A) f(x) is continuous at \[x=-1\]
B) f is differentiable at \[x=-1\]
C) f is differentiable \[\forall x\]
D) f is continuous everywhere
Correct Answer: A
Solution :
[a] \[\because f(x)\left\{ \begin{align} & 1,\,\,\,\,\,x\le -1 \\ & \left| x \right|,\,\,\,\,-1\le x<1 \\ & 0,\,\,\,\,\,\,\,\,\,x\ge 1 \\ \end{align} \right.\] Since, the doubtful points be \[x=-\text{ }1\]and \[x=1\] At \[x=-\text{ }1\] \[Lf'(-1)=\underset{h\to 0}{\mathop{lim}}\,\frac{f(-1-h)-f(1)}{-h}=\underset{h\to 0}{\mathop{lim}}\,\frac{1-1}{-h}=00\] \[Rf'(-1)=\underset{h\to 0}{\mathop{lim}}\,\frac{f(-1+h)-f(-1)}{+h}=\underset{h\to 0}{\mathop{lim}}\,\frac{\left| -1+h \right|-1}{h}\] \[=\underset{h\to 0}{\mathop{lim}}\,\frac{\left| -\left( 1+h \right) \right|-1}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-h-1}{h}=-1\] Thus, \[\therefore Lf'(-1)\ne R{f}'(-1)\] Hence, f (x) is not differentiable at \[x=-1\]but it is continuous at \[x=-1\] At \[x=-1\] \[\therefore Lf'(1)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1-h)-f(1)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left| 1-h \right|-0}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-h}{-h}=\underset{h\to 0}{\mathop{lim}}\,\left( \frac{-1}{h}+1 \right)\] i.e. Lf'(l) doesn't exists. Therefore, f (x) is not differentiable at \[x=1\]as well as, f (x) is not continuous at x = 1. Hence, option [a] is correct.
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