Functions Question 99
Question: Consider $ f(x)= \begin{cases} & \frac{x^{2}}{|x|},,x\ne 0 \\ & ,0,,x=0 \\ \end{cases} . $
[EAMCET 1994]
Options:
A) $ f(x) $ is discontinuous everywhere
B) $ f(x) $ is continuous everywhere
C) $ f’(x) $ exists in $ (-1,1) $
D) $ f’(x) $ exists in $ (-2,2) $
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Answer:
Correct Answer: B
Solution:
We have $ f(x)= \begin{cases} \frac{x^{2}}{|x|}, & x\ne 0 \\ 0, & x=0 \\ \end{cases} = \begin{cases} \frac{x^{2}}{x}=x, & x>0 \\ 0, & x=0 \\ \frac{x^{2}}{-x}=-x, & x<0 \\ \end{cases} . . $ We have $ \underset{x\to 0-}{\mathop{\lim }},f(x)=\underset{x\to 0}{\mathop{\lim }}-x=0,\underset{x\to 0+}{\mathop{\lim }},f(x)=\underset{x\to 0}{\mathop{\lim }},x=0 $ and $ f(0)=0. $ So $ f(x) $ is continuous at $ x=0. $ Also $ f(x) $ is continuous for all other values of x. Hence, $ f(x) $ is continuous everywhere. Clearly, $ Lf’(0)=-1 $ and $ Rf’(0)=1. $ therefore $ f(x) $ is not differentiable at $ x=0. $
BETA
