Q. A function $f(x)$ is defined as fallows for real $x$, $f(x) = \left\{ \begin{array}{r} 1 - x^{2},\ for\ x < 1 \\ 0,\ \ \ \ \ \ for\ x = 1 \\ 1 + x^{2},\ for\ x > 1 \end{array} \right.\$ Then,
A) $f(x)$, is not continuous at $x = 1$
B) $f(x)$ is continuous but not differentiable at $x = 1$
C) $f(x)$ is both continuous and differentiable at $x = 1$
D) None of the above
Explanation
(a)
LHL$= \lim_{x \rightarrow 1^{-}}{f(x)} = \lim_{h \rightarrow 0}\ \left\lbrack 1 - (1 - h)^{2} \right\rbrack = 0$
RHL$= \lim_{x \rightarrow 1^{+}}{f(x)} = \lim_{h \rightarrow 0}{\{ 1 + (1 + h)^{2}\}} = 2$
Also, $f(1) = 0$
$\Rightarrow \ \ RHL \neq LHL = f(1)$
Hence, $f(x)$ is not continuous at $x = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.