Correct option: A.દરેક $x$ , $0 \le x \le 2$ માટે સતત અને દરેક $x$ એ $(0,2)$ માં $x \ne {1}$ માટે વિકલનીય છે.
a
(a) $f(x) = \left\{ {\begin{array}{*{20}{l}}{x{\rm{ ,}}}&{0 \le x \le 1}\\{1{\rm{ ,}}}&{1 < x \le 2}\end{array}} \right.$
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(1 - h)$
$ = \mathop {\lim }\limits_{h \to 0} \,\,(1 - h) = 1$
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} \,f(1 + h) = 1$
Hence function is continuous in $(0, 2).$
Now $\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} \,(0 + h) = 0 = f(0)$
$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} \,(2 - h) = 1 = f(2)$
Hence function is continuous in $[0, 2]$
Clearly, from graph it is not differentiable at $x = 1.$
