Consider the following lists. \(\begin{array}{ll} \text { List I } & \text { List II } \\ \hline \text { (A)…
Consider the following lists.
\(\begin{array}{ll} \text { List I } & \text { List II } \\ \hline \text { (A) } f(x)=\frac{|x+2|}{x+2}, x \neq-2 & \text { 1. }\left[\frac{1}{3}, 1\right] \\ \hline \text { (B) } g(x)=\mid[x \mid x \in R & \text { 2. } Z \\ \hline \text { (C) } h(x)=|x-[x]|, x \in R & \text { 3. } W \\ \hline \text { (D) } f(x)=\frac{1}{2-\sin 3 x}, x \in R & \text { 4. }[0,1) \\ \hline & \text { 5. }\{-1,1\} \end{array}\)
\(\begin{array}{llll}A & B & C & D \\ 5 & 3 & 2 & 1\end{array}\)
\(\begin{array}{llll}A & B & C & D \\ 3 & 2 & 4 & 1\end{array}\)
\(\begin{array}{llll}A & B & C & D \\ 5 & 3 & 4 & 1\end{array}\)
\(\begin{array}{llll}A & B & C & D \\ 1 & 2 & 3 & 4\end{array}\)
Solution
\(\begin{aligned}
(\mathrm{A}) \because f(x) & =\frac{|x+2|}{x+2}, x \neq-2 \\
& =\left\{\begin{array}{ll}
\frac{x+2}{x+2}, & x > -2 \\
-\frac{x+2}{x+2}, & x < -2
\end{array}= \begin{cases}1, & x > -2 \\
-1, & x < -2\end{cases} \right.
\end{aligned}\)
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)
\(\begin{aligned}
& \because h(x)=|x-[x]|, x \in R=|\{x\}| \in[0,1) \\
& {[\because\{x\}=x-[x] \text { and }\{x\} \in[0,1)]}
\end{aligned}\)
So, range of \(h(x)\) is \([0,1)\).
\(\begin{array}{lrl}
\text { (D) } \because f(x)= & \frac{1}{2-\sin 3 x}, x \in R \\
\because & & -1 \leq \sin 3 x \leq 1, \forall x \in R \\
\Rightarrow & -1 \leq-\sin 3 x \leq 1 \\
\Rightarrow & 2-1 \leq 2-\sin 3 x \leq 2+1 \\
\Rightarrow & \frac{1}{3} \leq \frac{1}{2-\sin 3 x} \leq \frac{1}{1}
\end{array}\)
So, range of \(f(x)\) is \(\left[\frac{1}{3}, 1\right]\).
Hence, option (3) is correct.