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AP EAMCET · Maths · Functions

Consider the following lists.
\(\begin{array}{ll} \text { List I } & \text { List II } \\ \hline \text { (A) } f(x)=\frac{|x+2|}{x+2}, x \neq-2 & \text { 1. }\left[\frac{1}{3}, 1\right] \\ \hline \text { (B) } g(x)=\mid[x \mid x \in R & \text { 2. } Z \\ \hline \text { (C) } h(x)=|x-[x]|, x \in R & \text { 3. } W \\ \hline \text { (D) } f(x)=\frac{1}{2-\sin 3 x}, x \in R & \text { 4. }[0,1) \\ \hline & \text { 5. }\{-1,1\} \end{array}\)

  1. A \(\begin{array}{llll}A & B & C & D \\ 5 & 3 & 2 & 1\end{array}\)
  2. B \(\begin{array}{llll}A & B & C & D \\ 3 & 2 & 4 & 1\end{array}\)
  3. C \(\begin{array}{llll}A & B & C & D \\ 5 & 3 & 4 & 1\end{array}\)
  4. D \(\begin{array}{llll}A & B & C & D \\ 1 & 2 & 3 & 4\end{array}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\begin{array}{llll}A & B & C & D \\ 5 & 3 & 4 & 1\end{array}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} (\mathrm{A}) \because f(x) & =\frac{|x+2|}{x+2}, x \neq-2 \\ & =\left\{\begin{array}{ll} \frac{x+2}{x+2}, & x > -2 \\ -\frac{x+2}{x+2}, & x -2 \\ -1, & x < -2\end{cases} \right. \end{aligned}\) So, range of \(f(x)\) is \(\{-1,1\}\). (B) \(\because\)…