A) differentiable but not continuous
B) continuous but not differentiable
C) continuous as well as differentiable
D) neither continuous nor differentiable
Correct Answer: B
Solution :
Given\[x+|y|=2y\] \[\Rightarrow \]\[x+y=2y\]or\[x-y=2y\] \[\Rightarrow \]\[x=y\]or\[x=3y\] This represent a straight line which passes through origin. Hence, \[x+|y|=2y\]is continuous at x = 0. Now, we check differentiability at x = 0 \[x+|y|=2y\Rightarrow x+y=2y,y\ge 0\] \[x-y=2y,y<0\] Thus,\[f\left( x \right)=\left\{ \begin{matrix} x, & y<0 \\ {}^{x}/{}_{3}, & y\ge 0 \\ \end{matrix} \right\}\] Now, L.H.D.\[=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{-h}\] \[=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\frac{x+h-x}{-h}=-1\] R.H.D\[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}\] \[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\frac{x+h}{3}-\frac{x}{3}}{h}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{1}{3}=\frac{1}{3}\] Since, L.H.D \[\ne \] R.H.D. at x = 0 \[\therefore \]given function is not differentiable at x =0
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