Functions Question 564
Question: If $ f(x)= $ $ \begin{cases} & \frac{1-(x)}{1+x},,x\ne -1 \\ & 1,,,x=-1 \\ \end{cases} . $ , then the value of $ f(|2k|) $ will be (where
[ ] shows the greatest integer function) [DCE 2005]
Options:
A) Continuous at $ x=-1 $
B) Continuous at $ x=0 $
C) Discontinuous at $ x=\frac{1}{2} $
D) All of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)= \begin{cases} & \frac{1-|x|}{1+x},,x\ne -1 \\ & 1,,x=-1 \\ \end{cases} . $ and $ f(x)= \begin{cases} & 1,,x<0 \\ & \frac{1-x}{1+x},,x\ge 0 \\ \end{cases} . $ $ f(2x)= \begin{cases} & 1,,x<0 \\ & \frac{1-[2x]}{1+[2x]},x>0 \\ \end{cases} . $
Þ $ f(2x)= \begin{cases} & 1,x<0 \\ & 1,0\le x<\frac{1}{2} \\ & 0,\frac{1}{2}\le x\le 1 \\ & -\frac{1}{3},,,1\le x<\frac{3}{2} \\ \end{cases} . $
Þ $ f(x) $ , for all values of x where $ x<\frac{1}{2} $ a continous function and for $ x=\frac{1}{2} $ and $ x=1 $ $ f(x) $ be a discontinous function.
BETA
