A) Discontinuous at \[x=2\]
B) Not differentiable \[x=2\]
C) Differentiable at \[x=3\]
D) Continuous but not differentiable at \[x=3\]
Correct Answer: D
Solution :
\[\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(3-h)=\underset{h\to 0}{\mathop{\lim }}\,\,\,|3-h-3|\,\,=0\] \[\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(3+h)=\underset{h\to 0}{\mathop{\lim }}\,\,\,|3+h-3|\,\,=0\] \[\because \,\,\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\,f(x)=\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f(x)=f(3)\] Hence f is continuous at \[x=3\] Now \[L\,{f}'(3)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f(3-h)-f(3)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{|3-h-3|\,\,-0}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{h}{-h}=-1\] \[R\,{f}'(3)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f(3+h)-f(3)}{h}\]\[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{|3+h-3|\,\,-0}{h}=1\] \[\because L\,{f}'(3)\,\ne \,R\,{f}'(3)\]. Hence f is not differentiable at \[x=3\]. Trick : Can be seen by graph it is continuous but tangent is not defined at \[x=3\].You need to login to perform this action.
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