-(V), (B)-(III) , (C)-(II), (D)-(I)
• B. (A)-(III), (B)-(II) , (C)-(IV), (D)-(I)
• C. (A)-(V), (B)-(III) , (C)-(IV), (D)-(I)
• D. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Answer 1

The Correct Option is C

$f(x)=\frac{|x+2|}{x+2}, x \neq-2$

So, range of $f(x)$ is $\{-1,1\}$.
(B) $\because g(x)=|[x]|, x \in R$
As $[x] \in I \Rightarrow|[x]| \in W$
So, range of $g(x)$ is $W$.
(C) $\because h(x)=|x-[x]|, x \in R=|\{x\}| \in[0,1)$
$[\because\{x\}=x-[x]$ and $\{x\} \in[0,1)]$
So, range of $h(x)$ is $[0,1)$.
(D) $\because f(x)=\frac{1}{2-\sin 3 x}, x \in R$
$\because -1 \leq \sin 3 x \leq 1, \forall x \in R$
$\Rightarrow -1 \leq-\sin 3 x \leq 1$
$\Rightarrow 2-1 \leq 2-\sin 3 x \leq 2+1$
$\Rightarrow \frac{1}{3} \leq \frac{1}{2-\sin 3 x} \leq \frac{1}{1}$
So, range of $f(x) \text { is }\left[\frac{1}{3}, 1\right]$