Jee Main 2024 29 01 2024 Shift 1 - Question10
Question 10
Let $f(x)= \begin{cases}2+2 x, & x \in(-1,0) \\ 1-\frac{x}{3}, & x \in[0,3)\end{cases}$
$g(x)=\begin{array}{cc}x, & x \in[0,1) \\ -x, & x \in(-3,0)\end{array}.$
The range of $f \circ g(x)$ is
(1) $[0, 1]$
(2) $[-1,1]$
(3) $(0,1]$
(4) $(-1,1)$
Show Answer
Answer: (3)
Solution:
$f(x)= \begin{cases}2+2 x, & x \in(-1,0) \\ 1-\frac{x}{3}, & x \in[0,3)\end{cases}$
$g(x)=\begin{array}{cc}x, & x \in[0,1) \\ -x, & x \in(-3,0)\end{array} \Rightarrow g(x)=|x|, x \in(-3,0] \cup [0,1).$
$f(g(x))=\begin{array}{cc}2+2|x|, & |x| \in(-1,0) \Rightarrow x \in \emptyset \\ 1-\frac{|x|}{3}, & |x| \in[0,3) \Rightarrow x \in(-3,3)\end{array}.$
$f(g(x))= \begin{cases}1-\frac{x}{3}, & x \in[0,1) \\ 1+\frac{x}{3}, & x \in(-3,0)\end{cases}$
Range of $f \circ g(x)$ is $[0,1]$
BETA
