A) f (x) is discontinuous every where
B) f (x) is continuous every where
C) f(x) is continuous at x = 0 only
D) f(x) is discontinuous at x = 0 only
Correct Answer: B
Solution :
[b] \[f(x)=\left\{ \begin{matrix} \frac{{{x}^{2}}}{x}, & x\ne 0 \\ 0 & x=0 \\ \end{matrix} \right.\] \[=\left\{ \begin{matrix} \frac{{{x}^{2}}}{x}=x, & x>0 \\ 0, & x=0 \\ \frac{{{x}^{2}}}{-x}=-x, & x<0 \\ \end{matrix} \right.\] Now, \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,(-x)=0\] \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,(x)=0\] 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.
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