Functions Question 400
Question: Let $ f(x) $ be define on $ [-2,2] $ and is given by $ f(x)= \begin{cases} -1,,-2\le x\le 0 \\ x-1,,0\le x\le 2 \\ \end{cases} . $ , then $ f(| x |) $ is defined as
Options:
A) $ f(| x |)= \begin{cases} 1-2\le x\le 0 \\ 1-x,0<x\le 2 \\ \end{cases} . $
B) $ f(| x |)=x-1\forall x\in R $
C) $ f(| x |)= \begin{cases} -x-1,-2\le x\le 0 \\ x-1,0<x\le 2 \\ \end{cases} . $
D) None of these
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Answer:
Correct Answer: C
Solution:
[c] we have $ f(x)= \begin{cases} -1,-2\le x\le 0 \\ x-1,0\le x\le 2 \\ \end{cases} . $ $ f(| x |)= \begin{cases} -1,-2\le | x |\le 0 \\ | x |-1,0\le | x |\le 2 \\ \end{cases} .\Rightarrow f(| x |)=| x |-1,0\le | x |\le 2 $ ( $ (as-2\le | x |\le 0 $ is not possible)
$ \Rightarrow f(| x |)= \begin{cases} -x-1, \\ x-1, \\ \end{cases} ,\begin{matrix} -2\le x\le 0 \\ 0<x\le 2 \\ \end{cases} \begin{matrix} {} \\ {} \\ \end{cases} . $
BETA
